Thermal Radiation - III- Radn. energy exchange between gray surfaces

Lectures on Heat Transfer –
THERMAL RADIATION-III:
Radiation energy exchange
between gray surfaces
by
Dr. M. ThirumaleshwarDr. M. Thirumaleshwar
formerly:
Professor, Dept. of Mechanical Engineering,
St. Joseph Engg. College, Vamanjoor,
Mangalore,
India

Preface:
• This file contains slides on THERMAL
RADIATION-III: Radiation energy exchange
between gray surfaces.
• The slides were prepared while teaching
Heat Transfer course to the M.Tech.Heat Transfer course to the M.Tech.
students in Mechanical Engineering Dept.
of St. Joseph Engineering College,
Vamanjoor, Mangalore, India, during Sept.
– Dec. 2010.
Aug. 2016 2MT/SJEC/M.Tech.

• It is hoped that these Slides will be useful
to teachers, students, researchers and
professionals working in this field.
• For students, it should be particularly
useful to study, quickly review the subject,useful to study, quickly review the subject,
and to prepare for the examinations.
•

References:
• 1. M. Thirumaleshwar: Fundamentals of Heat &
Mass Transfer, Pearson Edu., 2006
• https://books.google.co.in/books?id=b2238B-
AsqcC&printsec=frontcover&source=gbs_atb#v=onepage&q&f=false
• 2. Cengel Y. A. Heat Transfer: A Practical
Approach, 2nd Ed. McGraw Hill Co., 2003
Aug. 2016 MT/SJEC/M.Tech. 4
Approach, 2nd Ed. McGraw Hill Co., 2003
• 3. Cengel, Y. A. and Ghajar, A. J., Heat and
Mass Transfer - Fundamentals and Applications,
5th Ed., McGraw-Hill, New York, NY, 2014.

References… contd.
• 4. Incropera , Dewitt, Bergman, Lavine:
Fundamentals of Heat and Mass Transfer, 6th
Ed., Wiley Intl.
• 5. M. Thirumaleshwar: Software Solutions to• 5. M. Thirumaleshwar: Software Solutions to
Problems on Heat Transfer – Radiation Heat
Transfer-Part-II, Bookboon, 2013
• http://bookboon.com/en/software-solutions-heat-transfer-
radiation-ii-ebook

Thermal Radiation – III:
Radiation energy exchange between gray
surfaces :
Outline…
Radiation heat exchange between gray
surfaces - electrical network method – two
surfaces - electrical network method – two
zone enclosures – Problems - three zone
enclosures – Problems - radiation
shielding – Problems - radiation error in
temperature measurement – Problems

Radiation energy exchange
between gray surfaces:
• Following assumptions are made to simplify the
solution:
• All the surfaces of the enclosure are opaque (τ = 0),
diffuse and gray
• Radiative properties such as ρ, ε and α are uniform
and independent of direction and frequency
• Radiative properties such as ρ, ε and α are uniform
and independent of direction and frequency
• Irradiation and heat flux leaving each surface are
uniform over the surface
• Each surface of the enclosure is isothermal, and
• The enclosure is filled with a non-participating medium
(such as vacuum or air)

The electrical network method:
• We shall discuss the ‘electrical network
method’, since it is simple to apply and
gives a physical ‘feel’ of the problem.
• This method, introduced by Oppenheim in
the 1950’s, is simple and direct; it
emphasizes on the physics of the problem,
emphasizes on the physics of the problem,
and is easy to apply.
• Before we introduce this method, let us
define two new quantities, viz. irradiation
and radiosity: (See Fig. 13.25)

• Irradiation, (G): is the total radiation incident upon a
surface per unit time, per unit area (W/m2).
• Radiosity, (J): is the total radiation leaving a surface,
with no regard for its origin (i.e., reflected plus emitted
from the surface) per unit time, per unit area (W/m2).

• Now, from Fig.(13.25), it is clear that total
radiation leaving the surface (i.e.
Radiosity, J) is:
• For a gray, opaque (τ = 0) surface, we
have:
J ρ G. ε E b
.
have:
• Therefore,
ρ 1 α( ) 1 ε( ) ...from Kirchoff's law
J 1 ε( ) G. ε E b
.
or, G
J ε E b
.
1 ε( )

• Now, net rate of radiation energy transfer
from the surface is given by: (rate of
radiation energy leaving the surface minus
the rate of radiation energy incident on the
surface), i.e. Q
A
J G
Q J ε E b
.
• Therefore,
i.e.
Q
A
J
J ε E b
.
1 ε
Q
ε A.
1 ε
E b J.
i.e. Q
E b J
1 ε( )
A ε.
W.....(13.48)

• By analogy with Ohm’s law, we can think of Q in
eqn.(13.48) as a current flowing through a
potential difference (Eb-J), and the factor (1-
ε)/A.ε as the resistance.
• Now, this resistance is the resistance to the flow
of radiant heat due to the nature of the surface
and is known as ‘surface resistance (R)’.
and is known as ‘surface resistance (R)’.
• i.e.
• Surface resistance for a surface ‘i’ is shown
schematically in Fig. (13.26,a).
R
1 ε( )
A ε.
....surface resistance

• For a black body emissivity ε = 1;
so, the surface resistance is zero,
and
• Also, many surfaces in numerous
applications are adiabatic, i.e. well
insulated, and net heat transfer
J i E bi ...for a black body....(13.49)
insulated, and net heat transfer
through such a surface is zero,
since in steady state, all the heat
incident on such a surface is re-
radiated. These are known as
re-radiating surfaces.

• Walls of a furnace is the familiar example
of a re-radiating surface. Obviously, for a
re-radiating surface, Qi = 0, and from eqn.
(13.48) we get:
• Note that the temperature of a re-radiating
J i E bi σ T i
4. ....for a re-radiating surface....(13.50)
• Note that the temperature of a re-radiating
surface can be calculated from the above
eqn; further, note that this temperature is
independent of the emissivity of the
surface.

• Again, consider two diffuse, gray and
opaque surfaces i and j, maintained at
uniform temperatures Ti and Tj,
exchanging heat with each other.
• Then, remembering the definitions of
radiosity and view factor, we can write for
the radiation leaving surface ‘i’ that strikes
the radiation leaving surface ‘i’ that strikes
surface ‘j’:
• Similarly, for surface j, we have:
Q i A i F ij
. J i
.
Q j A j F ji
. J j
.

• Therefore, net heat interchange between surfaces i and j
is:
Q ij A i F ij
. J i
. A j F ji
. J j
.
i.e. Q ij A i F ij
. J i J j
. W..(13.51)..since A i F ij
. A j F ji
. ...by reciprocity.
i.e. Q ij
J i J j
1
W....(13.52)
• Again, by analogy with Ohm’s law, we can write eqn.
(13.52) as:
1
A i F ij
.
Q ij
J i J j
R ij
W
where, R ij
1
A i F ij
.
...(13.53)

• Rij is known as ‘space
resistance’ and it represents
the resistance to radiative heat
flow between the radiosity
potentials of the two surfaces,
due to their relative orientation
and spacing.
• Space resistance is illustrated
• Space resistance is illustrated
in Fig. (13.26,b).
• Note from eqn. (13.52) that if
Ji > Jj, net heat transfer is from
surface i to surface j;
otherwise, the net heat transfer
is from surface j to surface i.

• Thus, for each diffuse, gray, opaque
surface, in radiant heat exchange with
other surfaces of an enclosure, there are
two resistances, viz. the surface
resistance, Ri = (1- εi)/(Ai.εi), and a space
resistance, Rij = 1/(Ai.Fij).
• For a N surface enclosure, net heat
transfer from surface ‘i’ should be equal to
the sum of net heat transfers from that
surface to the remaining surfaces. i.e.

• This situation is shown in Fig. (13.27).
Q i
1
N
j
Q ij
= 1
N
j
A i F ij
. J i J j
.
= 1
N
j
J i J j
R ij=
...W....(13.54)
i.e.
E bi J i
R i 1
N
j
J i J j
R ij=
W...(13.55)
J
Fig. 13.27 Radiation heat transfer from surface i
to other surfaces in a N-surface enclosure
Surface i
JiEbi
Qi
Ri
J1
J2
JN-1
JN
Ri1
Ri2
Ri(N-1)
RiN

• As can be seen from the above fig., rate of
radiation ‘current’ flow to surface ‘i’ through its
surface resistance must be equal to the sum of
all the radiation current flows from surface ‘i’ to
all other surfaces through the respective space
resistances.
• In general, there two types of radiation
problems:
problems:
• first (and most common), when the surface
temperature Ti, and therefore, the emissive
power Ebi is known;
• and, the second type is when the net radiation
heat transfer at the surface i is known.

• Eqn. (13.55) is useful in solving the first
type of problems, i.e. when the surface
temperature is known; instead, if the net
heat transfer rate at the surface is the
known quantity, eqn. (13.52) is the
applicable equation.
• Essentially, the problem is to solve for the
radiosities J1, J2,….Jn.
• Electrical network method is convenient to
use if the number of surfaces in an
enclosure is limited to about five.

• If the number of surfaces is more than five, the
direct approach is to apply eqn. (13.55) for each
surface whose temperature is known, and eqn.
(13.52) for each surface at which the net heat
transfer rate is known, and solve the resulting
set of N linear, algebraic equations for the N
unknowns, viz. J1, J2,…Jn by standard
mathematical methods.
• Once the radiosities are known, eqn. (13.48)
may be applied to determine either the heat
transfer rate or the temperature, as the case
may be.

Radiation heat exchange in two-
zone enclosures:
• Two-zone enclosure- simply means that the two
surfaces, together, make up the enclosure and
‘see’ only themselves and nothing else.
• Many, practically important geometries may be
classified as two-zone enclosures, eg. a small
body enclosed by a large body, a pipe passing
classified as two-zone enclosures, eg. a small
body enclosed by a large body, a pipe passing
through a large room, concentric spheres,
concentric, long cylinders , long, parallel plane
surfaces, etc.
• Fig. (13.28) shows a schematic of a typical two-
zone enclosure and the associated radiation (or,
thermal) network.

• Surfaces 1 and 2 forming the
enclosure are diffuse, gray and
opaque.
• Let their emissivities,
temperatures and areas be (ε1,
T1, A1) and (ε2, T2, A2)
respectively.
• The radiation network is shown
in Fig. 13.28.
in Fig. 13.28.
• Each surface has one surface
resistance associated with it and
there is one space resistance
between the two radiosity
potentials, and all the
resistances are in series, as
shown.

• The ‘heat current’ (Q12) in this circuit is calculated by
dividing the ‘total potential’ (Eb1 – Eb2) by the ‘total
resistance’ (R1 + R12 +R2). So, we write:
Q 12 Q 1 Q 2
E b1 E b2
R 1 R 12 R 2
i.e. Q 12
E b1 E b2
1 ε 1 1 1 ε 2
A 1 ε 1
. A 1 F 12
. A 2 ε 2
.
i.e. Q 12
σ T 1
4
T 2
4.
1 ε 1
A 1 ε 1
.
1
A 1 F 12
.
1 ε 2
A 2 ε 2
.
W......(13.56)
Eqn. (13.56) is an important equation, which gives net rate of heat
transfer between two gray, diffuse, opaque surfaces which form an
enclosure, i.e. which ‘see’ only each other and nothing else.

• Let us consider a few special cases of
two-surface enclosure:
• Case (i): Radiant heat exchange
between two black surfaces:
• For a black body, ε = 1, and J = Eb, as
explained earlier. i.e. surface resistance
[= (1 - ε)/(A.ε)] of a black body is zero.
[= (1 - ε)/(A.ε)] of a black body is zero.
Then, the radiation network will consist of
only a space resistance between the two
radiosity potentials, as shown in Fig.
(13.29):

• Then, from eqn. (13.56), we get:
Fig. 13.29 Radiation network for two black
surfaces forming an enclosure
Eb1= J1
R12
= 1/(A1
.F12
)
Q12
Eb2= J2
Q 12
σ T 1
4
T 2
4.
1
A 1 F 12
.
i.e. Q 12 A 1 F 12
. σ. T 1
4
T 2
4. W....for two black surfaces forming an
enclosure....(13.57)

• Next, we shall consider four cases of practical interest
where the view factor between the inner surface 1 and
the outer surface 2 (i.e. F12) is equal to 1:
• Case (ii): Radiant heat exchange for a
small object in a large cavity:
• See Fig. (13.30,a). A practical example of a small object
in a large cavity is the case of a steam pipe passing
through a large plant room.
through a large plant room.
• For this case, we have:
A 1
A 2
0
and, F 12 1

• And, eqn.(13.56) becomes:
• Case (iii): Radiant heat exchange
between infinitely large parallel plates:
• See Fig. (13.30,b). In this case, A1 = A2 =
A, say, and F12 = 1.
Q 12 A 1 σ. ε 1
. T 1
4
T 2
4. ....for small object in a large cavity.....(13.58)
A, say, and F12 = 1.
• Then, eqn. (13.56) becomes:
Q 12
A σ. T 1
4
T 2
4.
1
ε 1
1
ε 2
1
....for infinitely large parallel plates.....(13.59)

Fig. 13.30. Important ‘two surface enclosures’.

• Case (iv): Radiant heat exchange between infinitely
long concentric cylinders:
• See Fig. (13.30,c). In this case:
F 12 1
Q 12
A 1 σ. T 1
4
T 2
4.
1
ε 1
A 1
A 2
1
ε 2
1.
....for infinitely long concentric cylinders.....(13.60)
• Remember that A1 refers to the inner (or enclosed) surface.
• Eqn. (13.60) is known as ‘Christiansen’s equation’.
where
A 1
A 2
r 1
r 2

• Case (v): Radiant heat exchange between
concentric spheres:
• See Fig. (13.30,d). In this case:
F 12 1
Q 12
A 1 σ. T 1
4
T 2
4.
1
ε 1
A 1
A 2
1
ε 2
1.
....for concentric spheres.....(13.61)
• Remember, again, that A1 refers to the inner (or
enclosed) surface.
1 2 2
where
A 1
A 2
r 1
r 2
2

Problems on two-surface enclosures:
• Example 13.19 (M.U. 1991): A long pipe, 50
mm in diameter, passes through a room and is
exposed to air at 20 deg. C. Pipe surface
temperature is 93 deg.C. Emissivity of the
surface is 0.6. Calculate the net radiant heat losssurface is 0.6. Calculate the net radiant heat loss
per metre length of pipe.
• Solution:
• The pipe is enclosed by the room; so, it is two-surface enclosure
problem. Further, area of the pipe is very small, compared to the
area of the room. Therefore, this is a case of a small object
surrounded by a large area, and we have:

Aug. 2016 36
This is a two-zone enclosure problem.
Fig. shows the radiation network for
this problem.
MT/SJEC/M.Tech.

We have:

Radiation heat exchange in three
zone enclosures:
• Fig. (13.32,a) shows an enclosure made of three
opaque, diffuse, gray surfaces.
• Let the surfaces A1, A2, A3 be maintained at uniform
temperatures of T1, T2 and T3 respectively. Also, let the
emissivities be ε1, ε2 and ε3 respectively.
• The radiation network for this system of three surface
• The radiation network for this system of three surface
enclosure is shown in Fig. (13.32,b).
• While drawing the radiation network, the principle to be
followed is quite simple: first, draw the surface resistance
associated with each gray surface; then, connect the
radiosity potentials between surfaces by the respective
space resistances.

• It is considered that the temperature of
each surface is known; i.e. emissive power
Eb for each surface is known.
• Then, the problem reduces to determining
the radiosities J1, J2 and J3.
• This is done by applying Kirchoff’s law of
• This is done by applying Kirchoff’s law of
d.c. circuits to each node: i.e. sum of the
currents (or, rate of heat transfers)
entering into each node is zero.
• Doing this, we get the following three
algebraic equations:

Node J1:
E b1 J 1
R 1
J 2 J 1
R 12
J 3 J 1
R 13
0 .....(13.63,a)
Node J2:
E b2 J 2
R 2
J 1 J 2
R 12
J 3 J 2
R 23
0 .....(13.63,b)
Node J :
E b3 J 3 J 1 J 3 J 2 J 3
0 .....(13.63,c)
• Solving these three eqns. simultaneously, we get J1, J2
and J3.
• Remember to write each eqn. such that current flows
into the node; then, the magnitudes of the radiosities
would adjust themselves when all the three equations
are solved simultaneously.
Node J3:
b3 3
R 3
1 3
R 13
2 3
R 23
0 .....(13.63,c)

• Once the magnitudes of the radiosities are
known, expressions for net heat flows between
the surfaces are:
Q 12
J 1 J 2
R 12
J 1 J 2
1
A 1 F 12
.
....(13.64,a)
Q 13
J 1 J 3
R 13
J 1 J 3
1
A 1 F 13
.
....(13.64,b)
Q 23
J 2 J 3
R 23
J 1 J 2
1
A 2 F 23
.
....(13.64,c)

• And, net heat flow from each surface is:
Q 1
E b1 J 1
R 1
E b1 J 1
1 ε 1
A 1 ε 1
.
.....(13.65,a)
Q
E b2 J 2 E b2 J 2
.....(13.65,b)
Q 2
b2 2
R 2
b2 2
1 ε 2
A 2 ε 2
.
.....(13.65,b)
Q 3
E b3 J 3
R 3
E b3 J 3
1 ε 3
A 3 ε 3
.
.....(13.65,c)

• Eqn. set (13.64) is a set of general
equations for three diffuse, opaque, gray
surfaces.
• However, these equations will be modified
depending upon any constraint that may
be attached to any of the surfaces, i.e.
• say, if the surface is black or re-radiating:
• say, if the surface is black or re-radiating:
Ji = Ebi = σ.Ti
4.
• And, Qi = 0 for a re-radiating surface.
• If Qi at any surface is specified instead of
the temperature (i,.e. Ebi), then,
(Ebi – Ji)/Ri is replaced by Qi.

Few special cases of three-zone
enclosures:
• Case (i): Two black surfaces connected to a
third refractory surface:
• This is a three-zone enclosure, with two of the
surfaces being black and the third surface being
a re-radiating, insulated surface.
a re-radiating, insulated surface.
• Typical example is a furnace whose bottom is
the ‘source’ and the top is the ‘sink’ and the two
surfaces are connected by a refractory wall
which acts as a re-radiating surface.
• In effect, the source and sink exchange heat
through the re-radiating wall.

• However, in steady state, the re-radiating
wall radiates as much heat as it receives,
which means that net heat exchange
through the re-radiating wall (= Q) is zero,
i.e. Eb = J for the re-radiating wall.
• Therefore, once J (i.e. Eb) is calculated for
the re-radiating surface, its steady state
b
the re-radiating surface, its steady state
temperature can easily be calculated from:
Eb = σ. T4.

Eb1
= J1
E = J R
Eb2
= J2
R12
R12=1/(A1.F12 )
EbR= JR
R2RR1R
R1R=1/(A1.F1R )
R2R=1/(A2.F2R )
Fig.(a)
Fig. 13.33 Two black surfaces connected by a third re-radiating
surface and its radiation network
Eb1 = J1 Eb2 = J2
R12
R2RR1R JR
Fig.(b)

• The radiation network is drawn very easily by
remembering the usual principles, i.e.
• for a black surface, the surface resistance is
zero, i.e. Eb = J.
• For a re-radiating surface too, Eb = J, as already
explained;
• Further, for a re-radiating surface, Q = 0.
• Between two given surfaces, the radiosity
• Between two given surfaces, the radiosity
potentials are connected by the respective
space resistances, as shown.
• It may be observed that the system reduces to a
series-parallel circuit of resistances as shown in
Fig. (13.33,b).

• So, we write, for the total resistance of the circuit, Rtot:
1
R tot
1
R 12
1
R 1R R 2R
and, Q 12
E b1 E b2
R tot
E b1 E b2
1
R 12
1
R 1R R 2R
.
• Here, Q12 is the net radiant heat transferred between surfaces 1 and
2. Similar expressions can be written for heat transfer between
surfaces 2 and 3 (= Q23) and the heat transfer between surfaces 1
and 3 (= Q13).
i.e. Q 12 σ T 1
4
T 2
4. A 1 F 12
. 1
1
A 1 F 1R
.
1
A 2 F 2R
.
. .....(13.66)

Case (ii): Two gray surfaces surrounded
by a third re-radiating surface:

• In this case, there are two gray surfaces, and the
third surface is an insulated, re-radiating
surface.
• As already explained, the re-radiating surface
radiates as much energy as it receives;
therefore, net radiant heat transfer for that
surface is zero, i.e. Q 3 0
surface is zero, i.e. Q 3 0
i.e.
E b3 J 3
1 ε 3
A 3 ε 3
.
0
i.e. E b3 J 3

• i.e. once the radiosity of the re-radiating surface
is known, its temperature can easily be
calculated, since Eb3 = σ.T3
4.
• Further, note that T3 is independent of the
emissivity of surface 3.
• Now, the radiation network reduces to a simple
series-parallel circuit of the relevant resistances.
series-parallel circuit of the relevant resistances.
• Expression for heat flow rate is:
Q 1 Q 2
E b1 E b2
R tot
where, Rtot is the total resistance, given by:

R tot R 1
1
1
R 12
1
R 13 R 23
R 2
i.e. R tot
1 ε 1
A 1 ε 1
.
1
A 1 F 12
. 1
1 1
1 ε 2
A 2 ε 2
.
....(13.67)
1
A 1 F 13
.
1
A 2 F 23
.

Radiation shielding:
• One or more ‘radiation shields’ are used to
reduce radiant heat transfer between two given
surfaces.
• Radiation shield is, simply a thin, high reflectivity
• Radiation shield is, simply a thin, high reflectivity
surface placed in between the surfaces which
exchange heat between themselves.
• Radiation shields may be made of aluminium
foils, copper foils, or aluminized mylar sheets
etc.

• Fig. (13.36,a) shows to large parallel plates, 1
and 2 exchanging heat between themselves;
• Let their areas, temperatures (in Kelvin) and
emissivities be (A1, T1, ε1) and (A2, T2, ε2).
• Let a radiation shield 3, be placed between
these plates. Plate 3 is thin and made of a
material of high reflectivity.
• Let the emissivities of two sides of the radiation
shield be ε and ε as shown.
shield be ε3-1 and ε3-2 as shown.
• Radiation network for this system is shown in
Fig. (13.36,b).
• This is drawn, as usual, remembering that each
gray surface has a ‘surface resistance’
associated with it, and the two radiosity
potentials are connected by a ‘space resistance’.

• When there is no shield, the radiation heat
transfer between plates 1 and 2 is already
shown to be:
• With one shield placed between plates 1 and 2,
Q 12
A σ. T 1
4
T 2
4.
1
ε 1
1
ε 2
1
....for infinitely large parallel plates.....(13.59)
• With one shield placed between plates 1 and 2,
the radiation network will be as shown in Fig. (b)
above.
• Note that now all the relevant resistances are in
series.

• Net heat transfer between plates 1 and 2
is given as:
Q12-one shield = (Eb1-Eb2)/Rtot where Rtot is
the total resistance.
• i.e.
E b1 E b2
Q 12_one_shield
E b1 E b2
1 ε 1
A 1 ε 1
.
1
A 1 F 13
.
1 ε 3_1
A 3 ε 3_1
.
1 ε 3_2
A 3 ε 3_2
.
1
A 3 F 32
.
1 ε 2
A 2 ε 2
.
....for two gray surfaces with one radiation shield
placed in between.....(13.70)

• Now, for two large parallel plates, we note:
• Then, eqn. (13.70) simplifies to:
F 13 F 32 1 and, A 1 A 2 A 3 A
Q 12_one_shield
A σ. T 1
4
T 2
4.
1 1
1
1 1
1
....(13.71)
• Note that as compared to eqn. (13.59) for the
case of no-shield, we have, with one shield, an
additional term appearing in the denominator of
eqn. (13.71).
1
ε 1
1
ε 2
1
1
ε 3_1
1
ε 3_2
1

• Therefore, if there are N radiation shields, we
have, for net radiation heat transfer:
• If emissivities of all surfaces are equal, eqn.
(13.72) becomes:
Q 12N_shields
A σ. T 1
4
T 2
4.
1
ε 1
1
ε 2
1
1
ε3_1
1
ε 3_2
1 .....
1
ε N_1
1
ε N_2
1
...(13.72)
(13.72) becomes:
Q 12N_shields
A σ. T 1
4
T 2
4.
N 1( )
1
ε
1
ε
1.
1
N 1( )
Q 12no_shield
. ....(13.73)

• Note this important result, which implies that, when all
emissivities are equal, presence of one radiation shield
reduces the radiation heat transfer between the two
surfaces to one-half,
• Two radiation shields reduce the heat transfer to one-
third, 9 radiation shields reduce the heat transfer to one-
tenth etc.
• For a more practical case of the two surfaces having
ε ε
emissivities of ε1 and ε2, and all shields having the same
emissivity of εs, eqn. (13.72) becomes:
Q 12N_shields
A σ. T 1
4
T 2
4.
1
ε 1
1
ε 2
1 N
2
ε s
1.
....(13.74)

• To determine the equilibrium temperature of the
radiation shield:
• Once Q12 is determined from eqn. (13.71), the temperature of the
shield is easily found out by applying the condition that in steady
state:
• We can use either of the conditions: Q12 = Q13 or Q12 = Q32.
• Q13 or Q32 is determined by applying eqn. (13.59); i.e. we get:
Q 12 Q 13 Q 32 ...(13.75)
A σ. T 1
4
T 3
4.
• In both the above eqns., T3 is the only unknown, which can easily be
determined.
Q 12 Q 13
A σ. T 1 T 3
.
1
ε 1
1
ε 3
1
...(13.76,a)
or,
Q 12 Q 32
A σ. T 3
4
T 2
4.
1
ε 3
1
ε 2
1
....(13.76,b)

For a cylindrical radiation shield placed in
between two, long concentric cylinders:
• Consider the case of radiation heat transfer
between two long, concentric cylinders.
• The radiation heat transfer between two long,
concentric cylinders is already shown to be:

Q 12
A 1 σ. T 1
4
T 2
4.
1
ε 1
A 1
A 2
1
ε 2
1.
....for infinitely long concentric cylinders.....(13.60)
where
A 1
A 2
r 1
r 2
• Now, let a cylindrical radiation shield, 3, be
placed in between the inner cylinder (1)
and the outer cylinder (2), as shown in Fig.

• The radiation network for this system is exactly the same
as shown in Fig. (13.36,b):
• And, the radiation heat transfer between cylinders 1 and
2, when the shield is present, is given by:
2, when the shield is present, is given by:
• Now, for the cylindrical system, we have:
Q 12_one_shield
E b1 E b2
1 ε 1
A 1 ε 1
.
1
A 1 F 13
.
1 ε 3_1
A 3 ε 3_1
.
1 ε 3_2
A 3 ε 3_2
.
1
A 3 F 32
.
1 ε 2
A 2 ε 2
.
....for two gray surfaces with one radiation shield
placed in between.....(13.70)

F 13 F 32 1
A 1 2 π. r 1
. L.
A 2 2 π. r 2
. L.
and, A 3 2 π. r 3
. L.
• Then, eqn. (13.70) reduces to:
• In eqn. (13.77), we have: (A1/A2) = (r1/r2), and (A1/A3) = (r1/r3).
Q 12one_shield
A 1 σ. T 1
4
T 2
4.
1
ε 1
A 1
A 2
1
ε 2
1.
A 1
A 3
1
ε 3_1
1
ε 3_2
1.
...for concentric cylinders with one radiation shield...(13.77)

• Note that as compared to the relation for two
concentric cylinders with no shield (i.e. eqn.
13.60), an additional term appears in the
denominator of eqn. (13.77) (i.e. the third term)
when one radiation shield is introduced;
• If there is a second radiation shield, say (4), then
one more similar term will have to be added in
one more similar term will have to be added in
the denominator to take care of the resistance of
that shield.
• Equilibrium temperature of the shield is
determined by applying the principle that, in
steady state: Q 12 Q 13 Q 32

For a Spherical radiation shield placed in
between two concentric spheres:
• The radiation network for this system is shown in Fig.
(13.36,b).
• When there is no radiation shield, radiation heat transfer
between surfaces 1 and 2 is given by eqn. (13.61), viz.
Q 12
A 1 σ. T 1
4
T 2
4.
1
ε 1
A 1
A 2
1
ε 2
1.
....for concentric spheres.....(13.61)
where
A 1
A 2
r 1
r 2
2

• Again, when the radiation shield is present, the general relation for
radiation heat transfer between surfaces 1 and 2 is eqn. (13.70).
• Relation for radiant heat transfer between surfaces 1 and 2, is
exactly as eqn. (13.77), i.e.
Q 12one_shield
A 1 σ. T 1
4
T 2
4.
1
ε 1
A 1
A 2
1
ε 2
1.
A 1
A 3
1
ε 3_1
1
ε 3_2
1.
...for concentric spheres with one radiation shield...(13.78)
• In eqn. (13.78), we have:
• Equilibrium temperature of the shield is determined by applying the
principle that, in steady state:
A 1
A 2
r 1
r 2
2
and,
A 1
A 3
r 1
r 3
2
Q 12 Q 13 Q 32

Radiation error in temperature
measurement:
• An important application of radiation shields is in
reducing the radiation error in temperature
measurement.
• Consider the case of a hot fluid at a temperature
Tf, flowing through a channel, whose walls are at
a temperature T .
Tf, flowing through a channel, whose walls are at
a temperature Tw.
• Let the convective heat transfer coefficient
between the fluid and the thermometer bulb be
h.
• To measure the temperature of the fluid, a
thermometer (or a thermocouple) is introduced
into the stream, as shown in Fig. (13.39 a).

Tf , h
(a) Thermometer without radiation shield
Tw
qrad
qconv
Tc
Thermometer
Tc
Thermometer
Tw
Fig. 13.39 Radiation shielding of thermometers
Tw
Tc
Ts
Tw
Tf
, h
(b) Thermometer with radiation shield

• Let the reading shown by the thermometer be
Tc.
• This reading, however, does not represent the
true temperature of the fluid Tf, since the
thermometer bulb will lose heat by radiation to
the walls of the channel which are at a lower
temperature Tw (which is usually the case).
• So, in steady state, the thermometer bulb will
gain heat by convection from the flowing fluid
gain heat by convection from the flowing fluid
and will lose heat by radiation to the walls, and
as a result, the temperature Tc shown by the
thermometer will be some value in between Tf
and Tw .
• We wish to find out the true temperature of the
fluid Tf , by knowing the thermometer reading Tc.

• Making an energy balance on the thermometer bulb, in
steady state, we have:
• Without radiation shield:
• qconv to the bulb = qrad from the bulb
i.e. h A c
. T f T c
. ε c A c
. σ. T c
4
T w
4.
i.e. T f T c
ε c σ. T c
4
T w
4.
.....(13.79)
where Ac = surface area of thermometer bulb,
εc = emissivity of thermometer bulb surface
• Eqn. (13.79) gives the true temperature of the fluid Tf.
• Second term on the RHS of eqn. (13.79) represents the
error in temperature measurement due to radiation effect
i.e. T f T c
h
.....(13.79)

• Radiation error can be minimized by:
• (i) having low value of εc i.e. high reflectivity for the
bulb surface
• (ii) high value for h, the convective heat transfer
coefficient
• In practice, even if we start with a thermometer bulb
surface of high reflectivity, soon the emissivity value
rises to about 0.8 or 0.9 due to deposit formation,
rises to about 0.8 or 0.9 due to deposit formation,
corrosion or erosion of the bulb surface, etc.
• So, the most practical way to reduce the radiation error
in temperature measurement is to provide a cylindrical
radiation shield around the thermometer bulb, as shown
in Fig. (13.39,b) above.

• In steady state, the shield temperature (Ts) will stabilize
somewhere in between the fluid temperature Tf and the
wall temperature Tw.
• Then, in eqn. (13.79), Tw will be replaced by the effective
shield temperature Ts.
• Energy balance on the thermometer bulb:
• Heat transferred to the bulb from the fluid by convection =
Heat transferred from the bulb to the shield by radiation
Heat transferred from the bulb to the shield by radiation
• In eqn. (13.80), Fcs = view factor of thermometer bulb w.r.t the shield
and is, generally equal to 1.
i.e. h A c
. T f T c
.
σ T c
4
T s
4.
1 ε c
A c ε c
.
1
A c F cs
.
1 ε s
A s ε s
.
....(13.80)

• Making an energy balance on the shield:
heat transferred to shield from the fluid by convection +
heat transferred to shield from bulb by radiation =
heat transferred from shield to walls by radiation
i.e. 2 A s
. h. T f T s
.
σ T c
4
T s
4.
1 ε c
A c ε c
.
1
A c F cs
.
1 ε s
A s ε s
.
ε s A s
. σ. T s
4
T w
4. ....(13.81)
where, A s = area of shield on one side
ε s = emissivitty of shield surface
A c = area of bulb surface
ε c = emissivity of bulb surface
F cs = view factor of bulb w.r.t. shield

• In the first term of the above eqn. factor 2 appears since
convective heat transfer to the shield occurs on both
surfaces of the shield.
• Also, in writing the RHS, the inherent assumption is that:
F sw 1 ...view factor between the shield and the walls
and,
• Solving eqns. (13.80) and (13.81) simultaneously, we
obtain the shield temperature Ts and the thermometer
reading Tc, (if Tf is known), or Tf (if Tc is known).
A s
A w
0 ...i.e. surface area of shield is negligible compared to the area of the channel
walls

Thermal Radiation - III- Radn. energy exchange between gray surfaces

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