Infrared Heating Technical Manual for Salamander Ceramic Infrared Emitters

Comparing Different Forms of Infrared Heat

    Throughout the years many different forms of infrared heat sources have been developed. Some of the more familiar forms seen today are metal sheathed tubular heatersquartz tubesquartz lamps, gas-fired catalytic, flat faced panels, and ceramic emitters. Each source has its own distinctive set of properties:

Metal Sheath Quartz Tube Quartz Lamp Catalytic Flat Faced Panels Ceramic
Radiant Efficiency
Physical Strength
Very Low
Heat-Up Cool-Down
Very Fast
Very Slow
Max. Temp.
1400 °F
1600 °F
4000 °F
800 °F
1600 °F
1292 °F
Color Sensitivity Low Low High Low Low Low

Radiant Efficiency:  The total amount of infrared energy, relative to a “black body radiator,” that is emitted from the source.

Physical Strength:
   The physical strength of each source. A high rating indicates a very durable source that can withstand physical abuse such as dropping a wrench on the source.

Heat-Up/Cool Down:
   The amount of time required for the source to come up to operating temperature and cool back down to room temperature.

Maximum Temperature:
   Maximum operating temperature of the source.

Color Sensitivity:
    Refers to the ability of a typical load to absorb the spectral radiation emitted from a source based on the color of the load.  The shorter the wavelength emitted from a source the more color sensitive a load will be to the sources spectral radiation.

Radiant Emission Patterns of Ceramic Emitters
Radiant Emission Patterns
Concentrated                                   Uniform                                      Wide Area

       Salamander ceramic emitters are manufactured with three basic emitter faces: concave, flat, and convex. These emitter face styles will result in the specific radiant emission patterns as shown above. Note: Infrared radiation is emitted at right angles to the emission surface.

Concentrated: The concave surface will emit a “concentrated” radiant pattern which is highly effective when zone heating is desired as well as radiant heating in general.
Uniform: The flat surface will produce a “uniform” pattern for even heating at a close proximity between the emitter and the target being heated.
Wide Area: The convex shape gives off a “wide area” pattern which is desirable in comfort heating or other applications that require a dispersed radiant emission pattern.

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Salamander Radiant Emission Grid

         The Salamander radiant emission grid can be used to determine the proper ceramic emitter spacing when used in an application such as an infrared panel. In order to achieve an even heat pattern it is critical that the emitters are spaced so that their radiant emission patterns overlap when reaching the target. The more overlap that occurs, the more even the heat will be across the face of the product being heated. The area of highest radiant emission intensity for a single emitter is shown within the two dark crossed lines on the grid. In order for element emissions to overlap, the dashed line shows an intersection point at a distance of 7″ will occur if the emitters are placed a distance of 2″ apart from edge to edge. This same concept should be used to either determine the distance to place the product if using an existing panel, or placement of emitters if building a panel to guarantee radiant emission overlap.

Ceramic Infrared Panel Design

Typical Panel Configuration
Typical Ceramic Panel Design

Ceramic Emitter Mounting
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This drawing shows a ceramic heater and a reflector with an oblong hole for mounting.

LTE Ceramic Emitter Mounting
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Our larger LTE element has two mounting posts on the back side. This drawing shows an LTE and a reflector with two oblong holes for mounting.
LTE Mounting
This drawing shows an LTE installed in a reflector.

Wiring Specifications:

High temperature 842°F (450°C) MG or similar style wire (with a suitable temperature and amperage rating) should be used for all electrical connections made within the terminal area of the infrared panel. The high temperature wire must be run on top of (or above) the ceramic fiber insulation.
Ceramic terminal blocks are recommended to allow for quick emitter replacement, flexibility in zoning, and “touch safe” design.
The terminal cover for the infrared panel should be louvered or made out of expanded metal to minimize the temperature within the terminal area.

        The spacing of the emitters should be such that the resulting infrared emissions incident on the target will be even and maximized.
Emitters that are tightly spaced in an array will allow the target to be positioned close to the emitters and still result in even heating. The intensity and efficiency of the infrared radiation will be maximized and heat losses will be minimized.
Emitters that are loosely spaced in an array will force the target to be positioned further away in order to achieve even heating. This style of panel would typically result in a lower intensity infrared emission.
For more information about emitter mounting, using our two piece mounting clip design and warnings see our emitter mounting page.

Infrared Heating Basics

        This section of the technical manual is a summary of the physics involved in all infrared heating systems. The information can be used as an aid in calculating system power requirements as well as determining the feasibility of a given infrared heating application.

Infrared Energy:

        When infrared energy strikes an object it may be absorbed, transmitted, or reflected from the surface. The sum of the amount of energy absorbed, transmitted, and reflected must equal 100% of the total incident energy. An object is called a “blackbody” if it absorbs (or emits) 100% of incident infrared radiation.
1 = ρ + α + t

    ρ = reflectivity
    α = absorptivity
    t = transmissivity

Example: Infrared energy strikes an object that is 30% reflective, and 20% transparent, how much infrared energy is absorbed by the object?

1 = .30 + α + .20
α = 1 – .30 -.20 = .50 (or 50% )
        The term “blackbody radiation” was derived from an experiment in cavity radiation. A small hole was drilled into an object and light was focused into the hole. The hole (cavity) appeared to be black. Light that entered the cavity is trapped and absorbed into the object allowing no light to escape. Radiant energy emitted from a “blackbody” source is dependent only on the temperature of the cavity walls and is not at all dependent on any other characteristic of the source such as color.


        A true “blackbody” source for industrial applications has not yet been developed. However, various radiant heating elements are available with a wide range of radiant efficiencies. The efficiency of a radiant heater is given by its emissivity value. Emissivity is defined as the ratio of the radiant energy emitted by an object at a given temperature and the radiant energy emitted by a “blackbody” at the same temperature.

e =     _____________

    e = emissivity of source
    Ws = Total radiant energy emitted from a source at temperature T1
    Wbb = Total radiant energy emitted from a blackbody at temperature T1

        Infrared radiation is part of a broad electromagnetic spectrum. The relationship between electromagnetic radiation is as follows:
λ =   _______
    λ = Wavelength in meters
    c = Speed of light ( 3 x 10meters per second )
    f = Frequency in hertz ( cycles per second )

Infrared Spectrum:Infrared Spectrum

Stefan-Boltzmann Law:

        The Stefan-Boltzmann Law gives the total power radiated at a specific temperature from an infrared source. That is, the entire amount of infrared radiation (at a specific temperature) emitted from a given source at all associated wavelengths.
R = (e) x (σ) x ( T)    Watts / in2
    σ = Stefan-Boltzmann Constant
        [ 36.58072 x 10 -12 W/ in2 K]
    e = Emissivity Value of the Source
    T = Surface Temperature of the Source in K (Kelvin.)

        In order to understand the spectral distribution of infrared radiation from a source we must first understand Planck’s Law. Planck’s Law gives us the spectral distribution of radiation from a blackbody source. That is, a source that emits 100% infrared radiation at a given single temperature. It is important to understand at this point that in practice, infrared sources are made up of thousands of “point sources” that are all at different temperatures. Each point source will have a different spectral distribution and the combination of point sources will make up the entire spectral distribution. Therefore, we can only approximate the spectral distribution using an average surface temperature and emissivity value.

                (e) x ( 2.416069 x 10-25 )            Watts
R(λ) =    _______________________________________________
                ( λ ) 5 exp .014408/λT – 1 ]            in 2 . µm

    e = Emissivity of Source
    λ = Wavelength in Meters
    T = Temperature in K (Kelvin)
        K = (°F + 460) / 1.8

Spectral Distribution of a “Blackbody”

At Various Temperatures
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Wavelength, Microns (λm)

“A”….800 °F, λm = 4.14 mm             “B”….1000 °F, λm = 3.57 mm
“C”….1200 °F, λm = 3.14 mm           “D”….1400 °F, λm = 2.81 mm

        Notice in the Planck’s Law curves shown on the previous page that the spectral radiancy of the source increases proportionally with the source temperature. In other words, the radiant infrared output from a source increases as the temperature of the source increases. The overall infrared emissions from a given source is equal to the area under the associated Planck’s Law curve. By integrating Planck’s Law at a given temperature with respect to the wavelength we can calculate the amount of infrared emissions within a given range of wavelengths (See graph below).
        Also notice that as the temperature of the source increases, the peak wavelength of the source becomes shorter. When the temperature of the source becomes too high a noticeable amount of energy is emitted from the source as light. That is, a portion of the energy emitted from the source falls within the wavelengths associated with light. Referring back to the infrared spectrum chart shown on page 7, visible light occurs starting at .40µm and ends at .70µm. The infrared spectrum starts at .70µm and extends to 1000µm. Although the useful range of wavelengths for infrared heating applications occurs between .70µm to 10µm.
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        Wien’s Law gives the wavelength at which the spectral distribution (given by Planck’s Law) of the radiation emitted by a blackbody is at a maximum point. Note, however, that according to Plank’s Law a range of wavelengths is emitted from a source at a specific temperature! Wien’s Law simply gives the “peak wavelength”.

            2.898 x 10 -3       m K
λm =  __________________________
                          T k

    λm = Peak Wavelength in Meters
    T k = Temperature in K (Kelvin)
            K = (°F + 460)/1.8

Surface Temperature and Radiation Emissions:

        The curve shown below can be used as a quick reference to estimate the amount of infrared radiant energy emitted from a given source. The curves were derived using the Stefan-Boltzmann Law. For example, a 1000 °F (538 °C or 811 K) infrared source with an emissivity value of .80 (80%) will have an approximate radiant emission (from the curves below) of 12.5 Watt / in2. Using the Stefan-Boltzmann equation yields the following:

R = ( .80 )( 36.58072 x 10 -12 )[ (811)4] = 12.65 Watts / in2

Surface Temperature vs. Radiation Emission
at Various Emissivity Values
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Watts / in2

Emitter Surface Temperature

        The warm-up and cool-down curves shown below are based on the Salamander FTE style ceramic emitter. The curves for the Salamander HTE and LTE emitter can be approximated by using the following factors. If it is desired to know the time/temperature relationship for an HTE emitter, multiply the wattage of the desired HTE emitter by a factor of 2. That is, an HTE-500 will have the same temperature characteristics as an FTE-1000. If it is desired to know the time/temperature curves for an LTE emitter, multiply the wattage of the desired LTE emitter by a factor of .55. That is, an LTE-900 will have the same temperature characteristics as an FTE-500 (approximately). Note that the time/temperature curves are based on a single FTE emitter in a 70 °F (21 °C) ambient environment. When using the ceramic emitters in an array of multiple units the time/temperature curves can be significantly different.
Surface Temperature Warm-Up Time:
Single Emitter in 70 °F (21 °C) Ambient
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Time in Minutes

Surface Temperature Cool-Down Time:

Single Emitter in 70 °F (21 °C) Ambient
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Time in Minutes
Spectral Absorption Curves
        The following spectral absorption curves show the range of wavelengths that a particular material will absorb infrared radiation as well as the percentage of absorption. These curves are only representative of a particular sample of a given “virgin” material. In actual practice, coloring agents and other additives will change the look of the curves. However, the curves can be used to get a general idea of the range of infrared radiation in which the material will absorb.
Spectral Absorption Curve For Water
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Wavelength, µm

Spectral Absorption Curve For PVC
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Wavelength, µm

Spectral Absorption Curve For Polystyrene
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Wavelength, µm

    Spectral Absorption Curve For Polyethylene
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Wavelength, µm

Physical Properties of Materials

Material Densitylb. / ft3 Specific
Emissivity Thermal
Btu · in
Heat of
Heat of
Non-Metallic Solids:
Asphalt 65 0.40 0.93 1.20 40 250/121
Beeswax 60 1.67 75 144/62
Carbon 138 0.20 165
Cotton 92 0.31 0.77 0.41
Glass 165 0.20 5.4
Ice 57 0.53 32/0
Paper 58 0.45 0.93 0.82
Paraffin 56 0.70 1.56 63 133/56
Rubber 76 0.44 0.90 1.10
Wood, Oak 50 0.57 0.90 1.15
Wood, Pine 34 0.67 0.90 0.90
Plastics: Most Non-Metals Have An Emissivity of 0.90
ABS 69-76 0.3-0.4
Acrylic 69-74 0.34
Epoxy 66-88 0.25-0.3
Flouroplastic 131-150 0.28
Nylon 67-72 0.3-0.5
Phenolic 85-124 0.35
Polycarbonate 74-78 0.30
Polyester 66-92 0.2-0.35
Polyethylene 57-60 0.54
Polyamides 90 0.27-0.3
Polypropylene 55-57 0.46
Polystyrene 66 0.32
PVC 72-99 0.2-0.3
Aluminum 169 0.24 1536 1190/643
– Polished 0.09
– Med. Oxide 0.19
– Heavy Oxide 0.31
430 Stainless 475 0.11 150 2650/1454
– Polished 0.17
– Med. Oxide 0.57
– Heavy Oxide 0.85
Oil, Cottonseed 60 0.47 0.90
Oil, Vegetable 57.5 0.43 0.90 318/159
Paraffin 47.1 0.71 750/399
Water 62.4 1.0 0.93 4.08 965 212/100

Reference Data

 °C = 5/9(°F -32)        Or          °F = 9/5(°C) +32
                K = (°F +460)/1.8     Or          K = °C + 273
                °R = °F +460

Ohms Law:

E = Volts, I = Amps
R = Ohms, W = Watts

3 Phase Wye (Balanced Load) 3 Phase Delta (Balanced Load)
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Conversion Factors:

1 kW = 1000 Watt 1 mm = .03937 Inch 1 kg = 2.205 lb.
3412 BTU = 1 kW-HR 1 m = 39.37 Inch 1 g = .002205 lb.
1 HP = .746 kW 1 Inch = 2.54 cm 1 U.S. Gal. = 0.1337 Cu. Ft.
1 Boiler H.P. = 9.8 kW 1 km = .6214 Mile 1 U.S. Gal. = 3.785 liters

Estimating Power Requirements

        In a given heating system any or all of the three modes of heat transfer (convection, conduction, radiation) can be utilized. The intended purpose of the following examples is to focus on the infrared heating component only of each heating system. That is, it is assumed that 100% of the heat transfer in each example is by infrared radiation and any heat losses are considered to be negligible.


Given: A thermoforming application requires that an 1/8″ thick PVC sheet be heated to 350 °F (177 °C) in 1 minute.   Determine the power required using only infrared radiation.

Emissivity of the infrared source = .90
Emissivity of PVC Sheet = .90
Specific Heat of PVC = .30 BTU / lb. / °F
Density of PVC = 99 lbs. / ft3
Temperature Required = 350 °F (177 °C)
Ambient Temperature = 65 °F (18 °C)
Convert the PVC target into (lbs. / in2):
lbs./in2 = (99 lbs./ft3)( 1/1728 in3)(.125 in) = .007161  lbs./in2

The power required to heat the PVC sheet is given by:

Watt-Hour            (Weight)(Specific Heat)( T1 – T2 )
____________   =   ____________________________________________
       in2                                     3.412

=     _____________________________   =   .179

Thermoforming Example (cont.):

Warm-up Time:
                                Watt-Hour / in2
Warm-up Time =    __________________   x  60        Minutes
                                   Watt / in2

1 min. =    ________________   x  60
                      Watt / in2

Solve the “time” equation for Watt / in2:

Watt / in2 =  ____________    =  10.74
                        1 min
Watt / in= 10.74

This is the amount of infrared radiant energy that must be absorbed into the PVC sheet to heat the sheet to 350 °F (177 °C) in 1 minute.

At this point one can use Planck’s Law and the spectral absorption curve for PVC by superimposing these curves on each other and calculating the total area under the curves at which the two curves intersect (provided that accurate curves are available). This can be extremely time consuming.

A simplified method of estimating the power radiated and absorbed into the PVC sheet is given by the following:

The effective emissivity between two parallel plates is given by:

                    1                                  1
e =   __________________     =     __________________
        ( 1/e+ 1/e– 1)            ( 1/.9 + 1/.9 – 1)

   = .82

Thermoforming Example (cont.):
Two infrared heater panels will be used. One panel will heat the top of the PVC sheet, the other will heat the bottom of the PVC sheet. Heating both the top and bottom of the PVC sheet will minimize the temperature gradient within the sheet which could cause “part” deformation. Since two infrared panels will be used, the power required per panel is 1/2 of the 10.74 Watts / in2. Therefore, 5.37 Watts / in2 is required from each infrared panel.

Stefan-Boltzmann Law:

R = (.82)(36.58072 x 10 -12 )( ( T1 )– ( T2 )) = 5.37

T1= Source Temperature
T2 = Average PVC Temperature = (65+350)/2 = 208 °F
T2 = 371 K

Solve the equation for the source temperature.

        = 667 K          ( 741 °F or 394 °C)

Results: The surface temperature of the source must be at least 741°F (394 °C) to achieve a 350 °F (177 °C) PVC sheet temperature within 1 minute.

Water Evaporation:
Given: Estimate the amount of infrared radiation required to evaporate 4 grams of water per square foot every 5 seconds from a substrate material in a water based adhesive application. Assume the substrate to have a negligible mass.

Emissivity of the infrared source = .90
Emissivity of Water = .93
Specific Heat of Water = 1.0 BTU / lb. / °F
Latent Heat of Vaporization = 965 Btu / lb.
Boiling Point of Water = 212 °F (100 °C)
Ambient Temperature = 65 °F (18 °C)

Convert the grams of water per square foot to lbs. of water per square inch:(4 g/ft2)(.0022046 lb./g)( 1/144 ft2/in2) = 61.24 x 10-6 lbs./in2
The power required to heat the water is given by:
   Watt-Hour            (Weight)(Specific Heat)( T1 – T2 )
_______________   =    __________________________________________
          in2                                        3.412

                               (61.24 x 10-6)(1.0)(212 – 65)
                               =    ______________________________________    =  2.64 x 10-3

                                                 (965 Btu/lb.)(61.24 x 10-6 lbs./in2)
Latent Heat of Vaporization  =  ______________________________________

                                            = 17.32 x 10-3
Total Power Required  =  (2.64 x 10-3) + (17.32 x 10-3)

                                    = 22.60 x 10-3      _____________

Water Evaporation Example (continued):
Warm-up Time:
                                Watt-Hour / in2
Warm-up Time =    __________________   x  60        Minutes
                                   Watt / in2

                       (22.6 x 10-3)
5/60 min. =   ________________   x  60
                          Watt / in2

Solve the “time” equation for Watt / in2:

                     (22.6 x 10-3)(60)
Watt / in2 =    __________________    =  16.27
                          (5/60 min)

The effective emissivity between two parallel plates is given by:

                    1                                      1
e =   __________________     =        ___________________
        ( 1/e+ 1/e– 1)            ( 1/.9 + 1/.93 – 1)

   = .84

Stefan-Boltzmann Law:

R = (.84)(36.58072 x 10 -12 )( ( T1 )– ( T2 )) = 16.27

Where:  T1 = Source Temperature
             T2 = Average Water Temperature = (65+212)/2 = 138.5 °F (59 °C)
                   = 332.5 K
Solve the equation for the source temperature.

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        = 858 K = 1084 °F (584 °C)

The surface temperature of the source must be at least 1084 °F (584 °C) to evaporate 4 grams of water within 5 seconds.