UNIT-III

RADIATION

Thermal radiation is electromagnetic radiation generated by the thermal motion of charged

particles in matter. All matter with a temperature greater than absolute zero emits thermal

radiation. When the temperature of the body is greater than absolute zero, inter-atomic collisions

cause the kinetic energy of the atoms or molecules to change. This results in charge-

acceleration and/or dipole oscillation which produces electromagnetic radiation, and the wide

spectrum of radiation reflects the wide spectrum of energies and accelerations that occur even at

a single temperature.

Sunlight is part of thermal radiation generated by the hot plasma of the Sun. The Earth also

emits thermal radiation, but at a much lower intensity and different spectral distribution (infrared

rather than visible) because it is cooler. The Earth's absorption of solar radiation, followed by its

outgoing thermal radiation are the two most important processes that determine the temperature

and climate of the Earth.

A blackbody refers to an opaque object that emits thermal radiation. A perfect blackbody is one

that absorbs all incoming light and does not reflect any. At room temperature, such an object

would appear to be perfectly black (hence the term blackbody). However, if heated to a high

temperature, a blackbody will begin to glow with thermal radiation.

In fact, all objects emit thermal radiation (as long as their temperature is above Absolute Zero, or

-273.15 degrees Celsius), but no object emits thermal radiation perfectly; rather, they are better

at emitting/absorbing some wavelengths of light than others. These uneven efficiencies make it

difficult to study the interaction of light, heat and matter using normal objects.

At the beginning of the 20th century, scientists Lord Rayleigh, and Max Planck (among others)

studied the blackbody radiation using such a device. After much work, Planck was able to

empirically describe the intensity of light emitted by a blackbody as a function of wavelength.

Furthermore, he was able to describe how this spectrum would change as the temperature

changed. Planck's work on blackbody radiation is one of the areas of physics that led to the

foundation of the wonderful science of Quantum Mechanics, but that is unfortunately beyond the

scope of this article.

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Planck and the others found was that as the temperature of a blackbody increases, the total

amount of light emitted per second increases, and the wavelength of the spectrum's peak shifts

to bluer colors .

For example, an iron bar becomes orange-red when heated to high temperatures and its color

progressively shifts toward blue and white as it is heated further.

In 1893, German physicist Wilhelm Wien quantified the relationship between blackbody

temperature and the wavelength of the spectral peak with the following equation:

where T is the temperature in Kelvin. Wien's law (also known as Wien's displacement law) states

that the wavelength of maximum emission from a blackbody is inversely proportional to its

temperature. This makes sense; shorter-wavelength (higher-frequency) light corresponds to

higher-energy photons, which you would expect from a higher-temperature object.

For example, the sun has an average temperature of 5800 K, so its wavelength of maximum

emission is given by:

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This wavelengths falls in the green region of the visible light spectrum, but the sun's continuum

radiates photons both longer and shorter than lambda(max) and the human eyes perceives the

sun's color as yellow/white.

In 1879, Austrian physicist Stephan Josef Stefan showed that the luminosity, L, of a black body

is proportional to the 4th power of its temperature T.

where A is the surface area, alpha is a constant of proportionality, and T is the temperature in

Kelvin. That is, if we double the temperature (e.g. 1000 K to 2000 K) then the total energy

radiated from a blackbody increase by a factor of 24 or 16.

Five years later, Austrian physicist Ludwig Boltzman derived the same equation and is now

known as the Stefan-Boltzman law. If we assume a spherical star with radius R, then the

luminosity of such a star is

where R is the star radius in cm, and the alpha is the Stefan-Boltzman constant, which has the

value:

Blackbody Radiation Heat Transfer

The heat emitted by a blackbody (per unit time) at an absolute temperature of T is given by

the Stefan-Boltzmann Law of thermal radiation,

where has units of Watts, A is the total radiating area of the blackbody, and s is the Stefan-

Boltzmann constant.

A small blackbody at absolute temperature T enclosed by a much larger blackbody at absolute

temperatureTe will transfer a net heat flow of,

The small blackbody still emits a total heat flow given by the Stefan-Boltzmann law. However, the

small blackbody also receives and absorbs all the thermal energy emitted by the large enclosing

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blackbody, which is a function of its temperature Te. The difference in these two heat flows is the

net heat flow lost by the small blackbody.

Gray Body Radiation Heat Transfer

Bodies that emit less thermal radiation than a blackbody have surface emissivities e less than 1.

If the surface emissivity is independent of wavelength, then the body is called a "gray" body, in

that no particular wavelength (or color) is favored.

The net heat transfer from a small gray body at absolute temperature T with surface

emissivity e to a much larger enclosing gray (or black) body at absolute temperature Te is given

by,

Radiation Exchange between small gray bodies:

Consider two gray bodies 1 and 2 having emissivities 1 , 2 or absorptivities 1, 2.

They are said to be small if their size is very small compared to the distance between them. The

radiation emitted by 1 is partly absorbed by 2. The portion of radiation unabsorbed and thus

reflected on the first incidence is considered to be lost in space, that is, nothing returns again to

surface 1. The same can be said of surface 2 as well.

The energy emitted by body 1 = A11T14

The energy incident on body 2 = F12A121T14

The energy absorbed by body 2 = 2F12A11T14

The energy transfer from 1 to 2 is

Q1 = 12A1F12 T14

Similarly energy transfer from 2 to 1 is

Q2 = 12A2F21 T24

Net radiant heat exchange between the two bodies is

Q12 = 12AF ( T14

- T24)

Where AF = A1F12 = A2F21

4

Radiation exchange between large parallel gray planes Consider two very large parallel gray surfaces A1 and A2,a small distance apart and exchanging

radiation. All the radiation emitted by one plane must reach and be intercepted by the other

plane.

Radiation emitted by A1 = 1T14

Radiation absorbed by A2 = 21T14

(on first incidence)

Radiation reflected by A2 = 21T14

Radiation absorbed by A1 = 121T14

Radiation reflected by A1 = 121T14

Radiation absorbed by A2 = 2121T14 = 1212T1

4

The same applies to surface 2. The net exchange of energy for an area A is

Radiation shields Radiation shields are often used to reduce the heat transfer by radiation between surfaces by

effectively increasing the surface resistance without actually removing the heat from the overall

system. A very effective insulation can be provided by using many layers of radiation reflecting

films separated by a vacuum. Thin sheets of plastic coated with highly reflecting metallic films on

both sides serve as very effective radiation shields. A familiar application of radiation shields is

in the measurement of the temperature of a fluid by a thermometer or a thermocouple which is

shielded to minimize the radiation effects.

Consider use of a single shield in a two-surface enclosure, such as that associated with large

parallel plates:

5

This equation can be generalized for a system of two parallel plates separated by n screens of

emissivity as

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1. Calculate the following for an industrial furnace in the form of a black body and emitting

radiation at 2500 0C

(i) Monochromatic emissive power at 1.2 μm length

(ii) Wavelength at which the emissive is maximum

(iii) Maximum emissive powers

(iv) Total emissive power

Given Data:

Surface temperature, T = 2500 0C = 2773 K

To Find :

(i) Monochromatic emissive power, Eb at λ = 1.2 m length

λ = 1.2X10-6 m

(ii) Wavelength at maximum, λmax

(iii) Maximum emissive power, (Ebλ)max

(iv) Total emissive power, Eb

Solution:

(i) Ebλ = C1λ-5

C2

e λT -1 [HMT data book. Pg. 82]

Let C1 = 0.374177107 x10-15 W/m2 C2 = 0.014387752 mk

Ebλ = 0.374177107X10-15 X (1.2X10-6)-5 = 1.5037X1014

0.014387752 74.472

1.2X10-6X2773

e -1

Ebλ = 2.019X1012 W/m2

(ii) λmax T = 2898 μmk

λmax . 2773 = 2898 X 10-6 mk

λmax = 1.045 X10-6m

(iii) (Ebλ)max = 1.307X10-5 T5

= 1.307X10-5 (2773)5

(Ebλ) max = 2.143X 1012 W/m2

(iv) Eb = σT4

= 5.67X10-8 x (2773)4

Eb = 3.352 X106 W/m2

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2. Two parallel plates of size lm x lm spaced 0.5m apart are located in very large room, the walls are

maintained at a temperature of 27°C. One plate is maintained at a temperature of 900°C and the

other at 400°C. Their emissivities are 0.2 and 0.5 respectively. If the plates exchange heat between

themselves and surroundings, find the heat transfer to each plate and to them. Consider only the

plate surface facing each other. (16 marks) [ May/June 2012]

Given Data:

Plate size = 1m x 1m

Plate distance = 0.5 m

First plate temperature, T1 = 9000C = 1173k

Second plate temperature, T2 = 4000C = 673 k

Emissivity of first plate, ε1 = 0.2

Emissivity of second plate, ε2 = 0.5

To Find:

a) Heat transfer to each plate

b) Heat transfer to room.

Solution:

Heat transfer take place two plates and also room. So, this is three surface problem.

Area, A1 = 1 x 1 = 1m2 A1 = A2 =1m2

The room size is infinity. so A3 = ∞ [HMT data book pg.no.93]

X = L/D = 1/0.5 = 2

Y = B/D = 1/0.5 = 2

F12 = 0.41525

We know that, F11+F12+F13 = 1 But, F11 = 0 [Due to room]

We know that,

F11+F12+F13 = 1 But, F11 = 0

So, 0 +F12+F13 = 1

F13 = 1 – F12

= 1-0.41525

F13= 0.5847

Similarly, Stefan – Boltzmann law, Eb = σT4

F21+F22+F23 =1 Eb1 = σT14

But, F22 = 0 [Due to room] = 5.67X10-8 (1173)4

F21 + 0 + F23 = 1 Eb1= 107.34 X103 W/m2

F23 = 1-F21 = 1-0.41525

F23 = 0.5847

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Eb2 = σT24 = 5.67x10-8 (673)4

Eb2 = 11.63 x 103 W/m2

Eb3 = σT34 = 5.67x10-8 (300)4

Eb3 = 459.27 W/m2

At Node J1,

Eb1 – J1 + J2 – J1 + Eb3 –J1 = 0

4 1 1

A1F12 A1 F13

107.34 x103-J1 + J2 –J1 + 459.27 –J1 = 0

4 1 1

1x 0.41525 1x 0.5847

26835 - J1 + J2 - J1 + 268.54 - J1 = 0

4 2.408 2.408 1.102

26835 - 0.25 J1 + 0.415 J2 - 0.45 J1 + 268.54 - 0.5847 J1 = 0

-1.2497 J1 + 0.415 J2 = -27.10 x103 1

At Node J2,

(J1-J2) / (1/A1F12) + (Eb3 -J2) /1/A2F23 + (Eb2-J2)/2 =0

(J1-J2 ) / (1/1x 0.41525) + (459.27-J2) / (1/1x0.5847) + (11.63x103-J2) / 2 = 0

J1/2.408 - J2/2.408 + 268.54 - 0.5847 J2 + 5.815 x103 – 0.5 J2 = 0

0.415 J1 - 1.4997 J2 = - 6.08 x103 2

From solving Eqn. 1 & Eqn. 2

-1.2497 J1 + 0.415 J2 = - 27.10 x 103

0.415 J1 – 1.4997 J2 = - 6.08 x103

We get, J2 = 11.06 x 103 W/m2

J1 = 25.35 x 103 W/m2

Heat lost by plate 1, Q1 = Eb1 – J1

1-ε1 = 107.34X103 – 25.35 X103

A1ε1 (1-0.2) / (1X0.2)

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Q1 = 20.49 x 103 W

Heat lost by plate 2, Q2 = Eb2-J2 = 11.63X103 – 11.06 x 103

(1-ε2) /A2ε2 (1-0.5)/1 x 0.5

Q2 = 570W

Total heat lost, Q = Q1+Q2 = 20.49X103+570

Q = 21.06X103W

Total heat received/absorbed by room, Q = J1 – J3 + J2 - J3

1/ A1 F13 1/ A2F23

J3 = Eb3

= 25.35 x10-3 – 459.27 + 11.06 x103 – 459.27

1/1 x 0.5847 1/1 x 0.5847

Q =20.752 x 103 W

3. Derive Wien’s displacement law of radiation from plank’s law. (May/June 2012)

Wien’s displacement law states that the product of λmax and T is constant

(i.e.) λmax.T = Constant

From Planck’s law,

(Eλ)b = 2πC2hλ-5

e ch/λKT –1

(Eλ)b becomes maximum (if T = Constant) when

d(Eλ)b = 0

dλ.

d(Eλ)b = d c1 λ-5 = 0 (or)

d λ d λ c2

exp λt -1

exp C2/ λT -1 -5C1 λ-6 – C 1 λ-5 e C2/ λT C2/T -1/ λ2 = 0

e c2/ λT -1 2

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- exp c2/ λT + 1+(1/5) C2 (1/ λT) exp C2/ λT = 0

By Trial and error,

λmax T = C2 = 1.439X104 µmK

4.965 4.965

λmaxT =2898 µmk

4. (a) Find The energy emitted by a black body at 700°C [ Nov/Dec-2011]

GIVEN DATA:

Surface temperature, T = 7000C = 973k.

To FIND:

Energy emitted by a black body

SOLUTION :

Energy emitted, Eb = σT4

σ = 5.6 X 10-8 W/m2 K4 HMT data book Pg.no .82

Eb = 5.67X10-8 (973)4 = 50.82X103 W/m2

(b) A furnace is approximated as an equilateral triangular duct of sufficient length so that end effect

can be neglected. The hot wall of the furnace is maintained at 900k and has the same emissivity. Find

the net radiation heat blur leaving. The wall. Third wall of the furnace may be assured as a

reradiating surface.

Two of the surfaces of a long equilateral triangular furnace are maintained at uniform

temperature while the third wall is reradiating surface.

T2 = 900 K

ε2 = 0.8 2 3 Q1

Eb1

1

T1 = 400K

ε 1 = 0.8

The radiation network in This Case is a simple series parallel Connection.

Q1 = Eb1 – Eb2

A R1 + 1 + 1 + 1

R12 R13 R23

Q1 = Eb1 – Eb2

A 1-ε1 + F12 + 1 + 1-ε2

ε1 1/R13 + 1/R23 ε2

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F12 = F13 = F23 = 0.5 [Symmetry]

Eb1 = σT14 = 5.67 x 10-8 x (400)4 = 1451.52 W/m2

Eb2 = σT24 = 5.67x 10-8 x (900)4 = 37200.87 W/m2

Q = 37200.87-1451.52

A 1-0.8 + 0.5 + 1 + 1-0.8

0.8 0.5+0.5 0.8

Q = 17874.675 W/m2

A

1) a) Considering radiation in gases, obtain the exponential – decay Formula.

[Nov/Dec – 2011]

Consider a beam of monochromatic radiation with an intensity Iλ0 entering a gas layer of

Thickness L.

As the beam passes through The gas layer, its intensity gets reduce and the decrease is given by.

dIλx = -Kλ Iλx. dx

Where,

Iλx – Monochromatic intensity at a distance x

Kλ – Proportionality Constant

Integrating above equation between x=0 and X=L

∫X=Lx=0 dIλx = ∫x=L x=0 – Kλ . dx

Iλx

Ln IλL = -Kλ.L

Iλ0

IλL = Iλ0 e-Kλ.L

Where, IλL – radiation intensity at x = L

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5. b) Consider two concentric cylinders having diameters 10cm and 20cm and a length of 20cm.

Designating The open ends of the cylinders as surface 3 and 4, estimate the shape factore, F3-4 (10

marks) [Nov/Dec – 2011]

From data book pg.no 104

r3 = 5 cm

r4 = 10 cm; L=20cm

L/r4 = 20/10 = 2; r3/r4 = 5/10 = 0.5

F4-3 = 0.44

A3F3-4 = A4F4-3

F3-4 = A4/A3. F4-3 = π x102 x 0.44 = 4 x0.44 = 1.76

π x52

Shape factor, F3-4 = 1.76

6. a) Distinguish between irradiation and radiosity ? [ May/June 2013]

Radiation intensity is nothing but radiation heat transfer. The heat radiated per unit time per unit

area is The radiation intensity ( Q/A = q)

Irradiation: G = Total radiation incident upon a surface per unit time and per unit area.

Q/A = J- G (or) Q = Eb – J

(1-ε)/εA

b ) A double walled cylindrical vessel used for storing liquid oxygen at -18.c The inner wall is

coated with a point with ε = 0.02. The temperature of the inner surface of the outer wall is 2.c. It is

also coated with same painting. The inner wall surface area is 80% of outer the inner wall surface

area is 0.15 m2. Determine heat radiated. [Apr/May 2012 ]

SOLUTION :

A1 = 0.15 m2

0.15 = 0.8 A2

A2 = 0.15/0.8 = 0.1875 m2

R1 = 1-ε1 = 1-0.02 = 326.66 1/m2

A1ε2 0.15x0.02

R2 = 1-ε2 = 1-0.02 = 261.33 1/m2

A2ε2 0.1875x0.02

F12 = 1 [Since “L” is not given, L = , Refer page 84 of HMT data book ]

R = 1 = 1 = 6.666 /m2

A1F12 0.15x1

Eb1 – Eb2 = σ (T14 – T2

4 ) = 5.67X10-8 (2554 – 2934) = -178.14 W/m2

Q12 = Eb1-Eb2 = -178.14

R1+R+R2 326.66+6.66+261.33

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Q12 = -0.29956 W

“-ve sign indicates that heat is radiated in opposite direction.

7. Two very large parallel planes are respectively. 0.8 and 0.3. To minimize the radiation exchange

between the planes, a polished aluminum radiation shield is placed between them. If the emissivity of

the shield is 0.04 on both sides, find the percentage reduction in heat transfer rate. (10

marks)[Apr/May – 2011]

GIVEN DATA:

Emissivity, ε1 = 0.8, ε2 = 0.3, ε3 = 0.04

To find :

% Reduction in heat transfer rate.

Solution:

Heat transfer without radiation shield.

Q12 = ε σ A (T14 – T2

4)

Let ε = 1 = 1

1/ε1 + 1/ε2 – 1 1/0.8 + 1/0.3 -1

ε = 0.279

Q12 = 0.279 σ A (T14 – T2

4 ) 1

Heat Transfer with radiation shield

Qradiation shield = A σ (T14- T2

4)

1/ε1 + 1/ε2 + 2n/ε3 – (n+1)

Let n = No. of radiation shield. Qradiation shield

Qradiation shield = A σ (T14 – T2

4)

1/0.8+1/0.3+2X1/0.04 – (1+1)

= A σ[ T14 –T2

4 ]

52.58

Qradiation shield = 0.019 A σ (T14 – T2

4)

% Reduction in heat transfer rate = Qwithout shield – Q with shield

Qwithout shield

= 0.2790 σ A (T14 – T2

4 ) – 0.019A σ (T14 – T2

4)

0.279 σ A (T14 – T2

4)

= 0.279-0.019 = 93.18%

0.279

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8. Define the Following [ Nov/Dec – 2010 ]

(i) Black body (ii) Grey body (iii) Opaque body (iv) white body

(i) Black body :

This black body absorbs all the radiation falling on it surface.

For this body α = 1, So τ = 0, ρ = 0

(ii) Grey body :

For this grey body, absorptivity does not very with temperature and wave length of the incident

radiation. So, for a gray body α = (α) λ = Constant.

(iii) Opaque body :

This opaque body does not transmit any radiation falling on its surface. Here τ = 0, So, α+ ρ

= 1

(iv) White body:

This white body reflected all the radiation felling on its surfaces

9. The radiation shape factor of the circular surface of a thin hollow cylinder of 10cm diameter and

10cm length is 0.1716. What respect to itself ? [Nov/Dec – 2010]

Given data :

r1 =r2 = 10/2 = 5cm, L = 10cm, F1-2 = 0.1716

Here F1-2 = F2-1 as A1 = A2

The shape factor relation between all the three surfaces is given by,

F1-1 + F1-2 + F1-3 = 1 1

F3-3 + F3-2 + F3-1 = 1 2

But F3-1 = F3-2 and F1-1 = F2-2 = 0 3

Substituting (3) in (1)

F3-3 + F3-1 + F3-1 = 1

F3-3 = 1-2 F3-1 4

Also A1F1-3 = A3F3-1

F3-1 = F3-1 x A1/A3 = F1-3 x πR2

2πrL

F3-1 = F1-3 X r/2L

From equation (1):

F1-3 = 1-F1-2 (F1-1 = 0) = 1-0.1716 = 0.8284 5

F3-1 = 0.8284 x 5/(2x10) = 0.2071

Now substituting F3-1 in eqn. (4)

F3-3 = 1-2 (0.2071) = 0.5858

10. (a) Discuss briefly the variation of black body emissive power with wave length for different

temperatures [ Apr/May – 2008 ]

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A plot of (Eλ) b as a function of temperature and wavelength is given in

The plot shows the following characteristics of block body radiations.

1) The energy emitted at all wavelength increases with rise in temperature.

2) The peak spectral emissive power shift towards a smaller, wavelength at higher temperature.

3) The area under the monochromatic emissive power versus wavelength at any temperature. Given the rate

of radiation energy emitted within the wavelength interval dλ.

dEb = (Eλ)b . dλ.

b) The spectral emissivity function of an opaque surface at 800K is approximated as .

ε 1 = 0.3 0 ≤ λ 3µm

ε λ = ε 2 = 0.8 3 µm ≤λ 7µ m

ε 3 = 0.1 7 µm ≤λ

Calculate the average emissivity of the surface and its emissive power.

ε 1 = 0.3 0 ≤λ 3µm

ε λ = ε 2 = 0.8 3 µm ≤λ 7µ m

ε 3 = 0.1 7 µm ≤λ ∞

SOLUTION:

Average emissivity of the surface,

ε = ε 1 ε 2 + ε 2 ε 3 + ε 3 ε 1

= (0.3 x0.8) + (0.8 x 0.1) + (0.1 x 0.3) = 0.35

Emissive power, Eb = σ ε T4 (or) ε σ T4

= 0.35 x 5.6 7 x 10-8 x 8004

Eb = 8128.5 W/m2

11) Explain briefly the following [ April/May – 2008 ]

(i) Specular and diffuse reflection

(ii) Reflectivity & transmissivity

(iii) Reciprocity rule and summation rule

(i) Specular and diffuse reflection :

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Specular and reflection occurs from a surface such as a mirror, which is very smooth, an image of the

source of radiation is projected the angle of reflection is equal to angle of incidence. Diffuse

radiation occurs when the surface is rough and the reflection from the surface occurs partially in

discriminately in all directions.

(ii) Reflectivity and transmissivity :

ρ = Qρ/Q is the fraction of incident

Radiation reflected and is called reflectivity.

τ = Qτ/Q is the fraction of incident radiation transmitted and is called transmissivity.

(iii) Reciprocity rule and summation rule:

If Q be the rate at which a surface receives heat and of this amount Q is reflected, Qτ transmitted and

Qx absorbed. Then by the principle of conservation of energy, total sum must be equal to incident

radiation.

(i.e) Reflection + Transmission + Absorption = Incident radiation.

Qρ + Qτ + Qα = Q (or) Qρ/Q + Qτ/Q + Qα/Q = 1 ρ+τ+α = 1

1. A) Deduce the generalized equation for heat transfer to a system of two parallel plates separated by

“in” screens. Consider two parallel planes A1 and A2 as shown.

Let us consider two parallel planes, 1 and 2 each of area A at temperatures T1

and T2 respectively with a radiation shield placed between them as shown in fig. It is known that the net

heat exchange between parallel planes without any radiation shield placed between them is,

Q12 = A σ (T14 – T2

4 ) A

1 + 1 – 1

ε 1 ε 2

If the emissivity of the radiation shield is ε3, we can use this equation to find heat exchange between 1, 3

and 3,2.

Q13 = Aσ (T14 – T2

4) B

1 + 1 –1

ε 1 ε 3

Q32 = A σ (T34 – T2

4 ) C

1 + 1 – 1

ε 3 ε 2

Since the shield does not deliver or remove heat from the system:

Q13 = Q32

Eliminating T3 from equation (B) and (C).

Q13 = A σ (T14 – T2

4) D

(1/ ε 1 + 1/ ε 3 – 1 ) + (1/ ε 3 + 1/ ε 2 – 1)

Dividing equation (D) by (A), we get.

Q13 = (1/ ε 1 + 1/ ε 2 – 1) E

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Q12 (1/ ε 1 + 1/ ε 3 – 1) + (1/ ε 3 + 1/ ε 2 – 1)

If ε 1 = ε 2 = ε 3, Then right side of equation (E) reduces to ½ (or)

Q13 = Q32 = ½ Q12

Thus by inserting one shield between two parallel surface, the direct radiation heat

transfer between them is halved.

Sometime more than one shield is used, and in general case where there are in shields, all the

surface resistance would be the same, since the emissivities are equal. There will be two of these surface

resistance for each shield and one for each heat transfer surface, there will also be (n+1) shape resistance but

the shape factors are unity for all infinite parallel planes.

Total Resistance (Rn)= (2n+2) (1- ε / ε) + (n+1) (1) = (n+1) (2/ε–

1)

A A

The radiant heat transfer rate between two infinitely large parallel plates

separated by in shields is therefore.

Qn = 1 . A σ (T14 – T2

4 )

(n+1)(2/ε – 1 )

12.a)Emissivities of two parallel plates at 800.C and 300.C are 0.3 and 0.5 respectively. Find the net

energy transfer rate per square meter. [ Nov/Dec – 2007 ]

SOLUTION :

The heat exchange per unit area between two large parallel plates is given by.

Q12 = q12 = σ(T14 – T2

4 ) = 5.67 X 10-8 [ 10734 – 5734]

A 1/ε1 + 1/ε2 – 1 1/0.3 + 1/0.5 – 1

= 5.67 X 12352 = 16.2 KW/m2

4.33

b) Show from energy balance consideration that the radiation heat transfer from a plane Composite

surface area A4 and made up of plane surface area A2 and A3 to plane surface area A1 is given by,

A4F41 = A3F31 + A2 F21. and F14 + F12 +F13

PROOF :

If two surface A1 and A4

Are parallel and large, radiation

Occurs a gross the gap between them

so that A1 and A4 and all

radiation emitted by one falls on

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the other, then

F1-4 = F4-1 = 1

If one of the surface say A4 is divided into subareas A2 and A3 then A1F1-2 = A2 F12 + A3 F13

Thus it the radiant surface is subdivided, the shape factor gore that surface with respect to the

receiving surface is not equal to sum of individual shape factors.

A4F4-1 = A3F31 + A2F21 (or) F41 = F31 + F21

Hence the shape factor from a radiating surface to a subdivided receving surface is simply the sum

of individual shape factors.

c) Using the definition of radiosity and irradiation prove that the radiation heat exchange between

two grey bodies is given by the radiation

Qnet = σ(T14 – T24)

1-E1 + 1 + 1-ε2

A1E1 A1F1-2 A2ε2

The radiosity comprises meth original emit lance from the surface plus the reflected portion of any

radiation incident upon it.

J=E+PG (or) J=Eb +PG

Where Eb = Emissive power of a perfect black body at the same temperature.

Also α+ρ+τ=1

α+ρ = 1 (τ =0, the surface black opaque)

(or) ρ =1 – X

J = ε Eb + (1-α) G

But α = ε (by Kirchhoff’s law)

(or) J = ε Eb + (1-ε) G

(or) G = J-εEb

1-ε

The net energy leaving a surface is the difference between its radiosity and irradiation, Thus.

Qnet = J-G = J- εEb = J(1-ε) – (J-εEb)

A (1-ε) 1-ε

= J-Jε – J+εEb = ε[Eb –J]

1-ε 1-ε

Qnet = Aε ( Eb-J) = Eb-J

1-ε 1-ε/Aε

19

The quantity 1-ε/Aε is called surface resistance.

Now consider the exchange of radiant energy between two surface 1 and 2

The net interchange of heat between the surface (Q12) is given by Q12 = J1A1F1-2 – J2A2F2-1

But A1F1-2 = A2F2-1

Q12 = A1F1-2 (J1-J2)

(or) Q1-2 = J1-J2/1/A1F12)

1 is called space resistance.

A1F1-2

The net heat exchange between two gray surface is given by

1-ε1 1 1-ε2

A1ε1 A1F1-2 A2ε2

(Q12)net = Eb1 – Eb2 = σ(T14-T24)

1-ε1 + 1 + 1-ε2 1-ε1 + 1 +

1-ε2

A1ε1 A1F1-2 A2ε2 A1ε1 A1F1-2

A2ε2

13. A black body at 3000 K emits radiation. Calculate the following:

i) Monochromatic emissive power at 7 m wave length.

ii) Wave length at which emission is maximum.

iii) Maximum emissive power.

iv) Total emissive power,

v) Calculate the total emissive of the furnace if it is assumed as a real surface having emissivity

equal to 0.85.

Given: Surface temperature T = 3000K

Solution: 1. Monochromatic Emissive Power:

From Planck’s distribution law, we know

5

1b

2

e 1

CE

C

T

[From HMT data book, Page No.71]

Where

c1 = 0.374 10-15 W m2

c2 = 14.4 10-3 mK

20

= 1 10-6 m [Given]

15 6 5

b 3

6

1

12 2

b

0.374 10 [1 10 ] E

144 10

1 10 3000

E 3.10 10 W/m

2. Maximum wave length (max)

From Wien’s law, we know

3

max

3

max

-6

max

T 2.9 10 mK

2.9 10 =

3000

= 0.966 10 m

3. Maximum emissive power (Eb) max:

Maximum emissive power

(Eb)max = 1.307 10-5 T5

= 1.307 10-5 (3000)5

(Eb)max = 3.17 1012 W/m2

4. Total emissive power (Eb):

From Stefan – Boltzmann law, we know that

Eb = T4

[From HMT data book Page No.71]

Where = Stefan – Boltzmann constant

= 5.67 10-8 W/m2K4

Eb = (5.67 10-8) (3000)4

Eb = 4.59 106 W/m2

5. Total emissive power of a real surface:

(Eb)real = T4

Where = Emissivity = 0.85

(Eb)real = 8 40.85 5.67 10 (3000)

6 2

b real(E ) 3.90 10 W /m

21

14. Assuming sun to be black body emitting radiation at 6000 K at a mean distance of 12 1010

m from the earth. The diameter of the sun is 1.5 109 m and that of the earth is 13.2 106 m.

Calculation the following.

1. Total energy emitted by the sun.

2. The emission received per m2 just outside the earth’s atmosphere.

3. The total energy received by the earth if no radiation is blocked by the earth’s atmosphere.

4. The energy received by a 2 2 m solar collector whose normal is inclined at 45 to the sun. The

energy loss through the atmosphere is 50% and the diffuse radiation is 20% of direct radiation.

Given: Surface temperature T = 6000 K

Distance between earth and sun R = 12 1010 m

Diameter on the sun D1 = 1.5 109 m

Diameter of the earth D2 = 13.2 106 m

Solution:1. Energy emitted by sun Eb = T4

-8 4

b

-8 2 4

E = 5.67 10 (6000)

[ = Stefan - Boltzmann constant

= 5.67 10 W /m K ]

6 2

b

2

1 1

29

18 2

1

E = 73.4 10 W/m

Area of sun A 4 R

1.5 10 = 4

2

A 7 10 m

6 18

b

26

b

Energy emitted by the sun

E = 73.4 10 7 10

E 5.14 10 W

2. The emission received per m2 just outside the earth’s atmosphere:

The distance between earth and sun R = 12 1010 m

22

2

10 2

23 2

2

b

26

23

Area, A = 4 R

= 4 (12 10 )

A = 1.80 10 m

The radiation received outside the earth atmosphere per

m

E =

A

5.14 10 =

1.80 10

= 2855.

25 W/m

3. Energy received by the earth:

2

2

6 2

4 2

Earth area = (D )4

= [13.2 10 ]4

Earth area = 1.36 10 m

Energy received by the earth

4

17

2855.5 1.36 10

3.88 10 W

4. The energy received by a 2 2 m solar collector;

Energy loss through the atmosphere is 50%. So energy reaching the earth.

100 - 50 = 50%

= 0.50

Energy received by the earth

2

0.50 2855.5

1427.7 W/m ......(1)

Diffuse radiation is 20%

2

2

0.20 1427.7 = 285.5 W/m

Diffuse radiation = 285.5 W/m .........(2)

Total radiation reaching the collection

2

142.7 285.5

1713.2 W/m

2

Plate area = A cos

= 2 2 cos 45

= 2.82 m

Energy received by the collector

23

2.82 1713.2

4831.2 W

15. Two black square plates of size 2 by 2 m are placed parallel to each other at a distance of 0.5 m.

One plate is maintained at a temperature of 1000C and the other at 500C. Find the heat exchange

between the plates.

Given: Area A = 2 2 = 4 m2

T1 = 1000C + 273

= 1273 K

T2 = 500C + 273

= 773 K

Distance = 0.5 m

To find : Heat transfer (Q)

Solution : We know Heat transfer general equation is

where

4 4

1 2

121 2

1 1 1 12 1 2

T TQ

1 11

A A F A

[From equation No.(6)]

For black body 1 2 1

4 4

12 1 2 1 12

8 4 4 12

5

12 12

Q [T T ] A F

= 5.67 10 (1273) (773) 4 F

Q 5.14 10 F ......(1)

Where F12 – Shape factor for square plates

In order to find shape factor F12, refer HMT data book, Page No.76.

Smaller sideX axis =

Distance between planes

2 =

0.5

X axis = 4

Curve 2 [Since given is square plates]

X axis value is 4, curve is 2. So corresponding Y axis value is 0.62.

i.e., 12F 0.62

12 5

5

12

(1) Q 5.14 10 0.62

Q 3.18 10 W

24

16. Two parallel plates of size 3 m 2 m are placed parallel to each other at a distance of 1 m. One

plate is maintained at a temperature of 550C and the other at 250C and the emissivities are 0.35 and

0.55 respectively. The plates are located in a large room whose walls are at 35C. If the plates located

exchange heat with each other and with the room, calculate.

1. Heat lost by the plates.

2. Heat received by the room.

Given: Size of the plates = 3 m 2 m

Distance between plates = 1 m

First plate temperature T1 = 550C + 273 = 823 K

Second plate temperature T2 = 250C + 273 = 523 K

Emissivity of first plate 1 = 0.35

Emissivity of second plate 2 = 0.55

Room temperature T3 = 35C + 273 = 308 K

To find: 1. Heat lost by the plates

2. Heat received by the room.

Solution: In this problem, heat exchange takes place between two plates and the room. So this is

three surface problems and the corresponding radiation network is given below. Area A1 = 3 2 =

6 m2

2

1 2A A 6m

Since the room is large 3A

From electrical network diagram,

1

1 1

2

2 2

33

3 3

1 1 0.350.309

A 0.35 6

1 1 0.550.136

A 0.55 6

10 [ A ]

A

Apply

3 1 2

3 3 1 1 2 2

1 1- 10, 0.309, 0.136

A A A

values in electrical network diagram.

To find shape factor F12 refer HMT data book, Page No.78.

b 3X 3

c 1

a 2Y 2

c 1

X value is 3, Y value is 2, corresponding shape factor [From table]

F12 = 0.47

25

12F 0.47

We know that,

F11 + F12 + F13 = 1 But, F11 = 0

13 12

13

13

F 1 F

F 1 0.47

F 0.53

Similarly, F21 + F22 + F23 = 1 We know F22 = 0

23 21

23 12

13

23

F 1 F

F 1 F

F = 1 - 0.47

F 0.53

From electrical network diagram,

1 13

2 23

1 12

1 10.314 ....(1)

A F 6 0.53

1 10.314 ....(2)

A F 6 0.53

1 10.354 ....(3)

A F 6 0.47

From Stefan – Boltzmann law, we know

4

b

4

b1 1

4-8

3 2

b1

E T

E T

= 5.67 10 823

E 26.01 10 W /m .....(4)

4

b2 2

4-8

3 2

b2

4

b3 3

4-8

2

b3 3

E T

= 5.67 10 823

E 4.24 10 W /m .....(5)

E T

= 5.67 10 308

E J 510.25 W /m .....(6)

[From diagram]

The radiosities, J1 and J2 can be calculated by using Kirchoff’s law.

The sum of current entering the node J1 is zero.

At Node J1:

26

b1 1 b3 12 1

1 12 1 13

E J E JJ J0

1 10.309

A F A F

[From diagram]

3

1 2 1 1

3 1 2 1 1

3

1 2

26.01 10 J J J 510.25 J 0

0.309 0.354 0.314

J J J J 84.17 10 1625 0

0.309 0.354 0.354 0.354

-9.24J 2.82J 85.79 10 .....(7)

At node j2

b3 2 b2 21 2

1 12 2 23

E J E JJ J0

1 1 0.136

A F A F

3

1 2 2 2

3

1 2 2 2

3

1 2

J J 510.25 J 4.24 10 J0

0.354 0.314 0.136

J J J J510.25 4.24 100

0.354 0.354 0.314 0.314 0.136 0.136

2.82J 13.3J 32.8 10 ....(8)

Solving equation (7) and (8),

3

1 2

3

1 2

-9.24J 2.82J 85.79 10 .....(7)

2.82J 13.3J 32.8 10 .....(8)

3 2

2

3 2

1

J 4.73 10 W /m

J 10.73 10 W /m

Heat lost by plate (1) is given by

b1 11

1

1 1

E JQ

1

A

3 3

1

3

1

26.01 10 10.73 10Q

1 0.35

0.35 6

Q 49.36 10 W

Heat lost by plate 2 is given by

b2 22

2

2 2

E JQ

1

A

27

3 3

2

3

2

4.24 10 4.73 10Q

1 0.55

6 0.55

Q 3.59 10 W

Total heat lost by the plates

Q = Q1 + Q2

= 49.36 103 – 3.59 103

3Q 45.76 10 W ......(9)

Heat received by the room

1 3 2 3

1 13 1 12

J J J JQ

1 1

A F A F

3 3

b1 1

3

10.73 10 510.25 4.24 10 510.25

0.314 0.314

[ E J 512.9]

Q = 45.9 10 W .....(10)

From equation (9), (10), we came to know heat lost by the plates is equal to heat received

by the room.

17. A gas mixture contains 20% CO2 and 10% H2o by volume. The total pressure is 2 atm. The

temperature of the gas is 927C. The mean beam length is 0.3 m. Calculate the emissivity of the

mixture.

Given : Partial pressure of CO2, 2COP = 20% = 0.20 atm

Partial pressure of H2o,

2H 0P = 10% = 0.10 atm.

Total pressure P = 2 atm

Temperature T = 927C + 273

= 1200 K

Mean beam length Lm = 0.3 m

To find: Emissivity of mixture (mix).

Solution : To find emissivity of CO2

2

2

CO m

CO m

P L 0.2 0.3

P L 0.06 m - atm

From HMT data book, Page No.90, we can find emissivity of CO2.

From graph, Emissivity of CO2 = 0.09

28

2CO 0.09

To find correction factor for CO2

Total pressure, P = 2 atm

2CO mP L = 0.06 m - atm.

From HMT data book, Page No.91, we can find correction factor for CO2

From graph, correction factor for CO2 is 1.25

2COC 1.25

2 2

2 2

CO CO

CO CO

C 0.09 1.25

C 0.1125

To find emissivity of 2H o:

2H o mP L 0.1 0.3

2H o mP L 0.03 m - atm

From HMT data book, Page No.92, we can find emissivity of 2H o.

From graph Emissivity of 2H o= 0.048

2H o 0.048

To find correction factor for 2H o:

2

2

2

H o

H o

H o m

P P 0.1 21.05

2 2

P P1.05,

2

P L 0.03 m - atm

From HMT data book, Page No.92 we can find emission of H20

From graph,

Correction factor for H2O = 1.39

CH2O = 1.39

εH2O x CH2O = 0.048 x 1.39

εH2O x CH2O = 0.066

Correction factor for mixture of CO2 and H2o:

29

2

2 2

2

2 2

2 2

2 2

H o

H o CO

H o

H o CO

CO m H O m

CO m H O m

P 0.11.05

P P 0.1 0.2

P0.333

P P

P L P L 0.06 0.03

P L P L 0.09

From HMT data book, Page No.95, we can find correction factor for mixture of CO2 and 2H o.

From graph, Δε = 0.002

Total emissivity of gaseous mixture, εmix = εCo2 x Cco2 + εH2O x CH2O – Δε

= 0.1125 + 0.066 – 0.002

εmix = 0.1765

30