Radiation(Chapter14)

RADIATION

RADIATION

HEAT TRANSFER 

HEAT TRANSFER 

Radiation

Radiation

T

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 phenomenon. E

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Radiation

Radiation

T

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 phenomenon. E

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T

Types o

ypes of Radi

f Radiatio

ation

n

T

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1_{. The thermal energy of the hot source at T1 is}

converted into energy in the form of   electromagnetic waves.

2. These waves travel through intervening space in straight lines and strike a cold object at T₂. 3. The electromagnetic waves that stikes the

body are absorbed by the body and converted back to thermal energy or heat.

The amount of radiation emitted by a body depends on its temperature, and is proportional to T4. when this radiation strikes a surface, a portion of it is reflected , and the rest enters the surface. On the portion that enters, some are absorbed  by the material, and the remaining radiation is transmitted through.

Blackbody emissive power (W/m2) depends on temperature (T ) of surface

Mechanism of Radiation

The ratio of reflected energy to the incident energy is called reflectivity, . Similarly, transmitivity () and ab sorptivity () are defined as the fraction of the incident energy that is transmitted through or absorbed by the object.

Irradiation = the total amount of incident radiation that strikes a surface

Radiosity = the sum of the radiation emitted by a surface and the fraction of irradiation that is reflected by the surface.

e = emissive power G = total irradiation J = total radiosity

1

!





 V

X 

E

In general: Opaque material:

1

!

 V

E

E = absorptivity  V = reflectivity X = transmissivity I = emissivity E!I

Mechanism of Radiation

P eP T₁ T₂ T₃ Energy e Ideal Emitter Schematic T₃> T₂> T₁

A body that is assumed to absorb all radiant energy and does not reflect any is called a black body. Such a body also emits radiation. The ratio of the emissive power of a surface to that of a black body is called emissivity () and is equal to 1.0 for a black body. According to Kirchhoff¶s law, emissivity and absorptivity of a surface in surroundings at its own temperature are the same for both monochromatic and total radiation; thus for a given surface at thermal radiation

 = 

At thermal equilibrium

emissivity of surface = absorptivity

E

I 

transmissivity of solid surfaces = 0

emissivity is the only significant parameter emissivities vary from 0.1 _{(polished surfaces) to}

0.95 (blackboard)

=

The black body is an idealized surface having the following properties:

Perfect absorber: it absorbs all incident radiation of  wavelength and direction

Perfect emitter: for a prescribed temperature and wavelength, no surface can emit more than a black body.

Although the radiation emitted by a black body is a function of wavelength and temperature, it is independent of direction. That is, the black body is a diffuse emitter.

Black Body

INTENSITY FOR DIFFUSE BLACKBODY RADIATION

Black Body

 ± absorptivity = E_!

 ± emissivity = I!

 ± ideal emissive power = e _b

4 T  e_b ! W 

1

!

!

I 

E

 

P E { f 

 

P I  _{{ f}  4 T  e _gray ! W I  b  gray e e ! I  G _{ray Body}  ± absorptivity < 1  ± emissivity < 1  ± emissive power<1

Black Body

P eP Gray Body Black  Body Energy Real Body bl _ack   gray e e ! I  I  E !

 

I  { _f  Schematic

All Real Surfaces are ³

G

_rey´

IRRADIATION, INCIDENT RADIATION

The total energy emitted by a body, regardless of the wavelengths, is given by:

Where:

 = emissivity

A = surface area exposed T = absolute temperature  = Stefan-Boltzmann constant  = 5.67 x10-8 _W/m2_.K 4 _{= 0.1714 x 10}-8 _Btu/hr.ft2_.O_R 4

Emissive Power



4 4



1 1 1S   AT  T S  q !W I  

The total energy absorbed a body, regardless of  the wavelengths, is given by:

Where:

 = absorptivity

A = surface area exposed T = absolute temperature  = Stefan-Boltzmann constant  = 5.67 x10-8 _W/m2_.K 4 _{= 0.1714 x 10}-8 _Btu/hr.ft2_.O_R 4

Emissive Power

 

4 1S  a  AT S  q ! W 

If two surfaces are arranged so that radiant energy can be exchanged, a net flow of energy will occur  from the hotter surface to the colder surface. The size, shape and orientation of the two radiating surfaces or a system of surfaces are factors in determining the heat transfer rate between them.

View Factor, F₁₂= fraction of radiation leaving the surface 1 _{in all directions which is intercepted}

by surface 2.

For two black planes radiating to each other, the net radiation is expressed as

q1₂ = 1₂A1(T14-T₂4)

Where F1₂ is the view factor of surface 1 to surface

2, also

q₂1 = ₂1A₂(T14-T₂4)

For view factor cannot exceed unity. Such that A1F1₂ = A₂F₂1

and is independent of temperature

1

§

!  j  ij 

F 

F 

₁₁ 

F 

₁₂

...

F 

_1 j 

...

F 

_1n !

1

 ji   j  ij  i 

F 

 A

F 

 A

! _{Thermal Equilibrium}

V_{iew Factor: F}

ij = fraction of radiation from

surface i intercepted by surface j.

1 2

In the case of infinite parallel planes, F1₂=F₂1=1.0,

the geometric factor is omitted. q1₂ = A1(T14-T₂4)

When surfaces are connected by nonconducting but reradiating walls, the reradiating view factor  is 1₂, is used instead of 1₂ and is treated

similarly.

For two gray planes radiating to each other, the net radiation is expressed as

q1₂ = 1₂A1(T14-T₂4)

Where F1₂ is the new view factor and defined as

Radiant Transfer between Gray Bodies

1₂ = 1

1 _{+ A1} 1 _± 1 ₊ 1 _-1

1 b

e

1

 J 

 J 

 R

 _J 

₂ 1 1 1 1 I  I   A  2 2 2 1 I  I   A   R F   A_{1 1} 1 12 1 1 F   A  R F   A_{2 2} 1

No net heat flux wall

Analog circuit 12

Q



Q

12

!

q

12

A

1  Find: 2 b

e

X 1 2 R  R 

1 b

e

 J 

1

 J 

2

e

_b2  R

 J 

1 1 1 1 I  I   A  2 2 2 1 I  I   A   R F   A_{1 1} 1 12 1 1 F   A  R F   A_{2 2} 1  R  R  A F  F   A_{1 1 2 2} 1 1  1 b

e

 J 

1

 J 

2

e

_b2 1 1 1 1 I  I   A  2 2 2 1 I  I   A  12 1 1 F   A

1 b e  J 1  J 2 e_b2 1 1 1 1 I  I   A  2 2 2 1 I  I   A   R  R  A F  F   A_{1 1 2 2} 1 1  12 1 1 F   A 1 b e 1  J   J ₂ eb2 1 1 1 1 I  I   A  2 2 2 1 I  I   A   R  R  A F  F   A F   A 2 2 1 1 12 1 _{1 1} 1 1    R  R  A F  F   A F   A F   A 2 2 1 1 12 1 12 1 1 1 1   !

1 b

e

1

 J 

 J 

₂

e

b2 1 1 1 1 I  I   A  2 2 2 1 I  I   A  12 1 1 F   A  R  R  A F  F   A F   A F   A 2 2 1 1 12 1 12 1 1 1 1   ! 2 2 2 12 1 1 1 1 1 1 1 I  I  I  I   A F   A  A     1 b

e

e

_b2 2 2 2 12 1 1 1 1 12 1 1 1 1 1 I  I  I  I   A F   A  A  A     ! F     

1 b

e

e

_b2 12 1 1 F       A



4



2 4 1 1 12  A T  T  Q ! F     ₁₂  W  



_{1 2}



1 12  A eb eb Q ! F     12    12 1

tan

_ce

 _A

F     

Conduc

!

 R  R  A F  F   A F   A F   A 2 2 1 1 12 1 12 1 1 1 1   ! 2 2 2 12 1 1 1 1 12 1 1 1 1 1 I  I  I  I   A F   A  A  A     ! F     

1 2 net  q _{12 ,} 2 b

e

1 b

e

1  J   J ₂ 2 2 2 1 I  I   A  1 1 1 1 I  I   A  12 1 1 F   A 2 1 A  A !





2 2 2 12 1 1 1 1 2 1 1 12 1 1 1 I  I  I  I   A F   A  A e e  A q b b      !





1 1 1 2 1 2 1 12    ! I  I  b b e e q 1

Surface and View Factor Resistance

1

21 12 ! F  !

F 

1 S net   S  q _{1 ,}

e

_b1

e

_bS  1  J   J _S   S   S   S   A I  I   1 1 1 1 1 I  I   A   S  F   A_{1 1} 1 1 1 S  ! F  1 _{} 0}  S   A  A





 S   S  b S  b  S   A F   A  A e e  A q I  I  I  I  2 2 1 1 1 1 1 1 1 1 1 1 1      !



b b _S 



 S  e e q !  1 1 1 I  1 0 1 ! !  S   S  I  E



4 4



1 1 1S  T  T S  q _!W I  

Surface and View Factor Resistance

The total incident radiant energy upon a body which partially reflects, absorbs and transmits radiant energy is 2200 W/m2. of this amount, 450 W/m2 is reflected and 900 W/m2 is absorbed by the body. Find the transmitivity.

 = 1 _{±  ±  =} 1 _{± 450/2200 ± 900/2200 = 0.386}

Determine the total emissive power of a blackbody at

1₀₀₀O_C

Two black body rectangles, 1_{.8 m by 3.6 m are parallel}

and directly opposed and are 3.6 m apart. If surface 1 _{is at}

T1 = 95OC and surface 2 is at T₂ = 315OC, determine a)

the net rate of heat transfer b) the net energy loss rate from the 95OC surface if the surrounding other than surface 2 behave as black body at 295 K.