Thermal Radiant Absorption in Thin Semitransparent Plastics

Thermal Radiant Absorption in Thin Semitransparent Plastics Jim Throne, Dunedin, Florida

Raison d’etre

Energy input to plastic sheet is usually through a combination of radiant and convective sources. The model that is often used to predict the rate at which a specific plastic of a specific thickness heats is called the transient one-dimensional heat conduction model [7]. For thick- gauge sheet, being sheet greater than about 0.120 inch or 3.0 mm in thickness, the standard model assumes that the sheet is radiopaque, meaning that all the inbound energy is absorbed on the surface of the sheet. The conduction of this energy into the sheet from the surface is the controlling heat transfer factor. For thin-gauge sheet, conduction is no longer as major a factor. Instead, the ability for the sheet to absorb the inbound energy becomes significant. This paper focuses on energy uptake of thin-gauge semitransparent sheet.

Abstract

Infrared absorption characteristics of three thin semi-transparent plastics are examined, using their Fourier Transform Infrared or FTIR scans. Their heating characteristics are analyzed using the transient one-dimensional lumped parameter mathematical model.

Introduction

Thermoforming is the process by which shaped parts are made by heating and stretching plastic sheet [7]. Thin plastic sheet, being sheet 0.060 inch or 1.5 mm in thickness or less, is usually heated with radiant energy. Visual transparency is required for many thin- gauge parts such as rigid packaging. As noted below, most packaging plastics are semi- transparent, meaning they appear transparent in the visible wavelength range of 0.4 m to 0.7 m, but are semi-transparent in the far infrared region where radiant heating occurs. ‘Semi-transparent’ implies that, in addition to the sheet absorbing a portion of the inbound radiant energy, a portion of the energy is reflected from the sheet surface, and a portion is transmitted completely through the sheet.

In this note, the effect of plastic sheet material characteristics on the extent of radiant energy absorbed by the sheet is examined. For this paper, we consider energy interchange and radiant input only from one side of the sheet. At the end of the paper, we analyze energy absorption into thin semitransparent plastics using the lumped-parameter mathematical model.

Radiation Concepts

1 Radiation input to a substance is usually written as the sum of three characteristics – absorptivity, , being the amount of energy that is absorbed by the substance, transmissivity, , being the amount of energy that passes completely through the substance, and reflectivity1, , being the amount of energy that is reflected back toward the radiant source.

1  T  T  T (1)

Where the subscript T means the total amount of energy. Symbols without the subscript mean the local values of energy.

For a radiopaque plastic, transmissivity is zero [ = 0]. Reflectivity, , is given in terms of the index of refraction of the substance, as:

2 n1  n2      (2) n1  n2 

Where n1 is the index of refraction of the medium through which the radiant energy is passing and n2 is the index of refraction of the substance. For common glass in air, n1 = 1 and n2 = 1.5. Thus about 4% of light is reflected, or  ~ 0.04. The nature of the reflected energy in a radiopaque plastic only depends on the specularity of the surface of the plastic. In this note, specularity will not be considered as a technical issue2.

The mathematics for radiant heating of radiopaque solids is well-known and will not be discussed here. [1-4].

Semi-transparent Radiation Concepts

For semi-transparent solids such as optically transparent plastics – PVC, PS, PMMA, PC, PET, and others – transmissivity must be considered. For the first part of the discussion that follows, transmissivity is considered to be independent of wavelength, . There are two general methods for including transmissivity in radiant heating mathematics.

The ray-tracing method has been used extensively to describe the effect of transmission and reflection within a semi-transparent solid [5].

1 Please note that the symbol  is used herein for both local reflectivity and material density. These are not to be confused. 2 Specularity or molecular diffusion of incoming radiant energy will be the subject of a subsequent paper.

Figure 1 Ray-Tracing Method in a Semi-Transparent Solid [5, Fig. 18-2, p. 780]

Note that both inner and outer surfaces of the solid reflect radiant energy. The energy transmitted through point 1 is given as the inbound energy unit minus the reflectivity, or  The amount of that energy that is absorbed by the solid is given as the one minus the transmitted energy. But the transmitted energy is given as . Therefore  or That portion not absorbed,  reaches the rear surface of the solid where a portion of that is reflected, viz, , and the rest, , passes through the rear surface to the environment.

The fraction of the unit of unit inbound energy that is reflected is simply a progressive sum of the reflective energy from each bounce off the inner surfaces of the solid:

        T =          …)] (3a)

 1  2 2  T  1 2 2  (3b)  1   

The fraction of the energy transmitted is the progressive sum of all the energy rays that exited through the rear surface of the solid:

 1  2   T   2 2  (4) 1   

And finally, the most important term in the heating of the semi-transparent solid, the fraction of the energy absorbed:

T = 1 – T – T (5a)

1 1    (5b) T 1 

3 An alternate model, known as the net-radiation or heat flux model ([5], pp. 780-781) yields the same results.

As an example of the relationship between these three elements, consider the air-water arithmetic given above. If = 0.04, then

1 1   1 0.04  0.96 (5c) T 1 0.04 1 0.04

If the sheet is rather thick, the fraction of energy transmitted through the sheet will be small. As a result, the fraction of energy absorbed by the sheet will be:

T   d)

In other words, reflectivity for the air-water combination is essentially nil. What about plastics?

Radiation in Semi-transparent Plastics

In Table 1 are tabulated the refractive indices of many plastics. The common semi- transparent ones are highlighted.

Table 1 Refractive Indices of Plastics

Fluorcarbon (FEP) 1.34 - Polytetrafluoro--Ethylene (TFE) 1.35 Chlorotrifiuoro--Ethylene (CTFE) 1.42 - Cellulose Propionate 1.46 - Cellulose Acetate Butyrate 1.46 - 1.49 Cellulose Acetate 1.46 - 1.50 Methylpentene Polymer 1.485 - Ethyl Cellulose 1.47 - Acetal Homopolymer 1.48 - Acrylics 1.49 - Cellulose Nitrate 1.49 - 1.51 Polypropylene (Unmodified) 1.49 - Polyallomer 1.492 - Polybutylene 1.50 - Ionomers 1.51 - Polyethylene (Low Density) 1.51 Nylons (PA) Type II 1.52 - Acrylics Multipolymer 1.52 - Polyethylene (Medium Density) 1.52. - Styrene Butadiene Thermoplastic 1.52 - 1.55 Polyethylene Terephthalate (PET) 1.58

4 PVC (Rigid) 1.52. - 1.55 Nylons (Polyamide) Type 6/6 1.53 - Urea Formaldehyde 1.54 - 1.58 Polyethylene (High Density) 1.54 - Styrene Acrylonitrile Copolymer 1 .56 - 1.57 Polystyrene (Heat & Chemical) 1.57- 1.60 Polycarbonate (Unfilled) 1.586 - Polystyrene (General Purpose) 1.59 Polysulfone 1 .633

For the polymers listed in Table 1, the reflectivity range is 0.021 to 0.057. Although reflectivity is very small for plastics, it is considered in the analysis that follows.

Consider now transmissivity of semi-transparent plastics. It is generally accepted that the Beer-Lambert law holds for semi-transparent solids3:

  l  T   e (6)

Where  is the wavelength,  is the wavelength-dependent transmissivity,  is the wavelength-dependent absorption coefficient of the solid, and l is the thickness of the solid.

Radiant heating of most semi-transparent plastic solids occurs in the far infrared wavelength range of 2.5 m to perhaps 10 m. The primary range for thermoforming is about 2.5 m to about 7 m. This range corresponds to a temperature range of 350oF to about 1600oF. The Fourier Transform Infrared or FTIR device is ideal for determining the wavelength-dependent transmissivity of semi-transparent plastic solids. Several FTIR plots are shown in Figures 2-4. Note that the wavelength-dependent transmissivity is shown for two film thicknesses.

3 Although this theory was discovered by Pierre Bouguer, it is always mistakenly attributed to Johann Lambert who rediscovered it, and August Beer, who extended it to other materials, including atmospheric gases.

5 Figures 2-4. Infrared traces for three semi-transparent plastics

As is apparent from these plots, energy transmission, and therefore energy uptake, is highly wavelength-dependent. As an example, the plastics shown absorb 100% of inbound radiant energy in the 3.2 m to 3.7 m wavelength range. This is the fingerprint for all plastics having carbon-hydrogen bonds [C-H]. Certain plastics such as PVC absorb 100% of the inbound radiant energy around a wavelength of about 8 m, as well. This is not the case for others such as PE.

As noted earlier, the arithmetic was proposed for radiant properties that are independent of wavelength. As an example, average transmissivity values are shown in Table 2 for the two thicknesses of the three plastics shown in Figures 2-4. Keep in mind that the transmissivity values represent T.

Table 2 Average Transmissivity Values (3-7 m)

Transmissivity Transmissivity Plastic for thin film for thicker film PE ~0.88 @ 0.001 in ~0.56 @ 0.010 in PS ~0.76 @ 0.001 in ~0.35 @ 0.010 in PVC ~0.62 @ 0.003 in ~0.44 @ 0.012 in

Average and Mean Transmissivities

Because and l are known, local absorption coefficient values can be calculated from equation (6). And because these values are known for the three thicknesses of these plastics, the average absorption coefficient values, (thin + thick)/2, can be considered:

Table 3 Absorption Coefficient Values from Table 2

Absorption Absorption Mean Average Coefficient, in-1 Coefficient, in-1 Absorption Absorption Plastic for thin film for thicker film Coefficient Coefficient PE 128 58 86 93 PS 274 105 203 190 PVC 159 68 104 114

6 The mean absorption coefficient values can also be determined, from (thin x thick)1/2. Table 4 compares the average transmissivity values of Table 2 with those calculated using the mean and average absorption coefficient values of Table 3.

Table 4 Comparison of Table 2 Average Transmissivity Values with Mean and Average Calculated Values

Thin Film Table 2 Mean Mean Avg Avg Plastic value Calc’d Value Pct Error Calc’d Value Pct Error PE 0.88 0.93 + 5.7 0.91 + 3.5 PS 0.76 0.82 + 7.9 0.83 + 8.8 PVC 0.62 0.73 +17.7 0.71 +14.6

Thick Film Table 2 Mean Mean Avg Avg Plastic Value Calc’d Value Pct Error Calc’d Value Pct Error PE 0.56 0.42 -25.0 0.40 -28.6 PS 0.35 0.13 -63.0 0.15 -57.3 PVC 0.44 0.29 -34.1 0.26 -40.9

As is apparent in Figure 5, the absorption coefficient values calculated using the Beer- Lambert law, decrease with increasing sheet thickness for these three plastics. Although this effect has been observed for essentially all plastics for which there are infrared scans, there is no apparent reason for this anomaly4.

4 According to the literature on spectroscopy, for liquids, solutions that are not homogeneous can show deviations from the Beer-Lambert law because of the phenomenon of absorption flattening. The deviations will be most noticeable under conditions of low concentration and high absorbance. The deviations so noted tend to be minor, to a few percent, and not on the order of magnitude seen here. There seems to be no equivalent concept for plastic solids. There is also a strong indication that inclusions of different indices of refraction may act to alter the transmission of radiation through the solid [6]. Spectroscopic analyses of plastics may also lead to an understanding of the non-linear behavior of the Beer-Lambert law. This will be reviewed in a subsequent publication.

7 Beer-Lambert Exponential Transmissivity Equation 300 PS PE PVC 250

0 1 2 3 4 5 6 7 8 9 10 11 12 Sheet thickness, thousandths of inches

Figure 5. Calculated average absorption coefficient values for three semi- transparent plastics, noting the disparity between very thin films of 0.001 in and very thin thermoformable sheet.

Transmissivity Values for Thin-Gauge Plastic Sheet

Thermoformers rarely radiantly heat plastic films of thicknesses less than 0.010 inches or 250 m. Therefore, the transmissivity values for the thicker films of Table 3 are considered valid in further calculations. In addition, the Beer-Lambert law is assumed to be valid for all calculations for films thicker than 0.010 inches or 250 m.

Table 5 presents calculated values for absorptivity and reflectivity for the three plastics in Table 2 for original thick films and for sheets of 0.020 inches or 0.5 mm, 0.030 inches or 0.75 mm, and 0.040 inches or 1.00 mm in thickness.

Table 5 Calculated Absorptive and Reflectivity Values

Thick Films of Table 2 Absorption Avg Trans. Calc’d Calc’d Calc’d -1 Plastic Coefficient, in T trans,  Reflec.,  Absorb.,  PE 58 0.56 @ 0.010 in 0.61 0.045 0.062 0.395 PS 105 0.34 @ 0.010 in 0.38 0.052 0.059 0.608 PVC 68 0.44 @ 0.012 in 0.48 0.047 0.057 0.513

0.020 inch (0.5 mm) thick sheet Avg Calc’d Calc’d Calc’d

Plastic Trans. T trans,  Reflec.,  Absorb,  PE 0.31 0.34 0.045 0.050 0.645 PS 0.12 0.136 0.052 0.053 0.828

8 PVC 0.26 0.283 0.047 0.050 0.693

Plastic Trans, T trans,  Reflec.,  Absorb,  PE 0.176 0.193 0.045 0.047 0.778 PS 0.043 0.048 0.052 0.052 0.905 PVC 0.130 0.143 0.047 0.048 0.822

Plastic Trans, T trans,  Reflec.,  Absorb,  PE 0.106 0.115 0.045 0.046 0.848 PS 0.015 0.017 0.052 0.052 0.933 PVC 0.066 0.073 0.047 0.047 0.887

Note that even at 0.030 inch or 0.75 mm, PE and PVC still absorb only about 80% of the incident radiant energy. Figures 6 and 7 illustrate the effect of sheet thickness on transmissivity and absorptivity values for these three semi-transparent plastics.

Transmissivity of three semi-transparent plastics 0.6 PS PE PVC 0.5

0 10 20 30 40 Sheet Thickness, thousandths of inches

Figure 6 Thickness-dependent transmissivity values for three semi-transparent plastics

9 Absorptivity of three semi-transparent plastics

PS PE PVC 0 10 20 30 40 Sheet Thickness, thousandths of inches

Figure 7 Thickness-dependent absorptivity values for three semi-transparent plastics

The Development of the Sheet Heating Model

How can this information be used to determine the rate of heating of semi-transparent sheet? Consider the simple example of infinitely parallel plates. The radiating plate considered a black body with an emissivity =1 and its temperature is Tr*. The absorbing plate temperature is Tplastic*. [The asterisks mean that the temperatures are absolute.]

The inbound heat flux is given as:

*4 qinbound  Tr (7)

Where  is the Stefan-Boltzmann constant. The amount of energy absorbed by the sheet is given as:

*4 qinbound  T lTr (8)

The total energy interchange is thus:

*4 *4 qint erchange   T lTr Tplastic  (9)

It is apparent that the amount of energy absorbed is linearly dependent on the thickness- dependent absorptivity, T(l).

10 The arithmetic above assumes all the radiant energy is entering one surface of the plastic sheet and exiting the other. Most plastic sheets are heated from both sides. So the total energy uptake of the sheet is given as:

qint erchangex2  qint erchange1  qint erchange2 (10)

*4 *4 *4 qint erchangex2   T lTr1 Tr 2 Tplastic  (11)

For thin plastic sheets, where the conduction through the sheet is less significant than the energy exchange with the heating source, the sheet temperature can be approximated by the lumped-parameter model:

VcpdT  Qconvection  Qradiation

VcpdT  hATair Tplastic d  qint erchange Ad *4 *4 VcpdT  hATair Tplastic d  T lTr Tplastic Ad (12)

This model includes the effect of convection heat transfer at the surface of the sheet. V is the volume of the sheet, V=Al, where A is the sheet surface area and l is the sheet thickness in compatible units. Tair is the air temperature, h is the convection heat transfer coefficient, and  is the time.

The equation can be approximately solved in two ways – either analytically after a linearization of the radiation term or with finite difference analysis (FDE).

Approximate analytical solution

As a first step in linearization, a radiant heat transfer coefficient, hr, is defined as:

*4 *4 qint erchange T lTr Tplastic  hr      (13) Tr Tplastic   Tr Tplastic  

Now the lumped-parameter equation is approximated by:

dT lc plastic  hT T  h T T  (14) p d air plastic r radiation plastic

Replacing some of the terms:

hT T  h  h  C hT  h T  C  air plastic B  r  air r r lc p lc p B h  hr 

11 Where C, B, and C/B are (approximate) constants. (Keep in mind that hr is really dependent on plastic sheet temperature, as will be shown below in Table 9.)

dT plastic  BT  C  BC / B T  (15) d plastic plastic

dTplastic  Bd (16) C / B Tplastic 

If the initial sheet temperature, Tplastic=T0 when =0, then the first order equation can be integrated to yield an approximate analytical solution:

C / B Tplastic  ln   B  C / B T0  (17) C / B Tplastic     exp B   C / B T0 

Because the radiant heater temperature, Tr, is usually much greater than the ambient air temperature, Ta, the value of the term C/B is on the order of the radiant heater temperature value. Thus, from the last equation, it can be assumed that as time progresses, the plastic sheet temperature, Tplastic, approaches the radiant heater temperature in an exponential fashion.

Difference solution

Beginning with the one-dimensional equation, assume a forward difference:

*4 *4 lcpdT  hTair Tplastic d  T lTr Tplastic d (18)

dT  T  Ti Ti1 (19)

Where Ti is the temperature at  and Ti-1 is the previous temperature at . Again, T=T0 at =0. The difference equation now reads:

   *4 *4  hTair Ti1  T lTrad Ti1 Ti  Ti1       (20)  lc p   lcp 

Because of the strong influence of the fourth-power term, care must be taken in selecting a small enough value for the time step, .

Arithmetical values

12 Note that the terms in the brackets in equation (20) must have the units of temperature per h unit time. In the first bracket, the ratio must have the units of reciprocal time. lcp

Table 6 h Values for where l=0.030 inch and h = 1 Btu/ft2 h oF lcp o 3 -1 Plastic cp (Btu/lb F)  (lb/ft ) l  cp h/ l  cp (h ) PS 0.5 65.5 0.0819 12.2 PVC 0.5 84 0.105 9.52 PE 0.9 60 0.135 7.41

For the radiation term,  = 0.1714 x 10-8 Btu/ft2 h oR4, where oR = oF+460. By dividing the absolute temperature by 100, the arithmetic simplifies.

 4 4   T  460   Tplastic  460   l rad       *4 *4  T   T lTrad Ti1   100   100       0.1714 (21) lc  lc   p   p   

o As illustration, consider Trad = 700 F. The first parenthesis in the bracket isT(l) x 4 0.171411.6  3103T(l). For the three plastics above:

Table 7 Terms in radiant bracket for l = 0.030 inch

Plastic 3103/ l  cp T ( l) T ( l ) 3103/ l  cp 0.1714 / l  cp PS 37890 0.905 34290 2.093 PVC 29500 0.822 24249 1.634 PE 23000 0.778 17894 1.273

Table 8 Comparison of Inbound Radiant and Sheet Re-radiant Energy Terms [units = h-1] for l = 0.030 inch

Plastic Radiant @ 700 oF Sheet @ T= 80 oF Sheet @ 380 oF PS 34290 1780 10420 PVC 24249 1390 8135 PE 17894 1082 6338

Two observations can be made regarding the energy input to thin sheet. First, it is apparent, when comparing the relative values of convective and radiative heat transfer, that convection effects are negligible. Second, it is apparent that the major thermal

13 driving force is energy from the radiant heaters. For the calculations shown in Table 8, the energy give-back of the PS sheet at 380oF is only 30% of the total energy output of the heater. The value is less for lower sheet exit temperatures and the other plastics.

The radiation term can be linearized by creating a radiative heat transfer coefficient, hr, as shown in equation (13). The radiative heat transfer coefficient concept is important when comparing the relative influences of convective and radiative energy transfer rates on the heating rate of semi-transparent plastics. Table 9 gives calculated values of radiative heat transfer coefficients for the three semi-transparent plastics at the two temperatures in Table 8.

Table 9 2 o Radiative Heat Transfer Coefficient, hr, Btu/ft h F [Compare with Convective Value, h = 1]

o o Plastic T ( l) hr@Tplastic =80 F hr@Tplastic =380 F PS 0.905 4.29 6.11 PVC 0.882 4.18 5.89 PE 0.778 3.66 4.88

As is apparent and expected, radiant energy transfer rates are greater than convective energy transfer rates. (The reason the radiative heat transfer coefficient increases as the sheet temperature goes up is that the denominator of equation (13) decreases more rapidly with temperature than does the numerator.)

Calculated Heating Rates – An Example

Consider a comparison of the heating profiles for the three plastics in the above tables for sheet thickness, l = 0.030 inch, Figure 8. The radiant heater temperature is 700oF and the convective heat transfer coefficient, h = 1 Btu/ft2 h oF.

It appears that the high transmissivity and high specific heat of PE strongly influence its slow heating rate. The higher density and slightly higher transmissivity of PVC are the apparent reasons why it heats at a slower rate than PS.

14 One-dimensional Radiant Heating of Semi-Transparent Thin-Gauge Plastic Sheet - One Side Heating

Sheet thickness=0.030 in, Radiant Heater Temperature = 700F, Initial Sheet Temp = 80F, Air Temp = 80F

200 Polystyrene PVC PE

0 0 10 20 30 40 50 60 70 80 sheet heating, Aug2010.bas Heating time, seconds

Figure 8. One-dimensional radiant heating of semi-transparent thin-gauge plastic o sheet. Sheet thickness =0.030 inch, Radiant heater temperature, Tr=700 F, Initial o o sheet temperature, T0=80 F, Air temperature, Tair=80 F

Equation (20) and the arithmetic that follows it do not necessarily distinguish whether the energy in inbound from one or two energy sources, so long as Trad, h, and Tair are the same on both surfaces of the sheet. Heating uniformly from both sides increases the rate of heating, as is expected (Figure 9).

15 One-dimensional Radiant Heating of Semi-Transparent Thin-Gauge Plastic Sheet Thin Line - 2 Side Heating, Thick Line - 1 Side Heating Sheet thickness=0.030 in, Radiant Heater Temperature = 700F, Initial Sheet Temp = 80F, Air Temp = 80F 500 Polystyrene PVC PE PS PVC PE 400

0 0 10 20 30 40 50 60 70 80 Heating time, seconds Aug2010graphic3.bas

sheet heating, Aug2010.bas

Figure 9. One-dimensional radiant heating of semi-transparent thin-gauge plastic sheet. Thin lines indicate heating on 2 sides. Thick lines indicate heating on one side o (Figure 8). Sheet thickness =0.030 inch, Radiant heater temperature, Tr=700 F, o o Initial sheet temperature, T0=80 F, Air temperature, Tair=80 F

Limitations on the Lumped-Parameter Concept

The lumped-parameter model should be applied only when the dimensionless Biot number, Bi < 0.1 [7]. Bi = hl/k, where h is the heat transfer coefficient, l is a sheet thickness dimension, and k is the thermal conductivity of the plastic. If the sheet is heated on only one side, l equals the actual sheet thickness. If the sheet is heated on both sides, l equals the actual sheet half-thickness. In the technical literature, h is always assumed to be the convective heat transfer coefficient. In the analysis below, the combined heat transfer coefficient, hr + h, is also considered when determining the maximum allowable thickness for the lumped-parameter model.

For some bounds on the model, consider the three semi-transparent plastics used earlier. Note that the maximum allowable thickness decreases in proportion to the increase in heat transfer coefficient. For the illustration above, the convection heat transfer coefficient, h, was assumed to be constant at 1 Btu/ft2 h oF. As seen in Table 10, for convection only, the maximum allowable thickness values are quite large compared with the thicknesses used in earlier discussions on transmissivity and absorptivity. This implies that the use of the lumped-parameter model is applicable to the above analysis.

16 On the other hand, if the effect of radiation is included, using the radiative heat transfer coefficient, hr, given in equation (13), the lumped-parameter model may not be applicable at temperatures much above room temperature. Remember, however, that the maximum sheet thickness values in Table 10 represent half-thickness values when the sheet is heated equally on both sides5.

Having said that, the lumped-parameter model is the easiest to use when illustrating the role transparency plays in heating of semi-transparent plastic sheet.

Table 10 Thermal Conductivity and Maximum Sheet Thickness Values For Lumped-Parameter Model, Bi < 0.1

Thermal Maximum Maximum Maximum Conductivity, Thickness, l Thickness, l Thickness, l o o o Plastic Btu/ft h F inches, h in, h+hr (80 F) in, h+hr (380 F) PS 0.105 0.126 0.0218 0.0143 PVC 0.083-0.100 0.100-0.120 0.0207-0.0248 0.0119-0.0143 PE 0.183-0.292 0.220-0.350 0.0377-0.0600 0.0255-0.0405

Limitations on the Above Analysis

From even a casual glance at Figures 2 through 4, it is apparent that transmissivities are strongly wavelength-dependent. The analysis above assumes that it is proper to assume average values, at least through the wavelength range of 3 m through 7 m. Of course the proper method of analysis is to parse the values over a finite number of wavelength ranges [8]. Further, it was noted that the analysis above assumed that the emissivity values of the sheet and the heater were unity and that both the sheet and heater were planar and infinite in dimensions. These limitations can, of course, be removed, with the result being fine-tuning of the time-dependent heating curves depicted in Figure 9.

References

1. M.N. Ozisik, Boundary Value Problems of Heat Conduction, Dover Publications, Inc., New York, 1968. 2. M.F. Modest, Radiative Heat Transfer, McGraw-Hill, Inc., New York, 1993. 3. H.C. Hottel and A.F. Sarofim, Radiative Transfer, McGraw-Hill Book Company, New York, 1967. 4. S. Chandrasekhar, Radiative Transfer, Dover Publications, Inc., New York, 1960. 5. R. Siegel and J.R. Howell, Thermal Radiation Heat Transfer, 4/e, Taylor & Francis, New York, 2002.

5 It is important to keep in mind, however, that the lumped-parameter model contains no conduction component, meaning that thermal conductivity and thermal diffusivity are not included in the model. As a result, heating thin sheet from both sides doubles the energy input but does not affect the functional sheet thickness. For transient one-dimensional heat transfer in thick sheet, conduction into the sheet is important. As a result, with equal energy input to both sides of the sheet, the functional sheet thickness is one-half the actual sheet thickness [7].

17 6. J. Boulanger, O. Balima and A. Charette, Boundary Effects on the Efficiency of Direct Infra-Red Optical Tomography, 10th International Conference on Quantitative InfraRed Thermography, July 27-30, 2010, Québec (Canada) 7. J.L. Throne, Technology of Thermoforming, Carl Hanser Verlag, Munich, 1996, pp. 164-165. 8. Anon., Beer-Lambert Law, Department of Chemistry, The University of Adelaide,Adelaide, Australia, http://www.chemistry.adeleide.edu.au/external/soc- rel/content/beerslaw.htm.