CCS355 Neural Network & Deep Learning Unit II Notes with Question bank .pdf
Β
Heat transfer
1. HEAT TRANSFER
NAME: KEMKAR JAIMIN AJITKUMAR
EN. NO.: 160123119014
SEMESTER AND CLASS: 5TH D
BATCH: D2
SUBJECT: HEAT TRANSFER (2151909)
SUBJECT GUIDE: PROF. J. A. KATESHIA
2. CONTENT LAYOUT
βͺ Weinβs displacement
βͺ Kirchhoffβs law
βͺ solid angle
βͺ Lambertβs cosine law
βͺ Radiation heat exchange between black bodies
βͺ shape factor
3. WIENβS DISPLACEMENT LAW
βͺ Wienβs established a relationship between the wavelength at which the maximum value of
monochromatic emissive power occurs and absolute temperature of black body
βͺ Wienβs displacement law states that the product of T and π πππ₯ is constant i.e. π πππ₯ &
T=constant
βͺ The relationship between the true temperature of a blackbody (T) in degree kelvin and its
peak spectral existence or dominant wavelength (π πππ₯) is described by the
π πππ₯ =
π
π
=
2898ππ(Β°πΎ)
π
βͺ Where K is constant equaling 2898 ππΒ°πΎ
4. βͺ Wienβs law tells us that objects of
different temperature emit
spectra that peak at different
wavelength.
βͺ For example the average
temperature of the earth is
300Β°πΎ (80Β°πΉ).
π πππ₯ =
2898ππΒ°πΎ
π
π πππ₯ =
2898ππΒ°πΎ
300Β°πΎ
= 9.47ππ
5. Importance of Weinβs displacement law
βͺ The dominant wavelength provides valuable information about which part of the
thermal spectrum we might want to sense in. for example, if we are looking for
800Β°πΎ forest fires that have a dominant wavelength of approximately 3.62ππ then
the most appropriate remote sensing system might be a 3-5 ππ thermal infrared
detector.
βͺ If we are interested in soil, water and rock with ambient temperature on the
earthβs surface of 300Β°πΎ and a dominant wavelength of 9.66ππ, then a thermal
infrared detector operating in the 8 β 14ππ region might be most appropriate.
6. KIRCHHOFFβS LAW
βͺ At thermal equilibrium, the emissivity of a body (or surface) equals its absorptivity.
βͺ This law states that the ratio of total emissive power to absorptivity is constant for all
substances which are in thermal equilibrium with the surroundings. (books India)
βͺ This can be written in mathematical form for four bodies as follow:
πΈ1
πΌ1
=
πΈ2
πΌ2
=
πΈ3
πΌ3
=
πΈ4
πΌ4
βͺ Assume out of any four body, any one body, say fourth one is a black body. Then
πΈ1
πΌ1
=
πΈ2
πΌ2
=
πΈ3
πΌ3
=
πΈ π
πΌ π
[πΌ π = 1]
πΈ1
πΈ π
= πΌ1,
πΈ2
πΈ π
= πΌ2,
πΈ3
πΈ π
= πΌ3
7. βͺ But according to the definition of emissive
(π), we have
π1 = πΌ1, π2 = πΌ2, π3 = πΌ3
βͺ From equation, in general we can say
πΌ = π
8. SOLID ANGLE
βͺ A unit solid angle is defined as the angle covered by unit
area on a surface of sphere of unit radius when joined with
the center of the sphere and it is measured in the
steradians.
βͺ Or a solid angle is defined as a portion of the space inside
a sphere enclosed by a conical surface with the vertex of
the cone at the center of sphere. It is measured by the ratio
of the spherical surface enclosed by the cone to the square
of the radius of the sphere.
βͺ The solid angle subtended by the complete hemisphere is
given by
2ππ2
π2 = 2π and full sphere is
4ππ2
π2 = 4π.
ππ =
ππ΄π
π2 where, An = normal area
r = radius of sphere
9. LAMBERTβS COSINE LAW
βͺ Monochromic or spectral intensity of radiation is defined as the radiant energy emitted by a
black body at a temperature T, streaming through a unit area normal to the direction of
propagation per unit wavelength about a wavelength per unit solid angle about the
propagation of beam.
βͺ it is denoted by πΌ ππ and can be expressed as:
πΌ ππ =
ππππππ¦ ππππ‘π‘ππ
(πππππππ‘ππ ππππ)Γ(π€ππ£π πππππ‘β)Γ(π ππππ πππππ)
π
π2βππβππ
βͺ Lambert cosine law states that the total emissive power (E) from a surface in any direction is
directly proportional to the angle of emission.
πΈ = πΈ π β πππ π
where, E = total emissive power
En = total emission power in normal direction
10. RADIATION HEAT EXCHANGER BETWEEN TWO
BLACK BODIES
βͺ The radiant heat exchange between two bodies depends upon
1. The medium that intervenes the two bodies
2. The emitting characteristics of two bodies
3. The views the surface have of each other, i.e. how they βseeβ each other.
βͺ Let us assume that two bodies are black having non-absorbing medium between them. Let
us consider the area π π΄1 and π π΄2 of two surface. The distance between two bodies is βrβ and
the angles made by the normal β 1 and β 2 respectively. The energy leaving the surface πΌ with
π π΄1 area and absorbed by π π΄2is
(π12) πππ‘= π1β2 β π2β1
= ππ΄1 πΉ12(π1
4
β π2
4
)
11. SHAPE FACTOR
βͺ The shape factor may be defined as the fraction of radiative energy that is diffused from one
surface element and strikes the other surface directly with no intervening reflections.
π1β2 = πΉ1β2 β π β π΄1 β π1
4
π1β2 = π΄2 β πΉ1β2 β π β π2
4
βͺ Above equation is applicable to black surface only and must be used for surfaces having
emissivity other than one.