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 thingauge parts such as rigid packaging. As noted below, most packaging plastics are semitransparent, 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:
n  n 
  1 2
 n1  n2 
2
(2)
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].
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.
1
2
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 =  …)]

T   1 

1   2 2 

1   2 2 
(3a)
(3b)
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 
2 2
1    
T   
(4)
And finally, the most important term in the heating of the semi-transparent solid, the
fraction of the energy absorbed:
T = 1 – T – T
T 
1   1   
1  
(5a)
(5b)
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
 T  1  0.04
1 
1 
 0.96
1  0.04
1  0.04
(5c)

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 semitransparent ones are highlighted.
Table 1
Refractive Indices of Plastics
Fluorcarbon (FEP)
Polytetrafluoro--Ethylene (TFE)
Chlorotrifiuoro--Ethylene (CTFE)
Cellulose Propionate
Cellulose Acetate Butyrate
Cellulose Acetate
Methylpentene Polymer
Ethyl Cellulose
Acetal Homopolymer
Acrylics
Cellulose Nitrate
Polypropylene (Unmodified)
Polyallomer
Polybutylene
Ionomers
Polyethylene (Low Density)
Nylons (PA) Type II
Acrylics Multipolymer
Polyethylene (Medium Density)
Styrene Butadiene Thermoplastic
Polyethylene Terephthalate (PET)
1.34 1.35
1.42 1.46 1.46 - 1.49
1.46 - 1.50
1.485 1.47 1.48 1.49 1.49 - 1.51
1.49 1.492 1.50 1.51 1.51
1.52 1.52 1.52. 1.52 - 1.55
1.58
4
PVC (Rigid)
Nylons (Polyamide) Type 6/6
Urea Formaldehyde
Polyethylene (High Density)
Styrene Acrylonitrile Copolymer
Polystyrene (Heat & Chemical)
Polycarbonate (Unfilled)
Polystyrene (General Purpose)
Polysulfone
1.52. - 1.55
1.53 1.54 - 1.58
1.54 1 .56 - 1.57
1.57- 1.60
1.586 1.59
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:
 T    e   l
(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 350 oF 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)
Plastic
PE
PS
PVC
Transmissivity
for thin film
~0.88 @ 0.001 in
~0.76 @ 0.001 in
~0.62 @ 0.003 in
Transmissivity
for thicker film
~0.56 @ 0.010 in
~0.35 @ 0.010 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
6
Plastic
PE
PS
PVC
Absorption
Coefficient, in-1
for thin film
128
274
159
Absorption
Coefficient, in-1
for thicker film
58
105
68
Mean
Absorption
Coefficient
86
203
104
Average
Absorption
Coefficient
93
190
114
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
Plastic
PE
PS
PVC
Table 2
value
0.88
0.76
0.62
Mean
Calc’d Value
0.93
0.82
0.73
Mean
Pct Error
+ 5.7
+ 7.9
+17.7
Avg
Calc’d Value
0.91
0.83
0.71
Avg
Pct Error
+ 3.5
+ 8.8
+14.6
Table 2
Value
0.56
0.35
0.44
Mean
Calc’d Value
0.42
0.13
0.29
Mean
Pct Error
-25.0
-63.0
-34.1
Avg
Calc’d Value
0.40
0.15
0.26
Avg
Pct Error
-28.6
-57.3
-40.9
Thick Film
Plastic
PE
PS
PVC
As is apparent in Figure 5, the absorption coefficient values calculated using the BeerLambert 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
Absorption coefficient, 1/in
300
PS
PE
9
10
PVC
250
200
150
100
50
0
1
2
3
4
5
6
7
8
11
12
Sheet thickness, thousandths of inches
Figure 5. Calculated average absorption coefficient values for three semitransparent 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.
-1
Plastic Coefficient, in
T
PE
58
0.56 @ 0.010 in
PS
105
0.34 @ 0.010 in
PVC
68
0.44 @ 0.012 in
Calc’d
Calc’d
trans,  Reflec., 
0.61
0.045 0.062
0.38 0.052 0.059
0.48 0.047 0.057
0.020 inch (0.5 mm) thick sheet
Avg
Calc’d
Plastic
Trans.T
trans, 
PE
0.31
0.34
PS
0.12
0.136
Calc’d
Reflec., 
0.045 0.050
0.052 0.053
Calc’d
Absorb.,
0.395
0.608
0.513
Calc’d
Absorb, 
0.645
0.828
8
PVC
0.26
0.283
0.047 0.050
0.693
0.030 inch (0.75 mm) thick sheet
Avg
Calc’d
Plastic
Trans, T
trans, 
PE
0.176
0.193
PS
0.043
0.048
PVC
0.130
0.143
Calc’d
Reflec., 
0.045 0.047
0.052 0.052
0.047 0.048
Calc’d
Absorb, 
0.778
0.905
0.822
0.040 inch (1.00 mm) thick sheet
Avg
Calc’d
Plastic
Trans, T
trans, 
PE
0.106
0.115
PS
0.015
0.017
PVC
0.066
0.073
Calc’d
Reflec., 
0.045 0.046
0.052 0.052
0.047 0.047
Calc’d
Absorb, 
0.848
0.933
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
Transmissivity, tau
0.5
0.4
0.3
0.2
0.1
0
10
20
30
40
Sheet Thickness, thousandths of inches
Figure 6
plastics
Thickness-dependent transmissivity values for three semi-transparent
9
Absorptivity of three semi-transparent plastics
1
Absorptivity
0.8
0.6
0.4
0.2
PS
PE
PVC
0
10
20
30
40
Sheet Thickness, thousandths of inches
Figure 7
plastics
Thickness-dependent absorptivity values for three semi-transparent
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:
qinbound  Tr*4
(7)
Where  is the Stefan-Boltzmann constant. The amount of energy absorbed by the sheet
is given as:
qinbound   T l Tr*4
(8)
The total energy interchange is thus:

 

*4
qinterchange   T l  Tr*4  Tplastic
(9)
It is apparent that the amount of energy absorbed is linearly dependent on the thicknessdependent 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
qinterchangex2   T l  Tr*14  Tr*24  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c p dT  Qconvection  Qradiation
Vc p dT  hATair  Tplasticd  qint erchange Ad

 

*4
Vc p dT  hATair  Tplasticd   T l  Tr*4  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
  T l Tr*4  T plastic
hr 
Tr  Tplastic    Tr  Tplastic
qint erchange



(13)
Now the lumped-parameter equation is approximated by:
lc p
dTplastic
d
 hTair  Tplastic  hr Tradiation  Tplastic
(14)
Replacing some of the terms:
C
hT
air
 Tplastic
lc p

B
h  hr 
lc p
C hTair  hrTr 

h  hr 
B
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.)
dTplastic
d
  BT plastic  C  BC / B  Tplastic
dTplastic
C / B  T
plastic
(15)
  Bd
(16)
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  T plastic 
ln 
   B
 C / B  T0 
 C / B  T plastic 

  exp  B 
 C / B  T0 
(17)
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
lc p dT  hTair  Tplasticd   T l  Tr*4  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
 hTair  Ti 1  
 T l Trad
 Ti *41 
Ti  Ti 1  
    
 
lc p
 lc p



(20)
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c p
Plastic
PS
PVC
PE
Table 6
h
Values for
where l=0.030 inch and h = 1 Btu/ft2 h oF
lc p
o
cp (Btu/lb F)
 (lb/ft3)
lcp
h/lcp (h-1)
0.5
65.5
0.0819
12.2
0.5
84
0.105
9.52
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

 Trad  460   Tplastic  460  
 
 T l 
  
*4
 T l Trad
100
 Ti*41 
 100  

  (21)

  0.1714

lc p
lc p






As illustration, consider Trad = 700oF. The first parenthesis in the bracket isT(l) x
0.1714 11.6 4  3103 T(l). For the three plastics above:
Table 7
Terms in radiant bracket for l = 0.030 inch
Plastic
PS
PVC
PE
3103/ lcp
37890
29500
23000
T(l) T(l) 3103/ lcp
0.905
34290
0.822
24249
0.778
17894
0.1714 / lcp
2.093
1.634
1.273
Table 8
Comparison of Inbound Radiant and Sheet Re-radiant Energy Terms
[units = h-1] for l = 0.030 inch
Plastic
PS
PVC
PE
Radiant @ 700oF
34290
24249
17894
Sheet @ T= 80oF
1780
1390
1082
Sheet @ 380oF
10420
8135
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
Radiative Heat Transfer Coefficient, hr, Btu/ft2 h oF
[Compare with Convective Value, h = 1]
Plastic
PS
PVC
PE
T(l)
0.905
0.882
0.778
hr@Tplastic=80oF
4.29
4.18
3.66
hr@Tplastic=380oF
6.11
5.89
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
Sheet Temperature, oF
400
300
200
Polystyrene
PVC
PE
100
0
0
10
sheet heating, Aug2010.bas
20
30
40
50
60
70
80
Heating time, seconds
Figure 8. One-dimensional radiant heating of semi-transparent thin-gauge plastic
sheet. Sheet thickness =0.030 inch, Radiant heater temperature, Tr=700oF, Initial
sheet temperature, T0=80oF, Air temperature, Tair=80oF
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 T rad, 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
Sheet Temperature, oF
500
Polystyrene
PS
PVC
PVC
PE
PE
400
300
200
100
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
(Figure 8). Sheet thickness =0.030 inch, Radiant heater temperature, Tr=700oF,
Initial sheet temperature, T0=80oF, Air temperature, Tair=80oF
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
Plastic
PS
PVC
PE
Thermal
Conductivity,
Btu/ft h oF
0.105
0.083-0.100
0.183-0.292
Maximum
Thickness, l
inches, h
0.126
0.100-0.120
0.220-0.350
Maximum
Thickness, l
in, h+hr (80oF)
0.0218
0.0207-0.0248
0.0377-0.0600
Maximum
Thickness, l
in, h+hr (380oF)
0.0143
0.0119-0.0143
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
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
5. R. Siegel and J.R. Howell, Thermal Radiation Heat Transfer, 4/e, Taylor &
Francis, New York, 2002.
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/socrel/content/beerslaw.htm.
18