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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.

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.

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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.

	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.

Infrared Energy:

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

Emissivity:

                 W_s
e =     ^{_____________}
                W_bb

        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

Where:
    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

Wavelength, Microns (λm)

Surface Temperature and Radiation Emissions:

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

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

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)

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.
Calculations:

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
                                                 3.412

                                                 (965 Btu/lb.)(61.24 x 10^-6 lbs./in²)
Latent Heat of Vaporization  =  ^{______________________________________}
                                                                    3.412

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

                                                               Watt-Hour
                                    = 22.60 x 10^{-3      _____________}
                                                                       in²

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/e₁ + 1/e₂ – 1)            ( 1/.9 + 1/.93 – 1)

   = .84

Stefan-Boltzmann Law:
R = (.84)(36.58072 x 10 ^-12 )( ( T₁ )⁴ – ( T₂ )⁴ ) = 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.

        = 858 K = 1084 °F (584 °C)

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