2. RADIATION
• The radiative mode of heat transfer is
characterized by energy transported in the form
of electromagnetic waves. The waves travel at the
speed of light.
• Both the wave theory and the particle theory are
useful in helping to explain the behavior of
thermal radiation. The wave theory states that
radiation can be imagined to be a wave oscillating
with a frequency ϑand a wavelength(λ). The
product of the frequency and wavelength is the
velocity of propagation, which is the velocity of
light, c.
c= λϑ
3. • The particle theory assumes that radiant energy is
transported as packets of energy called photons. Each
photons travels with the speed of light with a distinct
energy level given by
e=hϑ
Where h is Planck’s constant.
• There are ways other than heating a surface that can
cause photons to be emitters from the body .at the
short-wavelength end of spectrum, for example, are
the x-rays, which can be produced by subjecting a
piece of metal to a stream of electrons. On the other
end of the spectrum are the radio waves, with long
wavelengths, that can be produced by electronic
equipments and crystals. The entire range of all
wavelengths is called the electromagnetic spectrum.
4. • Thermal radiation that is emitted from a surface
due solely to its temperature exists between
wavelengths of 10-7 and 10-4m .
• The human eye is able to detect electromagnetic
waves between wavelengths of approximately 3.8
x 10-7 m and 7.6 x 10-7 m, and radiation between
these wavelengths is called visible radiation.
• 1Å=1 angstrom=10-10m = 10-8 cm = 10-4 μm.
• 1 μm = 1 micrometer= 1 micron = 10-6 m= 10-4 cm
= 104 Å
5. RADIATION PROPERTIES
• Radiative properties are those properties
which quantitatively describe how radiant
energy interacts with the surface of the
material.
• In general, the radiative properties are
functions of wavelength. For example, a
surface may be good reflector in the visible
wavelength range and a poor reflector in the
infrared range.
6. TOTAL RADIATION PROPERTIES
• The total incident energy is referred to as the total
irradiation and is given the symbol G. when the
irradiation strikes a surface, a portion of energy is
absorbed within the material, a portion is reflected
from the surface, and the remainder is transmitted
through the body. Three of the radiative properties –
the absorptivity, reflectivity, and the transmittivity
describe how the incident energy is distributed into
these three categories.
• The absorptivity, α, of the surface is the fraction of
incident energy absorbed by the body . The
reflectivity, ρ of the surface is defined as the fraction
of incident energy reflected from the surface. The
transmissivity,τ,of the body is the fraction of energy
that is transmitted through the body
7.
8. • The transmissivity , τ, of the body is the fraction of
incident energy that is transmitted through the body.
• The energy balance may be expressed mathematically
as
• αG+ρG+τG=G
• Or simply α +ρ +τ=1
• If the surface is said to be a perfect reflector, all
irradiation is reflected, or
• ρ=1.0 and the energy balance for a perfectly
reflecting surface implies that
• τ-α=0
• A black body absorbs the maximum amount of
incident energy or α=1.0
• And therefore τ=ρ=0 for black body
9. • The emmissivity (ε) of a surface is defined as
the total emitted energy divided by the total
energy emitted by a black body at the same
temperature. The mathematical definition of
the total emissivity ɛ is then
• Since a black body emits the maximum
amount of radiation at a given
temperature, the emissivity of a surface is
always between zero and one. When a
surface is a black body, E(T)=Eb and ɛ=α=1.0
for a blackbody
10. PHYSICS OF RADIATION
CONCEPT OF A BLACK BODY
• A body that emits and absorbs the maximum
amount of energy at a given temperature is a
black surface or simply a blackbody. A blackbody
is a standard that can be approached in a practice
by coating the surface of the body or by
modifying the shape of the surface.
• Planck’s Law:
When a blackbody is heated to a temperature T,
photons are emitted from the surface of the
body. Max Planck showed that the energy
emitted at a wavelength λ from a blackbody at a
temperature T is Eb λ(T)=C1/ λ 5(eC2/ λ T-1)
11.
12. • Eb λ =monochromatic or spectral emissive power of a black
body at temperature T,W/m3.
• C1=first radiation constant=3.7418*10-16wm2
• C2=second radiation constant=1.4388*10-2m.k
• Wein’s displacement law:
The wavelength at which the black body emissive power
reaches a maximum value of a given temperature can be
determined from planck’s law by satisfying the condition
for a maximum value:( The product of maximum
wavelength and temperature is given by wein’s
displacement law)
λmaxT=2.898*10-3m.k
Where
13. λmax denotes the wavelength at which the maximum
monochromatic emissive power occurs for a black surface
with temperature T is called Wein’s displacement law.
(Ebλ)max =1.287*10-5T5 w/m3
As the temperture of the filament is increased, the
amount of radiant energy increases and more of the energy
is emitted at shorter wavelength. Above about 1000k ,a
small portion of the energy falls in the long wavelength or
red end of visible spectrum. Our eyes are able to detect this
radiation and the filaments appears to be a dull red color
above 1600k all visible wavelengths are included, so the
filament appears “white” hot at this temperature.
14. An example of an energy source that is at a high
temperature is the sun.The outer surface of the sun has
a temperature of approximately 5800k.According to
Wein’s law the value of λmax at this temperautre is
5.2*10-7m which lies in the range of visible wavelength.
Stefan boltzmann-law:
The total amount of radiative energy per unit area
leaving a surface with absolute temperature T overall
wavelength is called the total emissive power.Eb(T)=σT4
which is known as stefan boltzmann law.the symbol σ
is stefan boltzmann constant σ=5.67*10-8 w/m2k4
15. Non black surfaces: A simple example is a small body that is
non black surrounded by a black surface. Let the area of the
enclosed and surrounding surfaces be A1 and A2 respectively
and let their temperature beT1 and T2,respectively.the
radiation from surface A2 falling on surface A1 is Σa2f21t24.of
this the fraction α1,the absorptivity of area A1for radiation
from surface A2,is absorbed by surface A1.The remainder is
reflected back to the black surrounding and completely
reabsorbed by area A2.Surface A1 emits radiation in
amountA1ε1T14.Where ε1 is the emissivity of surface A1.All
this radiation is absorbed by the surface A2,and none is
returned by another reflection. The net energy loss by surface
A1 is
q12=σε1A1T14- σA2F21α1T24
16. But A2F21=A1
q12=σA1(ε1T14- α1T24 )
If surface A1 is gray, ε1= α1 and
q12=σA1ε1 (T14- T24 )
In general for gray surfaces,
q12=σA1F12 (T14- T24 )= σA2F21 (T14- T24 )
17. Relationship between absorptivity and
emissivity.
The fraction of incident radiation which is
absorbed by material is called the absorptivity.
Net thermal radiation from a gray body.
18. Q= σA1ε1 (T14- T24 )
q=hr(T1-T2).
hr =q/(T1-T2)= σε1 (T14- T24 )/(T1-T2)
Problem:
The tank is 0.5 m diameter and 1 m high and is
situated in a large space effectively forming black
surroundings.the estimation is based on a tank
surface temperature of 80 o C and an ambient
temperature of 25 o C .If the tank surface is
oxidized copper with an emissivity of 0.8 the
radiant heat flow is:
