Lesson 19 Radiation I

19.1 Nomenclature

Eλ,b = spectral emission from a blackbody surface [W/m2-μ]

Eb = emission from a blackbody surface [W/m2]

q = radiative heat flux [W/m2]

F = fraction of blackbody band emission

λ = wavelength in microns [μ=μm]

ε = total emissivity, between 0 and 1 (ε=1 → blackbody, ε < 1 → gray body)

α =absorptivity, between 0 and 1, α=ε if surface is blackbody or gray body

σ = Stefan-Boltzmann Constant = 5.67×10-8

W/m2-K

4

19.2 Fundamental Concepts

Radiation Radiation is the phenomenon of electromagnetic (EM) wave propagation. EM-waves

transfer heat energy through interactions with solid surfaces or fluids. Unlike conduction and

convention, no medium is required to transfer heat by radiation, and it becomes more

predominant as temperature increases. Figure 19.1 shows the range of EM waves that we

commonly experience in everyday life.

Figure 19.1 Electromagnetic Spectrum

Differences between Radiation and Convection

Radiation Convection

(a) Medium None needed Needed

(b) Distance Can reach very far Local

(c) Speed Speed of light Relatively low

(d) Math Integral equations Differential equations

(e) q" εσTs4 h(Ts –T∞)

(a) Radiation can travel through the vacuum of space from the sun to the earth, but also

through the air to warm the earth’s surface.

(b) We can see the light emitting from a candle from afar, but we cannot feel it even from

one foot away.

(c) Generally, conduction is the slowest mode of heat transfer.

Radiation = f(λ,ω) λ=wavelength, ω=solid angle. We will not consider ω.

Definitions Relating to λ and ω

Diffuse surface: emits same radiation intensities at different solid angles.

Gray surface: emits same radiation intensities at different wavelengths.

Spectral emissivity: function of wavelength.

Total emissivity: average of spectral emissivity; constant at a given temperature.

Angular emissivity: function of solid angle.

Hemispherical emissivity: average of angular emissivity; constant at a given temperature.

19.3 Blackbody Radiation

Blackbody Surfaces (a) Will absorb all incidental radiation, regardless of λ; nothing is reflected or transmitted.

(b) Emit the maximum amount of radiation for any T (i.e., ε=1).

(c) ≠ black surfaces. Even the sun can be approximated as blackbody.

Planck Blackbody Spectral Emissive Distribution

𝑬𝝀,𝒃 =𝑪𝟏

𝝀𝟓 (𝐞𝐱𝐩 (𝑪𝟐

𝝀𝑻) − 𝟏)

(1)

Where

C1=3.742×108 and

C2 = 1.439×104

Figure 19.2 Blackbody Spectral Emission as a function of λ, parameterized by T.

(a) λmaxT = 2898 (Wien’s Displacement) (1a)

(b) Through evolution, we have adapted to capture solar radiation (T=5800 K) around the

vicinity of the peak of the spectral emission.

(c) In order for our eyes to see emitted lighted (not merely reflected), temperatures must be

very high.

Stefan-Boltzmann Law

𝑬𝒃 = ∫ 𝑬𝝀,𝒃𝒅𝝀 = 𝝈𝑻𝟒∞

𝟎

(2)

Where σ = Stefan-Boltzmann Constant = 5.67×10-8

W/m2-K

4.

The law governing the relationship between the heat flux and the temperature, called the

Stefan-Boltzmann law, is given as

𝒒′′ = 𝜺𝝈𝑻𝟒 = 𝜺𝑬𝒃 and 𝒒 = 𝑨𝒒′′ = 𝑨𝜺𝝈𝑻𝟒 = 𝑨𝜺𝑬𝒃 (2a)

19.4 Fraction of Blackbody Emission

Fraction of Blackbody Emission Frequently we are interested in finding a portion of the emissive radiation between a range of

wavelengths from λ1 to λ2 (partial area under Eλ,b-λ curve). The fraction, F, can be derived as

𝐹𝜆1𝑇→𝜆2𝑇 =∫ 𝐸𝜆,𝑏𝑑𝜆

𝜆2

𝜆1

𝐸𝑏=

∫ 𝐸𝜆,𝑏𝑑𝜆𝜆2

0

𝜎𝑇4−

∫ 𝐸𝜆,𝑏𝑑𝜆𝜆1

0

𝜎𝑇4= 𝐹0→𝜆2𝑇 − 𝐹0→𝜆1𝑇

𝑭𝝀𝟏𝑻→𝝀𝟐𝑻 = 𝑭𝟎→𝝀𝟐𝑻 − 𝑭𝟎→𝝀𝟏𝑻 (3)

Values of F can be found from Table A-2 in Heat Transfer: Lessons with Examples Solved

by MATLAB, by Dr. Tien-Mo Shih.

λT is given in units of μ-K.

At λT=50000, F≈1.

Example Over all λ, a blackbody (ε=1) will emit a radiative flux 𝑞𝑏 = 𝜀𝜎𝑇4 = 𝜎𝑇4. Over a range of

λ, the flux will be 𝑞𝜆1→𝜆2= 𝑞𝑏 ∗ 𝐹𝜆1𝑇→𝜆2𝑇.

19.5 Summary of Equations

Planck Blackbody Spectral Emissive Distribution (1)

𝐸𝜆,𝑏 =𝐶1

𝜆5 (exp (𝐶2

𝜆𝑇) − 1)

C1=3.742×108

C2 = 1.439×104

Wien’s Displacement (1a) λmaxT = 2898 [μ-K]

Stefan-Boltzmann Law (2, 2a)

𝐸𝑏 = ∫ 𝐸𝜆,𝑏𝑑𝜆 = 𝜎𝑇4∞

0

𝑞′′ = 𝜀𝜎𝑇4 = 𝜀𝐸𝑏 and 𝑞 = 𝐴𝑞′′ = 𝐴𝜀𝜎𝑇4 = 𝐴𝜀𝐸𝑏

σ = Stefan-Boltzmann Constant = 5.67×10-8

W/m2-K

4

ε = total emissivity, between 0 and 1 (ε=1 → blackbody, ε < 1 → gray body)

α =absorptivity, between 0 and 1, α=ε if surface is blackbody or gray body

Fraction of Blackbody Emission (3) 𝐹𝜆1𝑇→𝜆2𝑇 = 𝐹0→𝜆2𝑇 − 𝐹0→𝜆1𝑇

Lesson 20

Radiation II

20.1 Nomenclature

E = radiative flux emitted by a plate

f1 = fraction of solar radiation received by earth = 0.0021%

G = incoming radiative flux arriving at the plate

m = plate mass

ελ = spectral emissivity

ε = total emissivity, between 0 and 1 (ε=1 → blackbody, ε < 1 → gray body)

α =absorptivity, between 0 and 1, α=ε if surface is blackbody or gray body

ρ = reflectivity (not density!)

τ = transmissivity (not shear stress!)

20.2 Emissivity

Spectral Emissivity Emissivity is the one radiative property that is unrelated to incoming fluxes; it is a function of

the λ and T of the emitting source. Let us define the spectral emissivity as

𝜺𝝀 =𝑬𝝀(𝝀, 𝑻)

𝑬𝝀,𝒃(𝝀, 𝑻)

(1a)

The total emissivity, ε, can be derived from this equation with integration over all λ and

making use of the fraction of blackbody emission, F:

𝜺 =𝑬(𝑻)

𝑬𝒃(𝑻)=

∫ 𝑬𝝀(𝝀, 𝑻)𝒅𝝀∞

𝟎

𝝈𝑻𝟒=

∫ 𝜺𝝀𝑬𝝀,𝒃(𝝀, 𝑻)𝒅𝝀∞

𝟎

𝝈𝑻𝟒

= 𝜺𝝀𝟏→𝝀𝟐𝑭𝝀𝟏𝑻→𝝀𝟐𝑻 + 𝜺𝝀𝟐→𝝀𝟑

𝑭𝝀𝟐𝑻→𝝀𝟑𝑻 + ⋯

(1b)

From Eq. (1b) we can also readily find the total radiative flux emitting from a body:

𝑬(𝑻) = 𝒒𝒔′′ = 𝜺𝝈𝑻𝒔

𝟒 (2)

20.3 Three Other Radiative Properties

Absorptivity, Reflectivity, and Transmissivity As with Eq. (1a) and Eq. (1b), the same can be applied for

1. absorptivity α (propensity to absorb radiation),

2. reflectivity ρ (propensity to reflect radiation),

3. transmissivity τ (propensity to let radiation transmit through).

𝜶𝝀 =𝑮𝝀,𝜶

𝑮𝝀 → 𝜶 =

𝑮𝒂(𝑻)

𝑮(𝑻)=

∫ 𝑮𝝀,𝒂𝒅𝝀∞

𝟎

∫ 𝑮𝝀𝒅𝝀∞

𝟎

=∫ 𝜶𝝀𝑮𝝀𝒅𝝀

∞

𝟎

∫ 𝑮𝝀𝒅𝝀∞

𝟎

(3a,b)

𝝆 =∫ 𝝆𝝀𝑮𝝀𝒅𝝀

∞

𝟎

∫ 𝑮𝝀𝒅𝝀∞

𝟎

(4)

𝝉 =∫ 𝝉𝝀𝑮𝝀𝒅𝝀

∞

𝟎

∫ 𝑮𝝀𝒅𝝀∞

𝟎

(5)

Note that these three properties all related to incoming radiation. If there is no mention of a

radiative heat source, then these values cannot be found. Even if incoming fluxes are the

same, the resulting α, ρ, and τ may be different depending on source temperature. See

example 20-5.

20.4 Gray Surfaces

Gray Surfaces Gray surfaces are defined as surfaces whose radiative properties are independent of

wavelength. Hence, for gray surfaces, all spectral properties are (1) constant and (2) equal

to the total properties.

Therefore, if G is the total incoming radiation, then it follows that

G = amount absorbed + amount reflected + amount transmitted = αG + ρG + τG

Dividing by through by G yields

𝟏 = 𝜶 + 𝝆 + 𝝉 (6a)

Special Cases Opaque surface: 𝜏 = 0 → 1 = 𝛼 + 𝜌

20.5 Kirchhoff’s Law

Kirchhoff’s law states that for most materials in most practical purposes,

𝜺𝝀 = 𝜶𝝀 (7)

We could argue this is not true, but consider the following example:

A material has ελ = 0.1 and αλ = 0.9

The material is placed in a 300K enclosure.

As time elapses, the material receives more energy than it emits.

The material will reach equilibrium when radiation in = radiation out:

𝜀𝜆𝑇𝑠4 = 𝛼𝜆𝑇𝑒𝑛𝑐𝑙

4 → 0.1𝑇𝑠4 = 0.9 ∗ 3004 → 𝑇𝑆 = 519.6 K

This violates the 2nd

Law of Thermodynamics. At equilibrium, Ts=300 K.

Eq. (7) by no means automatically implies that

𝜺 = 𝜶 (8)

Eq. (8) is only valid if

(a) the surface is gray, or

(b) the source is blackbody and Tsource = Tsurface, in which case Gλ = Eλ,b in Eq. (3b).

20.6 Energy Balance over a Typical Plate (Lumped Capacitance)

𝒎𝒄𝒗

𝒅𝑻

𝒅𝒕= 𝒉𝑨(𝑻∞ − 𝑻) + 𝑨(𝑮 − 𝝆𝑮 − 𝝉𝑮) − 𝑨𝜺𝝈𝑻𝟒

(9)

Notice that the term concerning incoming radiation is 𝐴(𝐺 − 𝜌𝐺 − 𝜏𝐺) = 𝐴𝛼𝐺. The amount

of energy absorbed is equal to the total incoming energy less the reflected and transmitted.

Hence, Eq. (9) can be rewritten as

𝒎𝒄𝒗

𝒅𝑻

𝒅𝒕= 𝒉𝑨(𝑻∞ − 𝑻) + 𝑨𝜶𝑮 − 𝑨𝜺𝝈𝑻𝟒

(10)

20.7 Steady-State Energy Balance with External Energy Supply

𝒒𝒔𝒖𝒑𝒑𝒍𝒊𝒆𝒅′′ = 𝒒𝒐𝒖𝒕

′′ − 𝒒𝒊𝒏′′ (11)

In equilibrium, the total heat flux externally supplied to a control volume is equal to the

difference in the heat flux out (emitted, reflected, transmitted) and the heat flux in

(absorbed).

Lesson 21

Radiation III

Introduction: This lesson focuses on radiation exchanges inside vacuum enclosures composed

by no more than 3 surfaces. These surfaces are considered flat, infinitely deep, and can be either

black or gray.

21.1 Nomenclature

F= View factor, subscript 12 = “from area A1 to area A2”

J = radiosity: radiation leaving a surface, including emitted and reflected radiation (ε, ρ)

21.2 View (or Shape/Configuration) Factors

Definition A view factor FA

B is defined as the fraction of all radiation leaving surface A which strikes

surface B. This fraction is related to the geometry and between the surfaces and the

geometry of the surfaces themselves.

Define a view factor between two surfaces of area A1 and A2, respectively:

F12 = (radiation from A1 striking A2) / (total radiation emitted from A1)

Under this definition, F12 > 1.

Reciprocity Rule

𝑨𝑨𝑭𝑨→𝑩 = 𝑨𝑩𝑭𝑩→𝑨 (1)

Energy Conservation Rule Because radiation leaving a surface is conserved, for any surface i with are Ai,

∑ 𝑭𝑨𝒊→𝑨𝒋= 𝟏 𝐬𝐨 𝑭𝟏𝟐 + 𝑭𝟏𝟑 + ⋯ + 𝑭𝟏𝒏

𝒏

𝒋=𝟏

= 𝟏

(2)

So for a triangular enclosure consisting of three flat surfaces of area A1, A2, and A3,

respectively, the following relations exist between the 3 areas and 6 viewing factors:

(a) 𝐹12 + 𝐹13 = 1 𝐴1𝐹12 = 𝐴2𝐹21 (d)

(b) 𝐹21 + 𝐹23 = 1 𝐴2𝐹23 = 𝐴3𝐹32 (e)

(c) 𝐹31 + 𝐹32 = 1 𝐴3𝐹31 = 𝐴1𝐹13 (f)

View Factor for a Triangle For a triangular enclosure, as shown in Figure 21.1, we can derive F12 in terms of the areas.

If we know F12, we can find the other 5 viewing factors.

Figure 21.1 A triangular enclosure consisting of 3 flat surfaces.

Derivation:

We want to find F12, and the only equation with this term is (a):

𝐹12 + 𝐹13 = 1

We want to get F13 in terms of F12. What other equations contain F13? The only other is (f).

𝐴3𝐹31 = 𝐴1𝐹13 so 𝐹13 =𝐴3

𝐴1𝐹31

Now what other equation has F31? The only other is (c):

𝐹31 + 𝐹32 = 1 so 𝐹31 = 1 − 𝐹32

Now what other equation has F32? The only other is (e):

𝐴2𝐹23 = 𝐴3𝐹32 so 𝐹32 =𝐴2

𝐴3𝐹23

Now what other equation has F23? The only other is (b):

𝐹21 + 𝐹23 = 1 so 𝐹23 = 1 − 𝐹21

Now we have a term related to F12! The equation relating F21 to F12 is (d), so

𝐹21 =𝐴1

𝐴2𝐹12

Now we take the long road back to getting F13 in terms of F12:

A3

θ1 θ2

A1

A2

𝐹23 = 1 − 𝐹21 = 1 −𝐴1

𝐴2𝐹12 → 𝐹32 =

𝐴2

𝐴3𝐹23 =

𝐴2

𝐴3(1 −

𝐴1

𝐴2𝐹12) =

𝐴2

𝐴3−

𝐴1

𝐴3𝐹12

𝐹31 = 1 − 𝐹32 = 1 −𝐴2

𝐴3−

𝐴1

𝐴3𝐹12 → 𝐹13 =

𝐴3

𝐴1𝐹31 =

𝐴3

𝐴1(1 −

𝐴2

𝐴3−

𝐴1

𝐴3𝐹12)

=𝐴3

𝐴1−

𝐴2

𝐴1− 𝐹12

Therefore,

𝐹12 + 𝐹13 = 𝐹12 +𝐴3

𝐴1−

𝐴2

𝐴1− 𝐹12 = 2𝐹12 +

𝐴3 − 𝐴2

𝐴1= 1

𝑭𝟏𝟐 =𝑨𝟏 + 𝑨𝟐 − 𝑨𝟑

𝟐𝑨𝟏

(3)

Areas Cannot Be Arbitrary Because the shape we are dealing with is a triangle, its dimensions cannot be arbitrary given.

In reference to Figure 21.1, we see that

A2 =A1cos(θ1) + A3cos(θ2) < A1 + A3

Always check to make sure this inequality holds true.

21.3 Black Triangle Enclosures

Energy Balance Consider the energy balance over surface 1 (ε=1). We have

𝑞1 = 𝑞𝑜𝑢𝑡 − 𝑞𝑖𝑛 = 𝐴1𝜎𝑇14 − 𝐴2𝐹21𝜎𝑇2

4 − 𝐴3𝐹31𝜎𝑇34

Using the reciprocity rule and dividing both sides by A1, we obtain

𝒒𝟏′′ = 𝝈𝑻𝟏

𝟒 − 𝑭𝟏𝟐𝝈𝑻𝟐𝟒 − 𝑭𝟏𝟑𝝈𝑻𝟑

𝟒 (4a)

The energy balances over surfaces 2 and 3 follow suit:

𝒒𝟐′′ = 𝝈𝑻𝟐

𝟒 − 𝑭𝟐𝟏𝝈𝑻𝟏𝟒 − 𝑭𝟐𝟑𝝈𝑻𝟑

𝟒 (4b)

𝒒𝟑′′ = 𝝈𝑻𝟑

𝟒 − 𝑭𝟑𝟏𝝈𝑻𝟏𝟒 − 𝑭𝟑𝟐𝝈𝑻𝟐

𝟒 (4c)

Assuming we have the view factors, there are 3 equations and 6 unknowns: q1”, q2”, q3”, T1,

T2, and T3. Three of these quantities must be given in order to solve a problem.

21.4 Gray Triangular Enclosures

Radiosity If an enclosure consists of gray surfaces, then we have to deal with a component of radiation

that is reflected. This is where we introduce radiosity, J. Radiosity is defined as all the

radiation leaving a surface, including emission and reflection. Consider the energy balance

over a control volume surrounding the surface shown in Figure 21.2.

Figure 21.2 Derivation of J1 and q1 relationship.

For gray bodies,

𝑱𝟏 = 𝜺𝟏𝝈𝑻𝟏𝟒 + 𝝆𝑮 = 𝜺𝟏𝝈𝑻𝟏

𝟒 + (𝟏 − 𝜺𝟏)𝑮𝟏 (5a,b)

Then the energy balance over A1 is

𝒒𝟏 = 𝒐𝒖𝒕 − 𝒊𝒏 = 𝑨𝟏(𝑱𝟏 − 𝑮𝟏) (6)

Then,

𝒒𝟏 =𝑨𝟏𝜺𝟏

𝟏−𝜺𝟏(𝝈𝑻𝟏

𝟒 − 𝑱𝟏) (7)

In a triangular enclosure, we have

𝒒𝟏′′ = 𝑱𝟏 − 𝑭𝟏𝟐𝑱𝟐 − 𝑭𝟏𝟑𝑱𝟑 (8a)

𝒒𝟐′′ = 𝑱𝟐 − 𝑭𝟐𝟏𝑱𝟏 − 𝑭𝟐𝟑𝑱𝟑 (8b)

𝒒𝟑′′ = 𝑱𝟑 − 𝑭𝟑𝟏𝑱𝟏 − 𝑭𝟑𝟐𝑱𝟐 (8c)

If T1 is given instead of q1”, then Eq. (8a) can be modified into

𝒂(𝟏, 𝟏)𝑱𝟏 + 𝒂(𝟏, 𝟐)𝑱𝟐 + 𝒂(𝟏, 𝟑)𝑱𝟑 = (𝜺𝟏

𝟏−𝜺𝟏) 𝝈𝑻𝟏

𝟒 (8d)

where

𝑎(1,1) =1

1 − 𝜀1, 𝑎(1,2) = −𝐹12, 𝑎(1,3) = −𝐹13

G1

q1”

A1

J1

21.5 Two Parallel Gray Plates with A1=A2

Special Case With two parallel gray plates of same area A1=A2, as shown in Figure 21.3, the situation

presented by Eqs. (8a-c) can be simplified to

𝒒𝟏′′ = 𝑱𝟏 − 𝒇𝑱𝟐 (9a)

𝒒𝟐′′ = 𝑱𝟏 − 𝒇𝑱𝟐 (9b)

where f is the viewing factor between the plates (because A1=A2, F12=F21=f). Then with Eq.

(7), we acquire two more equations:

𝒒𝟏′′ =

𝜺𝟏

𝟏−𝜺𝟏(𝝈𝑻𝟏

𝟒 − 𝑱𝟏) (9c)

𝒒𝟐′′ =

𝜺𝟐

𝟏−𝜺𝟐(𝝈𝑻𝟐

𝟒 − 𝑱𝟐) (9d)

This gives 4 equations and 6 unknowns. When 2 are given, the problem can be solved.

Figure 21.3 Energy Balance between two parallel plates.

q2”

q1” A1

A2

J2

fJ2

fJ1

J1