12B-Chapter 12, Sec 12-1.4 - 12.7 Black

7/27/2019 12B-Chapter 12, Sec 12-1.4 - 12.7 Black

1/20

Radiation: Processes and Properties

Surface Radiative Properties

Chapter 12

Sections 12.4 through 12.7


2/20

Emissivity

Surface Emissivity

Radiation emitted by a surface may be determined by introducing a property

(the emissivity) that contrasts its emission with the ideal behavior of a blackbody

at the same temperature.

The definition of the emissivity depends upon ones interest in resolving

directional and/or spectral features of the emitted radiation, in contrast

to averages over all directions (hemispherical and/or wavelengths (total).

The spectral, directional emissivity:

,

,

,

, , ,, , ,

,

e

b

I TT

I T

The spectral, hemispherical emissivity (a directional average):

2 2

0 0

2 2

0 0

/

,

/

, ,

, , , cos sin,,

, , cos sin

e

b b

I T d dE TT

E T I T d d


3/20

Emissivity (cont)

The total, hemispherical emissivity (a directional and spectral average):

0 b

b b

T E T d E TT

E T E T

,, , ,

To a reasonable approximation, the hemispherical emissivity is equal to

the normal emissivity.

n

Representative values of the total, normal emissivity:

Note:

Low emissivity of polished metals and increasing emissivity for unpolished

and oxidized surfaces.

Comparatively large emissivities of nonconductors.


4/20

Emissivity (cont)

Representative spectral variations:

Note decreasing with increasing for metals and different behavior for nonmetals.,n

Representative temperature variations:

nWhy does increase with increasing for tungsten and not for aluminum oxide?


5/20

Abs, Ref & Trans

Response to Surface Irradiation: Absorption, Reflection

and Transmission

There may be three responses of a semitransparentmedium to irradiation:

Reflection from the medium refG, .

Absorption within the medium absG, .

Transmission through the medium trG, .

Radiation balance

ref abs tr G G G G , , ,

In contrast to the foregoing volumetric effects, the response of an opaque material

to irradiation is governed by surface phenomena and 0, .trG

, ,ref tr G G G

The wavelength of the incident radiation, as well as the nature of the material,

determine whether the material is semitransparent or opaque.

Are glass and water semitransparent or opaque?


6/20

Abs, Ref & Trans (cont)

Unless an opaque material is at a sufficiently high temperature to emit visible

radiation, its coloris determined by the spectral dependence of reflection in

response to visible irradiation.

What may be said about reflection for a white surface? Ablacksurface?

Why are leaves green?


7/20

Absorptivity

Absorptivity of an Opaque Material The spectral, directional absorptivity:

, ,

,

,

, ,, ,

, ,

i abs

i

I

I

The spectral, hemispherical absorptivity:

2 2

0 0

2 2

0 0

/

, , ,

/

,

, , , , cos sin

, , cos sin

abs i

i

G I d d

G I d d

To what does the foregoing result simplify, if the irradiation is diffuse?

If the surface is diffuse?

The total, hemispherical absorptivity:

0

oabsG dG

G G d

If the irradiation corresponds to emission from a blackbody, how may the

above expression be rewritten?

Assuming negligible temperature dependence,

The absorptivity is approximately independent of the surface temperature,

but if the irradiation corresponds to emission from a blackbody, why does

depend on the temperature of the blackbody?


8/20

Reflectivity

Reflectivity of an Opaque Material

The spectral, directional reflectivity: Assuming negligible temperature

dependence:

, ,

,

,

, ,, ,, ,

i ref

i

II

The spectral, hemispherical reflectivity:

22

0 0

/

, , ,

,

, , , , cos sin

, ,

ref i

i

G I d d

G I

To what does the foregoing result simplify if the irradiation is diffuse?

If the surface is diffuse?

The total, hemispherical reflectivity:

0

0

abs G dGG G d

Limiting conditions of diffuse and

spectral reflection. Polished and rough surfaces.


9/20

Reflectivity (cont)

Note strong dependence of and 1 on .

Is snow a highly reflective substance? White paint?


10/20

TransmissivityTransmissivity

The spectral, hemispherical transmissivity: Assuming negligible temperature

dependence,

,trG

G

Note shift from semitransparent to opaque conditions at large and small wavelengths. The total, hemispherical transmissivity:

0

0

trtrG dG

G G d

,

For a semitransparent medium,

11


11/20

Kirchhoffs Law

Kirchhoffs law equates the total, hemispherical emissivity of a surface to its

total, hemispherical absorptivity:

Kirchhoffs Law

However, conditions associated with its derivation are highly restrictive:

Irradiation of the surface corresponds to emission from a blackbody at the

same temperature as the surface.

However, Kirchhoffs law may be applied to the spectral, directional properties

without restriction:

, ,

Why are there no restrictions on use of the foregoing equation?


12/20

Diffuse/Gray Surfaces

Diffuse/Gray Surfaces With 2 2

0 0

2 2

0 0

/

,

/

cos sin

cos sin

d d

d d

and 2 20 0

2 2

0 0

/

, ,

/

,

cos sin

cos sin

i

i

I d d

I d d

Under what conditions may we equate to ?

With

0 b

b

E d

E T

,

and 0 G d

G

Under what conditions may we equate to ?

Conditions associated with

assuming a gray surface:


13/20

Problem: Surface Emissivity and Absorptivity

Problem 12.49: Determination of the solar absorptivity and total emissivity

of a diffuse surface from knowledge of the spectral

distribution of and the surface temperature.

KNOWN: Spectral, hemispherical absorptivity of an opaque surface.

FIND: (a) Solar absorptivity, (b) Total, hemispherical emissivity for T s = 340K.


14/20

Problem: Surface Emissivity and Absorptivity (cont)

SCHEMATIC:

ASSUMPTIONS: (1) Surface is opaque, (2) = , (3) Solar spectrum has G = G,Sproportional to E,b (, 5800K).

ANALYSIS: (a) The solar absorptivity may be expressed as

S ,b ,b0 0E ,5800K d / E ,5800K d .

The integral can be written in three parts using F(0 ) terms.

S 1 2 30 0.3 0 1.5 0 0.3 0 1.5F F F 1 F .

From Table 12.1,

T = 0.3 5800 = 1740 mK F(0 0.3 m) = 0.0335T = 1.5 5800 = 8700 mK F(0 1.5 m) = 0.8805.


15/20

Problem: Surface Emissivity and Absorptivity (cont)

Hence,

S 0 0.0355 0.9 0.8805 0.0335 0.1 1 0.8805 0.774.

(b) The total, hemispherical emissivity for the surface at 340K may be expressed as

,b b0E ,340K d / E 340K .

With = , the integral can be written in terms of the F(0 ) function. However, it isreadily recognized that since

0 1.5 m,340K F 0.000 at T 1.5 340 510 m K

there is negligible emission below 1.5 m.

It follows that

0.1

COMMENTS: The assumption = is satisfied if the surface is irradiated diffusely or ifthe surface itself is diffuse. Note that for this surface under the specified conditions of solar

irradiation and surface temperature, S. Such a surface isspectrally selective.

P bl E B l f I di t d S f


16/20

Problem: Energy Balance for an Irradiated Surface

Problem 12.90: Determination of the emissivity and absorptivity of a coated

vertical plate exposed to solar-simulation lamps and the magnitude

of the irradiation required to maintain a prescribed plate

temperature.

KNOWN: Vertical plate of height L 2 m suspended in quiescent air. Exposed surface withdiffuse coating of prescribed spectral absorptivity distribution subjected to simulated solar

irradiation, GS,. Plate steady-state temperature Ts 400 K

P bl E B l f I di t d S f ( t)


17/20

Problem: Energy Balance for an Irradiated Surface (cont)

FIND: Plate emissivity, , plate absorptivity, , and plate irradiation, G.

ASSUMPTIONS: (1) Steady-state conditions, (2) Ambient air is extensive, quiescent, (3)

Spectral distribution of the simulated solar irradiation, GS, , proportional to that of a blackbody

at 5800 K, (4) Coating is opaque, diffuse, and (5) Plate is perfectly insulated on the edges and the

back side, and (6) Plate is isothermal.

PROPERTIES: Table A.4, Air (Tf 350 K, 1 atm): 20.92 10-6 m2s, k = 0.030 W/mK,

= 29.90 10-6 m2/s, Pr 0.700.

SCHEMATIC:

P bl E B l f I di t d S f ( t)


18/20


ANALYSIS: (a) Perform an energy balance on the plate as shown in the schematic on a per unitplate width basis,

in outE E 0

4s sG T h T T L 0

where and are determined from knowledge of and h is estimated from an appropriatecorrelation.

Plate total emissivity: Expressing the emissivity in terms of the band emission factor, F (0 - T),

1 s 1 s1 20 T 0 TF 1 F

0.9 0 0.1 1 0 0.1


19/20


Plate absorptivity: With the spectral distribution of simulated solar irradiation proportional to

emission from a blackbody at 5800 K,

1 s 1 s1 20 T 0 TF 1 F

0.9 0.7202 0.1 1 0.7202 0.676


20/20


Irradiation on the Plate: Substituting numerical values into Eq. (1),

4 48 2

0.676G 0.1 5.67 10 W m K 400 K 25.66W m K 400 300 K 0

2G 1052 W m

12B-Chapter 12, Sec 12-1.4 - 12.7 Black

Documents

Transcript of 12B-Chapter 12, Sec 12-1.4 - 12.7 Black