Chapter 13: Radiation Heat
Transfer
Yoav Peles
Department of Mechanical, Aerospace and Nuclear Engineering
Rensselaer Polytechnic Institute
Objectives
•
•
•
•
•
•
When you finish studying this chapter, you should be able to:
Define view factor, and understand its importance in radiation
heat transfer calculations,
Develop view factor relations, and calculate the unknown view
factors in an enclosure by using these relations,
Calculate radiation heat transfer between black surfaces,
Determine radiation heat transfer between diffuse and gray
surfaces in an enclosure using the concept of radiosity,
Obtain relations for net rate of radiation heat transfer between
the surfaces of a two-zone enclosure, including two large
parallel plates, two long concentric cylinders, and two
concentric spheres,
Quantify the effect of radiation shields on the reduction of
radiation heat transfer between two surfaces, and become
aware of the importance of radiation effect in temperature
measurements.
The View Factor
• Radiation heat transfer between surfaces depends on the
orientation of the surfaces relative to
each other as well as their radiation
properties and temperatures.
• View factor is defined to account for the
effects of orientation on radiation heat
transfer between two surfaces.
• View factor is a purely geometric
quantity and is independent of the surface properties and
temperature.
• Diffuse view factor ─ view factor based on the assumption that
the surfaces are diffuse emitters and diffuse reflectors.
• Specular view factor ─ view factor based on the assumption that
the surfaces are specular reflectors.
• Here we consider radiation exchange between diffuse surfaces
only, and thus the term view factor simply means diffuse view
factor.
• The view factor from a surface i to a surface j is
denoted by Fi→j or just Fij, and is defined as
• Fij=the fraction of the radiation leaving surface i
that strikes surface j directly.
• Consider two differential surfaces dA1 and dA2 on two
arbitrarily oriented surfaces A1 and A2, respectively.
• The rate at which radiation leaves
dA1 in the direction of q1 is:
I1cos q1dA1
• Noting that
dw21=dA2cos q2/r2,
• the portion of this radiation that strikes dA2 is
dA2 cos q 2
(13-1)
QdA dA  I1 cos q1dA1dw21  I1 cos q1dA1
2
1
2
r
• The total rate at which radiation leaves dA1 (via
emission and reflection) in all directions is the
radiosity (J1=pI1) times the surface area:
dQdA1  J1dA1  p I1dA1
(13-2)
• Then the differential view factor dFdA1→dA2 (the
fraction of radiation leaving dA1 that strikes
dA2)
QdA dA
cos q1 cos q 2
dFdA1 dA2 
1
QdA1
2

pr
2
dA2
(13-3)
• The view factor from a differential area dA1 to a
cos q1 cos q 2
finite area A2Fis:
dA2
(13-4)
dA  A  
2
1
2
A2
pr
• The total rate at which radiation leaves the
entire A1 in all directions is
QA1  J1 A1  p I1 A1
(13-5)
• considering the radiation that leaves dA1 and
strikes dA2, and integrating it over A1,
QA1 dA2   QdA1 dA2
A1
I1 cos q1 cos q 2 dA2

dA1
2
r
A1
(13-6)
• Integration of this relation over A2 gives the
radiation that strikes the entire A2,
QA1  A2   QA1 dA2
A2
I1 cos q1 cos q 2

dA1dA2
2
r
A2 A1
(13-7)
• Dividing this by the total radiation leaving A1 (from Eq. 13–
5) gives the fraction of radiation leaving A1 that strikes A2,
which is the view factor F12,
QA  A
cos q1 cos q 2
1
F12  FA  A 

dA1dA2 (13-8)
2


QA
A1 A A
pr
1
1
2
2
1
2
1
• The view factor F21 is readily determined from Eq. 13–8
by interchanging the subscripts 1 and 2,
QA  A
cos q1 cos q 2
1
F21  FA  A 

dA1dA2 (13-9)
2


QA
A2 A A
pr
2
2
1
1
2
2
1
• Combining Eqs. 13–8 and 13–9 after multiplying the former
by A1 and the latter by A2 gives the reciprocity relation
A1F12  A2 F21
(13-10)
• When j=i:
Fii=the fraction of radiation leaving
surface i that strikes itself directly.
– Fii=0: for plane or convex surfaces and
– Fii≠0: for concave surfaces
• The value of the view factor ranges
between zero and one.
– Fij=0 ─ the two surfaces do not have a
direct view of each other,
– Fij=1─ surface j completely surrounds
surface.
View Factors Tables for Selected
Geometries (analytical form)
View Factors Figures for Selected
Geometries (graphical form)
View Factor Relations
• Radiation analysis on an enclosure consisting
of N surfaces requires the evaluation of N2
view factors.
• Fundamental relations for view factors:
–
–
–
–
the reciprocity relation,
the summation rule,
the superposition rule,
the symmetry rule.
The Reciprocity Relation
• We have shown earlier that the pair of view
factors Fij and Fji are related to each other
by
Ai Fi  j  Aj Fj i
(13-11)
• This relation is referred to as the reciprocity
relation or the reciprocity rule.
• Note that:
Fj i  Fi  j
when
Ai  Aj
Fj i  Fi  j
when
Ai  Aj
The Summation Rule
• The conservation of energy
principle requires that the entire
radiation leaving any surface i of
an enclosure be intercepted by the
surfaces of the enclosure.
• Summation rule ─ the sum of the
view factors from surface i of an
enclosure to all surfaces of the
enclosure, including to itself, must
equal unity. N
F
j 1
i j
1
(13-12)
• The summation rule can be applied to each surface of
an enclosure by varying i from 1 to N.
• The summation rule applied to each of the N surfaces
of an enclosure gives N relations for the determination
of the view factors.
• The reciprocity rule gives 1/2N(N-1) additional
relations.
• The total number of view factors that need to be
evaluated directly for an N-surface enclosure becomes
1

 1
N   N  N  N  1   N  N  1
2

 2
2
The Superposition Rule
• Sometimes the view factor associated with a given
geometry is not available in standard tables and charts.
• Superposition rule ─ the view factor from a surface i
to a surface j is equal to the sum of the view factors
from surface i to the parts of surface j.
• Consider the geometry shown in the figure below.
• The view factor from surface 1 to the combined
surfaces of 2 and 3 is
F1 2,3  F12  F13
(13-13)
• From the chart in Table 13–2:
– F12 and F1(2,3)
and then from Eq. 13-13:
– F13
The Symmetry Rule
• Symmetry rule ─ two (or
more) surfaces that possess
symmetry about a third
surface will have identical
view factors from that surface.
• If the surfaces j and k are symmetric about the surface i
then
Fi  j  Fi k
• Using the reciprocity rule, it can be shown that
Fj i  Fk i
View Factors between Infinitely Long
Surfaces: The Crossed-Strings Method
• The view factor between two-dimensional surfaces
can be determined by the simple crossed-strings
method developed by H. C. Hottel in the 1950s.
• Consider the geometry shown in the figure.
• Hottel has shown that the view factor
F1 2 can be expressed in terms of
the lengths of the stretched strings as
F12
L5  L6    L3  L4 


2 L1
(13-16)
Radiation Heat Transfer: Black Surfaces
• Consider two black surfaces of arbitrary shape
maintained at uniform temperatures T1 and T2.
• The net rate of radiation heat transfer
from surface 1 to surface 2 can be expressed as
Q12 =
Radiation leaving
the entire surface 1
that strikes surface 2
-
Radiation leaving
the entire surface 2
that strikes surface 1
 A1Eb1F12  A2 Eb 2 F21
(W)
(13-18)
• Applying the reciprocity relation A1F12=A2F21 yields


Q12  A1 F12 T14  T24
(W)
• For enclosure consisting of N black surfaces
N
N

Qi   Qi j   Ai Fi  j Ti 4  T j4
j 1
j 1

(13-19)
(W) (13-20)
Radiation Heat Transfer: Diffuse, Gray
Surfaces
• To make a simple radiation analysis possible, it
is common to assume the surfaces of an
enclosure are:
–
–
–
–
–
opaque (nontransparent),
diffuse (diffuse emitters and diffuse reflectors),
gray (independent of wavelength),
isothermal, and
both the incoming and outgoing radiation are
uniform over each surface.
• For a surface i that is gray and
opaque (ei=ai and ai+ri=1), the
radiosity can be expressed as
J i  e i Ebi  ri Gi
(13-21)
 e i Ebi  1  e i  Gi
(W/m2 )
where:
Ebi   Ti 4
• For a surface that can be approximated as a blackbody
(ei=1), the radiosity relation reduces to:
J i  Ebi   Ti 4
(blackbody)
(13-22)
Net Radiation Heat Transfer to or from
a Surface
• The net rate of radiation heat transfer from a
surface i of surface area Ai is expressed as
Qi =
Radiation leaving
entire surface i
 Ai  Ji  Gi 
-
Radiation incident
on entire surface i
(W)
(13-23)
• Solving for Gi from Eq. 13–21 and substituting
into Eq. 13–23 yields

J i  e i Ebi
Qi  Ai  J i 
1- e i

 Aie i
 Ebi  J i 

 1- e i
(W) (13-24)
• In an electrical analogy to Ohm’s law, Eq.13-24 can
be rearranged as
Ebi  J i
Qi 
Ri
(W)
(13-25)
where surface resistance to
radiation is
1- e i
(13-26)
Ri 
Aie i
• For a blackbody Ri=0 and the net rate of radiation heat
transfer in this case is determined directly from Eq.
13–23.
• Reradiating surface ─ an adiabatic surface:
– when convection effects is negligible,
– under steady-state conditions.
• Reradiating surface must lose as much radiation
energy as it gains, thus:
Qi  0
Eq. 13-25
J i  Ebi   Ti 4
(W/m 2 ) (13-27)
• The temperature of a reradiating surface is
independent of its emissivity.
Net Radiation Heat Transfer
between Any Two Surfaces
• Consider two diffuse, gray, and opaque
surfaces of arbitrary shape maintained at
uniform temperatures.
• The net rate of radiation heat transfer from
surface i to surface j can be expressed as
Qi j =
Radiation leaving
the entire surface i
that strikes surface j
-
 Ai J i Fi  j  Aj J j Fj i
Radiation leaving
the entire surface j
that strikes surface i
(W)
(13-28)
• Applying the reciprocity relation AiFij= AjFji
yields
Qi  j  Ai Fi  j  J i  J j 
(W)
(13-29)
• In analogy to Ohm’s law
Qi  j 
Ji  J j
Ri  j
(W)
(13-30)
where space resistance to radiation is
Ri j
1

Ai Fi j
(13-31)
• In an N-surface enclosure, the conservation of energy
principle requires
Qi   Qi  j   Ai Fi  j  J i  J j = 
N
j 1
N
N
j 1
j 1
Ji  J j
Ri  j
• Combining Eqs. 13–25 and 13–32 gives
N J J
Ebi  J i
i
j

Ri
j 1 Ri  j
(W)
(13-33)
(W) (13-32)
Methods of Solving Radiation
Problems
• In the radiation analysis of an enclosure, either
– the temperature or
– the net rate of heat transfer
must be given for each of the surfaces.
• Two methods commonly used:
– surfaces with specified net heat transfer rate (Eq. 13-32)
Qi  Ai  Fi  j  J i  J j 
N
(13-34)
j 1
– surfaces with specified temperature (Eq. 13-33)
 Ti 4  J i 
1 ei
ei
F J
N
j 1
i j
i
Jj
(13-35)
• The equations above give N linear algebraic equations
for the determination of the N unknown radiosities for
an N-surface enclosure.
• Once the radiosities J1, J2, . . . , JN are available, the
unknown heat transfer rates can be determined from
Eq. 13–34.
• The unknown surface temperatures can be determined
from Eq. 13–35.
• Two methods to solve the system of N equations:
– direct method
• very suitable for use with today’s popular equation solvers
– network method
• not practical for enclosures with more than three or four surfaces
• simple and emphasis on the physics of the problem.
Radiation Heat Transfer in TwoSurface Enclosures
• Consider an enclosure consisting of two opaque surfaces at
specified temperatures.
• Need to determine the net rate of
radiation heat transfer.
• Known: T1, T2, e1, e2, A1, A2, F12.
• Surface resistances:
– two surface resistances,
– one space resistance.
• The net rate of radiation transfer is
expressed as
Q12 
Q12 
1  e1 
 T14  T24 
Eb1  Eb 2
 Q1  Q2
R1  R12  R2
A1e1  1 A1F12  1  e 2  A2e 2
(13-36)
Simplified forms of Eq. 13–36 for
some familiar arrangements
Radiation Heat Transfer
in Three-Surface Enclosures
• Consider an enclosure consisting of three opaque,
diffuse, and gray surfaces.
• Known: T1, T2, T3, e1, e2, e3, A1, A2, A3, F12.
• Since the temperatures are known Eb1, Eb2, and Eb3 are
considered known.
• Need to determine
the radiosities
J1, J2, and J3.
• The three equations for the determination of these
three unknowns are obtained from the requirement
that the algebraic sum of the currents (net radiation
heat transfer) at each node must equal zero.
Eb1  J1 J 2  J1 J 3  J1


0
R1
R12
R13
J1  J 2 Eb 2  J 2 J 3  J 2


0
R12
R2
R23
(13-41)
J1  J 3 J 2  J 3 Eb 3  J 3


0
R13
R23
R3
• Once the radiosities J1, J2, and J3 are available, the net
rate of radiation heat transfers at each surface can be
determined from Eq. 13–32.
Radiation Shields and the Radiation
Effects
• Radiation heat transfer between two surfaces
can be reduced greatly by inserting a thin, highreflectivity (low-emissivity) sheet of material
(radiation shields) between the two surfaces.
• Radiation heat transfer between two large
parallel plates of emissivities e1 and e2
maintained at uniform temperatures T1 and T2 is
given by Eq. 13–38:
A T14  T24 
Q12,no shield 
1
e1

1
e2
1
• Consider a radiation shield placed between these two
plates.
Radiation
network:
• The rate of radiation heat transfer is
Q12,one shield
(13-42)
Eb1  Eb 2

1  e 3,1 1  e 3,2
1  e1
1 e2
1
1





A1e1 A1 F13 A3e 3,1 A3e 3,2 A3 F32 A2e 2
• Noting that F13=F23=1 and A1=A2=A3=A for
infinite parallel plates, Eq. 13–42 simplifies to
Q12,one shield 

A T14  T24


1 1
  1
1

 1
   1  

 e1 e 2   e 3,1 e 3,2 
No shield resistance
(13-43)
shield resistance
• The radiation heat transfer through large
parallel plates separated by N radiation shields
Q12,N shield 

A T14  T24

1 1
  1
1

 1 
   1  

e
e
e
e
2
3,2
 1
  3,1

No shield resistance
shield 1 resistance

(13-44)
 1

1


 1
e

e
N ,2
 N ,1

shield N resistance
Radiation Exchange with Emitting and
Absorbing Gases
• Nonparticipating medium ─ medium that is
completely transparent to thermal radiation (no
emission, absorption, or scattering).
• Examples of nonparticipating medium:
– air at ordinary temperatures and pressures,
– gases that consist of monatomic molecules (e.g., Ar and He).
– gases that consist of symmetric diatomic molecules (e.g., N2
and O2).
• Examples of participating medium
– asymmetric molecules such as H2O, CO2, CO, SO2, and
hydrocarbons HmCn
• by absorption at moderate temperatures, and
• by absorption and emission at high temperatures.
The presence of a participating medium complicates the
radiation analysis considerably:
• A participating medium is a volumetric phenomena,
• Gases emit and absorb radiation at a number of narrow
wavelength bands, and the gray assumption may not
always be appropriate,
• The emission and absorption characteristics of the
constituents of a gas mixture also depends on the
temperature, pressure, and composition of the gas
mixture,
• Scattering ─ the change of direction of radiation due to
reflection, refraction, and diffraction.
• We limit our consideration to gases that emit and
absorb radiation.
Radiation Properties of a Participating
Medium
• Consider a participating medium of thickness L.
• A spectral radiation beam of intensity Il,0 is incident
on the medium, which is attenuated as it propagates
due to absorption.
• The decrease in the intensity of
radiation is proportional to the
– intensity Il,
– thickness dx.
• Beer’s law
dI l  x   kl I l  x  dx (13-47)
• kl is the spectral absorption coefficient (units m-1).
• Separating the variables and integrating from
x=0 to x=L gives (assuming kl =constant)
Il ,L
I l ,0
 e  kl L
(13-48)
• The spectral transmissivity
tl 
Il ,L
I l ,0
 e  kl L
(13-49)
• Radiation passing through a nonscattering (and
thus nonreflecting) medium is either absorbed
or transmitted  al+tl=1
• spectral absorptivity of a medium of
thickness L
a l  1  t l  1  e  kl L
(13-50)
• From Kirchoff’s law, the spectral emissivity of
the medium is
e l  a l  1  e  kl L
(13-51)
• The spectral absorption coefficient of a medium
(and thus el, al, tl), in general, vary with:
–
–
–
–
wavelength,
temperature,
pressure, and
composition.
• Optically thick medium ─ a medium with a
large value of kl L.
• For optically thick medium el≈al≈1.
Emissivity and Absorptivity
of Gases and Gas Mixtures
• The band nature of absorption of most gases are strongly
nongray.
• The nongray nature of properties should be considered in
radiation calculations for high accuracy.
• Satisfactory results can be
obtained by assuming the
gas to be gray, and using an
effective total absorptivity
and emissivity determined
by some averaging process.
• An approach which assumes
the gas to be gray, was
Spectral absorptivity of CO2 at 830 K
developed by Hottel (1954) and 10 atm for a path length of 38.8 cm.
and is presented below.
• The emissivity and absorptivity of
a gas component in a mixture
depends primarily on its density.
• Emissivity at a total pressure P
other than P=1 atm is determined
by multiplying the emissivity
value at 1 atm by a pressure
correction factor Cw obtained
from Figure 13–37a for water
vapor
e w  Cwe w, 1 atm
Emissivity of H2O in a mixture of
nonparticipating gases at a total pressure of 1
atm for a mean beam length of L.
(13-52)
Fig. 13-37 Correction factors for the emissivities
of H2O at pressures other than 1 atm
• When CO2 and H2O gases exist together in a mixture
with nonparticipating gases
e g  e c  e w  De
 Cce c , 1 atm  Cwe w, 1 atm  De
•
(13-53)
De is the emissivity correction factor.
Emissivity correction De for use in eg=ew+ec-De when both
CO2 and H2O vapor are present in a gas mixture.
• Hottel and his coworkers considered the emission of
radiation from a hemispherical gas body to a small
surface element located at the center of the base of the
hemisphere.
• It is certainly desirable to extend the reported
emissivity data to gas bodies of other geometries.
• Mean beam length L ─ represents the radius of an
equivalent hemisphere.
• The mean beam lengths
for various gas
geometries are listed in
Table 13–4.
• The absorptivity of a gas that contains CO2 and
H2O gases for radiation emitted by a source at
temperature Ts can be determined from
a g  a c  a w  Da

Da  De
(13-54)
• The absorptivities of CO2 and H2O can be
determined from the emissivity charts (Figs.
12–36 and 12–37) as
0.65
(13-55)
• CO2: a c  Cc  Tg Ts   e c Ts , Pc LTs / Tg 
• H2O: a w  Cw  Tg
Ts 
0.45
 e w Ts , Pw LTs / Tg 
(13-56)
• The rate of radiation energy emitted by a gas to a
bounding surface of area As
Qg ,e  e g As Tg4
(13-57)
• The net rate of radiation heat transfer between the gas
and a black surface surrounding it becomes

Qnet  e g As Tg4  a g As Ts4  As e g Tg4  a g Ts4
energy emitted
by the gas to
the surface

(13-58)
energy emitted
by the surface
to the gas
• For surfaces that are nearly black with an emissivity
es>0.7 (Hottel)
Qnet , gray 
es 1
2
Qnet ,black 
es 1
2

As e g Tg4  a g Ts4

(13-59)