Radiation(Chapter14)
Short Description
Download Radiation(Chapter14)...
Description
RADIATION HEAT TRANSFER
Radiation Thermal radiation is the energy emitted by matter as a result of its finite temperature. Any matter with temperature above absolute zero (0 K) emits electromagnetic radiation. Electromagnetic radiation can be visualized as waves traveling at the speed of light, thus, radiation is a surface phenomenon. Electromagnetic radiation is categorized into type ypes by their wavelengths ths.
Radiation Thermal radiation is the energy emitted by matter as a result of its finite temperature. Any matter with temperature above absolute zero (0 K) emits electromagnetic radiation. Electromagnetic radiation can be visualized as waves traveling at the speed of light, thus, radiation is a surface phenomenon. Electromagnetic radiation is categorized into type ypes by their wavelengths ths.
Types of of Radiatio Radiation n
The type of radiation emitted by a body depends on its temperature. The hotter the object is, the shorter the wavel aveleength and and the great eater its amount.
Mechanism of Radiation 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.
Mechanism of Radiation 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 4
Eb !W T
Mechanism of Radiation
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.
Mechanism of Radiation e = emissive power G = total irradiation J = total radiosity
I = emissivity E!I In general:
E = absorptivity V = reflectivity X = transmissivity
E V X ! 1 Opaque material:
E V !1
Energy
Schematic
Ideal Emitter T3> T2> T1
T3
eP
T2 T1
e P
Black Body 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 =
Black Body At thermal equilibrium emissivity of surface = absorptivity
I
=
E
transmissivity of solid surfaces = 0 emissivity is the only significant parameter
emissivities vary from 0.1 (polished surfaces) to 0.95 (blackboard)
Black Body 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. INTENSITY FOR DIFFUSE BLACKBODY RADIATION
Black Body Black Body ± absorptivity = E!
E ! I ! 1
± emissivity = I!
e b ! W T 4
± ideal emissive power = e b G
ray Body
± absorptivity < 1 ± emissivity < 1 ± emissive power< 1
E { f P I {
f P
e gray
! I e b
e gray
! W I T
4
Energy
Schematic
I !
e gray ebl ack
Black Body
eP
Gray Body Real Body
P
E
! I
I {
f
All Real Surfaces are ³Grey´
IRRADIATION, INCIDENT RADIATION
Emissive Power The total energy emitted by a body, regardless of the wavelengths, is given by:
q1S !W I1 AT 14 T S 4 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.OR 4
Emissive Power The total energy absorbed a body, regardless of the wavelengths, is given by:
q1S ! aW AT S 4 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.OR 4
Radiant Transfer between Surfaces 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.
Surface and View Factor Resistance
Surface and View Factor Resistance
Radiant Transfer between Black Bodies 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
Radiant Transfer between Black Bodies View
Factor: Fij = fraction of radiation from surface i intercepted by surface j. 2
1
§ F ij !1
F 11 F 12 ... F 1 j ...F 1n
!1
j
Ai F ij ! A j F ji
Thermal Equilibrium
Radiant Transfer between Black Bodies 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.
Radiant Transfer between Gray Bodies 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 12
=
1 1
F12
+ A1
1
A2 2
± 1 +
1
1
-1
Surface and View Factor Resistance R
Find:
Q12
1
2
No net heat flux wall
R 1
1 I 1
Analog circuit
Q12 ! q12 A1
A1 F 1 R
A1I 1
J 1
X
1 A2 F 2 R
J R
eb1
1 I 2 A2I 2
J 2
eb 2 1 A1 F 12
Surface and View Factor Resistance 1 I 1
1
1
A1 F 1 R
A2 F 2 R
A1I 1
J 1
eb1
J R
A2I 2
1 A1 F 12
J 2
1
1 I 1
1 I 2
A1 F 1 R
eb 2 1
A2 F 2 R
A1I 1
eb1
1 I 2 A2I 2
J 1
1 A1 F 12
J 2
eb 2
Surface and View Factor Resistance 1
1 I 1
1
A1 F 1 R
A2 F 2 R
1 I 2
A1I 1 1
J 1
e b1 A1 F 12
A2I 2
! A1 F 12
J 2
A1 F 12
eb 2
1 1 A1 F 1 R
1 1
A2 F 2 R
1 I 1
A1 F 12
1 A1 F 1 R
A1I 1
e b1
1
J 1
1
1 I 2
A2 F 2 R
A2I 2
J 2
eb 2
Surface and View Factor Resistance 1 I 1
1
1 I 2
A1I 1
A1 F 12
A2I 2
e b1 A1F 12
J 2
J 1 !
1 I 1 A1I 1
1 1
A1 F 12 1 I 1 A1I 1
eb1
1 I 2
A1 F 12 ! A1 F 12
A1 F 12
1 1 A1 F 1 R
A2I 2 1
eb 2
1 I 2 A2I 2
eb 2
1 A2 F 2 R
Surface and View Factor Resistance 1 A1F 12
eb1
eb 2
Conductance ! A1F 12 A1F 12 !
1 I 1 A1I 1
1 1 A1 F 12
1 I 2
A1 F 12
! A1 F 12
1 1 A1 F 1 R
A2I 2
eb1 eb 2 Q12 ! A1F 12 4 4 W T 1 T 2 Q12 ! A1F 12
1 A2 F 2 R
Surface and View Factor Resistance Radiation heat transfer between two infinite parallel plates
A1
F 12 ! F 21 ! 1 1 I 1 1 A1I 1
eb 1 1
2
q12 A1 !
q 12 , net
A2
1 I 2 A2I 2
A1 F 12
J 1
!
J 2
eb 2
eb1 eb 2 1 I 1 A1I 1 q12
!
1 1 1 I 2 A1 F 12 A2I 2
e 1 e 2 b
1 I 1
b
1 I 2
1
Surface and View Factor Resistance Radiation heat transfer between small objects and infinite surrounding
F 1 S ! 1 E S
S
q 1 S , net
1
! I S ! 1
1
1 I S
A1I 1
A1 F 1 S
A S I S
J 1
q1 S A1 !
1 I 1 A1I 1
4
q1S !W I1 T 1 T S
A S
}0
1 I 1
eb1
4
A1
ebS
J S eb1 eb S 0 1 1 1 I 2 A1 F 1 S A2I S
q1 S ! I 1 eb1 eb S
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 1000OC E b = T4 = 5.67 x 10-8 (1273.15K) = 149 kW/m2
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.
View more...
Comments