HHeeaatt aanndd MMaassss TTrraannssffeerr …brijrbedu.org/Brij Data/Brij HMT/SM/Chapter-9 Radiation... · · 2017-09-02For more information log on Brij Bhooshan Asst. Professor

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Brij Bhooshan Asst. Professor B.S.A College of Engg. & Technology, Mathura (India)

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In This Chapter We Cover the Following Topics

Art. Content Page

9.1 Prevost’s Theory 3

9.2 Radiation Property 3

9.3 Stefan’s Law 5

9.4 Emissive Power 5

9.5 Emissivity 6

9.6 Kirchhoff's Law 7

9.7 Concept of Black Body 9

9.8 Planck's Law of Radiation 9

9.9 Rayleigh-Jeans' Law 10

9.10 Wien's Law 10

9.11 Wien's Displacement Law 11

9.12 Planck's Law in Dimensionless Form 11

9.13 Black Body Radiation in a Certain Range of Wavelength 12

9.14 Radiation from Real Surfaces 13

9.15 Intensity of Radiation

Lambert's cosine law

14

15

9.16 Radiation Heat Transfer Exchange Between Two Black Bodies 15

9.17 Shape Factor Algebra 17

9.18 Electrical Analogy of Radiant Exchange in Black Enclosure 22

9.19 Radiant Heat Exchange in Black Body with Surface 23

9.20 Radiation Heat Transfer between Gray Bodies

Radiant Exchange between Two Small Gray Bodies


Infinite Long Concentric Cylinder

Heat Exchange Small Body in a Large Enclosure

24

24

25

26

27




2 Chapter 9: Radiation Heat Transfer

9.21 Radiosity and Irradiation 28

9.22 Radiation Network for Gray Surfaces Exchanging Energy 29

9.23 Hottel’s Crossed String Method 33

9.24 Radiation Shields 34

9.25 Radiation Error in High Temperature Measurement 36

9.26 Radiation from Cavities 37

9.27 Apparent Emissivity of a Cavity 38

9.28 The Radiation Heat-Transfer Coefficient 39

9.29 Radiation from Gases and Vapours 40

9.30 Absorptivity of Gases 41

9.31 Radiant Heat Exchange Between a Gas Volume and a Black Enclosure 42

References:

1- J. P. Holman, Heat Transfer, 9th Edn, MaGraw-Hill, New York, 2002.

2- James R. Welty, Charles E. Wicks, Robert E. Wilson, Gregory L. Rorrer

Fundamentals of Momentum, Heat, and Mass Transfer, 5th Edn, John Wiley & Sons,

Inc., 2008.

3- F. Kreith and M. S. Bohn, Principal of Heat Transfer, 5th Edn, PWS Publishing Co.,

Boston, 1997.

4- P. K. Nag, Heat and Mass Transfer, 2nd Edn, MaGraw-Hill, New Delhi 2005.

Please welcome for any correction or misprint in the entire manuscript and your

valuable suggestions kindly mail us [email protected].




3 Heat and Mass Transfer By Brij Bhooshan

Thermal radiation is that electromagnetic radiation emitted by a body as a result of its

temperature. Thermal radiation refers to the radiant energy emitted by bodies by virtue

of their own temperatures, resulting from the thermal excitation of the molecules.

Thermal radiation is generally described in terms of electromagnetic waves, all of which

propagated at the speed of light 3 108 m/s.

The wavelength and frequency of radiation propagating in a medium are related by

C =

where C is the velocity of light in the medium, is the frequency and, is the

wavelength.

The propagation of thermal radiation takes place in the form of discrete quanta, each

quantum having energy of

E = h

where h is Planck’s constant and has the value h = 6.625 × 10−34 J·s.

Radiation is electromagnetic wave belonging mainly to the infrared of the

electromagnetic spectrum. The radiation being are electromagnetic do not require any

medium for transfer.

The factors on which the radiation emitted by a body depends are the following

1. Temperature of the body,

2. Nature of the surface of the body,

3. Surface area of the body.

If the intervening space between two bodies is vacuum, the radiative energy transfer

between them occurs with the speed of light. If an absorbing medium intervenes the

bodies, the net radiative transfer is reduced, since the medium itself absorbs some

energy. The study of radiation energy transfer is thus separated into two distinct parts:

1. Radiation energy transfer in a nonparticipating medium where the medium is

either vacuum or does not interfere with the energy propagation.

2. Radiation energy transfer in a participating medium, which intervenes with its

propagation by absorbing, emitting or scattering energy. Layers of certain gases

and vapours which absorb and emit radiation at certain ranges of wavelengths

would be considered as the participating medium.

9.1 PREVOST’S THEORY

All bodies emit thermal radiation, unless the body is at absolute zero temperature. If a

body is placed in a surrounding at the same temperature as itself, its temperature does

not change, since the rate at which energy is radiated is equal to the rate at which

energy is received from the surroundings. When a hot body is placed in cooler

surroundings, the rate at which it radiates is faster than the rate at which it absorbs the

incident radiation, as a result of which its temperature decreases till the state of

thermal equilibrium is reached. This is known as Prevost's theory of heat exchange.

9.2 RADIATION PROPERTY

Matter can emit, absorb, reflect and transmit radiant energy. If Q is the total radiant

energy incident upon the surface of a body, some part of it (Qa) will be absorbed, some

part (Qr) reflected and some part (Qt) transmitted through the body (Diagram 9.1). By

energy balance

Q = Qa + Qr + Qt





or

+ ρ + τ = 1 (9.1)

where α is the fraction of incident radiation which is absorbed, called absorptivity, ρ is

the fraction which is reflected, called reflectivity and τ is the fraction which is

transmitted through the body, called transmissivity or transmittance.

Diagram 9.1 Radiation incident on a surface

There are two types of reflection that can occur, specular reflection and diffuse

reflection. In specular reflection, the angle of incidence of the radiation is equal to the

angle of reflection. The reflection shown in Diagram 9.1 is specular reflection. Most

bodies do not reflect in a specular manner, they reflect radiation in all directions. Diffuse

reflection is sometimes likened to a situation in which the incident thermal radiation is

absorbed and then reemitted from the surface, still retaining its initial wavelength.

For black body: = 1, ρ = 0, τ = 0.

For opaque body: + ρ = 1, τ = 0.

For white body: = 0, ρ = 1, τ = 0.

For gray body: , ρ, τ = uniform.

For transparent body: = 0, ρ = 0, τ = 1.

Most solids do not transmit any radiation and are opaque. By increasing ρ with high

surface polishing, or can be decreased. If ρ is reduced, increases. The reflectivity

depends on the character of the surface. Therefore, the absorptivity of an opaque body

can be increased or decreased by appropriate surface treatment.

When the surface is highly polished, the angle of incidence θi, is equal to the angle of

reflection θr, and the reflection is said to be specular. When the surface is rough, the

incident radiation is distributed in all directions, and the reflection is said to be diffuse

(Diagram 9.2).

Diagram 9.2 Types of reflection

Most gases have high values of and low values of and ρ. Air at atmospheric pressure

and temperature is transparent to thermal radiation for which τ = 1 and = 0, ρ = 0. For

example gases like CO2 and H2O vapour are highly absorptive at certain ranges of

wavelengths.

(a) Specular or

mirror like

Mirror image

of source

Reflected

ray

Normal

Incident

ray

Source

(b) Diffuse reflection

Reflected

ray

Normal

Incident

ray

Source

(c) Actual or irregular reflection

Reflected

ray

Normal Incident

ray

Metal

Qa Absorb radiation

Incident radiation

Q

Qr

Qt Transmitted radiation

Reflected radiation





9.3 STEFAN’S LAW

It’s sate that “The rate at which an objects radiates is directly proportional to the fourth

power of its absolute temperature.”

E = T4 [9.2]

By considering the radiation as such a gas, the principles of quantum-statistical

thermodynamics can be applied to derive an expression for the energy density of

radiation per unit volume and per unit wavelength as

If we taken a dimensionless variable, then

x = h/kT or = c/

Integrating above equation

The value of definite integral is

where

Then

Eb = T4 [9.3]

Equation (9.3) is called the Stefan-Boltzmann law, Eb is the energy radiated per unit

time and per unit area by the ideal radiator, and σ is the Stefan-Boltzmann constant,

which has the value

σ = 5.669 × 10−8 W/m2 ·K4.

It is the depend of total hemispherical radiation on temperature.

The total emissive power of a black body is given by

After integration we get

Eb = T4

Above equation is same as the equation (9.3).

9.4 EMISSIVE POWER

If the radiation from a heated body is dispersed into a spectrum by a prism, it is found

that the radiant energy is distributed among various wavelengths.

The total emissive power of a body, E, is defined as the total radiant energy emitted by

the body at a certain temperature per unit time and per unit surface area at all

wavelengths.





The monochromatic emissive power of a body, E, is defined as the radiant energy

emitted by the body per unit time and per unit surface area at a particular wavelength

and temperature. The radiation energy emitted by a black body per unit time and per

unit surface area is given by

Eb = T4

Eb is called the total emissive power of a black body.

The variation of Eb with and T is shown in Diagram 9.3. The total radiation, Eb, is

distributed among wavelengths varying from 0 to . Therefore, at a particular

temperature,

Diagram 9.3 Variation of monochromatic emissive power of a black body with and T

For wavelengths ranging from to + d, dEb is the change of total emissive power.

dEb = dEb d

which is also called spectral or radiation intensity of a black body. Since the area under

each curve for a certain temperature represents the total emissive power Eb, as the

temperature increases, the area increases and therefore Eb increases.

9.5 EMISSIVITY

The emissivity of a surface is defined as the ratio of radiation emitted by the surface to

the radiation emitted by a black body at the same temperature. It is denoted by , and it

varies between 0 and 1.

The emissivity of a real surface varies with the temperature of the surface as well as the

wavelength and direction of emitted radiation. The emissivity of a surface at a certain

wavelength is called spectral emissivity . The emissivity in a certain direction is called

directional emissivity θ, where θ is the angle between the direction of radiation and the

normal to the surface. The emissivity of a surface averaged over all directions is called

the hemispherical emissivity, and the emissivity averaged over all wavelengths is called

the total emissivity.

E(T) is the total emissivity of a real surface at temperature (T).

Wave length

(m)

Total emittance

Visible light Locus of max

1200C

Eb Eb

d

Monochromatic

emissive power

Eb

0 2 4 6 8 10 12 106

500C





E(T) = ε(T)∙T4

Thus, Spectral emissivity or monochromatic emissivity is defined in a similar manner

where E(T) is the spectral or monochromatic emissive power of the real surface.

When the emissivity of the material does not change with temperature, it is called a

gray body. A gray body has a constant monochromatic emissivity , with respect to

wavelength as shown in Diagram (9.4).

Diagram 9.4 Emmissivities of real surface, gray body and black body

Thus, a gray body has a characteristic emissivity value (< 1) which does not vary with

temperature.

The monochromatic emissive power, , of a real surface can vary significantly from

black body emissive power, as illustrated in Diagram 9.5.

This is due to the fact that E = Eb. Since for the real surface is a characteristic of the

surface and not the temperature of the surface, the area under the real surface Eb curve

will not be a constant fraction of the area under the Eb curve as the temperature varies.

Diagram 9.5 comparison of emissive power of a black body, gray body and a real surface varying

with

9.6 KIRCHHOFF'S LAW

It’s sate that “The ratio of emissive power to absorptive power same for all surface at the

same temperature.”

Diagram 9.6 Energy exchange between a body and the enclosing wall

A small body of surface area A1 is placed in a hollow evacuated space kept at a constant

uniform temperature T (Diagram 9.6). After some time, the body will attain at steady

Hollow evacuated space kept at

constant uniform temperature

Small body of surface area A1 1

Monochromatic

emissive power

Eb

40

30

20

10

0 1 2 3 4 5 6

0 Wave length ()()

Real

surface

ε = ε = 0.61 (Gray body)

ε = ε = 1 (Black body)

Real surface

Gray body

0

1.0 Black body ( = 1.0)





state the same temperature as that of the interior of the space and thereafter will

radiate as much energy as it receives.

Let I be the radiant energy falling upon the body per unit time per unit surface area, E1

be the total emissive power of the body and, 1 be the absorptivity of the body.

At steady state,

Energy absorbed = Energy emitted

IA1 1 = E1 A1

Otherwise there would be an energy flow in to or out of the body which would raise or

lower its temperature.

If we replace the body in the enclosure with black body of the same shape and surface

area, and allow it to come to equilibrium which enclosure at the same temperature.

IA1 B = Eb A1

I = Eb/B [9.8]

From Eqns. (9.7) and (9.8), we have

ε = [9.9]

Equation (9.9) is Kirchhoff’s identity.

This is Kirchhoff’s law which states that the emissivity of the surface of a body is equal to

its absorptivity when the body is in thermal equilibrium with its surroundings.

A black body in addition to being a perfect absorber (B = 1) is also a perfect emitter (B

= 1) of radiant energy.

Kirchhoff’s law also holds for monochromatic radiation, for which

Therefore,

= ε [9.10]

Eq. (9.10) shows, the monochromatic absorptivity at the same wavelength.

Therefore, the monochromatic emissivity of a body is equal to the monochromatic

absorptivity at the same wavelength.

Now we know that

Then





9.7 CONCEPT OF BLACK BODY

Black body is a substance which absorbs all the radiation incident on its surface. Ex.

Charcole, lamp black, etc.

Good absorbers are also good emitters in the state of thermal equilibrium.

A black body has the property of absorbing photons of all wave length. The radiation

incident of the body appears to black.

If a light ray is incident of the hole, then its fails to come out of the hole because it

suffers the multiple reflections in state that the enclosure hole is the black body

(Diagram 9.7).

Diagram 9.7 Black body cavity

Suppose the energy emitted from a portion of the surface in the cavity is Eb. After one

reflection, this becomes ρEb, after two reflections, ρ2Eb, etc. Let us suppose the

radiation leaving the hole of this cavity to be composed of rays which have been directly

radiated, reflected once, twice, etc.

Then the energy emitted from the hole is

E = εEb + ρεEb + ρ2 εEb + = εEb (1 + ρ + ρ2 + )

The energy streaming out from the hole is black body radiation.

9.8 PLANCK'S LAW OF RADIATION

Planck’s introduced the quantum concept in 1900 and with it the idea that radiation is

emitted not in a continuous energy state but in discrete amounts or quanta. The

intensity of radiation emitted by a black body, derived by Planck, is

where Ib is the intensity of radiation from a black body between wavelengths and +

d.

The total emissive power between wavelengths and +d is

Suppose

C1 = 2hc2 = 3.742 108 [Wμm4/m2]; C2 = hc/k = 1.4388 108 [μmk]

Then Eq. (9.14) will be

Isothermal

enclosure

3rd reflection and

partial absorption

2nd reflection and

partial absorption

1st reflection and

partial absorption

Hole Incident ray





Planck's law is the basic law of thermal radiation. From this law it follows that the

emissive power Eb by individual isotherms passes through a maximum, as shown in

Diagram 9.8. With wavelengths = 0 and = it vanishes.

Diagram 9.8 Energy distribution of a black body varying with and T

The amount of energy radiated from a black body Eb as a function of wave length, now

dividing both side T5 in Eq. (9.14), then we get

The fraction of total energy radiated between 1 and 2 is given by

Similarly we get

The value of Eb0(T) are given as a function of product of T.

9.9 RAYLEIGH-JEANS' LAW

Planck's law has two extreme cases. One of the extreme cases is the condition T ≫ C2,

then C2/T ≪ 1

This equation is known as Rayleigh-Jeans law which is found to be valid for large

wavelengths.

9.10 WIEN'S LAW

The second extreme case corresponds to the condition

T ≪ C2, then C2/T ≫ 1

m1 T1 = m2 T2 = m3 T3 = m4 T4 = 0.2898 cmK Eb

m4 T4

m3 T3

m2 T2

m1 T1

=

T4 = 600 K

T3 = 800 K

T2 = 1000 K

T1 = 1400 K





The unity present in the denominator of Eq. (9.15) can be neglected, and the

relationship becomes

This equation is known as Wien's law, which is found to be valid for short wavelengths.

9.11 WIEN'S DISPLACEMENT LAW

The maximum values of the emissive power Eb can be obtained by differentiating

Planck's equation (9.15) with respect to and equating it to zero.

Put C2 /T = x and rearranging

The solution of above equation is

max T = 2.898 103 mK [9.21]

This above equation is termed as Wien's displacement law. Here, max is the wavelength

at which Eb is the maximum at a particular temperature. The value of (Eb)max shifts

towards the shorter wavelengths with increasing temperature.

Diagram 9.9

The maximum emissive power of a black body can be found from Planck's law; = max is

replaced from Eq. (9.15).

(Eb)max = C3 T5 W/m3 [9.22]

where the constant C3 = 1.287 105 W/m3 K5. From Eq. (9.22) it follows that the

magnitude of (Eb)max proportional to the fifth power of the absolute temperature of the

body.

9.12 PLANCK'S LAW IN DIMENSIONLESS FORM

Its expressed in dimensionless form by using Eq. (9.22). Therefore,

Eb

2m 1m

T2

T1

Eb

T2 > T1

T2 T1





From Eqs. (9.23) and (9.24), the maximum of this relationship corresponds to the values

of

This is also equal to the ratio of area A1 to the total area under the curves at T (Diagram

9.11). Similarly, for the wavelength range between and 1 and 2, the fraction of

radiation at temperature T, as shown in Diagram 9.11 would be

Diagram 9.10 fraction of radiation in a range of wavelength expressed as area ratio

9.13 BLACK BODY RADIATION IN A CERTAIN RANGE OF WAVELENGTH

The fraction of total radiation f from a black body in the wavelength range of 0 at a

certain temperature T is given by Eq. (9.25).

From Eqns. (9.15) and (9.16),

This gives Eb/T5 are given as a function of product of T.

Again

Now using equation (9.27), then we have

Diagram 9.11 shows the graph of Eb/T5 vs T.

1 2

Eb

Eb

d

T

A1

A2





Diagram 9.11 Total energy of black body radiation below T

At T = 1 K, the curve represents Eb vs . The area under the curve is equal to the

Stefan-Boltzmann constant. The fractional energy in the range of 1T and 2T given by

which also represents the area ratio which can be read from the scale at the top of

Diagram 9.11.

9.14 RADIATION FROM REAL SURFACES

A gray body is one the monochromatic emissivity of which has the same value at all

wavelengths. For a gray body, is independent of so that

= = ε =

even though the temperatures of the incident radiation and of the receiving surface are

not the same.

In general, however, the emissivity of real surfaces varies with wavelength. If the

variation of monochromatic emissivity with is known, the emissive power of the real

body can be found by plotting the product in Eb vs (Diagram 9.12).

Diagram 9.12 Variation of E with

An average emissivity of a real body can then be obtained from

Then

Since = E/Eb. If the real body is gray body, then the ratio Eg/Eb is constant for all

wavelengths.

Real surface,

Gray body = const.

T = const.

Black body ( = 1.0) T = const.

E

Eb

E

Eg

Real surface

E = εEb

E = εEb

(Gray body)

Black

body

Total energy

found below T

T d(T)

Total

area =

At T = 1

K

0.2898

cmK

Eb/T5





The total average absorptivity can be similarly obtained from the distribution of

monochromatic absorptivity .

If a plot of with is known, can be obtained from the above equation.

9.15 INTENSITY OF RADIATION

The intensity of radiation, I is defined as the rate of heat radiation in a given direction

from a surface per unit solid angle per unit area of the projection of the surface on a

plane normal to the direction of radiation (Diagram 9.13).

Diagram 9.14

A solid angle is defined as the ratio of the spherical surface enclosed by a cone, with its

vertex at the centre of the sphere, to the square of the radius of the sphere (Diagram

9.13). The solid angle subtended by the spherical surface dA at the centre would be

dA/r2, where r is the radius of the sphere. If dA = r2, the solid angle subtended is 1

steradian. The whole spherical surface subtends at the centre an angle of 4 steradian

or 1 steragon.

Let us consider the radiation emitted from an elemental black surface of area dA1 at the

temperature T (Diagram 9.14).

Diagram 9.14 Radiation from differential area in hemisphere

Energy will be radiated in all direction in the entire hemisphere.

Suppose that the spherical strip area dA of radius r sin θ and thickness r dθ which

subtends at the centre a solid angle of dA/r2.

where dq is the total energy emitts.

The amount of radiation dQ directed towards this area is

0 ≤ ≤ 2

0 ≤ θ ≤ /2

dA1 r

η

dθ

θ

d

r sin θ

dA

(b) Intensity of radiation

(a) Differential solid angle

r

dω = dA/r2 θ

η

dA

Emitting area

Intensity of

radiation I (θ, )

Solid angle dω

A1





Total radiation in the entire hemisphere is

The emissive power of a black body is times the intensity of the emitted radiation.

Thus, the intensity I depends only on temperature T.

The total emissive power per unit area will be, if the surface does not radiate diffusely,

then

Lambert's Cosine Law

The law states that the total emissive power Eθ from a radiating plane surface in any

direction is directly proportional to the cosine of angle of emission.

Eθ = E cos θ [9.36]

9.16 RADIATION HEAT TRANSFER EXCHANGE BETWEEN TWO BLACK

BODIES

The radiant heat exchange between two bodies depends upon

1. The views the surfaces have of each other, i.e. how they "see" each other.

2. Their emitting and absorbing characteristics and

3. The medium that intervenes the two bodies.

Let us assume that the two bodies are black and the medium in nonparticipating in the

energy exchange.

Suppose that the area elements dA1 and dA2 on the two surfaces (Diagram 9.15). The

distance between them is r and the angles made by the normals to the two area

elements with the line joining them are ϕ1 and ϕ2 respectively. The projected area of dA1

in the direction of radiation is dA1 cos ϕ1.

Diagram 9.15 Radiant heat exchange between two surfaces

r

A1

N

1

N2

A2

dA1

dA2





Now solid angle is

Energy leaving dA1 and intercepted (and absorbed, since the surface is black) by dA2 is

Since I is independent of the direction the energy emitted per unit of area dA1, per unit

solid angle is proportional to the cosine of the angle ϕ1. This is named as Lambert's

cosine law.

From Eq. (9.34)

The total radiation leaving A1 and being absorbed by A2 is

where

Energy leaving dA2 and intercepted by dA1 is

The total radiation leaving A1 and being absorbed by A2 is

where

The net energy exchange between A1 and A2, then

(Q12)net = Q12 Q21 = A1 l2 (

) [9.42]

From equation (9.39) and (9.41),

Here, 12 is called the shape factor of A1 with respect to A2. This is the fraction of energy

leaving A1 that strikes A2 (and is absorbed, because the surface is black).

Similarly, 21 is the shape factor of A2 with respect to A1 which is also the fraction of

energy leaving A2 that strikes A1 (and is absorbed).

Since the values of 12 and 21 depend on how the two surfaces are exposed to each other

or "see" each other, these are also called view factors, geometry factors or configuration

factors. Unless A1 = A2, 12 = 21. But

A1 12 = A2 21 [9.43]

This is known as the reciprocity theorem.





Application 9.1: Consider the radiation from the small area dA1 to the flat disk A2, as

shown in Diagram 9.16. The element of area dA2 is chosen as the circular ring of radius

x.

Diagram 9.16 Radiation from a small-area element to a disk.

Solution: From Diagram 9.16, dA2 = 2 x.dx, and 1 = 2

Now shape factor is

Put

9.17 SHAPE FACTOR ALGEBRA

Consider an enclosure consisting of n surface as shown in Diagram 9.17. The energy

leaving the surface i must be intercepted by some surface of the enclosure, including i

itself if it is concave. That is

Diagram 9.17 Enclosure of black surface

If the surface j is subdivided in to m subsurface as shown in diagram, then

ii is the self-viewing factor and it is non-zero only for concave surfaces.

1

2

3

1

2

L

dA1 R =

R x





In addition, the reciprocity theorem holds true for any two surfaces of the enclosure.

The decomposition of one or both the surfaces into subdivisions produces a combination

of geometrical configurations for which the shape factor is easily determined. Of the two

surfaces A1 and A2, if A1 is subdivided into two parts A3 and A4, then the radiant heat

exchange between A1 and A2 is

If the surfaces are black,

Since, T1 = T2 = T3, then

The radiant exchange from A2 to Al

For two large plate are placed at parallel then

If surfaces are flat or convex the shape factor with respect to itself is zero.

Application 9.2: Suppose a sphere of surface area A1 is completely enclosed by an

irregular surface of area A2 as shown in Diagram 9.18. Find the shape factor.

Diagram 9.18

Solution: The energy conversation

The spherical surface is convex and hence itself factor is given 11 = 0.

From reciprocity theorem

Energy conservation for surface A2 is

Application 9.3: Suppose an enclosure of three non-concave surfaces as shown in

Diagram 9.19. Find the shape factor.

Diagram 9.19

Solution: From energy conversation to the surfaces 1, 2 and 3 gives

1 2

3





12 + 13 = 1 (a)

21 + 23 = 1 (b)

31 + 32 = 1 (c)


After solving equation (a) – (f), we get

Application 9.4: If two concentric block cylinder, if 1 represents the outer surface of the

inner cylinder and 2 represents the inner surface of the outer cylinder as shown in

Diagram 9.20. Find the shape factor.

Diagram 9.20 Radiant heat exchange between concentric cylinders

Solution: From energy conversation

11 + 12 = 1

11 = 0, 12 = 1


Energy conservation for surface A2 is

If d1 = 2d2, 50% of radiation from surface 2 falls on itself.

Application 9.5: Let us consider 1 hemispherical black cavity with a flat black plate

over it (Diagram 9.21). If surface of the cavity is denoted by 1 and that of the plate by 2.

Diagram 9.21 Radiation in a hemispherical cavity

Solution: From energy conversation

11 + 12 = 1; 21 + 22 = 1

22 = 0, 21 = 1


2

1

1

2





Therefore 50% of radiation emitted from the hemispherical surface is striking the

surface itself, and is absorbed.

Application 9.6: Consider the surface 1, 2, 3 as shown in Diagram 9.22. Find the shape

factor 13 in terms of the known shape factor for perpendicular rectangles with a

common edge.

Diagram 9.22

Solution: Suppose that the shape factor for radiation from A3 to the combined area A1,2

is desired. This shape factor must be given very simply as

3−1,2 = 3−1 + 3−2 [a]

that is, the total shape factor is the sum of its parts. We could also write Equation (a) as

A3 3−1,2 = A3 3−1 + A3 3−2 [b]

and making use of the reciprocity relations

A3 3−1,2 = A1,2 1,2−3

A3 3−1 = A1 1−3

A3 3−2 = A2 2−3

the expression could be rewritten

A1,2 1,2−3 = A1 1−3 + A2 2−3 [c]

which simply states that the total radiation arriving at surface 3 is the sum of the

radiations from surfaces 1 and 2.

Application 9.7: Suppose we wish to determine the shape factor 1−3 for the surfaces in

Diagram 9.23 in terms of known shape factors for perpendicular rectangles with a

common edge.

Diagram 9.23

Solution: We may write

1−2,3 = 1−2 + 1−3 [a]

1−3 = 1−2,3 + 1−2 [b]

in accordance with Equation (a)

A1 1,2−3 = A1 1−2 + A1 1−3 [c]





From reciprocity relations

A1 1−2,3 = A2,3 2,3−1 [d]

A1 1−2 = A2 2−1 [e]

A1 1−3 = A3 3−1 [f]

After solving we get

A2,3 2,3−1 = A2 2−1 + A3 3−1

Application 9.8: Determine the shape factor 1−4 for the surfaces in Diagram 9.24 in

terms of known shape factors for perpendicular rectangles with a common edge.

Diagram 9.24

Solution: Let us call the combined surface A1 and A2 as B and the combined surface A3

and A4 as C.

Then AB = A1-2, AC = A3-4.

AB B−C = A1 1−C + A2 2−C [a]

Since radiation leaving the surface B is the sum of radiation leaving the surface A1 and

A2, A1 1−C will be

A1 1−C = A1 1−3 + A1 1−4 [b]

AB B−3 = A1 1−3 + A2 2−3 [c]

A1 1−3 = AB B−3 – A2 2−3 = A1–2 1,2−3 – A2 2−3 [d]

From equation (b) and (d), then we have

A1 1−C = A1 1−3,4 = A1,2 1,2−3 – A2 2−3 + A1 1−4 [e]

From equation (e) and (a), then we have

AB B−C = A1–2 1,2−3,4 = A1,2 1,2−3 – A2 2−3 + A1 1−4 + A2 2−3,4

Application 9.9: Find the shape factor A1,2 by using the Hottels cross strings method

for surface as shown in Diagram 9.25.

Diagram 9.25

Solution: From geometry

D

D

C

B A

C

B





Using the Hottel’s cross-link strings formula

2L1 12 = Sum of length of the two crossed link Sum of length of the uncrossed link [9.47]

9.18 ELECTRICAL ANALOGY OF RADIANT EXCHANGE IN BLACK

ENCLOSURE

We know that

Between any two black surface the radiant heat exchange rate given the or rate of

radiant flux between two black surface

For surface 1 and any other surface, designated i, in a black enclosure the radiant

exchange

For surface 1 views n other surfaces,

From equation (9.48)

Diagram 9.26 Electrical analogy of radiative energy transfer between two black surfaces 1 & 2.

where (Eb1 – Eb2) is the driving force or potential between the two nodes 1 and 2 for

radiative energy transfer and 1/A1 1–2 is the resistance. The corresponding network is

given in Diagram 9.26.

Diagram 9.27 Radiative flux between the walls of a four wall black enclosure

4

3

2

1

2 1





For radiative flux between the four walls of a black enclosure (Diagram 9.27),

The equivalent network is shown in Diagram 9.27.

From an analogous Kirchhoff s law, the net heat flux leaving node point 1 is

which is identical with Eq. (9.52).

9.19 RADIANT HEAT EXCHANGE IN BLACK BODY WITH SURFACE

Let us consider an enclosure having a single source surface 1 a single sink surface 2 and

refractory surface R each of the surfaces maintained at uniform temperatures T1, T2 and

T3 respectively (Diagram 9.28).

Diagram 9.28 Enclosure of black surfaces with a no net flux refractory surface R

The net radiant heat transfer from hot surface 1

Since refractory walls store no energy

From equation (9.55)

Let

Then

1–R = 1– 1–2, and 2–R = 1– 2–1. Diagram 9.28 shows the electrical network equivalent

to the furnace enclosure of Diagram 9.28. The node R has no potential of its own.

R

2 1

R

2 1





Now

Using equation (9.57)

If A1 = A2, then

If the surface 2 does not "see" the surface 1, there is no direct incident radiation on

surface 2 from surface 1, then

If A1 = A2, and 1–2 = 0 then

Equations (9.57) and (9.58) solve many furnace problems. But these are in error because

the temperatures of the surfaces are not uniform in practice. Hottel extended this type of

analysis to multisurface cases, each surface being at a uniform temperature.

9.20 RADIATION HEAT TRANSFER BETWEEN GRAY BODIES

Calculation of radiant heat exchange between black surfaces is relatively easy, since all

the radiant energy which strikes a surface is absorbed. Once the geometrical shape

factor has been determined, the calculation of heat transfer is quite simple. This is,

however, not the case with nonblack bodies, since all the radiation striking a surface is

not absorbed, some part is reflected. Also the emissivities and absorptivities are not

uniform in all directions and for all wavelengths. The problem is somewhat simplified if

the bodies are considered gray in which case and are constant over the entire spectrum

of wavelength and so their average values are equal irrespective of temperatures, i.e.

= .

Since the transmissivity is zero for solid bodies, + ρ = 1 and for a gray body, the

reflectivity of the surface is considered as an additional parameter.


Let us consider two gray bodies, represented by suffix 1 and 2, having emissivity’s 1 and

2. Let us suppose that the bodies are small compared with the distance between them.

It may thus be assumed that of the radiation unabsorbed and reflected diffusely at each

surface, a negligible proportion returns to the original emitting body.

The energy emitted by the body 1 is 1A1(T1)4 of which 12[1A1(T1)4] is incident on the

second body, of which 2 12[1A1(T1)4] is absorbed. Since the bodies are gray, 2 = 2 and

energy transfer from 1 to 2 is





Q12 = A1 12 12

Similarly

Q21 = A2 12 21

Therefore, the net radiant heat transfer between the two bodies is

(Q12)net = A1 12 12 (

) [9.62]

The equivalent emissivity eq of two small gray bodies is

eq = 1 2 [9.63]

This is also called the view factor 12 for gray bodies

12 = 1 2 [9.64]

In practice, of course, a portion of the reflected radiation from each is returned to the

other body and reradiated so that the true equivalent emissivity eq or l2 is greater than

that given by Eq. (9.63), which gives, in fact, the least possible value.


The radiant interchange between two infinite parallel gray planes involves no geometry

factor, since l2 = 21 = 1.0.

Let us consider two gray planes, as shown in Diagram 9.29. For gray surfaces, according

to Kirchhoff’s law = and ρ = 1–. Surface 1 emits E1 per unit time and area. Surface

2 absorbs the fraction 2E1 and reflects (1– 2)E1 back towards A1. The net heat

transferred per unit of surface 1 to 2 is the emission E1 minus the fraction of E1 and E2

which is ultimately absorbed by surface 1 after successive reflections. Therefore,

Diagram 9.29 Radiant heat exchange between two infinite parallel gray planes

Suppose z = (1– 1)(1– 2), then

Put = , and 2 = 2, then we have





Similarly we get the radiant which leaves surface 2 is

The net heat transferred per unit time of surface 1 to 2 is

Now we know that

where

where 12 is the view factor for gray bodies.

Infinite Long Concentric Cylinder

Consider a infinite long concentric cylinder as shown in Diagram 9.30.

Diagram 9.30 Infinite long concentric cylinders

Suppose the energy emitted per unit area by inner cylinder.

Now we know that 12 = 1, and

Inner cylinder emits energy = E1,

Outer cylinder absorb energy is

Outer cylinder reflect energy is

Inner cylinder absorb energy is

Inner cylinder reflects energy is

Energy absorbed by inner cylinder on the second reflection is

Heat lost by the inner cylinder per unit area is given by

Outer cylinder A2, ε2, T2

Inner cylinder A1, ε1, T1





Similarly heat lost by the outer cylinder per unit area is given by

Net radiation heat transfer between the inner and outer concentric cylinder is

Now we know that

where

Heat Exchange Small Body in a Large Enclosure

Suppose that all the radiations emitted by the small body would be intercepted by the

outer enclosure. Then 12 = 1.

Energy emitted by small body 1 and absorbed by the outer large enclosure 2 is

Energy emitted by enclosure 2 is

Energy incident upon the small body 1 is

Energy absorbed by the small body 1 is

Net radiation heat transfer between the small body 1 and outer large enclosure 2 is





If T1 = T2, and Q1–2, then

For gray surface

9.21 RADIOSITY AND IRRADIATION

In calculating radiant heat transfer from gray surfaces, two terms will now be

introduced. Radiosity is the term used to indicate the total radiant energy leaving a

surface per unit time and per unit surface area. It is denoted by the symbol J. This

quantity differs from the emissive power in that the radiosity includes reflected energy

as well as the original emission, regardless of any directional dependence or spectral

preference.

Irradiation is the term used to denote the total radiation incident on a surface per unit

time and per unit surface area. It is denoted by the symbol G.

Diagram 9.31 (a) Surface energy balance for opaque material; (b) Radiosity and irradiation in a

gray body; (c) element representing “surface resistance” in the radiation-network method

Now the term radiosity is the sum of the energy emitted and the energy reflected, when

no energy is transmitted as shown in Diagram 9.31(b), so that

The net energy leaving the surface is the difference between the radiosity and

irradiation

From Eq. (9.71),

Substituting in Eq. (9.72),

Equation (9.73) provides a basis for network representation of the gray enclosure. If the

numerator on the right side is considered as the potential difference, the denominator as

the "surface resistance” to radiation heat transfer and the heat flow as current, then a

network element could be drawn, as shown in Diagram 9.31(c).

Let us now consider the exchange of radiant energy by two surfaces A1 and A2 as shown

in Diagram 9.32(a).

(a) (b)

(c)

Gray

body





Diagram 9.32 (a) Radiant interchange between two gray surfaces; (b) Element representing

"space resistance" in radiation network method

The total radiation which leaves the surface 1, the amount that reaches surface 2 is

J1A1 12, and of the total energy leaving surface 2, the amount that reaches surface 1 is

J2A2 21.

The net energy interchange between two surfaces is

Since A1 12 = A2 21,

We may thus construct a network element (Diagram 32(b)) which represents Eq. (9.74).

The denominator 1/(A1 12) is called the "space resistance" and the numerator (J1 – J2) is

the potential difference.

9.22 RADIATION NETWORK FOR GRAY SURFACES EXCHANGING ENERGY

To construct a network for a particular radiation heat transfer problem, we only have to

connect a "surface resistance", (1–)/A, to each surface and a "space resistance",

1/(Ax xy), between the radiosity potentials.

Diagram 9.33 shows a network which represents two surfaces exchanging radiative

energy with each other.

Diagram 9.33 Radiation network for two surfaces which see each other and nothing else

The net heat transfer would be given by dividing the overall potential difference with

the sum of the resistances.

If 1/(A1 12) is used to represent the overall resistance or A1 12 as the overall inductance,

then

where

or

(b) (a)

1 2





For a three-body problem, the radiation network is shown in Diagram 9.34.

Diagram 9.34 Radiation network for three surfaces which see each other and nothing else

The radiation heat exchange between body 1 and body 2 would be

and that between body 1 and body 3,

The values of radiosities have to be calculated for determining the heat flows in a

problem of this type. Kirchhoff s law on electrical dc network theory which states that

the algebraic sum of the currents entering a node is zero is applied.

The network method can be conveniently used to solve the problem of energy exchange

between two gray surfaces connected by a nonconducting and reradiating surface to

form an enclosure (Diagram 9.28). Diagram 9.28 will get modified, since unlike black

surfaces, gray surfaces have surface resistance. The corresponding radiation network for

the system is shown in Diagram 9.35.

Diagram 9.35 Radiation network for two gray surfaces enclosed by a third surface which is

nonconducting and reradiating

Node JR is not connected to any surface resistance, since the surface 1 has no source of

its own and it only reflects energy. It is termed as floating node. Now

Similarly, 2R = 1 – 21,

The network above (Diagram 9.35) is a simple series parallel system and may be solved

to compute the heat flow.

Let R’ is the sum of the resistances 1/[A1(1 – 21)] and 1/[A2(1 – 21)], which are in series.





Let Req be the equivalent resistance of the parallel resistances of R’ and 1/[A1 12], so that

The total resistance Rt offered to heat flow is

Therefore, the heat flow

where

which is the same as Eq. (9.59).

(i) If the surfaces are black, 1 = 1, 2 = l, for which

(ii) For infinite parallel planes, A1 and A2 are equal and radiation shape factor 12 =

21 = 1, since all the radiation leaving one plane reaches the other. Equation

(9.82) reduces to

where

This can also be obtained from Eq. (9.78).

(iii) For two long concentric cylinders as shown in Diagram 9.36, 12 = 1, Eq. (9.82)

becomes





Diagram 9.36 Radiation interchange between two concentric surfaces

where

This can also be obtained directly from Eq. (9.78).

If A1 = d1L and A2 = d2L,

where d1 and d2 are the diameters of the inner and outer cylinder respectively,

(iv) Equations (9.85) and (9.86) would also hold good for concentric spheres, for

which

where r1 and r2 are the radii of the inner and outer spheres.

(v) When a small body is enclosed by a large body, A1/A2 ~0, and 12 = 1, where

suffix 1 stands for the small body, Eq. (9.78) reduces to

which can be used to estimate the radiation energy loss from a hot object in a large

room.

For gray enclosures, any of the three network circuits may represent them Diagram

9.37.

The overall conductance is A1 12 and the overall resistance is

So that

If no radiating surfaces exist, then ’12 = 12 = 1.





Diagram 9.37 Electrical analog of enclosure of gray surfaces with a refractory surface

For the network (Diagram 9.37), it is clear that the following equation is an alternative

form of the equation

where

If A1 and A2 do not see each other, 12 = 0. Therefore, from Eq. (9.84)

From Eq. (9.91),

Even if the two surfaces do not see each other, still they receive energy via the

reradiating surface connected between them.

If the temperature of each type of enclosure is not uniform, the surfaces are then to be

divided into a number of parts, each of which may be considered to have a uniform

temperature. This would result in a number of source and sink surfaces and of no-net-

flux or reradiating surfaces.

9.23 HOTTEL’S CROSSED STRING METHOD

Let us consider the geometry as shown in Diagram 9.38 and find the shape factor 12

between surfaces 1 and 2. Let us connect the end points with tightly stretched strings

indicated by dashed lines.

If we consider the triangle ABD,

By reciprocity theorem,

R

2 1





Diagram 9.38 Determination of shape factor 12 by using the crossed-string method

Using equations (9.93) and (9.94)

Similarly

Now

In terms of lengths of strings,

instead of areas, Hottel’s cross-string method can thus be expressed as

9.24 RADIATION SHIELDS

Radiation heat transfer between two surfaces may be reduced either by using the

materials which are highly reflective or by introducing radiation shields between them.

Diagram 9.39(a) shows two infinite parallel gray planes interchanging radiative energy

between them with and without a radiation shield.

6

5

4

3 2

1

A

C

B

D





Diagram 9.39 (a) Radiation between parallel infinite planes with and without radiation shield

(b) Radiation network for two parallel planes separated by one radiation shield

For the case with a radiation shield between the surfaces, at equilibrium

where T3 is the equilibrium temperature of the shield. If T3 is known, the heat transfer

rate can easily be calculated. The radiation network with one shield is shown in

Diagram 9.39(b).

If the two parallel planes are of equal emissivity ,

If the third plane placed between them also has the same emissivity, at equilibrium

At thermal equilibrium Q13/A = Q32/A, therefore

Now using equations (9.100) and (9.101), then we have

Now using equations (9.99) and (9.102), then we have

By the use of one radiation shield, the net radiant heat transfer is reduced by 50%. The

position of the shield so long as it does not touch either of the planes does not alter its

effectiveness.

Generalization:

If N shields are placed between the two planes 1 and 2, there would be (2N + 2) "surface

resistances", two for each shield and one for each heat transfer surface, and (N + 1)

"space resistances" (which would all be unity).

The total assistance would thus be

(b)

With

shield

Without

shield

1 2 3 2 1 (a)

Q/A





The resistance when no shield is present is

The resistance with the shields in place is (N+1) times as large as when the shields are

absent. Thus

For cylindrical radiation shield

9.25 RADIATION ERROR IN HIGH TEMPERATURE MEASUREMENT

If the temperature of a high temperature gas stream is measured by the insertion of a

thermometer or thermocouple, the effects of the radiant exchange between the pipe

walls and the temperature sensing element introduce considerable error.

Diagram 9.40 Gas temperature measurement with (a) a bare thermocouple and (b) a shielded

thermocouple

If Tg is the gas temperature to be measured and Tc is the measured temperature at

steady state, the heat transfer by convection from gas to the thermocouple is equal to

the heat transfer by radiation from thermocouple to the wall (Diagram 9.40(a)), so that

where cw = c, the emissivity of the thermocouple which is very small compared to the

enclosing wall, and Ac is the area of the thermocouple. Here, (Tg – Tc) is the

thermocouple error.

When the couple is shielded, this error is considerably reduced. At steady state, (i) the

heat transfer by convection from gas to couple is equal to that by radiation from couple

to shield and (ii) the heat transfer by convection from gas to shield and that by radiation

from couple to shield are equal to heat transfer by radiation from shield to wall

(Diagram 9.40(b)).

where

Ts is the equilibrium temperature of the shield, As is the shield surface area and s is the

emissivity of the shield. Since heat is transferred by convection from gas to both sides of

the shield, the total surface area of the shield is 2As, as provided in the first term of Eq.

(a)

Gas

Gas

Shield (b)





(9.107). Assuming a value of Ts and solving for Tg from Eqs. (9.106) and (9.107) by trial

and error, the true gas temperature can be computed.

9.26 RADIATION FROM CAVITIES

The emission of radiant energy from cavities of regular geometrical shapes (Diagram

9.41) can be estimated.

Diagram 9.41 Radiant emission from cavities

Let us consider the conical cavity, as shown in Diagram 9.41(a), which is of diameter D,

height H, lateral length L, semi-vertex angle and surface area A1. The temperature T1

of the surface is uniform. Part of the radiation from the surface falls on itself, of which a

portion is absorbed and the remainder is reflected.

Rate of emission from the surface

Of this, the amount falling on A1 and absorbed by it

where 11 is the shape factor of the conical surface with respect to itself.

The amount reflected

Of this reflected energy, quantity falling on A1 and absorbed

Reflected

Absorbed

Reflected

and so on

Net rate of emission from the surface

Q = Total emission rate – Total absorption rate

(c) Cylindrical cavity (b) Hemispherical cavity

(a) Conical cavity

D

H

D

R

H

D





Let us consider an imaginary flat surface A2 closing the cavity (Diagram 9.42). Since A1

and A2 together form an enclosure,

Diagram 9.42 Conical cavity with a flat plate on top

Since 22 = 0, 21 = 1.

Now

From Diagram 9.42,

Surface area of cone

Now using equation (9.108)

Put for

Similarly, expressions for radiant emissions from cylindrical and hemispherical cavities

can be obtained.

9.27 APPARENT EMISSIVITY OF A CAVITY

Consider the cavity shown in Diagram 9.43 having an internal concave surface area A i

and emissivity i radiating out through the opening with area A0. The cavity exchanges

radiant energy with a surrounding at Ts having an area that is large compared to the

area of the opening.

If one considers the imaginary surface A0 covering the opening and exchanging heat

with Ai we have 0i = 1.0

D

H L

dx

x

r





Diagram 9.43 Apparent emissivity of cavity

and, from reciprocity,

But, i0 = is so that

The net radiant exchange of surface Ai with the large enclosure As is given by

and the net radiant energy exchange of an imaginary surface A0 having an apparent

emissivity a with the large surroundings is given by

for A0 at the same temperature at the cavity surface Ai.

Now using equations (9.111), (9.112) and (9.113), we have

We can observe the following behavior for a in limiting cases:

a = i for A0 =Ai or no cavity at all

And a →1.0 for Ai >> A0.

9.28 THE RADIATION HEAT-TRANSFER COEFFICIENT

Frequently in engineering analysis, convection and radiation occur simultaneously

rather than as isolated phenomena. An important approximation in such cases is the

linearization of the radiation contribution so that

Qtotal = Qconvection + Qradiation

htotal = hconvection + hradiation

where

In terms of reference temperature TR,

Here TR is a reference temperature, and T1 and T2 are the respective surface

temperatures. In effect, equation (9.116) represents a straight-line approximation to the

radiant heat transfer as illustrated in Diagram 9.44.

≫ Surrounding Ts





Diagram 9.44 Tangent approximation for hr

The factor, , accounts for geometry and surface condition of the radiating and

absorbing surface.

By constructing a tangent to the relation curve at T = T1, the following relations are

obtained for hr and TR:

and

9.29 RADIATION FROM GASES AND VAPOURS

Gases such as oxygen, nitrogen, hydrogen, dry air, etc. are transparent to thermal

radiation, i.e. these gases neither emit nor absorb radiant energy at the temperature of

interest. On the other hand, some gases and vapours such as CO2, CO, H2O, SO2, NH3,

hyrocarbons, etc. absorb and emit radiant energy significantly. Of these gases, CO2 and

H2O vapour are the most important so far as the atmosphere and the industrial furnaces

are concerned.

Diagram 9.45 Emission and absorption bands of gas radiation

There are, however, certain differences between radiation from gases and that from

solids.

(1) Solids radiate and absorb energy at all wavelengths over the entire spectrum

(from = 0 to = ). Gases like CO2 and H2O emit and absorb radiation only

between narrow ranges of wavelengths termed as bands (Diagram 9.45). Gases

are, therefore, named as selective radiators.

(2) For a solid body, the emission and absorption of radiation are essentially surface

phenomena. But in calculating the radiation emitted or absorbed by a gas layer,

its thickness, pressure and shape as well as surface area must be taken into

account. Incident radiation is slowly absorbed by the gas layer, the intensity

decreasing with the thickness of the layer.

Transparent

Emitted and absorbed





9.30 ABSORPTIVITY OF GASES

When monochromatic radiation at an intensity I0 passes through a gas layer of

thickness L as shown in Diagram 9.46, the radiant energy absorption in a differential

distance dx is governed by the equation

Diagram 9.46 Absorption of radiant energy by a gas layer

where Ix is the radiation intensity at a distance x from the wall where the intensity is

I0 and k is the monochromatic absorption coefficient. The decrease in intensity is

proportional to Ix as well as dx, and the constant of proportionality is k.

The integration of Eq. (9.119) gives

The radiation intensity IL decreases exponentially with the thickness of gas layer L.

Thus is known as Beer’s law.

Now according to our definition, the monochromatic transmissivity will be

If the gas is non reflecting, then

Amount of energy absorbed in the gas layer

where G = 1– exp(–kL) is called the monochromatic absorptivity of the gas. By

Kirchhoff’s law,

For large values of L, G = 1. Thus, for thick gas layers, gas approaches black body

radiation within the wavelength bands.

To find the effective absorptivity G or emissivity G of a gas volume over all wavelength

bands, the total intensity of radiation at a distance x is

and at length L,

Energy absorbed by the gas layer in all bands

Gas layer of thickness dx





where G = 1– exp(–kL) and k is the total absorption coefficient. It may be noted that k

is not generally equal to k.

9.31 RADIANT HEAT EXCHANGE BETWEEN A GAS VOLUME AND A BLACK

ENCLOSURE

Let us consider a gas volume at temperature TG enclosed by a black surface at

temperature Tw. Rate of radiation emitted by the gas and falling on the black enclosure

walls

where G is the emissivity of the gas at TG. All this energy is absorbed by the black

enclosure walls.

Rate of energy emitted by the black wall

But all this energy is not absorbed by the gas.

Rate of absorption of energy by the gas

where ’G is the absorptivity of the gas at TG for incident radiation at Tw.

Hence, the net rate of radiant heat exchange between the gas and the black enclosure is

where ’G is not known, an approximation can be made by using G, i.e. the absorptivity

of the gas at Tw for incident radiation from a black body at Tw, instead of G.

For a gray enclosure having an emissivity w for its wall, the net heat exchange may

approximately be obtained by

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HHeeaatt aanndd MMaassss TTrraannssffeerr …brijrbedu.org/Brij Data/Brij HMT/SM/Chapter-9 Radiation... · · 2017-09-02For more information log on Brij Bhooshan Asst. Professor