thermal energy emitted by matter as a result of vibrational

8/14/2019 Thermal Energy Emitted by Matter as a Result of Vibrational

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Thermal energy emitted by matter as a result of vibrational androtational movements of molecules, atoms and electrons. The energyis transported by electromagnetic waves (or photons). Radiationrequires no medium for its propagation, therefore, can take place alsoin vacuum. All matters emit radiation as long as they have a finite(greater than absolute zero) temperature. The rate at which radiationenergy is emitted is usually quantified by the modified Stefan-Bolzmann law:

where the emissivity , , is a property of the surface characterizinghow effectively the surface radiates compared to a "blackbody"(0


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G = GAbsorbed irradiation


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Electromagnetic radiation spectrumThermal radiation spectrum range: 0.1 to 100 mmIt includes some ultraviolet (UV) radiation and all visible (0.4-0.76

mm) and infrared radiation (IR).


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The Planck Distribution

The Planck law describes theoretical spectraldistribution for the emissive power of a black body. Itcan be written as

where C 1=3.742x10 8 (W. m4/m2) and C 2=1.439x10 4(m.K) are two constants. The Planck distribution isshown in the following figure as a function of wavelength for different body temperatures.


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Spectral blackbody emissive power


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Planck Distribution At given wavelength, the emissive power increaseswith increasing temperature

As the temperature increases,more emissive energyappear at shorter wavelengths

For low temperature (>800 K), all radiant energy fallsin the infrared region and is not visible to the human

eyes. That is why only very high temperature objects,such as molten iron, can glow.

Sun can be approximated as a blackbody at 5800 K


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Solar Irradiation


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L = r

LAngles and Arc LengthWe are well accustomed tothinking of an angle as a twodimensional object. It may

be used to find an arc length:

A = r 2d

r

Solid Angle

We generalize the idea of an angle andan arc length to three dimensions anddefine a solid angle, , which like thestandard angle has no dimensions.The solid angle, when multiplied bythe radius squared will havedimensions of length squared, or area,and will have the magnitude of theencompassed area.


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Projected AreaThe area, dA 1, as seen from the

prospective of a viewer, situated at an

angle from the normal to thesurface, will appear somewhatsmaller, as cos dA 1. This smaller area is termed the projected area.A projected = cos Anormal

dA 1 dA 1cos

Intensity

The ideal intensity, I b, may now be defined as the energyemitted from an ideal body, per unit projected area, per unittime, per unit solid angle.

= d dAdq

I 1cos


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Rsin

R

dA 1

dA 2

Spherical GeometrySince any surface will emit radiation outwardin all directions above the surface, thespherical coordinate system provides a

convenient tool for analysis. The three basiccoordinates shown are R, , and ,representing the radial, azimuthal and zenithdirections.

In general dA 1 will correspond to the emittingsurface or the source. The surface dA 2 willcorrespond to the receiving surface or the

target. Note that the area proscribed on thehemisphere, dA 2, may be written as: ][])sin[(2 d Rd RdA =

]sin22 d d RdA =

Recalling the definition of the solid angle, dA = R 2dwe find that: d = R 2sin dd


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Real SurfacesThus far we have spoken of ideal surfaces, i.e. those that emit energy according tothe Stefan-Boltzman law: E b = Tabs 4

Real surfaces have emissive powers, E, which are somewhat less than that obtained

theoretically by Boltzman. To account for this reduction, we introduce theemissivity, .

b E E

Emissive power from any real surface is given by: E = Tabs

4

IncidentRadiation,G

ReflectedRadiation

AbsorbedRadiation

TransmittedRadiation

Receiving PropertiesTargets receive radiation in one of threeways; they absorption, reflection or

transmission.Absorptivity, , the fraction of incidentradiation absorbed .Reflectivity, , the fraction of incidentradiation reflected . Transmissivity, , the fraction of incidentradiation transmitted .


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We see, from Conservation of Energy, that: + + = 1

In this course, we will deal with only opaque surfaces, = 0, so that:

+ = 1Relationship Between Absorptivity, , and Emissivity, Consider two flat, infinite planes, surface A and surface B, both emitting radiationtoward one another. Surface B is assumed to be an ideal emitter, i.e. B = 1.0.

Surface A will emit radiation according to the Stefan-Boltzman law as:

EA = ATA4

and will receive radiation as:

GA = ATB4The net heat flow from surface A will be:q = ATA4 - ATB4

Now suppose that the two surfaces are at exactly the same temperature. The heatflow must be zero according to the 2 nd law. If follows then that: A = A

SurfaceA, T A

SurfaceB, T B


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Thermodynamic properties of the material, and may depend on temperature. Ingeneral, this will be the case as radiative properties will depend on wavelength, . Thewave length of radiation will, in turn, depend on the temperature of the source of radiation.

The emissivity, , of surface A will depend on the material of which surface A iscomposed, i.e. aluminum, brass, steel, etc. and on the temperature of surface A.

The absorptivity, , of surface A will depend on the material of which surface A iscomposed, i.e. aluminum, brass, steel, etc. and on the temperature of surface B.

Black Surfaces

Within the visual band of radiation, any material, which absorbs all visible light,appears as black. Extending this concept to the much broader thermal band, we speak of surfaces with = 1 as also being black or thermally black. It follows that for such a surface, = 1 and the surface will behave as an ideal emitter. The terms idealsurface and black surface are used interchangeably.


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Diffuse Surface: Refers to directionalindependence of the intensity associatedwith emitted,reflected ,or incidentradiation.

Grey Surface: A surface for which the spectralabsorptivity and the emissivity are independentof wavelength over the spectral regions of

surface irradiation and emission.


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Relationship Between Emissive Power and IntensityBy definition of the two terms, emissive power for an ideal surface, E b, andintensity for an ideal surface, I b.

= hemispherebb d I E cosReplacing the solid angle by its equivalent in spherical angles :

= 20 20 sincos d d I E bb

Integrate once, holding I b constant: =2

0 sincos2

d I E bb

bbb I I E ==

2

0

2

2

sin2

Integrate a second time. (Note thatthe derivative of sin is cos d.)

E b = I b


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= d dAdq

I 1cos = d dA I dq 1cos Next we will project the receiving

surface onto the hemispheresurrounding the source. First findthe projected area of surface dA 2,dA 2cos 2. (2 is the angle between

the normal to surface 2 and the position vector, R.) Then find thesolid angle, , which encompassesthis area.

dA 2

dA 2cos 2

R

To obtain the entire heat transferred from a finite area, dA 1, to a finite area, dA 2, weintegrate over both surfaces:

= 2 1 22211

21coscos

A A R

dAdA I q


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Total energy emitted from surface 1 : q emitted = E 1A1 = I1A1View Factors-Integral MethodDefine the view factor, F 1-2 , as the fraction of energy emitted from surface 1,which directly strikes surface 2.

1

22211

2121

2 1

coscos

A I R

dAdA I

qq

F A A

emitted

==

=2 1

22121

121

coscos1 A A R

dAdA A

F

Example Consider a diffuse circular disk of diameter D and area A j and a

plane diffuse surface of area A i


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= 2 1 2 2121121coscos1

A A RdAdA

A F

Since dA 1 is a differential area = 2 2 22121 coscos A R dA F ( )

=

2

2

2

21

2

A

dr r R L

F

=

24

2

212

A R

dr r L F

Let 2 L2 + r 2 = R 2. Then 2 d = 2rdr. =2

4

2

212

A

d L F

2

022

22

221

12

22

D

A L L L F

+==

22

22

0222

221 4

14

4 D L

D L D L

L F D

+=

+

=


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EnclosuresIn order that we might apply conservation of energy to the radiation process, we mustaccount for all energy leaving a surface. Weimagine that the surrounding surfaces act as anenclosure about the heat source which receiveall emitted energy. For an N surfaced enclosure,we can then see that:

11

, ==

N

j ji F

This relationship is known asthe Conservation Rule.

Reciprocity i j j jii F A F A =This relationship is

known as Reciprocity. Example: Consider two concentric spheres shown tothe right. All radiation leaving the outside of surface 1will strike surface 2. Part of the radiant energy leavingthe inside surface of object 2 will strike surface 1, partwill return to surface 2. Find F 2,1 . Apply reciprocity.

2

2

1

2

12,1

2

11,22,111,22

==== D D

A A

F A A

F F A F A

1 2


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Associative RuleConsider the set of surfaces shown to the right:Clearly, from conservation of energy, the fractionof energy leaving surface i and striking the

combined surface j+k will equal the fraction of energy emitted from i and striking j plus thefraction leaving surface i and striking k.

i

jk

k i jik ji F F F + +=)( This relationship is knownas the Associative Rule.

RadiosityRadiosity, J, is defined as the total energy leavinga surface per unit area and per unit time.

J E b + G

E b G G


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Net Exchange Between SurfacesConsider the two surfaces shown.Radiation will travel from surface i tosurface j and will also travel from j to i.

qi j = JiAi Fi j

likewise,qj i = JjAj Fj j

The net heat transfer is then:

qj i (net) = JiAi Fi j - JjAj Fj j

From reciprocity we note that F1 2A1 =F21A2 so that

qj i (net) = JiAi Fi j - Jj Ai Fi j = AiFi j(Ji Jj)

J j

Ji


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Net Energy Leaving a SurfaceThe net energy leaving a surface will be the difference between the energyleaving a surface and the energy received by a surface:

q1 = [Eb G]A1

Combine this relationship with the definition of Radiosity to eliminate G.J Eb + G G = [J - Eb]/

q1 = {Eb [J - Eb]/ }A1Assume opaque surfaces so that + = 1 = 1 , and substitute for .q1 = {Eb [J - Eb]/(1 )}A1

Put the equation over a common denominator:

( )111 11

1 A

J E A

E J E q bbb

=

+=

assume that = ( ) J E A A J E q bb

= = 11

111


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Electrical Analogy for RadiationWe may develop an electrical analogy for radiation, similar to that

produced for conduction. The two analogies should not be mixed: theyhave different dimensions on the potential differences, resistance andcurrent flows.

EquivalentCurrent

EquivalentR e s i s t a n c e

Poten t i a l Di ffe rence

Ohms Law I R V

Net Energy Leav ingSur face

q1?

A 1 E b - J

N e t E x c h a n g e

B e t w e e n Su r f a c e s

qi? j211

1

F A

J1 J 2


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Example: Consider a grate fed boiler.Coal is fed at the bottom, moves across thegrate as it burns and radiates to the walls andtop of the furnace. The walls are cooled byflowing water through tubes placed inside of the walls. Saturated water is introduced at the

bottom of the walls and leaves at the top at aquality of about 70%. After the vapour isseparated from the water, it is circulatedthrough the superheat tubes at the top of the

boiler. Since the steam is undergoing a

sensible heat addition, its temperature willrise. It is common practice to subdivide thesuper-heater tubes into sections, each havingnearly uniform temperature. In our case we

will use only one superheat section using anaverage temperature for the entire region.

SHsection

CoalChute

Surface 1Burning Coal

Surface 2Water Walls

Surface 3

SuperheatTubes


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Energy will leave the coal bed, Surface 1, as

described by the equation for the net energyleaving a surface. We draw the equivalentelectrical network as seen to the right :

E b1 T14

J1

R 1

1

1 A

The heat leaving from the surface of the coal may proceed to either the water walls or to the super-heater section. That part of the circuit is represented by a

potential difference between Radiosity:

J1

R 1

E b1

J2 J3

R 12 R 13

311

1 F A

211

1

F A

It should be noted that

surfaces 2 and 3 will alsoradiate to one another.

R 1

E b1

J2 J3

R 12 R 13

J1 311

1

F A

322

1

F A

211

1

F A


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It remains to evaluate the net heat flowleaving (entering) nodes 2 and 3.

1

1 A

311

1

F A

322

1

F A

211

1

F A

R 1

J2 J3

R 12 R 13

J1

R 2

E b1

E b2 E b3

1

1 A

22

21 A

Insulated surfaces. In steady state heat transfer, a surface cannotreceive net energy if it is insulated. Because the energy cannot be stored by asurface in steady state, all energy must be re-radiated back into the enclosure.

Insulated surfaces are often termed as re-radiating surfaces . Black surfaces: A black, or ideal surface, will have no surfaceresistance:

01

111 =

=

A A In this case the nodal Radiosity

and emissive power will be equal.


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Large surfaces: Surfaces having a large surface areawill behave as black surfaces, irrespective of the actual surface

properties:

011

=

=

AConsider the case of an object, 1, placed

inside a large enclosure, 2. The system willconsist of two objects, so we proceed toconstruct a circuit with two radiosity nodes.

1/(A 1F12)

J1 J2

Now we ground both Radiosity nodesthrough a surface resistance.

1/(A 1F12)J1 J2

E b2 T24

(1-2)/(2A2)

R 1 R 12 R 2

(1-1)/(1A1)

E b1 T14


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Since A 2 is large, R 2 = 0. The view factor, F 12 = 11/(A 1F12)

J1 J2

E b1 T14 E b2 T24R 1 R 12

(1-1)/(1A1)

Sum the series resistances:R Series = (1- 1)/(1A1) + 1/A 1 = 1/( 1A1)

Ohms law:i = V/R

or by analogy:q = E b/R Series = 1A1(T14 T 24)

Returning for a moment to the coal grate furnace, let us assume that we know

(a) the total heat being produced by the coal bed, (b) the temperatures of thewater walls and (c) the temperature of the super heater sections.


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Apply Kirchoffs law about node 1, for the coal bed:

013

13

12

12113121 =

+

+=++ R

J J R

J J qqqq

Similarly, for node 2:

023

23

12

21

2

2223212 =++=++ R

J J R

J J R

J E qqq b

And for node 3:0

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32

13

31

3

3332313 =++=++ R

J J R

J J R

J E qqq b

The three equationsmust be solvedsimultaneously. Sincethey are each linear inJ, matrix methods may

be used:

=

3

3

2

2

1

3

2

1

231332313

231312212

13121312

11111

11111

1111

R

E R E

q

J

J

J

R R R R R

R R R R R

R R R R

b

b


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Surface 1: Find the coal bed temperature, given the heat flow:

25.0111

1

1

14

1

1

111

+===

J RqT

R

J T

R

J E q b

Surface 2: Find the water wall heat input, given the water wall temperature:

2

24

2

2

222 R

J T R

J E q b ==

Surface 3: (Similar to surface 2) Find the water wall heat input,given the water wall temperature:

3

34

3

3

333

R

J T

R

J E q b

==

thermal energy emitted by matter as a result of vibrational

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