Black Body Radiation - A black body is an idealized object that absorbs all electromagnetic radiation. - It emits thermal radiation according to Planck's law, where the spectrum is determined solely by temperature. - Real materials emit at a fraction of black body levels, defined by their emissivity. Near-perfect black bodies can be approximated but not perfectly realized.

1. RADIATION
PREPARED UNDER THE GUIDANCE OF
Prof. Dr. Manish Mehta
BVM Engineering College , V.V Nagar

2. NAME ENROLLMENT NO.
MARMIK PATEL 140080125025
REJOICE PARACKAL 140080125026
MANAN SHAH 140080125027
SHREYANSH AGHERA 140080125028
Compiled by

3. CONCEPT OF DIFFERENT BODIES
 BLACK BODY:- A black body is an object that absorbs all the radiant energy reaching its surface from
all the
directions with all the wavelengths.It is perfect absorbing body.
so , for black body;
The black body is a hypothetical body . However its concept is very important . When
the black
body
absorbs heat , its temperature raises.
 WHITE BODY:- If all the incident radiation falling on the body are reflected , it is called a white body.
for white body,
 GREY BODY:- A grey body is defined as a body whose absorptivity of a surface does not vary with
variation in
temperature and wavelength of the incident radiation.
 OPAQUE BODY:- When no radiation is transmitted through the body , it is called as opaque body.
for opaque body;

4. KIRCHOFF’S LAW
• This law states that the ratio of total power to absorptivity is constant for al the substances which are in thermal
equilibrium with the surroundings. This can be written inn mathematical form for bodies as follow,
Assume out of any four body , any one body , say fourth one ,is black body .Then;
But according to the definition of emissivity , we have;
From equation (2) in general , we can say that;

5. Let us consider a small body with area A1 and absorptivity a1.The small body (1) is covered by a large radiation body (2)
with area A . If the large body is a black body than the total emissive power of body (2) is Eb. The energy fall on the bod
from body (2) is Eb . Of this energy , body (1) will absorbed the energy a1*A1*Eb . When thermal equilibrium is arrieved
the heat absorbed by the body (1) must be eual to the heat emitted (E1) from body 1 . So at equilibrium,
………………….(1)
Now remove body 1 and put body 3 of area A3 and absorptivity a3 .repeat the same procedure again.
…………………(2)

6. from equation 1 and 2, we have;
According to the definition of emissivity,
Above equation represents the proof of the kirchoff’s law.

7. The Bernoulli Equation
• It is an approximate relation between pressure,
velocity and elevation
• It is valid in regions of steady, incompressible flow
where net frictional forces are negligible
• Viscous effects are negligible compared to inertial,
gravitational and pressure effects.
• Applicable to inviscid regions of flow (flow regions
outside of boundary layers)
• Steady flow (no change with time at a specified
location)

8. • The value of a quantity may change from one location to another. In the case of
a garden hose nozzle, the velocity of water remains constant at a specified
location but it changes from the inlet to the exit (water accelerates along the
nozzle).
Steady flow

9. Acceleration of a Fluid Particle
• Motion of a particle in terms of
distance “s” along a streamline
• Velocity of the particle, V = ds/dt,
which may vary along the streamline
• In 2-D flow, the acceleration is
decomposed into two components,
streamwise acceleration as, and
normal acceleration, an.
2
n
V
a
R
=
• For particles that move along a straight path, an =0

10. Fluid Particle Acceleration
• Velocity of a particle, V
(s, t) = function of s, t
• Total differential
• In steady flow,
• Acceleration,
V V
dV ds dt
s t
∂ ∂
= +
∂ ∂
or
dV V ds V
dt s dt t
∂ ∂
= +
∂ ∂
0;and ( )
V
V V s
t
∂
= =
∂
s
dV V ds V dV
a V V
dt s dt s ds
∂ ∂
= = = =
∂ ∂

11. Derivation of the Bernoulli Equation (1)
• Applying Newton’s second law of conservation of linear momentum relation in
the flow field
( ) sin
dV
PdA P dP dA W mV
ds
θ− + − =
ds is the massm V dAρ ρ= =
W=mg= g ds is the weight of the fluiddAρ
sin =dz/dsθ
- -
dz dV
dpdA gdAds dAdsV
ds ds
ρ ρ=
,dp gdz VdVρ ρ− − =
21
Note V dV= ( ), and divding by
2
d V ρ
21
( ) 0
2
dp
d V gdz
ρ
+ + =
Substituting,
Canceling dA from each term and simplifying,

12. Derivation of the Bernoulli Equation (2)
 Integrating
2
constant (along a streamline)
2
dp V
gz
ρ
+ + =∫
2
constant (along a streamline)
2
p V
gz
ρ
+ + =
For steady flow
For steady incompressible flow,

13. Bernoulli Equation
• Bernoulli Equation states
that the sum of kinetic,
potential and flow (pressure)
energies of a fluid particle is
constant along a streamline
during steady flow.
• Between two points:
2 2
1 1 2 2
1 2 or,
2 2
p V p V
gz gz
ρ ρ
+ + = + +
2 2
1 1 2 2
1 2
2 2
p V p V
z z
g g
+ + = + +
γ γ
2
pressure head; velocity head, z=elevation head
2
p V
g
= =
γ

14. Example 1
Figure E3.4 (p. 105) Flow of
water from a syringe

15. • Water is flowing from a hose attached to a water main at
400 kPa (g). If the hose is held upward, what is the
maximum height that the jet could achieve?
Example 2

16. • Water discharge from a large tank. Determine the water
velocity at the outlet.
Example 3

17.  change in flow conditions
• Frictional effects can not be neglected in long and narrow flow passage,
diverging flow sections, flow separations
• No shaft work
Limitations on the use of Bernoulli Equation

18.  A black body (also blackbody) is an idealized physical body that absorbs all
incident electromagnetic radiation, regardless of frequency or angle of incidence.
A white body is one with a "rough surface [that] reflects all incident rays completely
and uniformly in all directions."[1]
 A black body in thermal equilibrium (that is, at a constant temperature) emits
electromagnetic radiation called black-body radiation. The radiation is emitted
according to Planck's law, meaning that it has a spectrum that is determined by the
temperature alone (see figure at right), not by the body's shape or composition.
 A black body in thermal equilibrium has two notable properties:
 It is an ideal emitter: at every frequency, it emits as much energy as – or more energy
than – any other body at the same temperature.
Black Body Radiation

19.  It is a diffuse emitter: the energy is radiated isotropically, independent of direction.
 An approximate realization of a black surface is a hole in the wall of a large enclosure (see
below). Any light entering the hole is reflected indefinitely or absorbed inside and is unlikely to
re-emerge, making the hole a nearly perfect absorber. The radiation confined in such an
enclosure may or may not be in thermal equilibrium, depending upon the nature of the walls and
the other contents of the enclosure.[3][4]
 Real materials emit energy at a fraction—called the emissivity—of black-body energy levels. By
definition, a black body in thermal equilibrium has an emissivity of ε = 1.0. A source with lower
emissivity independent of frequency often is referred to as a gray body.[5][6] Construction of
black bodies with emissivity as close to one as possible remains a topic of current interest.[7]

20.  Kirchhoff in 1860 introduced the theoretical concept of a perfect black body
with a completely absorbing surface layer of infinitely small thickness, but
Planck noted some severe restrictions upon this idea. Planck noted three
requirements upon a black body: the body must (i) allow radiation to enter but
not reflect; (ii) possess a minimum thickness adequate to absorb the incident
radiation and prevent its re-emission; (iii) satisfy severe limitations upon
scattering to prevent radiation from entering and bouncing back out. As a
consequence, Kirchhoff's perfect black bodies that absorb all the radiation
that falls on them cannot be realized in an infinitely thin surface layer, and
impose conditions upon scattering of the light within the black body that are
difficult to satisfy
Kirchhoff's perfect black bodies

21. Black-body Radiation
λ peak =
2.9 x 10-3 m
T(Kelvin)
Lightintensity
UV IR

22. λpeak vs Temperature
λ peak =
2.9 x 10-3 m
T(Kelvin)
T
3100K
(body temp)
2.9 x 10-3 m
3100
=9x10-6m
58000K
(Sun’s surface)
2.9 x 10-3 m
58000 =0.5x10-6m
infrared light
visible light

23. “Room temperature” radiation

24. The Planck Distribution
A. A. Michelson (late 1900s): “The grand underlying principles (of physics) have been firmly established...
...the future truths of physics are to be looked for in the sixth place of decimals.”
Planck credited with the birth of quantum mechanics (1900)
- developed the modern theory of black-body radiation

25. uantum nature of radiation
1st evidence from spectrum emitted by a black-body
What is a black-body?
An object that absorbs all incident radiation, i.e. no reflection
A small hole cut into a cavity is the most popular and realistic example.
⇒Νονε οφ τηε ινχιδεντ ραδιατιον εσχαπεσ
What happens to this radiation?
• The radiation is absorbed in the walls of the cavity
• This causes a heating of the cavity walls
• Atoms in the walls of the cavity will vibrate at frequencies characteristic of the temperature
of the walls
• These atoms then re-radiate the energy at this new characteristic frequency
The emitted "thermal" radiation characterizes the equilibrium
temperature of the black-body

26. Black-body spectrum

27. lack-body spectrum
• Black-bodies do not "reflect" any incident radiation
They may re-radiate, but the emission characterizes the black-body only
• The emission from a black-body depends only on its temperature
We (at 300 K) radiate in the infrared
Objects at 600 - 700 K start to glow
At high T, objects may become white hot
Wien's displacement Law
λµ Τ = χονσταντ = 2.898 ×10−3 µ.Κ, ορ λµ ∝ Τ−1
Found empirically by Joseph Stefan (1879); later calculated by Boltzmann
σ = 5.6705 ×10−8 Ω.µ−2.Κ−4.
Α βλαχκ−βοδψ ρεαχηεσ τηερµαλ εθυιλιβριυµ ωηεν τηε ινχιδεντ ραδιατιον ποωερ ισ βαλανχεδ βψ
τηε ποωερ ρε−ραδιατεδ, ι.ε. ιφ ψου εξποσε α βλαχκ−βοδψ το ραδιατιον, ιτσ τεµπερατυρε ρισεσ
υντιλ τηε ινχιδεντ ανδ ραδιατεδ ποωερσ βαλανχε.
Stefan-Boltzmann Law
Power per unit area radiated by black-body R = σ Τ 4

28. ayleigh-Jeans equation
Consider the cavity as it emits blackbody radiation
The power emitted from the blackbody is proportional to the radiation energy density in the cavity.
One can define a spectral energy distribution such that u(λ)dλ is the fraction of energy per unit volume
in the cavity with wavelengths in the range λ to λ + dλ.
Then, the power emitted at a given wavelength, R(λ) ∝ u(λ)
u(λ) may be calculated in a straightforward way from classical statistical physics.
= (# modes in cavity in range dλ) × (average energy of modes)
# of modes in cavity in range dλ, ν(λ)δλ = 8πλ−4 δλ
Αϖεραγε ενεργψ περ µοδε ισ κΒΤ, αχχορδινγ το κινετιχ τηεορψ
⇒ υ(λ) = κΒΤ ν(λ) = 8πκΒΤ λ−4

29. Wien, Rayleigh-Jeans and Planck distributions
( ) ( ) ( )
( )
/
RJ W P /4 5 5
8 8
; ;
1B
T
B
hc k T
k T e hc
u u u
e
β λ
λ
π π
λ λ λ
λ λ λ
−
= µ =
−
Wilhelm Carl Werner Otto Fritz Franz Wien

30. he ultraviolet catastrophe
There are serious flaws in the reasoning by Rayleigh and Jeans
Furthermore, the result does not agree with experiment
Even worse, it predicts an infinite energy density as λ → 0!
(This was termed the ultraviolet catastrophe at the time by Paul Ehrenfest)
Agreement between theory and
experiment is only to be found at
very long wavelengths.
The problem is that statistics
predict an infinite number of
modes as λ→0; classical kinetic
theory ascribes an energy kBT to
each of these modes!

31. nck's law (quantization of light energy)
In fact, no classical physical law could have accounted for measured blackbody spectra
The problem is clearly connected with u(λ) → ∞, ασ λ → 0
Planck found an empirical formula that fit the data, and then made appropriate changes to the classical calculation
so as to obtain the desired result, which was non-classical.
Max Planck, and others, had no way of knowing whether the calculation of the number of modes in the
cavity, or the average energy per mode (i.e. kinetic theory), was the problem. It turned out to be the latter.
The problem boils down to the fact that there is no connection between the energy and the
frequency of an oscillator in classical physics, i.e. there exists a continuum of energy states that are
available for a harmonic oscillator of any given frequency. Classically, one can think of such an
oscillator as performing larger and larger amplitude oscillations as its energy increases.

32. axwell-Boltzmann statistics
Define an energy distribution function ( ) ( ) ( )0
exp / ,such that 1Bf E A E k T f E
∞
= − =∫
Then, ( )0 0
exp( / )B BE E f E dE EA E k T dE k T
∞ ∞
= = − =∫ ∫
This is simply the result that Rayleigh and others used, i.e. the average energy of a classical
harmonic oscillator is kBT, regardless of its frequency.
Planck postulated that the energies of harmonic oscillators could only take on discrete values equal to
multiples of a fundamental energy ε = ηφ, ωηερε φ ισ τηε φρεθυενχψ οφ τηε ηαρµονιχ οσχιλλατορ, ι.ε. 0, ε, 2ε,
3ε, ετχ....
Then, En = nε = νηφ ν = 0, 1, 2...
Here, h is a fundamental constant, now known as Planck's constant. Although Planck knew of no physical
reason for doing this, he is credited with the birth of quantum mechanics.

33. he new quantum statistics
( ) ( )exp / exp /n n B Bf A E k T A nhf k T= − = −
Replace the continuous integrals with a discrete sums:
( )
0 0
exp /n n B
n n
E E f nhf A nhf k T
∞ ∞
= =
= = × −∑ ∑
( )
0 0
exp / 1n B
n n
f A nhf k T
∞ ∞
= =
= − =∑ ∑
Solving these equations together, one obtains:
( ) ( ) ( )
/
exp / 1 exp / 1 exp / 1B B B
hf hc
E
k T hf k T hc k T
ε λ
ε λ
= = =
− − −
Multiplying by D(λ), το γιϖε....
( )
5
( )
exp / 1B
hc
u
hc k T
λ
λ
λ
−
µ
−
This is Planck's law

34. Grey Body Radiation

35.  The calculation of the radiation heat transfer between black surfaces is relatively easy
because all the radiant energy that strikes a surface is absorbed.
 The main problem is one of determining the geometric shape factor, but once this is
accomplished, the calculation of the heat exchange is very simple.
 When nonblack bodies are involved, the situation is much more complex, for all the energy
striking a surface will not be absorbed; part will be reflected back to another heat-transfer
surface, and part may be reflected out of the system entirely.
 The problem can become complicated because the radiant energy can be reflected back and
forth between the heat-transfer surfaces several times.
 The analysis of the problem must take into consideration these multiple reflections if correct
conclusions are to be drawn.
 We shall assume that all surfaces considered in our analysis are diffuse, gray, and uniform in
temperature and that the reflective and emissive properties are constant over the entire
surface. Two new terms may be defined:

37. As shown in Figure 8-24, the radiosity is the sum of the energy
emitted and the
energy reflected when no energy is transmitted, or