chromalox radiant infrared heating - source evaluations

8/10/2019 Chromalox Radiant Infrared Heating - Source Evaluations

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Technical

Infrared heating is frequently missapplied andcapacity requirements underestimated due toa lack of understanding of the basic principlesof radiant heat transfer. When infrared energyfrom a source falls upon an object or workpiece, not all the energy is absorbed. Some ofthe infrared energy may be reflected ortransmitted. Energy that is reflected ortransmitted does not directly heat the workpiece and may be lost completely from theprocess (Figure 2).

Another important factor to consider inevaluating infrared applications is that theamount of energy that is absorbed, reflectedor transmitted varies with the wave length of

the infrared energy and with different materialsand surfaces. These and other importantvariables have a significant impact on heatenergy requirements and performance.

Infrared Emitters & Source Temperatures The amount of radiant energy emitted from aheat source is proportional to the surfacetemperature and the emissivity of the material.This is described by the Stefan-Boltzmann Lawwhich states that radiant output of an idealblack body is proportional to the fourth powerof its absolute temperature. The higher thetemperature, the greater the output and moreefficient the source.

Figure 2

Tec h n ica l Info r m atio nRadiant Infra re d Heating - Theo ry & Principles

I nfrared Theory

Infrared energy is radiant energy which passesthrough space in the form of electromagneticwaves (Figure 1). Like light, it can be reflectedand focused. Infrared energy does not dependon air for transmission and is converted toheat upon absorption by the work piece. Infact, air and gases absorb very little infrared.As a result, infrared energy provides forefficient heat transfer without contact betweenthe heat source and the work piece.

Figure 1

Reflectivity Materials with poor emissivityfrequently make good reflectors. Polished gold

with an emissivity of 0.018 is an excellentinfrared reflector that does not oxidize easily.Polished aluminum with an emissivity of 0.04is an excellent second choice. However, oncethe surface of any metal starts to oxidize orcollect dirt, its emissivity increases and itseffectiveness as an infrared reflector decreases.

Table 1 Approximate Emissi vi ti es

Spectr al D istr ibuti on of a Blackbody

at Var ious Temperatures

RadiantEnerg

y

Wavelength (M icrons)

Note As the temperature increases, thepeak output of the source shifts to the left ofthe electromagnetic spectrum with a greaterpercentage of the output in the near infrared

range. This is referred to as the WienDisplacement Curve and is an important factorin equipment selection.

Emissivity In practice, most materials andsurfaces are gray bodies having anemissivity or absorption factor of less than1.0. For practical purposes, it can be assumedthat a poor emitter is usually a poor absorber.For example, polished aluminum has anemissivity of 0.04 and is a very poor emitter. Itis highly reflective and is difficult to heat withinfrared energy. If the aluminum surface ispainted with an enamel, emissivity increasesto 0.85 - 0.91 and is easily heated withinfrared energy. Table 1 lists the emissivity ofsome common materials and surfaces.

Absorption Once the infrared energy isconverted into heat at the surface, the heattravels into the work by conduction. Materialssuch as metals have high thermal conductivityand will quickly distribute the heat uniformlythroughout. Conversely, plastics, wood andother materials have low thermal conductivityand may develop high surface temperatureslong before internal temperatures increaseappreciably. This can be an advantage whenusing infrared heating for drying paint, curingcoatings or evaporating solvents on non-metalsubstrates.

Wien Displacement Curve

Peak Wavelength

1400F

1200F

1000F

80 0F

40 0F

Emissivity and an Ideal Infrared Source The ability of a surface to emit radiation is

defined by the term emissivity.The same termis used to define the ability of a surface toabsorb radiation. An ideal infrared sourcewould radiate or absorb 100% of all radiantenergy. This ideal is referred to as a perfectblack body with anemissivityof unity or 1.0.The spectral distribution of an ideal infraredemitter is below.

1 2 3 4 5 6 7 8 9 10

Aluminum 0.04 0.055 0.11-0.19Brass 0.03 0.06-0.2 0.60Copper 0.018-0.02 0.57Gold 0.018-0.035 Steel 0.12-0.40 0.75 0.80-0.95Stainless 0.11 0.57 0.80-0.95

Lead 0.057-0.075 0.28 0.63Nickel 0.45-0.087 0.37-0.48Silver 0.02-0.035 Tin 0.04-0.065 Zinc 0.045-0.053 0.11Galv. Iron 0.228 0.276

Miscellaneous Materials

Asbestos 0.93-0.96Brick 0.75-0.93Carbon 0.927-0.967Glass, Smooth 0.937Oak, Planed 0.895Paper 0.924-0.944Plastics 0.86-0.95Porcelain, Glazed 0.924Quartz, Rough, Fused 0.932Refractory Materials 0.65-0.91Rubber 0.86-0.95

Water 0.95-0.963Paints, Lacquers, Varnishes

Black/White Lacquer 0.8-0.95Enamel (any color) 0.85-0.91Oil Paints (any color) 0.92-.096Aluminum Paint 0.27-0.67

Metals Polished Rough Oxidized

Transmission Most materials, with theexception of glass and some plastics, areopaque to infrared and the energy is eitherabsorbed or reflected. Transmission lossescan usually be ignored. A few materials, suchas glass, clear plastic films and open fabrics,may transmit significant portions of theincident radiation and should be carefullyevaluated.

Controlling Infrared Energy Losses Onlythe energy absorbed is usable in heating thework product. In an unenclosed application,losses from reflection and re-radiation can beexcessive. Enclosing the work product in anoven or a tunnel with high reflective surfaceswill cause the reflected and re-radiated energyto be reflected back to the work product,eventually converting most of the originalinfrared energy to useful heat on the workproduct.


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Technical

A

B

C

D

10 0

90

80

70

60

50

40

30

20

10

0

Tec h n ica l Info r m atio nRadiant Infra re d Heating - Sou rc e Evaluation s

Evaluati ng Infrared Sources

Commonly available infrared sources includeheat lamps, quartz lamps, quartz tubes, metalsheath elements, ceramic elements andceramic, glass or metal panels. Each of thesesources has unique physical characteristics,operating temperature ranges and peak energywavelengths. (See characteristics chartbelow.)

Source Temperature & Wave LengthDistribution All heat sources radiateinfrared energy over a wide spectrum ofwavelengths. As the temperature increases forany given source:

1.The total infrared energy output increases

with more energy being radiated at allwavelengths.

2. A higher percentage of the infrared energyis concentrated in the peak wavelengths.

3.The energy output peak shifts toward theshorter (near infrared) wavelengths.

The peak energy wavelength can be deter-mined using Wiens Displacement Law.

Peak Energy= 5269 microns/R(Microns) Source Temp. (F) +460

Source= 5269 microns/R =2.83 microns1400F 1400F +460

Source= 5269 microns/R =5.49 microns500F 500F +460

Absorption by Work Product Materia ls inProcess Applications While most materialsabsorb long (far) infrared wavelengthsuniformly, many materials selectively absorbshort (near) infrared energy in bands. Inprocess heating applications this selectiveabsorption could be very critical to uniformand effective heating.

For process heating, it is recommended thatthe infrared source have a peak output

wavelength that best matches the selectiveabsorption band of the material being heated.When the major absorption wavelengths of thematerial being heated are known, the chartbelow provides guidance in selecting the mostefficient heat source. The relative percentageof radiant energy emitted by specific sourceand falling in a particular wavelength rangecan be determined from the chart.

Example Plastic materials are known tohave high infrared absorption rates inwavelengths between 3 and 4 microns. Selecta source which provides the most effectiveoutput to heat plastics in the 3 and 4 micronrange.

1. Enter Bottom of Chart at 3 and 4 microns,read up to corresponding points onselected element curve (use 1400F metalsheath in this example).

%R

adiantEn

ergyBelowWavelength

2. From These Points, move left to read thecorresponding percentages (29% and

51%).

3. The Di fferencebetween these two values(22%) is the percentage of radiant energyemitted by the element within selectedwavelengths limits.

4 . To Obta inthe maximum percentage of theenergy emitted by a given element in thedesired wavelength band, multiply thepercentage in 3 above by the conversionefficiency for the selected element(comparison chart 56% x 22% =12.2%).

In this example, a high temperature source(quartz lamp 4000F) with a peak in the 1.16

micron range, while more energy conversionefficient, would not be as effective as a lowertemperature metal sheath or panel heaterswith a peak in the 2.8 to 3.6 micron range.Quartz tubes (1600F) would provide similarpeak wavelengths.

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

Wavelength - Mi crons

3900 2150 1400 1000 740 550 410 310

Temperature (T) degrees F = Radiation Pea ks

Percentage Increment of Radiant EnergyFalling Below any Wavelengthfor a Black Body atTemperature T

A - 4000F Source TemperatureB - 1600F Source TemperatureC - 1400F Source TemperatureD - 1000F Source Temperature

Source Temperature (F) 3000 - 4000F 3000 - 4000F Up to 1600F Up to 1500F Up to 1600F 200 - 1600F Up to 1700FBrightness 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

Mechanical Ruggedness Poor Fair Good Excellent Good Good Fair

Chromalox Model QR QRT RAD, URAD RCH CPL, CPLI, CPH CPHI

Infrared Source Glass Bulb T3 Quartz Lamp Quartz Tube M etal Sheath Ceramic Ceramic Coated Quartz Face

Character isti cs of Commerciall y Used I nfr ared Heat Source

Tungsten Filament Nickel Chrome Resistance Wire Wide Area Panels


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Technical

Alkyd Paint Steel 320 3.9 3Epoxy Paint Steel 356 8.1 5Acrylic Paint Steel 392 8.1 2Powder Coat Steel 400 13 6

Tec h n ica l Info r m atio nRadiant Infra re d Heating - Pr oc ess App lications

Appli cati on Parameters

Typical industrial applications of radiantheating include curingor baking(powders,paints, epoxies, adhesives, etc.), drying(water, solvents, inks, adhesives, etc.) andproduct heating(preheating, soldering, shrinkfitting, forming, molding, gelling, softening,and incubating). The following are generalguidelines that can be used in evaluating andresolving most radiant heating problems.Unfortunately, the process is so versatile andits applications so varied that it is not feasibleto list solutions to every problem.

To determine heat energy requirements andselect the best Chromalox infrared equipmentfor your application, it is suggested theproblem be defined using a check list similarto below. Several of the key factors on the listare discussed on this and following pages:

1. Product to be heated2. Physical dimensions and weight/piece3. Surface coating or solvents, if any4. Infrared absorption characteristics5. Production rate (lbs/hr, pieces/hr, etc.)6. Work handling method during heating

(continuous, batch or other)7. Element response time (if critical)8. Power level requirements in kW/ft2based

on Time/Temperature relationship (ifknown)

9. Starting work temperature10 . Final work temperature11 . Ventilation (if present or required)12 . Available power supply13 . Space limitations

Infrared Absorption Characteristics Aspreviously discussed, many materials,particularly plastics, selectively absorbinfrared radiation. The following chart providesdata on some common plastic materials andthe recommended source temperatures forthermoforming applications.

Element Response Time Some applica-tions, such as continuous web heating of

paper or plastic film, require quick shutdownof heaters in case of work stoppage. In theseapplications, residual radiation from theinfrared heaters and associated equipmentmust be considered. Residual radiation fromthe element is a function of the operatingtemperature and mass. Quartz lamps andtubes have relatively low mass and theinfrared radiation from the resistance wiredrops significantly within seconds aftershutdown. However, the surrounding quartzenvelope acts as a secondary source ofradiation and continues to radiate considerableenergy. Metal sheathed elements have moremass and slightly slower response time. Wide

area panels have the most mass and theslowest response time for both heat up andcool down. The following chart shows theaverage cool down rate of various sourcesafter shutdown. Actual cool down of thesource and work product will vary withequipment design, product temperature,ambient temperature and ventilation.

Glass Bottles 104 6.4 30Adhesives Paper 3.2 30

Heating

PVC Shrinking 300 3.2 60ABS Forming 340 9.7

Drying Surface& Temp Time

Heating Substrate (F) W/In2 (sec)

Deriving Time-Temperature Information fromEmpirical Testing If specific information isnot readily available for a particular workproduct, a simple but effective test will usuallyprovide enough preliminary data to proceedwith a design. Place one or more radiantheaters in a position with the radiationdirected at a work product sample. Thedistance between the face of the heater and

the sample should approximate the expectedspacing in the final application. Position thesample so that it is totally within the radiatedarea. Energize the heater(s) and record thetime necessary to reach desired temperature.Calculate the W/in2falling on the work pieceusing the exposed area of the work productand the maximum kW/ft2at the face of theheater as listed in the product catalog page. Ifthe data is not available and a sample test cannot be performed, the following table providesa few suggested watt densities as guidance.

LPDE 3.3 - 3.9 877 - 1170HDPE 3.2 - 3.7 950 - 1170

PS 3.2 - 3.7 950 - 1170(6.4 - 7.4) 245 - 355

PVC 1.65 - 1.8 2440 - 2700(2.2 - 2.5) 1625 - 1910

PMMA 1.4 - 2.2 1910 - 3265PA-66 1.9 - 2.8 1405 - 2285

(3.4 - 5) 585 - 1075

Cellulose 2.2 - 3.6 990 - 1910

Acetate (5.2 - 6) 440 - 545

Absorption Ideal SourceBand(s) Temperature

Plastic (microns) (

F)

Paint Baking 4-6 1 - 2Metal Dry Off 15 8Thermoforming 10 - 15 Fusing or Embossing(plastic films) 5-6 Silk Screen Drying 5-6

Application Heat Up Hold

W/In2 on Work

Contact your Local Chromalox Salesoffice for further information or assis-tance in determining time/temperaturerequirements for a particular application.

Power Level or Radiation Intensity Inmost process applications, more than oneradiant heater is needed to produce thedesired results. When heaters are mounted

together as close as possible, the net radiantoutput of the array is defined as the maximumpower level or radiation intensity. The catalogpages for radiant heaters indicate themaximum kW/ft2at the face of each heater.Typical ranges for radiation intensity (powerlevel) are as follows:

Source Temperature Vs. T ime

A

1600

1500

1400

1300

1200

11001000

90 0

80 0

70 0

60 0

50 0

40 0

30 0

SheathorSurfaceTemperature(F)

0 30 60 90 120 150 180

Elapsed Time (Sec.)

Time-Temperature Rela tionship A criticalstep in the evaluation of a radiant heatingapplication is to determine the time necessary

to develop work piece temperature and theelapsed time needed to hold temperature inorder to obtain the desired results (curing ordrying). The following chart shows time/temperature relationships for several typicalinfrared applications and materials.

A. Quartz Lamp 3/8 Dia.B. Quartz Tube 1/2 Dia.C. Metal SheathD. Wide Area Panel

E. Ceramic Heater

Surface TimeCuring Substrate Temp (F) W/In2 (min)

Source TemperatureAfter Shutdown

B

E

CD

Low 1 - 2Medium 2 - 3High Over 3

Radiant Intensity Heater Outputor Power Level (kW/Ft2)


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Technical

Tec h n ica l Info r m atio nRadiant Infra re d Heating - Pro cess App licat ion sDetermining kW Required It is difficultto develop simple calculations for radiant

heating applications because of the manyvariables and process unknowns. Designdata gained from previous installations orfrom empirical tests is frequently the mostreliable way of determining installed kWrequirements. Total energy requirementscan be estimated with conventional heatloss equations. The results of conventionalequations will provide a check against dataobtained from nomographs or empiricaltesting. As a minimum, conventionalequations should include the following.

1. Calculate the Sensible Heatrequired tobring work to final temperature. Basecalculations on specific heat and pounds ofmaterial per hour.

2. Determine Latent Heat of Vaporization(when applicable). Latent heat ofvaporization is normally small for solventsin paints and is frequently ignored.However, when water is being evaporated,the kilowatt hours required may be quitesignificant.

3. Ventilation Air (when applicable). The risein air temperature for work temperatures,350F or less, can usually be estimated as50% of final work temperature rise. Forhigher work temperatures, assume air andwork temperature are the same.

4. Conveyor Belt or Chain HeatRequirements.Assume temperature riseof conveyor to be the same as worktemperature rise.

5. Wall, Floor and Ceiling Losses for

Enclosed Ovens. For uninsulated metalsurfaces, refer to Graph G-125S. Forinsulated walls, refer to Graph G-126S.

6. Oven End Losses.For enclosed ovens, thiswill depend on shape of end area and

whether or not air seals are used. Ifsilhouette shrouds are used, a safety factorof 10% is acceptable.

7. The Sum of The Lossescalculated in 1-6above will be the minimum total heatenergy requirement based on conventionalheat loss equations.

Infrared He ating Equations Infrared energyrequirements can also be estimated by using

equations and nomographs developedspecifically for infrared applications.

Product Heating For product heating, thefollowing equation can be used

kW = Lbs/hr x CpxT F3412 Btu/kW x Efficiency(RE) x VF x

Where:Lbs/hr =Pounds of work product per hourCP=Specific heat in Btu/lb/FT=Temperature rise in FEfficiency (RE) =Combined efficiency of the

source and reflector

VF =View Factor is the ratio of theinfrared energy intercepted by the workproduct to the total energy radiated bythe source. For enclosed ovens, use afactor of 0.9. For other applications, referto the view factor table.

=Absorption (emissivity) factor of the workproduct

Drying & Solvent Evaporation Removingsolvent or water from a product requiresraising the product temperature to thevaporization temperature of the solvent andadding sufficient heat to evaporate it. Tocalculate heat requirements for solvent

evaporation, the following information must beknown.

1. Pounds of solvent to be evaporated perhour

2. Pounds of work product per hour3. Initial temperature of product and solvent4. Specific heat of product5. Specific heat of solvent6. Vaporization temperature of solvent

(ie: water =212F)7. Heat of vaporization of solvent8. Source/reflector efficiency9. View factor

10 . Absorption factor (emissivity)WARNING Hazard of Fire.Flammablesolvents in the atmosphere constitute a firehazard. When flammable volatiles are releasedin continuous process ovens, the National FirePrevention Association recommends not lessthan 10,000 ft3of air be removed from theoven per gallon of solvent evaporated.Reference NFPA Bulletin 86 "Ovens andFurnaces", available from NFPA, P.O. Box9101, Quincy MA 02269.

For drying, use the following equation.

kW =QWP+QS+QLH

3412 Btu/kW x Efficiency(RE) x VF x

Where:QWP=Btu required by work product to raise

the temperature from initial to vaporiza-tion temperature

Qs = Btu required by solvent to raise thetemperature from initial to vaporizationtemperature

QLH=Btu required for the latent heat of thevaporization of the solvent

Efficiency (RE) =Combined efficiency of thesource and reflector

VF =View Factor for enclosed ovens, use a

factor of 0.9. For other applications, referto the view factor table.

=Absorption (emissivity) factor of the workproduct

Controls Most control systems for infraredprocess heating can be divided into twocategories, open loop or manual systems andclosed loop, fully automatic systems.

Open Loops or M anual Systems Thesimplest and most cost effective controlsystem is an input controllers (percentagetimer) such as the Chromalox VCF Controlleroperating a magnetic contactor. The timer

cycles the radiant heaters on and off for shortperiods of time (typically 15 - 30 seconds).This control system works best with metalsheath heaters, which have sufficient thermalmass to provide uniform radiation. It can beused with quartz tube or quartz lamp heatersby using special circuitry to switch from full tohalf voltage rather than full on and full off.

Closed Loop or Automatic Systems Sinceinfrared energy heats the work product bydirect radiation, closed loop control systemsthat depend on sensing and maintaining airtemperature are relatively ineffective (except in

totally enclosed ovens). In critical applicationswhere temperature tolerances must be closelyheld, non-contact temperature sensorsoperating SCR control panels are recom-mended. Non-contact temperature sensorscan be positioned to measure only the workproduct temperature. Properly positioned,non-contact temperature sensors and SCRcontrol panels can provide very accurateradiation and product temperature control.


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Technical

Tec h n ica l Info r m atio nRadiant Infra re d Heating - Pr oc ess App licat ion s

Nomograph for Product Heating Forproduct heating, the nomograph at the right

can be used. The nomograph does not takeinto account heat energy requirements for airventilation. To estimate the kW for totalproduct heating:

1 . D ete rm inepounds of material per hour tobe heated (A)

2 . Read acrossto the specific heat of thematerial (B)

3. Read upto desired temperature rise in F(C)

4 . Read acrossto overall efficiency (D).

Overall efficiency =Product AbsorptionFactor x View Factor x Source Efficiency.Determine Product Absorption Factor(surface emissivity) of the work product(ie: =0.85 for enamel sheet metal).Determine View Factor (use 0.9 as a viewfactor for well designed or enclosedovens). Determine Source efficiency.Typical Source/Reflector efficiencies are:

Quartz Lamps 0.70 to 0.80Quartz Tubes 0.60 to 0.70Metal Sheath 0.55 to 0.65

5 . R ea d dow nto Kilowatts required (E).

Estimati ng Total Ki lowatts for Product Heating

200

175

150

125

100

75

50

25

0. 80. 6

0. 4

0. 30. 2

0. 1

250 300 350 400 100 90 80 70 60 50 40

10 30 50 70 90 110 130 150

20 0

30 0

50 0

80 01000

2000

3000

5000800010000

PoundsofMaterialPerHour

Kilowatts of Infrared Required

Overall EfficiencyTemperature Rise F

Specific Heatof Material

.0 .05

AB

CD

E

1. 0

30

20

10

60 55 50 40% 100 90 80 70 60 50% 40

30

20

10

100 80 60 40 20 0 20 40 60 80 100 120 140

Pounds of Water Evaporated Per Hour Kil owatts of I nfrared Energy Required

Source/ReflectorEfficiency

Work Product Absorption Factor(Emissivity x View Factor)

C

DA

B

65

70

80

Nomograph for Drying The nomograph tothe right can be used to estimate Kilowattsrequired to evaporate water from the surfacesof work product. Graph is based on an initialstarting product temperature of 70F. It doesnot take into account heat energy require-ments for air circulation or ventilation.

1. Determine poundsof water (solvent) perhour to be evaporated (A)

2. Read upto Source/Reflector efficiency (B).Depending on the configuration andcleanliness of the reflector, typical Source/

Reflector efficiencies are:Quartz Lamps 0.70 to 0.80Quartz Tubes 0.60 to 0.70Metal Sheath 0.55 to 0.65

3 . Read acrossto Work Product AbsorptionFactor (C). This value is based on theemissivity of the work product surface(ie: =0.85 for enameled sheet metal) andthe view factor of the oven or space. Use0.9 as a view factor for well designed orenclosed ovens.

4 . R ea d dow nto Kilowatts required (D).

Estimating Infr ared Ki lowatts for D rying

Note To evaporate solvents other than water, calculate the energy required to heat the solventto vaporization temperature using the weight, specific heat and temperature rise. Calculate thelatent heat of vaporization and add to the energy required to heat the solvent to vaporizationtemperature.


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Technical

Baking & Curing The nomograph to theright can be used to determine the watt

density required on the work product forbaking and curing of paints and coating.Lacquers are cured primarily by evaporation ofthe solvent and can be cured by infrared in2 - 15 minutes. Enamels are cured primarilyby polymerization and require a longer time(15 - 20 minutes). Varnishes, japans andhouse paints cure mainly by oxidation but canusually can be accelerated by infrared heating.To find approximate watt density needed forbaking:

1 . Loca tetemperature product is to reach infive minutes (A)

2 . Read acrossto line representing gauge ofthe material being heated (B)

3 . Re ad upto ventilation air in feet perminute over surface of the product (C). Ifnot known, estimate feet per minute basedon cubic feet per minute of ventilation orcirculating air divided by the the approxi-mate cross sectional area of the oven. Inapplications with no forced ventilation, use2 - 5 fpm.

4 . Re ad ri ghtto the absorption factor for thework product surface or coating (ie: =0.85 for enameled sheet metal) (D)

5 . Re ad dow nto watt density required onthe product surface (E).

Tec h n ica l Info r m atio nRadiant Infra re d Heating - Pro cess App licat ion s

0 1 2 3 4 5 6 7

10 0

20 0

30 0

40 0

50 0

60 0Temperature(F)TobeAttainedin

FiveMinutes

Ventilation AirFeet Per Minute

(Air at 80F)

0

5015 0

30 0

12 Gauge Sheet Steel

B

16 Gauge

20 Gauge

24 Gauge

28 Gauge

A

Wood

C

D

E

0.90

AbsorptionFactor(Emissivity)

Esti mati ng Watt Densi ty for Cur ing or Baking

Determining Heater Fixture Spacing

Having determined the total required kilowattsand the desired W/in2on the work product, thenext step is to deterine the spacing and thenumber of heaters. In most conveyor typeoven applications, a 12" spacing from the faceof the heater to the work product producesuniform distribution of the radiation. Thegraph to the right shows centerline tocenterline spacing of Chromalox radiantheaters to obtain various intensities on thework based on a spacing of 12" from the faceof the heater to the work product. Specificapplications may require the distance to beincreased or decreased.

The graph is applicable to line or point infraredsources installed in reflectors. Refer to viewfactor charts for ceramic heaters and flat panelinfrared sources.

I ntensi ty Vs. Spacing Point & Li ne Infrared Sources

L20

18

16

14

12

10

8

6

4

2

AverageW/In2o

nWorkProduct

2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36

Heater Center to Center Distance = L (In. )

A B

C

D

E

A = QRT(3/8" Dia. Quartz Tube)

B = RAD/QRT(1/2" Dia. Quartz Tube)

C = S-RADD = RADD/U-RAD/QR

(3/8" Dia. Quartz Tube)E = U-RP

3-11/16"12"

Watt Density Required on Product (W/In 2)

0.750.65

0.50

0.35


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Technical

Tec h n ica l Info r m atio nRadiant Infra re d Heating - Pr oc ess App licat ion sView Factor for Flat Panels While theradiation pattern from line and point infrared

sources can be controlled by reflectors, theradiation pattern from flat panels is diffusedand the infrared energy is emitted from a largearea. Consequently, the shape of the sourceand the target are a significant factor indetermining the Watt density falling on thework product. For parallel surfaces inapplications such as thermoforming or webheating, the incident energy falling on the workproduct is determined by a View Factor. Viewfactor is defined as the percentage or fractionof infrared energy leaving the surface of a flatpanel (source) which is intercepted by thesurface of the work product (target). The viewfactor for parallel surfaces (rectangles) can bedetermined from the graph. Example Findthe view factor for a 12 by 24" panel heatermounted 4" from a continuous web infrareddrying application. X/L =24" 4" =6,Y/L =12"4" =3. Read left from the interceptof X/L =6 and Y/L =3 with a view factor of 0.7.

View Factor for Two Parall el Sur faces

Radiant Oven Heating Example Amanufacturer of 66 gallon electric waterheaters wishes to bake the paint on sheetmetal jackets (open top and bottom) at 350F.The jackets weigh 33 lbs, are 26" in diameterby 45" high with an outside area of 25.5 ft2.The process requires 20 jackets be painted per

hour. The jackets will be suspended from aconveyor chain on 9 ft centers and will berotated as they move. The chain weighs 12lbs/ft. The heaters will be installed in a tunneloven with 2 inches of insulation and reflectivewalls. The oven is 8 ft long, 4 ft wide and 7 fthigh and has end openings 3 ft by 6 ft.Preliminary test results show the jackets mustbe baked for six minutes for a satisfactoryfinish. The paint weighs 7.25 lbs/gal, contains50% volatiles and covers 212 ft2per gallon.Assume a room temperature of 70FSpecific heat of steel =0.12 Btu/lb/FBoiling point of solvent =170FSpecific heat of solvent =0.34 Btu/lb/F

Latent heat of vaporization =156 Btu/lbHeat Required for Operation

1. Heat Absorbed by Jackets (20 jackets/hr x 33 lbs =660 lbs/hr)

660 lbs/hr x0.12 Btu/lb/F x (350 - 70F) =6.5 kW3412 Btu/kW

2. Heat Absorbed by Solvent Solventvolume

25.5 ft2 x 20 jackets/hr x 50% =1.20 gal/hr212 ft2/gal

Heat required to heat solvent to 70F

1.2 gph x 7.25 lb/gal x 0.34 Btu/lb x (170-70F)=0.1 kW3412 Btu/kW

Heat required to vaporize solvent

1.20 gph x 7.25 lb/gal x 156 Btu/lb =0.4 kW

3412 Btu/kWHeat absorbed by solvent =0.1 +0.4 =0.5 kW

3. Heat Required by Ventilation Air (NFPArecommendation is a minimum of 10,000cubic feet per gallon of solvent evaporated.)Densiity of air =0.080 lbs/ft3

Specific heat of air =0.240 Btu/lb/F

Note Ventilation air is heated by re-radiation and convection from the work, ovenwalls, etc. Air temperature is always less thanthe work temperature. Assume a 200F airtemperature.

Volume =1.20 gph x 10,000 ft3

=12,000 ft3

/hr12,000 ft3/h x 0.08 lb/ft3x 0.24 Btu/lb/F x (200-70F)

3412 Btu/kW

Heat absorbed by ventilation air =8.78 kW

4. Conveyor Chain & Hangers Normallythe conveyor chain is outside theradiation pattern of the heaters and isheated by convection from air in the tunnel.Since the heat absorbed by the air hasalready been accounted for, the heatabsorbed by the conveyor may be ignored.(Conveyor speed should provide 6 minutesin the 8 foot heated area.)

Total He at Absorbed

6.5 kW +0.5 kW +8.8 kW =15.8 kW

Heat Losses Heat losses from oven surfacewith 2 inches of insulation (Graph G-126S) =12 W/ft2. Assume inside surface temperatureof wall and ceiling =250F, T=180F)

Wall area 7 ft x 8 ft x 2 ft =112 ft2Ceiling and floor area 8 ft x 4 ft x 2 ft =64 ft2

Open tunnel ends =3 ft x 6 ft x 2 ft =36 ft2

Heat loss from outside surfaces of oven

176 ft2 x 12 W/ft2=2.1 kW/hr1000 W/kW

Heat loss from open oven ends (assume theopen ends are equal to an uninsulated metalsurface under the same conditions as the ovensurfaces) (See Graph G-125S.)

36 ft2 x 0.6 W/ft2x 180F =3.89 kW/hr1000 W/kW

Total Heat Losses 2.1 kW +3.98 kW =5.99kW

Total Heat Capacity Required for Operation

15.8 kW +5.99 kW =21.8 kW/hr

As with any process heat calculation, it is notpossible to account for all the variables andunknowns in the application. A safety factor isrecommended. For radiant heating applications,a safety factor of 1.4 is suggested.

Total Heat Required = 21.8 x 1.4 =30.5 kWh

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Technical

FixtureLocation

Tec h n ica l Info r m atio nRadiant Infra re d Heating - Com for t HeatingI ndoor Spot H eati ng

Infrared spot heating of work stations andpersonnel in large unheated structures orareas has proven to be economical andsatisfactory. The following guidelines may beused for spot heating applications (areas withlength or width less than 50 feet).

1. Determinethe coldest anticipated insideambient temperature the system mustovercome. If freeze protection is providedby another heating system, thistemperature will be 40F.

2. Determinethe equivalent ambienttemperature desired (normally 70F is thenominal average).

3. Subtract 1 from 2 to determine thetheoretical increase in ambient temperature(T) expected from the infrared system. Ifdrafts are present in the occupied area (airmovement over 44 feet per minute(0.5 mph) velocity), wind shielding orprotection from drafts should beconsidered.

4. Determinethe area to be heated in ft2. Thisis termed the design or work area (AD)(Fig. 1).

5 . Mul ti pl ythe design area by one watt persquare foot times the theoretical

temperature increase (T) desired asdetermined in Step 3 (minimum of 12

watts per square foot). The design factor ofone watt per square foot density assumes afixture mounting height of 10 feet. Add 5%for each foot greater than 10 feet inmounting height. Avoid mounting fixturesbelow 8 feet.

6. Determinefixture mounting locationsa) In areas where the width dimension is25 feet or less, use at least two fixturesmounted opposite each other atthe perimeter of the area and tilted at anangle. This provides a greater area ofexposure to the infrared energy by

personnel in the work area. Tilt thefixtures so that the upper limit of thefixture pattern is at approximately six feetabove the center of the work station area(Figure 2).b) When locating fixtures, be sure toallow adequate height clearance for largemoving equipment such as cranes and lifttrucks.c)Avoid directing infrared onto outsidewalls.

7. Est imate(tentatively) the radiated patternarea. Add length of fixture to the fixturepattern width (W) to establish pattern

length (L). Pattern Area =L x W (Fig. 3).

Figure 2 Til ted I nfrared

Fixtur es for Spot Heati ng

Figure 1 Design Area

WWidth of Radiation

Pattern Figure 3 Patt ern Area

Angle ofUniform

Radiation

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L

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W/ 2

11.5'

10'

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8. Divide the design area (Step 4) into thepattern area (Step 7).

Q =Pattern AreaDesign Area

If the pattern area is equal to or greaterthan the design area, quotient (Q) will beequal to or greater than 1 and coverage isadequate. If Q is less than 1, the design

area exceeds the pattern area of individualfixtures. Adjust the heater locations andpatterns or add additional fixtures withpatterns overlapping as necessary, toensure adequate coverage.

9. Multiplyquotient (Q in Step 8) by theincrease in theoretical temperature (TofStep 3) by the design area (ADof Step 4) todetermine the amount of radiation to beinstalled.

Radiation (Watts) =Q x T x AD

10 . Many Typesof radiant heaters areavailable for comfort heating applicationsincluding ceiling, wall and portable floorstanding models. Choose specific fixturesfrom the product pages. It is preferred thathalf the wattage requirements be installedon each side of the work station in thedesign area.

Controls Manual control by percentagetimers may be adequate for a small installation.To provide better control of comfort levels invarying ambient temperatures, divide the totalheat required into two or three circuits so thateach fixture or heating element circuit can beswitched on in sequence. Staging can beaccomplished by using multistage air thermo-

stats set at different temperatures.

Indoor Area Heati ng

In many industrial environments, area heating(areas with length or width greater than 50 ft)can be accomplished economically withmultiple infrared heaters. For quick estimates,determine the minimum inside temperatureand use a factor of 0.5 watts per square foot ofdesign area for each degree of theoretical

temperature. If the calculated heat loss of thestructure, including infiltration or ventilationair, is less than the quick estimate, select thelower value. Locate heaters uniformlythroughout the area with at least a 30% overlain radiation pattern.

Outdoor Spot Heati ng

The same guidelines outlined under IndoorSpot Heating should be followed except thatwatts per square foot for each degree oftheoretical ambient temperature increaseshould be doubled (approximately 2 watts persquare foot for each 1F). This factor applies toutdoor heating applications with little or nowind chill effect on personnel. If wind velocitieare a factor in the application, determine theequivalent air temperature from the Wind ChillChart in NEMA publication HE3-1971 or otherinformation source.

Note Increasing the infrared radiation tomassive levels to offset wind chill can creatediscomfort and thermal stress. In outdoorexposed applications, a wind break or shieldinis usually more effective.

60Reflector22 Tilt

chromalox radiant infrared heating - source evaluations

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