Course Name: Heat Transfer
Course Code: UME 720
Topic: Thermal Radiation
by
Dr. Madhup Kumar Mittal
Associate Professor
Department of Mechanical Engineering
Thapar Institute of Engineering & Technology
Fundamentals of Thermal Radiation
1. Radiation is the energy emitted by a body in the form
of energy waves which are also called
electromagnetic waves because they consist of both
electric and magnetic field waves. That is why
radiation emitted by a body is also called
electromagnetic radiation.
2. Electromagnetic radiation can be described as a stream
of photons, which are massless particles each travelling
in a wave-like pattern and moving at the speed of light.
Electric field wave
Magnetic field wave
Flashlight
Photons
3. There are many types of electromagnetic radiation such as gamma ray radiation, X-ray radiation,
microwave radiation, thermal radiation etc.
4. Different types of electromagnetic radiations are produced through various mechanisms. For example,
gamma rays are produced by nuclear reactions, X-rays by the bombardment of metals with highenergy electrons, microwaves by special types of electron tubes such as klystrons and magnetrons.
continued….
Fundamentals of Thermal Radiation
5. In 1864, James Maxwell observed an interesting mechanism which also produced electromagnetic radiations.
He observed that electromagnetic radiations are produced whenever electric charges accelerate, decelerate or
change the direction of their movement.
accelerates
or decelerate
electromagnetic
radiation
electromagnetic
radiation
change of direction
6. The observations of Maxwell can be used as an explanation for emission of radiation from any hot material.
➢ In a hot solid material, the molecules continuously vibrate. The
continuous vibration of molecules in solids results in production
of electromagnetic radiations.
➢ In hot liquid or gas, the molecules collide with each other
randomly, send each other off in different directions at different
speeds. The collision of molecules in liquid and gases result in
production of electromagnetic radiations.
vibrating
charge
electromagnetic
radiation
electromagnetic
radiation
Solid material
Liquid or gas
➢The amount of radiation emitted from a hot material depends on the temperature of the material producing
the radiation. The radiation emitted by a body as a result of its temperature is called Thermal Radiation.
Fundamentals of Thermal Radiation
Thermal Radiation
• Thermal radiation is the thermal energy emitted by a body in the form of electromagnetic waves as a
result of its temperature.
• Electromagnetic waves can be described as a stream of massless energy particles, called photons
(proposed by Max Planck), each moving at the speed of light in a wave like pattern whose wavelength is
given by 𝝀 = 𝒄Τ𝝂.
• Each photon is considered to have an energy of 𝑬 = 𝒉. 𝒄Τ𝝀, where h = 6.62×10-34 J.s is Planck’s constant.
• The strength or energy of thermal radiation emitted by a body depends upon the temperature of body.
As E ∝ 1/λ, thermal radiation emitted by bodies at low temperature (or at room temperature) has high
wavelength and vice-versa.
Flashlight
Photons
Continued…
Fundamentals of Thermal Radiation
Thermal Radiation
• All bodies at a temperature above absolute zero emit thermal radiations over a wide range of wavelength. However,
most of the thermal radiation (≈ 99%) emitted by a body lies in the wavelength range of 0.1 to 100 μm, hence the
thermal radiation is also defined as the portion of electromagnetic spectrum that extends from 0.1 to 100 μm.
• The radiation emitted by bodies at room temperature lies in the infrared range of spectrum.
• Bodies start emitting noticeable radiation (radiation in visible range) at temperatures above 527°C (800 K).
visible
𝑬 = 𝒉. 𝒄Τ𝝀
Spectrum of Electromagnetic Radiation
A surface that reflects red while
absorbing the remaining parts of the
incident light appears red to the eye
Fundamentals of Thermal Radiation
Concept of Black Body
• A Black body is an idealized body which is used as a standard body against which the radiative
properties of real bodies are compared.
• A Black body is a perfect absorber,
i.e., it absorbs all the incident radiation.
Black body
All radiation absorbed
Real body
Partial radiation absorbed
• A Black body is also a perfect emitter, i.e., at a specified temperature, no surface can emit more
energy than a black body.
Eblackbody > Erealbody
Both bodies at the same temperature T
Fundamentals of Thermal Radiation
A body emits radiation
in all directions
1012
Red
1014
Visible light region
5800 K
4000 K
Locus of max. power
λT = 2898 μm-k
Eb (λ), W/m2-μm
A body emits radiation in all directions
over a wide range of wavelength. In 1901,
Max Planck developed the following
relation for the amount of radiation energy
emitted per unit time, per unit surface area,
at a wavelength λ by a black body at an
absolute temperature T.
Violet
Planck’s law for black body radiation
1010
2000 K
1000 K
108
500 K
300 K
106
100 K
104
102
Eb(λ, T) is called monochromatic (single wavelength)
emissive power of black body.
1
0.01
0.1
1
10
λ (μm)
100
1000
Fundamentals of Thermal Radiation
Total emissive power of black body (Stefan-Boltzman Law)
The total emissive power of a black body [Eb(T)] is defined as the total radiant energy emitted in all directions
over the entire wavelength range, per unit time, per unit surface area, by a black body at an absolute temperature T.
𝜆=∞
𝐸𝑏 𝑇 = න 𝐸𝑏 𝜆, 𝑇 𝑑𝜆
𝜆=0
= න
𝜆=0
𝐶1
𝑑𝜆
𝜆5 𝑒𝑥𝑝 𝐶2 Τ𝜆𝑇 − 1
= 5.67 ×
𝐸𝑏 𝑇 = 𝜎𝑏 𝑇 4
10−8
𝑇4
= 𝜎𝑏
(W/m2)
𝐸𝑏 𝜆, 𝑇
𝐸𝑏 𝜆, 𝑇
𝜆=∞
𝐸𝑏 𝑇
𝑇4
dλ
λ
where, σb = Stefan-Boltzman constant = 5.67×10−8 W/m2-K4
Fundamentals of Thermal Radiation
Monochromatic Emissivity (ελ) of non-black body (real body)
The ratio of the monochromatic emissive power of a real body [E(λ,T)] to
the monochromatic emissive power of black body [Eb(λ,T)] at the same
wavelength and temperature is called the monochromatic emissivity of
real body.
𝐸(𝜆, 𝑇)
𝜀𝜆 =
𝐸𝑏 (𝜆. 𝑇)
Emissivity (ε) of non-black body (real body)
The ratio of the total emissive power of a real body [E(T)] to the total
emissive power of black body [Eb(T)], both being at the same temperature,
is called the average emissivity or simply the emissivity of non-black body.
T = const.
E (λ)
λ
ελ
1
Blackbody,
ε =1
ε = average
emissivity for
non-black body
𝜆=∞
‫=𝜆׬‬0 𝐸 𝜆, 𝑇 𝑑𝜆
𝐸(𝑇)
𝐸(𝑇)
𝜀=
= 𝜆=∞
=
𝐸𝑏 (𝑇) ‫׬‬
𝜎𝑏 𝑇 4
𝐸
𝜆,
𝑇
𝑑𝜆
𝑏
𝜆=0
0
ελ for non-black body
0
∴ 𝐸(𝑇) = 𝜀 𝐸𝑏 (𝑇) = 𝜀 𝜎𝑏 𝑇 4 = Emissive power of any real body at temperature T
λ
Fundamentals of Thermal Radiation
Absorptivity, reflectivity and transmissivity of a body
When a radiant energy emitted by a body impinges upon a semi-transparent body, a portion of it is absorbed, a part
of it is reflected back and the remaining portion is transmitted as shown in the Fig. below:
……(1)
Absorbed radiation
semi-transparent
(Ga)
body
Transmitted
radiation (Gt)
• The absorptivity, reflectivity and transmissivity of real body are not constant.
Rather, they vary with the wavelength of radiation impinging on it.
1
αλ
0
• The absorptivity (α), reflectivity (ρ) and transmissivity (τ) in Eq. (1) are the
average values of monochromatic absorptivity (αλ), monochromatic
reflectivity (ρλ) and monochromatic transmissivity (τλ) respectively, averaged
over the entire wavelength range of impinging radiation on the body.
1
ρλ
0
1
τλ
0
αavg
λ
ρavg
λ
τavg
λ
Fundamentals of Thermal Radiation
Types of radiative bodies (radiative surfaces)
Incident radiation
(1) Black Body (Black surface)
• A black body is one which absorbs all the radiation impinging on it.
• For black body: α = 1; ρ = τ = 0
• Black body is also a perfect emitter, i.e., ε = 1 for black body. It means, at a
specified temperature, no surface can emit more energy than a black body.
• There does not exists a perfectly black body which can absorb all the incident
radiation. Hence, black body is a hypothetical ideal body which is used as a
standard body against which the radiative properties of real bodies are
compared.
(2) Real Body (Non-black body, Non-black surface, Real surface)
• A real body is a non-black body which absorbs only a certain portion of total
radiation striking on it. Thus the absorptivity of a real body is blow unity (α < 1).
• The absorptivity of a real body is wavelength dependent, i.e., the absorptivity
of a real body does not remain constant over the entire range of wavelength of
incident radiation. Thus for real body: α < 1 and α = f (λ).
• The other radiation properties such as ρ, τ and ε are also less than one
and wavelength dependent.
α =1
𝐸(𝑇) = 𝜀 𝐸𝑏 (𝑇) = 𝜀 𝜎𝑏 𝑇 4
Both bodies at the same temperature T
1
αλ
0
αavg
λ
1
ρλ
0
1
τλ
0
ρavg
λ
τavg
λ
continued….
Fundamentals of Thermal Radiation
(3) Gray Body (Gray surface)
• For the purpose of simplification of analysis, all real bodies are
assumed to have constant values of radiation properties
irrespective of wavelength of incident radiation.
• Thus for gray body:
(α, ρ, τ & ε) < 1 and (α, ρ, τ & ε) ≠ f (λ).
εreal surface
1
ε
α real surface
τ real surface
τ
0
ρ gray surface
λ
0
1
α gray surface
λ
ρ real surface
1
ρ
• The condition of constant radiation properties too is not satisfied
by the real body and as such gray body is also a hypothetical
concept like a black body.
λ
0
1
α
0
εgray surface
τ gray surface
λ
(4) Opaque Body (Opaque surface)
• An opaque body is one which does not allow any amount of radiation
to transmit through it. Thus for opaque body: τ = 0 and α + ρ =1.
• Since most of the solids and liquids encountered in engineering are
almost non-transparent (opaque) to thermal radiation, all radiation
thermal analysis is done by assuming bodies involved to be opaque.
τ=0
Opaque body
& α + ρ =1
No radiation transmitted
Fundamentals of Thermal Radiation
Large enclosure behave like a black body
A large enclosure at temperature ‘T’ behave like a black body
because radiation emitted by an interior surface will undergo
multiple reflections and some amount of radiation will be
absorbed by inner surface upon each reflection. Hence entire
radiation will be eventually absorbed by the enclosure.
Kirchoff’s Law of Radiation
Kirchoff’s law of radiation states that the ratio of the emissive power ‘E’ to the absorptivity ‘α’ is same
for all the bodies which are in thermal equilibrium with their surrounding enclosure.
1
Large enclosure at
temp. T
Small body 1 (α1)
at temp. T
2
Large enclosure at
temp. T
Small body 2 (α2)
at temp. T
3
Large enclosure at
temp. T
𝐸1
𝐸2
𝐸3
𝐸𝑛
=
=
=
𝛼1
𝛼2
𝛼3
𝛼𝑛
Small body 3 (α3)
at temp. T
continued….
Proof of Kirchoff’s Law of Radiation
Let us consider a large enclosure which encloses a small body ‘1’ of surface area A1.
Let us assume both the enclosure and body are at the same temperature T. Under the
steady state condition, thermal equilibrium must exist between the body and the
enclosure, i.e.
Radiation emitted by the body (1) = radiation absorbed by the body (1)
𝐸 (𝑇) …(1)
𝐴1 𝐸1 𝑇 = 𝛼1 𝐴1 𝐸𝑏 (𝑇)
or, 𝐸𝑏 (𝑇) = 1
𝛼1
Now we remove body ‘1’ and replace it by body ‘2’ having surface
area A2, absorptivity α2 and emissivity ε2. In this case, we may write:
𝐸 (𝑇)
…(2)
𝐴2 𝐸2 𝑇 = 𝛼2 𝐴2 𝐸𝑏 (𝑇) or, 𝐸𝑏 (𝑇) = 2
𝛼2
𝐸1 (𝑇)
𝐸2 (𝑇)
𝐸3 (𝑇)
=
=
= … … … 𝐸𝑏 (𝑇)
𝛼1
𝛼2
𝛼3
From above Eq. (3), we can also write that for any radiating body (having
absorptivity α) which is in thermal equilibrium with the surrounding :
Thus in general we can write:
…(3)
𝐸(𝑇)
= 𝛼 …(4)
𝐸𝑏 (𝑇)
Large enclosure at
temp. T
1
Small body 1 (A1,
α1, ε1) at temp. T
2
Large enclosure at
temp. T
Small body 2 (A2,
α2, ε2) at temp. T
We also know that the ratio of the emissive power of a real body (E) to the emissive power of black body (Eb), both being at
the same temperature, is called the emissivity of body, i.e.,
𝐸(𝑇)
…(5) ⸫ from Eq.(4) and Eq.(5),
𝜀=
𝐸𝑏 (𝑇)
we have α = ε
Thus, Kirchoff’s law of radiation also states that the absorptivity of a body is equal to its emissivity when the body is in
thermal equilibrium with its surroundings.
Radiation Heat Transfer Between Surfaces
Radiation heat transfer between surfaces depends on:
Surface-3
Surface-2
• Temperatures of the surfaces
• Radiation properties of the surfaces
• Orientation of surfaces relative to each other
Surface-1
Shape factor (View factor, geometric factor, configuration factor)
• Shape factor is defined as the fraction of radiative energy that is radiated from one surface and strikes the
other surface directly.
F12
Direct radiation from surface "1" incident upon surface "𝟐"
=
Total radiation radiated from surface"1"
• Similarly F21 =
Direct radiation from surface "2" incident upon surface "𝟏"
Total radiation radiated from surface"2"
• In general, Fij represents the fraction of the radiation leaving surface ‘i’ that strikes surface ‘j’ directly
Radiation Heat Transfer Between Surfaces
Salient features of shape factor
When two surfaces 1 & 2, of area A1 and A2
respectively, exchange radiant energy with each
other, the shape factor follows reciprocity
theorem, i.e.,
2 ,A
2
A1F12 = A2F21
1 , A1
A concave surface has a shape factor with respect
to itself because the radiant energy coming out
from one part of the surface is intercepted by the
another part of the same surface. The shape
factor with respect to itself for surface ‘1’ is
denoted as F11.
1
F11 ≠ 0
For a flat surface or convex surface, shape factor with respect to itself
is zero as no radiation leaving from their surfaces strikes themselves.
F11 = 0
F11 = 0
If a enclosure consists of many surfaces,
then each surfaces of enclosure follows
the following relation for shape factor.
F11 + F12 + F13 +………..+F1N = 1
F21 + F22 + F23 +………..+F2N = 1
…………………………………..
N
1
2
3
This is known as conservation
principle or summation rule.
FN1 + FN2 + FN3 +………..+FNN = 1
Shape factor for a body inside a
enclosure is 1 as all radiation leaving
from the exterior surface of inside
body strikes the enclosure surface.
Enclosure
2
1 Inner body
(sphere)
F12 = 1
Radiation Heat Transfer Between Surfaces
Determine the shape factor from any one
surface to any other surface for the following
enclosure configuration:
2
3
2
1
3
1
A tube with cross-section
of an equilateral triangle
Thus, for surface-1
For surface-1
F11 + F12 + F13 = 1
F11 = 0
F11 = 0 (because it is a flat surface)
F12 = 0.5
⸫ F12 + F13 = 1
F13 = 0.5
and, F12 = F13 (By symmetry)
Similarly for surface-2
⸫ F12 = F13 = 0.5
F22 = 0
F23 = 0.5
F21 = 0.5
Similarly for surface-3
F33 = 0
F31 = 0.5
F32 = 0.5
Radiation Heat Transfer Between Surfaces
Determine the shape factor from any one
surface to any other surface for the following
enclosure configuration:
2
For plane surface-1
F11 + F12 = 1
F11 = 0 (because it is a flat surface)
⸫ F12 = 1
r
1
Plane surface
For hemispherical surface-2
F22 + F21 = 1
r
Hemispherical sphere of radius
r closed by a plane surface
We know that:
A1F12 = A2F21 (By reciprocity theorem)
∴ F21
A1
=
F
A2 12
π𝑟 2
=
× 1 = 0.5
1Τ2(4𝜋𝑟 2 )
⸫ F22 = 0.5
Radiation Heat Transfer Between Surfaces
Determine the shape factor from any one
surface to any other surface for the following
enclosure configuration:
Spherical enclosure
For surface-1 of inner sphere
F11 + F12 = 1
F11 = 0 (because it is a convex surface)
⸫ F12 = 1
2
For spherical enclosure surface-2
1 (sphere)
F22 + F21 = 1
r
2r
A1F12 = A2F21 (By reciprocity theorem)
∴ F21 =
Sphere of radius r inside spherical
enclosure of radius 2r
A1
F12
A2
4π𝑟 2
=
× 1 = 0.25
2
4π(2𝑟)
⸫ F22 = 0.75
Radiation Heat Transfer Between Surfaces
Net radiation heat exchange between black surfaces
• Consider two black
opaque
surfaces
of
arbitrary shape of area
A1 & A2, and maintained
at uniform temperatures
T1 and T2 respectively, as
shown in Figure.
T1 > T2
α=1
ε=1
ρ=0
τ=0
• Amount of radiation energy leaving surface ‘1’ per unit
time per unit area is given by:
Eb1 = σb T14
WΤm2
• Total amount of radiation energy leaving from entire
surface ‘1’ per unit time is given by:
Q1 = 𝐴1 σb T14
W
• Amount of radiation energy leaving surface ‘1’
and striking surface ‘2’ is given by:
Q12 = 𝐴1 𝐹12 σb T14
W
• Similarly, the amount of radiation energy leaving
surface ‘2’ and striking surface ‘1 ‘is given by:
Q21 = 𝐴2 𝐹21 σb T24
W
• The net rate of radiation heat transfer from
surface ‘1’ to surface ‘2’ can be calculated as:
Q12
𝑛𝑒𝑡
= Q12 − Q21
= 𝐴1 𝐹12 σb T14 − 𝐴2 𝐹21 σb T24
• By applying reciprocity relation (i.e., A1F12 = A2F21),
the above equation can be written as:
Q12
𝑛𝑒𝑡
= 𝐴1 𝐹12 σb T14 − T24
Question
A small sphere (outside diameter = 60 mm) with a
surface temperature of 300°C is located at the
geometric centre of a large hollow sphere (inside
diameter = 360 mm) with an inner surface
temperature of 15°C. Calculate how much of
emission from the inner surface of the large sphere
is incident upon the outer surface of the small
sphere; assume that both sides approach black body
behaviour. Also determine the net interchange of
radiation heat between the two spheres.
Solution
The net interchange of heat between the two sphere is:
2
𝑄𝑛𝑒𝑡 = 𝐴1 𝐹12 σb T14 − T24
r1
𝑄𝑛𝑒𝑡 = 4𝜋 × 0.032 × 1 × 5.67 × 10−8 300 + 273
1
r2
𝑄𝑛𝑒𝑡 = 0.0113 × 5.67 ×
573
100
4
288
−
100
4
− 15 + 273
4
= 64.66 𝑊 (𝑨𝒏𝒔)
4
Radiation Heat Transfer Between Surfaces
Radiation Heat Transfer Between Gray (Non-Black) Surfaces
Irradiation (G)
It denotes the total radiant energy incident upon
unit surface area per unit time. Its unit is W/m2.
Radiosity (J)
It denotes the total radiant energy leaving from
unit surface area per unit time. Its unit is W/m2.
J = εEb + ρ𝐺 = εσb T 4 + ρ𝐺
For black surface, ε = 1 & ρ = 0
⸫ For black surface, J = Eb, i.e., the radiosity of
black surface is equal to its emissive power.
Opaque gray surface
with emissivity ε &
at temperature T
Radiation Heat Transfer Between Surfaces
Net Radiation Heat Transfer from or to a Gray Surface
Let us consider a opaque gray surface i of surface area
Ai and at a temperature Ti. Let Gi be the total radiant
energy incident upon the unit surface area per unit time.
Ji
For a surface i, the radiosity (Ji) can be expressed as:
𝐽𝑖 = 𝜀𝑖 𝐸𝑏𝑖 + 𝜌𝑖 𝐺𝑖
(W/m2)
Gi
....(1)
For opaque surface i: 𝛼𝑖 +𝜌𝑖 = 1
𝑜𝑟,
since, 𝛼𝑖 = 𝜀𝑖 (from Kirchoff ′ s law)
𝜌𝑖 = 1 − 𝛼𝑖
Surface i
ρi G i
εiEbi
Ai, εi
∴ 𝜌𝑖 = 1 − 𝜀𝑖
⸫ Eq (1) can be written as:
𝐽𝑖 = 𝜀𝑖 𝐸𝑏𝑖 + 1 − 𝜀𝑖 𝐺𝑖
𝑜𝑟,
𝐽𝑖 − 𝜀𝑖 𝐸𝑏𝑖
𝐺𝑖 =
1 − 𝜀𝑖
....(2)
Continued…
Radiation Heat Transfer Between Surfaces
Net Radiation Heat Transfer from or to a Gray Surface
𝐽𝑖 − 𝜀𝑖 𝐸𝑏𝑖
𝐺𝑖 =
1 − 𝜀𝑖
....(2)
ρiGi
Gi
If, Ji > Gi
The net rate of heat transfer from
surface i of surface area Ai is
denoted by Qi and is expressed as:
𝑄𝑖 = 𝐴𝑖
(W)
𝐽𝑖 − 𝜀𝑖 𝐸𝑏𝑖
𝐽𝑖 −
1 − 𝜀𝑖
εiEbi
𝑄𝑖 = 𝐴𝑖 ( 𝐺𝑖 − 𝐽𝑖 )
Ai, εi
Surface i
𝑄𝑖 = 𝐴𝑖 ( 𝐽𝑖 − 𝐺𝑖 )
If, Gi > Ji
The net rate of heat transfer to surface
i can be expressed as:
Ji
⇒ 𝑄𝑖 =
Eq (3) can be represented
in the form of electrical
network as given below:
I
V1
Ji
V2
R
e 𝑉
𝑉1 −
2
𝐼=
𝑅𝑒
𝐴𝑖 𝜀𝑖 𝐸𝑏𝑖 − 𝐽𝑖
𝐸𝑏𝑖 − 𝐽𝑖 𝐸𝑏𝑖 − 𝐽𝑖
....(3)
⇒ 𝑄𝑖 =
=
=
1 − 𝜀𝑖
1 − 𝜀𝑖
𝑅𝑖
𝐴𝑖 𝜀𝑖
𝐽𝑖 − 𝐸𝑏𝑖 𝐽𝑖 − 𝐸𝑏𝑖
=
1 − 𝜀𝑖
𝑅𝑖
𝐴𝑖 𝜀𝑖
Surface i
Ji
1−ε
=
Aε
Qi
Ebi
The factor 𝑅𝑖 = (1 − 𝜀𝑖 )Τ𝐴𝑖 𝜀𝑖 is called surface resistance because this
resistance is related to surface properties of the radiating body.
....(4)
Eq (4) can be represented in the form
of electrical network as given below:
1−ε
=
Aε
Qi
(W)
Surface i
Ebi
Radiation Heat Transfer Between Surfaces
Radiation Heat Exchange Between Any Two Gray Surface
Surface j
Let us consider two opaque gray surfaces i & j of surface area Ai & Aj,
and maintained at uniform temperatures Ti & Tj respectively.
Jj
Radiation heat exchange between surface i and
surface j can be expressed as:
Ji
Surface i
Eq (1) can be represented in the form of
electrical network as given below:
𝑄𝑖𝑗 = 𝐴𝑖 𝐽𝑖 𝐹𝑖𝑗 − 𝐴𝑗 𝐽𝑗 𝐹𝑗𝑖
Surface j
By using reciprocity relation, i.e., Ai Fij = Aj Fji, above Eq. becomes
𝑄𝑖𝑗 = 𝐴𝑖 𝐹𝑖𝑗 𝐽𝑖 − 𝐽𝑗
𝑜𝑟,
𝐽𝑖 − 𝐽𝑗 𝐽𝑖 − 𝐽𝑗
𝑄𝑖𝑗 =
=
1
𝑅𝑖𝑗
𝐴𝑖 𝐹𝑖𝑗
Jj
....(1)
The factor 𝑅𝑖𝑗 = 1Τ𝐴𝑖 𝐹𝑖𝑗 is called space resistance, because this resistance
is related to the gap and orientation between the radiating surfaces .
Ji
Surface i
Radiation Heat Transfer in Two-Surface Enclosures
Let us consider an enclosure consisting of two opaque gray surfaces 1 & 2 of surface area A1 & A2, and maintained
at uniform temperatures T1 & T2 respectively. Surfaces 1 & 2 have emissivities ε1 & ε2 respectively.
Ji
T1 > T2
Gi
Q12
Q1
1
Q2
ρiGi
εiEbi
1 Q
1
Surface i
2
𝑄𝑖 = 𝐴𝑖 ( 𝐽𝑖 − 𝐺𝑖 )
Ai, εi
Eb1
1 − ε1
A1 ε1
(W)
As the surface 1 and surface 2 making an enclosure, so we can write:
Net rate of radiation heat transfer from surface 1 (Q1) = Net radiation heat exchange
between surface 1 and surface 2 (Q12) = Net rate of radiation heat transfer to surface 2 (Q2)
1
Q1
J1
Q12
J1
J2
1
𝐴1 𝐹12
⸫ we can also write for an enclosure consisting of surfaces 1 & 2 that:
Net rate of radiation heat transfer from surface 1 to surface 2, (Q12)net = Q1 = Q12 = Q2
𝑄12
𝑛𝑒𝑡
𝐸𝑏1 − 𝐽1 𝐽1 − 𝐽2 𝐽2 − 𝐸𝑏2
=
=
=
1 − 𝜀1
1
1 − 𝜀2
𝐴1 𝜀1
𝐴1 𝐹12
𝐴2 𝜀2
2
Q2
Q2
J2
2
Eb2
...(1)
continued….
Radiation Heat Transfer in Two-Surface Enclosures
𝑄12
𝑛𝑒𝑡
𝐸𝑏1 − 𝐽1 𝐽1 − 𝐽2 𝐽2 − 𝐸𝑏2
=
=
=
1 − 𝜀1
1 − 𝜀2
1
𝐴1 𝜀1
𝐴1 𝐹12
𝐴2 𝜀2
...(1)
Q12
Q1
1
From above Eq. (1), we can write following three separate equations:
1
1 − 𝜀1
..(2)
𝐽1 − 𝐽2 = 𝑄12 𝑛𝑒𝑡
𝐸𝑏1 − 𝐽1 = 𝑄12 𝑛𝑒𝑡
𝐴1 𝐹12
𝐴1 𝜀1
..(3)
Q2
2
𝐽2 − 𝐸𝑏2 = 𝑄12
1 − 𝜀2
𝐴2 𝜀2
𝑛𝑒𝑡
..(4)
Adding Eqs. (2), (3) & (4), we get:
𝐸𝑏1 − 𝐸𝑏2 = 𝑄12
⇒ 𝑄12
⇒ 𝑄12
𝑛𝑒𝑡
𝑛𝑒𝑡
1 − 𝜀1
1
1 − 𝜀2
+
+
𝐴1 𝜀1 𝐴1 𝐹12 𝐴2 𝜀2
𝐸𝑏1 − 𝐸𝑏2
𝐸𝑏1 − 𝐸𝑏2
=
=
1 − 𝜀1
1 − 𝜀2 𝑅1 + 𝑅12 + 𝑅2
1
𝐴1 𝜀1 + 𝐴1 𝐹12 + 𝐴2 𝜀2
𝑛𝑒𝑡
𝜎𝑏 𝑇14 − 𝑇24
=
1 − 𝜀1
1 − 𝜀2
1
+
+
𝐴1 𝜀1 𝐴1 𝐹12 𝐴2 𝜀2
..(5)
Eq (5) can be
represented in the
form of electrical E
b1
network as shown
in the fig.
Q1
J1
Q12
R12
R1
1
J2
Q2
R2
2
If both the surfaces are black, i.e., ε1 = ε2 = 1, then this Eq. is reduced to
𝑄12
𝑛𝑒𝑡
= 𝐴1 𝐹12 𝜎𝑏 𝑇14 − 𝑇24
Eb2
Radiation Heat Transfer in Two-Surface Enclosures
The net radiation heat flow between two gray surfaces, forming an
enclosure, can be represented by an electric circuit as shown in Figure :
𝑄12
𝑛𝑒𝑡
𝐸𝑏1 − 𝐸𝑏2
𝐴1 𝜎𝑏 𝑇14 − 𝑇24
=
=
1 − 𝜀1
1 − 𝜀2 1 − 𝜀1
1 − 𝜀2 𝐴1
1
1
+
+
+
+
𝐴1 𝜀1 𝐴1 𝐹12 𝐴2 𝜀2
𝜀1
𝐹12
𝜀2 . 𝐴2
Net radiation heat flow (Q12) for the common cases of two-surface enclosures of practical interest are given below:
Concentric spheres
2
1
∴ 𝑄12
𝑄12
𝑛𝑒𝑡
=
𝐴1 𝜎𝑏 𝑇14
𝑇24
−
1 − 𝜀1
1 − 𝜀2 𝐴1
1
+
+
𝜀1
𝐹12
𝜀2 . 𝐴2
𝐴1 4𝜋𝑟12 𝑟12
Here, F12 = 1 and
=
=
𝐴2 4𝜋𝑟22 𝑟22
𝐴1 𝜎𝑏 𝑇14 − 𝑇24
=
𝑛𝑒𝑡 =
1 − 𝜀1
1 − 𝜀2 𝑟12
1 1 − 𝜀2 𝑟12
𝜀1 + 1 + 𝜀2 . 𝑟22 𝜀1 + 𝜀2 . 𝑟22
Very long Concentric cylinders
𝑄12
1
𝑛𝑒𝑡
𝐴1 𝜎𝑏 𝑇14 − 𝑇24
=
1 − 𝜀1
1 − 𝜀2 𝐴1
1
+
+
.
𝜀1
𝐹12
𝜀2 𝐴2
2
Here, F12 = 1 and
𝐴1 𝜎𝑏 𝑇14 − 𝑇24
∴ 𝑄12
𝐴1 2𝜋𝑟1 𝐿 𝑟1
=
=
𝐴2 2𝜋𝑟2 𝐿 𝑟2
𝐴1 𝜎𝑏 𝑇14 − 𝑇24
𝐴1 𝜎𝑏 𝑇14 − 𝑇24
=
𝑛𝑒𝑡 = 1 − 𝜀
1 − 𝜀2 𝑟1
1 1 − 𝜀2 𝑟1
1
+
1
+
.
𝜀1
𝜀2 𝑟2
𝜀1 + 𝜀2 . 𝑟2
continued….
Radiation Heat Transfer in Two-Surface Enclosures
Very large parallel plates or surfaces close to each other
𝑄12
𝑛𝑒𝑡
𝐴1 𝜎𝑏 𝑇14 − 𝑇24
=
1 − 𝜀1
1
1 − 𝜀2 𝐴1
+
+
.
𝜀1
𝐹12
𝜀2 𝐴2
Here, F12 = 1 and A1 = A2 = A
∴ 𝑄12
𝐴 𝜎𝑏 𝑇14 − 𝑇24
𝐴 𝜎𝑏 𝑇14 − 𝑇24
=
𝑛𝑒𝑡 = 1 − 𝜀
1
−
𝜀
1
1
1
2
+1+
+ −1
𝜀1
𝜀2
𝜀1 𝜀2
Small object in a large enclosure
𝑄12
𝑛𝑒𝑡
𝐴1 𝜎𝑏 𝑇14 − 𝑇24
=
1 − 𝜀1
1
1 − 𝜀2 𝐴1
+
+
.
𝜀1
𝐹12
𝜀2 𝐴2
Here, F12 = 1 and
∴ 𝑄12
𝐴1
→0
𝐴2
𝐴1 𝜎𝑏 𝑇14 − 𝑇24
= 𝜀1 𝐴1 𝜎𝑏 𝑇14 − 𝑇24
𝑛𝑒𝑡 =
1 − 𝜀1
+1
𝜀1
Question
Determine the heat lost by radiation per meter length of 80 mm diameter pipe at 300°C, if pipe is
Case-1: located in a large room with red brick walls at a temperature of 27°C;
Case-2: enclosed in a 160 mm diameter red brick conduit at a temperature of 27°C.
Take ε for pipe and brick conduit = 0.79 and 0.73 respectively.
Solution Case-1
Given:
• emissivity of pipe = ε1 = 0.79;
• radius of pipe = r1 = 80/2 mm = 40 mm = 0.04 m
• temperature of pipe = T1 = 300 + 273 = 573 K
• temperature of room = T2 = 27 + 273 = 300 K
Pipe (1)
Large room (2)
Radiation heat flow between the pipe and large room is given by:
𝑄12
𝐸𝑏1 − 𝐸𝑏2
𝐴1 𝜎𝑏 𝑇14 − 𝑇24
=
=
1 − 𝜀1
1
1 − 𝜀2 1 − 𝜀1
1
1 − 𝜀2 𝐴1
+
+
+
+
𝐴1 𝜀1 𝐴1 𝐹12 𝐴2 𝜀2
𝜀1
𝐹12
𝜀2 . 𝐴2
In this case, F12 = 1 and
….(1)
𝐴1
→0
𝐴2
𝐴1 𝜎𝑏 𝑇14 − 𝑇24
𝐴1 𝜎𝑏 𝑇14 − 𝑇24
∴ 𝑄12 =
=
= 𝜀1 𝐴1 𝜎𝑏 𝑇14 − 𝑇24
1 − 𝜀1
1
+1
−1+1
𝜀1
𝜀1
∴ 𝑄12 = 0.79 × 2𝜋 × 0.04 × 1 × 5.67 × 10−8 573
4
− 300
4
= 1122.6 𝑊 Τ𝑚
(𝑨𝒏𝒔)
continued
Question
Determine the heat lost by radiation per meter length of 80 mm diameter pipe at 300°C, if pipe is
Case-1: located in a large room with red brick walls at a temperature of 27°C;
Case-2: enclosed in a 160 mm diameter red brick conduit at a temperature of 27°C.
Take ε for pipe and brick conduit = 0.79 and 0.73 respectively.
Solution Case-2
Given:
• emissivity of pipe and brick conduit = ε1 & ε2 = 0.79 & 0.73
• radius of pipe = r1 = 80/2 mm = 40 mm = 0.04 m
• radius of brick conduit = r2 = 160/2 mm = 80 mm = 0.08 m
• temperature of pipe = T1 = 300 + 273 = 573 K
• temperature of brick conduit = T2 = 27 + 273 = 300 K
Radiation heat flow between the pipe and brick conduit is given by:
𝑄12
𝐸𝑏1 − 𝐸𝑏2
𝐴1 𝜎𝑏 𝑇14 − 𝑇24
=
=
1 − 𝜀1
1
1 − 𝜀2 1 − 𝜀1
1
1 − 𝜀2 𝐴1
+
+
+
+
.
𝐴1 𝜀1 𝐴1 𝐹12 𝐴2 𝜀2
𝜀1
𝐹12
𝜀2 𝐴2
In this case: 𝐹12 = 1 and
….(1)
𝐴1 2𝜋𝑟1 𝐿 𝑟1 0.04
=
= =
= 0.5
𝐴2 2𝜋𝑟2 𝐿 𝑟2 0.08
𝐴1 𝜎𝑏 𝑇14 − 𝑇24
𝐴1 𝜎𝑏 𝑇14 − 𝑇24
∴ 𝑄12 =
=
1 − 𝜀1
1 − 𝜀2 𝑟1
1 1 − 𝜀2 𝑟1
+1+
.
+
.
𝜀1
𝜀2 𝑟2
𝜀1
𝜀2 𝑟2
2𝜋 × 0.04 × 1 × 5.67 × 10−8 573
∴ 𝑄12 =
1
1 − 0.73
+
× 0.5
0.79
0.73
4
− 300
4
= 979.05 𝑊 Τ𝑚 (𝑨𝒏𝒔)
Pipe (1)
Brick conduit (2)
Radiation Shield
Radiation heat transfer between two surfaces can be reduced greatly by inserting a thin, opaque and highly reflective
sheet between the two surfaces. Such highly reflective thin sheet is called radiation shield.
𝑄12
Eb1
Eb2
𝑤𝑖𝑡ℎ𝑜𝑢𝑡 𝑠ℎ𝑖𝑒𝑙𝑑
=
𝐸𝑏1 − 𝐸𝑏2
1 − 𝜀1
1
1 − 𝜀2
+
+
𝐴1 𝜀1 𝐴1 𝐹12 𝐴2 𝜀2
∴ 𝑄12
𝐴 𝜎𝑏 𝑇14 − 𝑇24
𝑤𝑖𝑡ℎ𝑜𝑢𝑡 𝑠ℎ𝑖𝑒𝑙𝑑 = 1 − 𝜀
1 − 𝜀2
1
+1+
𝜀1
𝜀2
∴ 𝑄12
𝐴 𝜎𝑏 𝑇14 − 𝑇24
𝑤𝑖𝑡ℎ𝑜𝑢𝑡 𝑠ℎ𝑖𝑒𝑙𝑑 =
1
1
+ −1
𝜀1 𝜀2
𝑄12
𝑤𝑖𝑡ℎ 𝑜𝑛𝑒 𝑠ℎ𝑖𝑒𝑙𝑑
=
We know that, for very large parallel
surfaces: F12 = 1 and A1 = A2 = A
….(1)
𝐸𝑏1 − 𝐸𝑏2
1 − 𝜀1
1
1 − 𝜀3
1 − 𝜀3
1
1 − 𝜀2
+
+
+
+
+
𝐴1 𝜀1 𝐴1 𝐹13 𝐴3 𝜀3
𝐴3 𝜀3 𝐴3 𝐹32 𝐴2 𝜀2
For very large parallel surfaces: F13 = F32 = 1 and A1 = A2 = A3 = A
∴ 𝑄12
Eb1
Eb2
Eb3
Under steady state
condition
Q13 = Q32 = Q12
∴ 𝑄12
𝐴 𝜎𝑏 𝑇14 − 𝑇24
𝑤𝑖𝑡ℎ 𝑜𝑛𝑒 𝑠ℎ𝑖𝑒𝑙𝑑 = 1 − 𝜀
1 − 𝜀3
1 − 𝜀3
1 − 𝜀2
1
+1+
+
+1+
𝜀1
𝜀3
𝜀3
𝜀2
𝐴 𝜎𝑏 𝑇14 − 𝑇24
𝑤𝑖𝑡ℎ 𝑜𝑛𝑒 𝑠ℎ𝑖𝑒𝑙𝑑 = 1
1
1
1
+ −1 +
+ −1
𝜀1 𝜀3
𝜀3 𝜀2
….(2)
Radiation Shield
Let us consider a special case when emissivities of radiating surfaces and shield are same, i.e., ε1 = ε2 = ε3 = ε
2
1
T1 T2
ε1 ε2
Q12
1
𝑄12
shield
2
𝑄12
𝑤𝑖𝑡ℎ𝑜𝑢𝑡 𝑠ℎ𝑖𝑒𝑙𝑑
𝐴 𝜎𝑏 𝑇14 − 𝑇24
𝐴 𝜎𝑏 𝑇14 − 𝑇24
=
=
1
1
2
+ −1
−1
𝜀1 𝜀2
𝜀
𝑤𝑖𝑡ℎ 𝑜𝑛𝑒 𝑠ℎ𝑖𝑒𝑙𝑑
𝐴 𝜎𝑏 𝑇14 − 𝑇24
=
1
1
1
1
+ −1 +
+ −1
𝜀1 𝜀3
𝜀3 𝜀2
T1 T3 T3 T2
ε1 ε3 ε3 ε2
Q12
1
shield
shield
T1 T3 T3 T4 T4 T2
ε1 ε3 ε3 ε4 ε4 ε2
Q12
𝐴 𝜎𝑏 𝑇14 − 𝑇24
1 𝐴 𝜎𝑏 𝑇14 − 𝑇24
1
=
= .
= 𝑄12
2
2
2
2
2
−1 +
−1
−1
𝜀
𝜀
𝜀
2
𝑄12
𝑤𝑖𝑡ℎ 𝑡𝑤𝑜 𝑠ℎ𝑖𝑒𝑙𝑑𝑠
If there are N shields with same surface emissivities
as that of radiating surfaces, then:
𝑤𝑖𝑡ℎ𝑜𝑢𝑡 𝑠ℎ𝑖𝑒𝑙𝑑
𝐴 𝜎𝑏 𝑇14 − 𝑇24
=
1
1
1
1
1
1
+ −1 +
+ −1 +
+ −1
𝜀1 𝜀3
𝜀3 𝜀4
𝜀4 𝜀2
𝐴 𝜎𝑏 𝑇14 − 𝑇24
1 𝐴 𝜎𝑏 𝑇14 − 𝑇24
1
=
= .
= 𝑄12
2
2
2
2
3
3
−1 +
−1 + −1
−1
𝜀
𝜀
𝜀
𝜀
𝑄12
𝑤𝑖𝑡ℎ 𝑁 𝑠ℎ𝑖𝑒𝑙𝑑𝑠
1
𝐴 𝜎𝑏 𝑇14 − 𝑇24
1
=
.
=
𝑄12
2
𝑁+1
𝑁
+
1
𝜀−1
𝑤𝑖𝑡ℎ𝑜𝑢𝑡 𝑠ℎ𝑖𝑒𝑙𝑑
𝑤𝑖𝑡ℎ𝑜𝑢𝑡 𝑠ℎ𝑖𝑒𝑙𝑑
Radiation Shield
Temperature of shield in steady state condition
Eb1
Eb2
Eb3
Under steady state condition
Net heat transfer between surfaces 1 & 3 (Q13) = Net heat transfer between surfaces 3 & 2 (Q32), i.e.,
𝐴 𝜎𝑏 𝑇14 − 𝑇34
𝐴 𝜎𝑏 𝑇34 − 𝑇24
=
1
1
1
1
+
−
1
+
𝜀1 𝜀3
𝜀3 𝜀2 − 1
⇒
𝑇14 − 𝑇34
𝑇34 − 𝑇24
=
1
1
1
1
+
−
1
+
𝜀1 𝜀3
𝜀3 𝜀2 − 1
….(1)
Above Eq. (1) is used to determine temperature of shield under steady state condition.
1 4
4
For special case when ε1 = ε2= ε3= ε, Eq. (1) is reduced to: 𝑇3 = 𝑇1 + 𝑇24
2
Radiation Heat Transfer in Three-Surface Enclosures
On applying heat balance for each node, we get:
Eb2
For node J1:
1 − 𝜀2
𝐴2 𝜀2
Q2
2
J2
ε2, A2, T2
1
𝐴2 𝐹23
Q23
1
𝐴1 𝐹12
Q13
Q12
3
1
𝐴1 𝐹13
J1
ε1, A1, T1
1
Q1
1 − 𝜀1
𝐴1 𝜀1
Eb1
ε3, A3, T3
𝐸𝑏1 − 𝐽1
𝐽1 − 𝐽2
𝐽1 − 𝐽3
=
+
1 − 𝜀1
1
1
𝐴1 𝜀1
𝐴1 𝐹12
𝐴1 𝐹13
or,
𝐸𝑏1 − 𝐽1 𝐽1 − 𝐽2 𝐽1 − 𝐽3
−
−
=0
1 − 𝜀1
1
1
𝐴1 𝜀1
𝐴1 𝐹12
𝐴1 𝐹13
……Eq. (1)
Similarly, for node J2:
𝐸𝑏2 − 𝐽2 𝐽1 − 𝐽2 𝐽2 − 𝐽3
+
−
=0
1 − 𝜀2
1
1
𝐴2 𝜀2
𝐴1 𝐹12
𝐴2 𝐹23
…Eq. (2)
and, for node J3:
𝐸𝑏3 − 𝐽3
𝐽2 − 𝐽3
𝐽1 − 𝐽3
+
+
=0
1 − 𝜀3
1
1
𝐴2 𝐹23
𝐴1 𝐹13
𝐴3 𝜀3
…Eq. (3)
Question
Consider two large parallel plates one at
temperature of 727°C with emissivity 0.8 and other
at temperature of 227°C with emissivity 0.4. An
aluminium radiation shield with an emissivity 0.05
on both side is placed between the plates. Calculate:
(1) percentage reduction in heat transfer rate
between the plates.
(2) Equilibrium temperature of the shield
Solution
plate
1
Given:
• temperature of plate 1
= T1 = 727 + 273 = 1000 K
• emissivity of plate 1= ε1 = 0.8
• temperature of plate 2
= T2 = 227 + 273 = 500 K
• emissivity of plate 2 = ε2 = 0.4
• emissivity of shield = ε3 = 0.05
shield
3
plate
2
T1 T3 T3 T2
ε1 ε3 ε3 ε2
(1)
𝑄12
𝐴 𝜎𝑏 𝑇14 − 𝑇24
𝐶
=
=
= 0.363 𝐶
𝑤𝑖𝑡ℎ𝑜𝑢𝑡 𝑠ℎ𝑖𝑒𝑙𝑑
1
1
1
1
𝜀1 + 𝜀2 − 1
0.8 + 0.4 − 1
where,
𝐶 = 𝐴 𝜎𝑏 𝑇14 − 𝑇24
𝐴 𝜎𝑏 𝑇14 − 𝑇24
𝑄12 𝑤𝑖𝑡ℎ 𝑜𝑛𝑒 𝑠ℎ𝑖𝑒𝑙𝑑 =
1
1
1
1
+
−
1
+
+
𝜀1 𝜀3
𝜀3 𝜀2 − 1
𝐴 𝜎𝑏 𝑇14 − 𝑇24
=
= 0.024 𝐶
1
1
1
1
0.8 + 0.05 − 1 + 0.05 + 0.4 − 1
⸫ percentage reduction in heat
0.363 𝐶 − 0.024 𝐶
=
× 100 = 93.4 % 𝑨𝒏𝒔
transfer rate between the plates
0.363 𝐶
(2) Under steady state condition:
Q13 = Q32
𝐴 𝜎𝑏 𝑇14 − 𝑇34
𝐴 𝜎𝑏 𝑇34 − 𝑇24
=
1
1
1
1
+
−
1
+
𝜀1 𝜀3
𝜀3 𝜀2 − 1
⇒
10004 − 𝑇34
𝑇34 − 5004
=
1
1
1
1
+
−1
+
−1
0.8 0.05
0.05 0.4
⇒ 𝑇34 = 5452.84 × 108
∴ 𝑇3 = 859.32 𝐾 (𝑨𝒏𝒔)
END