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AndreaFerrari(LTCM-SGM-EPFL) Heat&MassTransfer–NaturalConvec@oninsingle-phasefluids 22/05/17 1

Thermalradia+onheattransfer

Andrea Ferrari John Richard Thome


Theproblemofradia+veheatexchange. •  Thermal radia4on is energy emi8ed by ma8er that is at non-zero temperature. •  Allbodiesconstantlyemitenergy which is transported by electromagne4c waves.

•  In contrast to conduc4on and convec4on in which the heat transfer requires the presence of a temperature difference in some material medium, radia+ondoesnotrequireanymedium(radia4on heat transfer is much more efficient in a vacuum).

•  In contrast to conduc4on and convec4on in which the heat only goes from warmer bodies to colder bodies, inthermalradia+ontheheatcangofromcoldsurfacestohotsurfaces(beside of course going from hot surfaces to cold surfaces). Themagnitudeofheatemi;edbyasurfacedependsontemperature(a cold surface emits less energy than a hot surface)

Ø  Key ideas from Chapter 1

Radia@veheattransfer10. Radiative heat transfer

The sun that shines from Heaven shines but warm,And, lo, I lie between that sun and thee:The heat I have from thence doth little harm,Thine eye darts forth the fire that burneth me:

And were I not immortal, life were doneBetween this heavenly and earthly sun.

Venus and Adonis, Wm. Shakespeare, 1593

10.1 The problem of radiative exchange

Chapter 1 described the elementary mechanisms of heat radiation. Be-fore we proceed, you should reflect upon what you remember about thefollowing key ideas from Chapter 1:

• Electromagnetic wave spectrum • The Stefan-Boltzmann law• Heat radiation & infrared radiation • Wien’s law & Planck’s law• Black body • Radiant heat exchange• Absorptance, α • Configuration factor, F1–2

• Reflectance, ρ • Emittance, ε• Transmittance, τ • Transfer factor, F1–2

• α+ ρ + τ = 1 • Radiation shielding• e(T) and eλ(T) for black bodies

The additional concept of a radiation heat transfer coefficient was devel-oped in Section 2.3. We presume that all these concepts are understood.

The heat exchange problem

Figure 10.1 shows two arbitrary surfaces radiating energy to one another.The net heat exchange, Qnet, from the hotter surface (1) to the cooler

525


Theelectromagne+cspectrum. Radia4on originates due to emission of energy by ma8er and it is transported in a par4cular medium by electromagne4c waves of frequency ν and wavelength λ. The two proper4es are related by λ = c/ν, where c is the speed of light in the medium. For propaga4on in a vacuum, c=c0= 3 108 m/s.

Radia@veheattransferintroduc@on

10-5 10-4

10-3 10-2 10-1 1 10 102 103

Thermalradia@on

Infrared

MicrowaveUVX-rays

Gammarays

violet

blue

green

yellow

red

Visible

λ(µm)0.4

0.7

Spectrumofelectromagne@cradia@on


Absorptance,reflectanceandtransmi;ance. When a radiant heat flux q (usually referred to as Irradia&on, G) hits a semitransparent medium that is not black a frac4on 0 ≤ α ≤ 1 (absorptance) of the total incident energy is absorbed; a frac4on 0 ≤ ρ ≤ 1 (reflectance) is reflected by the surface; and a frac4on 0 ≤ τ ≤ 1 (transmi;ance) is transmi8ed through it. An opaque medium experiences no transmission (τ = 0). In general, liquids can be considered opaque to radia4on heat transfer and gases can be considered transparent to it (ρ = 0). Solid can be opaque (as in the case of metals) or semitransparent (as in the case of some polymers and some semiconductor materials).

It is evident that: α + ρ + τ = 1

Since ma8ers absorb and emit energy differently at different wavelengths, this rela4on can be also wri8en for the energy carried by each wavelength λ

αλ + ρλ + τλ = 1

Radia@veheattransfer:introduc@on

G=αG+ρGG=αG+ρG+τG

ρG

αG

τG

ρG

αGsemitransparent

mediumopaquemedium


Blackbodies. The model for the perfect thermal radiator is a so-called black body. •  A black body absorbsallenergythatreachesitregardless of wavelength and direc4on,

reflec4ng and transmi`ng nothing (ρ = τ = 0 and α = 1).

•  For a prescribed temperature and wavelength, nosurfacecanemitmoreenergythanablackbody.

•  The radia4on emi8ed by a black body is a func4on of the temperature and the wavelength but not of the direc4on (isotropicradia+on).

Ø  Stephan-Boltzmannlaw. The flux of energy radia4ng from a body is commonly designated e(T) [W/m2]. The symbol eλ(λ,T) designates the distribu4on of radia4ve flux in λ (monochroma.c emissive power). Thus e(T) is the integral over all the wavelength of eλ(λ,T).

For a black body at absolute temperature T the total emissive power is:


e(�, T ) =

Z �

0e�(�, T )d� e(T ) =

Z 1

0e�(�, T )d�

eb(T ) = �T 4

� = 5.67 10�8 W/m2 K4

Althoughcloselyapproximatedbysomesurfaces,nosurfacehaspreciselytheproper@esofablackbody(notevenablackhole)T:absolutetemperature


Blackbodyspectralemissivepower(Planckdistribu+on). The predic4on of how the monochroma4c emissive power of a black body, eλ(λ,T), depends on λ and T has been formulated by Max Planck in 1901 and it poses the founda4ons of modern quantum mechanics:


e�b =

2⇡hc20�5

[exp(hc0/kBT�)� 1]

Thermalradia@on

•  The emi8ed radia4on varies con+nuously with λ

•  At any λ the magnitude of emi8ed radia4on increaseswithincreasingtemperature

•  The spectral region in which the radia4on is concentrated depends on T with moreradia+onappearingatshorterwavelengthsforincreasingT

•  A significant frac4on of the radia4on emi8ed by the sun (which can be approximated as a black body at T = 5800 K) is in the visible region of the spectrum, while for T≤800 K emission is predominant in the infrared region of the spectrum (not visible)

Sun

Earth


Wien’sdisplacementlaw. The maximum wavelength at a given temperature can be found by taking the deriva4ve of the Planck distribu4on and set it equal to zero. λmax and T are related by:

By integra4ng Planck distribu4on over all the wavelengths, Stephan-Boltzmann law can be found (e(T) = σT4).

Radiantheatexchange. If a heated black-body (1) radiates only to some other black-body(2) it means that all the heat leaving object 1 arrives at object 2 and all the heat arriving at object 1 comes from object 2. The net heat transferred from object 1 to 2 is the difference between Q1 to 2 = A1eb(T1) and Q2 to 1 = A1eb(T2).

If object 1 “sees” other object than object 2 then a viewfactor, F1-2 must accounted for. F1-2 is the frac4on of energy leaving object 1 that is intercepted by object 2


�max

T = 2898µmK

Qnet = A1eb(T1)�A1eb(T2) = A1�(T41 � T 4

2 )

Qnet = A1F1�2�(T41 � T 4

2 )


Realbodies. The black body radia4on represents an upperlimitof the radia4on that can be absorbed or emi8ed at a certain temperature. Realbodiesabsorbandemitlessradia+onthanablackbody.Moreover the spectral radia4on emi8ed by a real surface differs from Planck distribu4on and may dependonthedirec+on.We can characterize the emissive power of a non-black body using a property called emi;ance, 0 < ε ≤ 1

When radia4on is exchanged between two bodies that are not black we have:

where Ƒ1-2 depends on the emi8ance of both bodies as well as the geometrical “view”

Iftheobject1ismuchsmallerthanobject2(a small object in a much larger isothermal environment) the view factor can be neglected and Ƒ1-2isequaltotheemi;anceofobject1


enon�black

= ✏eb

(T ) = ✏�T 4

Qnet = A1F1�2�(T41 � T 4

2 )

F1�2 = ✏1 if A1 ⌧ A2


Theheatexchangeproblem. When two arbitrary surfaces radiate energy one to each other, the net heat exchange, Qnet, from the ho8er surface (1) to the cooler surface (2) depends on the following influences:

Ø  If surfaces (1) and (2) are black, if they are surrounded by air, and if noheatflowsbetween them by conduc4on or convec4on, then onlythefirst3considera+onsareinvolvedindeterminingQnet. The last 3 considera4ons complicate the problem considerably and in this case atransferfactor, Ƒ1-2, must be included in determining Qnet.

Theproblemofradia@veheatexchange

526 Radiative heat transfer §10.1

Figure 10.1 Thermal radiation between two arbitrary surfaces.

surface (2) depends on the following influences:

• T1 and T2.

• The areas of (1) and (2), A1 and A2.

• The shape, orientation, and spacing of (1) and (2).

• The radiative properties of the surfaces.

• Additional surfaces in the environment, whose radiation may bereflected by one surface to the other.

• The medium between (1) and (2) if it absorbs, emits, or “reflects”radiation. (When the medium is air, we can usually neglect theseeffects.)

If surfaces (1) and (2) are black, if they are surrounded by air, and ifno heat flows between them by conduction or convection, then only the

1)  T1 and T2 2)  The areas A1 and A2 3)  The shape, orienta4on and spacing of (1) and

(2) 4)  The radia4ve proper4es of the surfaces 5)  Addi4onal surfaces in the environment,

whose radia4on may be reflected by one surface to the other

6)  If the medium between (1) and (2) it absorbs, emits, or reflects radia4on (if the medium is air usually these effects can be neglected)


Somedefini+ons1.   Emi;ance.A real body emits at some frac4on, ε, of a black body at the same temperature.

Both the total power emi8ed and the monochroma4c emissive power are always lower for a real body than the black body value given by Planck’s law. Thus, the monochroma4c emi8ance, ελ, and the total emi8ance, ε, are defined as:

For real bodies 0 < ε, ελ < 1, while for black bodies ε=ελ=1. The emi8ance depends en4rely on the proper4es of the surface and its temperature (it is not a func4on of the evironment of the body).


✏� =e�(�, T )

e�b(�, T )✏� =

e(T )

eb(T )=

R10 ✏�e�b(�, T )d�

�T 4

Table 10.1 Total emittances for a variety of surfaces [10.1]

Metals Nonmetals

Surface Temp. (◦C) ε Surface Temp. (◦C) ε

Aluminum Asbestos 40 0.93–0.97Polished, 98% pure 200−600 0.04–0.06 BrickCommercial sheet 90 0.09 Red, rough 40 0.93Heavily oxidized 90−540 0.20–0.33 Silica 980 0.80–0.85

Brass Fireclay 980 0.75Highly polished 260 0.03 Ordinary refractory 1090 0.59Dull plate 40−260 0.22 Magnesite refractory 980 0.38Oxidized 40−260 0.46–0.56 White refractory 1090 0.29

Copper CarbonHighly polished electrolytic 90 0.02 Filament 1040−1430 0.53Slightly polished to dull 40 0.12–0.15 Lampsoot 40 0.95Black oxidized 40 0.76 Concrete, rough 40 0.94

Gold: pure, polished 90−600 0.02–0.035 GlassIron and steel Smooth 40 0.94

Mild steel, polished 150−480 0.14–0.32 Quartz glass (2 mm) 260−540 0.96–0.66Steel, polished 40−260 0.07–0.10 Pyrex 260−540 0.94–0.74Sheet steel, rolled 40 0.66 Gypsum 40 0.80–0.90Sheet steel, strong 40 0.80 Ice 0 0.97–0.98

rough oxideCast iron, oxidized 40−260 0.57–0.66 Limestone 400−260 0.95–0.83Iron, rusted 40 0.61–0.85 Marble 40 0.93–0.95Wrought iron, smooth 40 0.35 Mica 40 0.75Wrought iron, dull oxidized 20−360 0.94 PaintsStainless, polished 40 0.07–0.17 Black gloss 40 0.90

Stainless, after repeated 230−900 0.50–0.70 White paint 40 0.89–0.97heating Lacquer 40 0.80–0.95

Lead Various oil paints 40 0.92–0.96Polished 40−260 0.05–0.08 Red lead 90 0.93Oxidized 40−200 0.63 Paper

Mercury: pure, clean 40−90 0.10–0.12 White 40 0.95–0.98Platinum Other colors 40 0.92–0.94

Pure, polished plate 200−590 0.05–0.10 Roofing 40 0.91Oxidized at 590◦C 260−590 0.07–0.11 Plaster, rough lime 40−260 0.92Drawn wire and strips 40−1370 0.04–0.19 Quartz 100−1000 0.89–0.58

Silver 200 0.01–0.04 Rubber 40 0.86–0.94Tin 40−90 0.05 Snow 10−20 0.82Tungsten Water, thickness ≥0.1 mm 40 0.96

Filament 540−1090 0.11–0.16 Wood 40 0.80–0.90Filament 2760 0.39 Oak, planed 20 0.90

528

Table 10.1 Total emittances for a variety of surfaces [10.1]

Metals Nonmetals

Surface Temp. (◦C) ε Surface Temp. (◦C) ε

Aluminum Asbestos 40 0.93–0.97Polished, 98% pure 200−600 0.04–0.06 BrickCommercial sheet 90 0.09 Red, rough 40 0.93Heavily oxidized 90−540 0.20–0.33 Silica 980 0.80–0.85

Brass Fireclay 980 0.75Highly polished 260 0.03 Ordinary refractory 1090 0.59Dull plate 40−260 0.22 Magnesite refractory 980 0.38Oxidized 40−260 0.46–0.56 White refractory 1090 0.29

Copper CarbonHighly polished electrolytic 90 0.02 Filament 1040−1430 0.53Slightly polished to dull 40 0.12–0.15 Lampsoot 40 0.95Black oxidized 40 0.76 Concrete, rough 40 0.94

Gold: pure, polished 90−600 0.02–0.035 GlassIron and steel Smooth 40 0.94

Mild steel, polished 150−480 0.14–0.32 Quartz glass (2 mm) 260−540 0.96–0.66Steel, polished 40−260 0.07–0.10 Pyrex 260−540 0.94–0.74Sheet steel, rolled 40 0.66 Gypsum 40 0.80–0.90Sheet steel, strong 40 0.80 Ice 0 0.97–0.98

rough oxideCast iron, oxidized 40−260 0.57–0.66 Limestone 400−260 0.95–0.83Iron, rusted 40 0.61–0.85 Marble 40 0.93–0.95Wrought iron, smooth 40 0.35 Mica 40 0.75Wrought iron, dull oxidized 20−360 0.94 PaintsStainless, polished 40 0.07–0.17 Black gloss 40 0.90

Stainless, after repeated 230−900 0.50–0.70 White paint 40 0.89–0.97heating Lacquer 40 0.80–0.95

Lead Various oil paints 40 0.92–0.96Polished 40−260 0.05–0.08 Red lead 90 0.93Oxidized 40−200 0.63 Paper

Mercury: pure, clean 40−90 0.10–0.12 White 40 0.95–0.98Platinum Other colors 40 0.92–0.94

Pure, polished plate 200−590 0.05–0.10 Roofing 40 0.91Oxidized at 590◦C 260−590 0.07–0.11 Plaster, rough lime 40−260 0.92Drawn wire and strips 40−1370 0.04–0.19 Quartz 100−1000 0.89–0.58

Silver 200 0.01–0.04 Rubber 40 0.86–0.94Tin 40−90 0.05 Snow 10−20 0.82Tungsten Water, thickness ≥0.1 mm 40 0.96

Filament 540−1090 0.11–0.16 Wood 40 0.80–0.90Filament 2760 0.39 Oak, planed 20 0.90

528

✏


Somedefini+ons1.   Emi;ance. One par4cular kind of surface behavior is that for which ελ is independent of λ.

Such a surface is called graybody. The monochroma4c emissive power, eλ(T), for a gray body is a constant frac4on, ε, of the black body emission (ε=ελforgraybodies)


§10.1 The problem of radiative exchange 529

Figure 10.2 Comparison of the sun’s energy as typically seenthrough the earth’s atmosphere with that of a black body hav-ing the same mean temperature, size, and distance from theearth. (Notice that eλ, just outside the earth’s atmosphere, isfar less than on the surface of the sun because the radiationhas spread out over a much greater area.)

No real body is gray, but many exhibit approximately gray behavior. Wesee in Fig. 10.2, for example, that the sun appears to us on earth as anapproximately gray body with an emittance of approximately 0.6. Somematerials—for example, copper, aluminum oxide, and certain paints—areactually pretty close to being gray surfaces at normal temperatures.

Yet the emittance of most common materials and coatings varies withwavelength in the thermal range. The total emittance accounts for thisbehavior at a particular temperature. By using it, we can write the emis-sive power as if the body were gray, without integrating over wavelength:

e(T) = ε σT 4 (10.5)

We shall use this type of “gray body approximation” often in this chapter.

Comparisonofthesun’senergyastypicallyseenthroughtheearth’satmosphere(con@nuousline)withthatofablackbodyhavingthesamemeantemperature,sizeanddistancefromtheearth(dashedline).Inthiscaseεisafunc@onofλ.Inset:poweremiiedbyagraybodywithε=½comparedwiththatofablackbodyatthesametemperature(εisanumberandnotafunc@onofλ).


Somedefini+ons2.   Diffuseandspecularemi;anceandreflec+on.The energy emi8ed or reflected by a non-

black surface may leave the body diffusively or specularly. o  When the reflec4on of emission is diffuse, there is no preferred direc4on for outgoing rays o  Black body emission is always diffuse o  Real body emission is between diffuse and specular. However, for most surfaces we can assume

diffuse behavior. This approxima4on works well for many engineering problems, in part because most tabulated spectral and total emi8ance have been averaged over all angles.


530 Radiative heat transfer §10.1

Specular or mirror-likereflection of incoming ray.

Reflection which isbetween diffuse andspecular (a real surface).

Diffuse radiation in whichdirections of departure areuninfluenced by incomingray angle, θ.

Figure 10.3 Specular and diffuse reflection of radiation.(Arrows indicate magnitude of the heat flux in the directionsindicated.)

In situations where surfaces at very different temperatures are in-volved, the wavelength dependence of ελ must be dealt with explicitly.This occurs, for example, when sunlight heats objects here on earth. So-lar radiation (from a high temperature source) is on visible wavelengths,whereas radiation from low temperature objects on earth is mainly in theinfrared range. We look at this issue further in the next section.

Diffuse and specular emittance and reflection. The energy emitted bya non-black surface, together with that portion of an incoming ray ofenergy that is reflected by the surface, may leave the body diffusely orspecularly, as shown in Fig. 10.3. That energy may also be emitted orreflected in a way that lies between these limits. A mirror reflects visibleradiation in an almost perfectly specular fashion. (The “reflection” of abilliard ball as it rebounds from the side of a pool table is also specular.)When reflection or emission is diffuse, there is no preferred direction foroutgoing rays. Black body emission is always diffuse.

The character of the emittance or reflectance of a surface will nor-mally change with the wavelength of the radiation. If we take account ofboth directional and spectral characteristics, then properties like emit-tance and reflectance depend on wavelength, temperature, and anglesof incidence and/or departure. In this chapter, we shall assume diffuse



Somedefini+ons3.   Intensityofradia+on. Radia4on is emi8ed by all parts of a plane surface in all direc4ons into

the hemisphere above the surface, and the direc4onal distribu4on of emi8ed (or incident) radia4on is usually not uniform. The quan4ty that describes the magnitude of radia4on emi8ed (or incident) in a specified direc4on in space is called radia&onintensity, denoted by I(λ,θ,ϕ)

Ø  Plane angle and solid angle

dω r

dA

dα dlr

d! =dA

r2steradiand↵ =

dl

rradian



Somedefini+ons3.   Intensityofradia+on,I(λ,θ,ϕ). We consider the emission in a par4cular direc4on from an

element of surface dA1 (orange surface). The direc4on of radia4on is specified in terms of the zenith (0<ϑ<π/2) and azimuthal (0<φ<2π) angle, of a spherical coordinate system. The area dAn (pink surface) through which the radia4on passes subtends a differen4al solid angle dω when viewed from a point on dA1. As shown in the picture, the area dAn is a rectangle with dimension r dθ x r sinθ dφ

d! =dAn

r2=

r2 sin ✓d✓d�

r2= sin ✓d✓d�

Z

hd! =

Z 2⇡

0

Z ⇡/2

0sin ✓d✓d� = 2⇡

Z ⇡/2

0sin ✓d✓ = 2⇡

Thesolidangleassociatedwiththeen@rehemispheremaybeobtainedbyintegra@ngbetweenφ=0toφ=2πandθ=0andθ=π/2



Somedefini+ons3.   Intensityofradia+on,I(λ,θ,ϕ). The intensity of radia4on, I(λ,θ,ϕ) [W/m2] , is defined as the

rate at which radiant energy is emi8ed at wavelength λ and in the (θ,φ) direc4on, per unit area of emi`ng surface normal to this direc4on and per unit solid angle about this direc4on. Note that theareausedtodefineIisthecomponentofdA1perpendiculartothedirec+onofradia+on(dA1cosθ, see picture).

o  The total (or hemispherical) emissive power [W/m2] of a diffuse surface can be obtained by integra4ng over all angles o  If the surface is also blackwe have

I(�, ✓,�) =dQ

dA1 cos ✓ d!

q = dQ/dA1 =

Z 2⇡

0

Z ⇡/2

0I cos ✓ sin ✓d✓d� = ⇡I

Ib =eb⇡

=�T 4

⇡I�b =

e�b

⇡= fn(T,�)(10.7b) (10.7c)


Theproblemofpredic+ngα. The total emi8ance, ε, of a surface is determined only by the physical proper4es of the surface, whereas thetotalabsorptancedependsonthesourcefromwhichthesurfaceabsorbsradia+on, as well as the surface’s own characteris4cs. This happens because the surface may absorb some wavelengths be8er than others and the incident radia4on may be a func4on of λ. This λ distribu4on in turn depends on the temperature and the physicalproper+esof allthesurfacesinvolved in the heat exchange process (very complicated problem).

Kirchhoff’slaw

✏�(T, ✓,�) = ↵�(T, ✓,�)

✏�(T ) = ↵�(T )

✏(T ) = ↵(T )

(10.8a)exactform(alwaysvalid)

(10.8b)diffuseform(validfordiffusebodies)

(10.8c)diffuse,grayform(validfordiffuse,graybodies)

Ingeneral,thesimplestfirstes@mateforthetotalabsorptanceisthediffusegraybodyapproxima@on(Eq.10.8c).Itwillbeaccurateeitherifthemonochroma@cemiiancedoesnotvarymuchwithλorifthebodiesexchangingenergyareatsimilarsurfacetemperature.

Kirchhoff’slaw.It is an expression that allows α to be (easily) determined under certain condi4ons. The general form of Kirchhoff’s law states thatabodyinthermodynamicequilibrium(that is, at a constant temperature) emitsasmuchenergyasitabsorbsineachdirec+onandineachwavelength. If, for example, a body emits more energy than it absorbs in one direc4on, θ1, it will be refrigerated along this direc4on. However, this would violate the Second Law of Thermodynamic which requires that there can be no net heat transfer between bodies at the same temperatures. Thus we have:

§1

0.2

Kirch

hoff

’sla

w5

35

Figu

re10.5

Heat

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eentw

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ivided

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!"#

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2 )σT

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ur

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us

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and

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§10.2 Kirchhoff’s law 535

Figure 10.5 Heat transfer between twoinfinite parallel plates.

ond integral divided by σT 42 . Hence,

qnet = ε1(T1)σT 41! "# $

emitted by plate 1

− α1(T1, T2)σT 42! "# $

absorbed by plate 1

(10.10)

We see that the total absorptance depends on T2, as well as T1.Why does total absorptance depend on both temperatures? The de-

pendence on T1 is simply because αλ1 is a property of plate 1 that maybe temperature dependent. The dependence on T2 is because the spec-trum of radiation from plate 2 depends on the temperature of plate 2according to Planck’s law, as was shown in Fig. 1.15.

As a typical example, consider solar radiation incident on a warmroof, painted black. From Table 10.1, we see that ε is on the order of0.94. It turns out that α is just about the same. If we repaint the roofwhite, ε will not change noticeably. However, much of the energy ar-riving from the sun is carried in visible wavelengths, owing to the sun’svery high temperature (about 5800 K).3 Our eyes tell us that white paintreflects sunlight very strongly in these wavelengths, and indeed this isthe case — 80 to 90% of the sunlight is reflected. The absorptance of

3Ninety percent of the sun’s energy is on wavelengths between 0.33 and 2.2 µm (seeFigure 10.2). For a black object at 300 K, 90% of the radiant energy is between 6.3 and42 µm, in the infrared.


Solarradia+onandgreenhouseeffect. The Sun is a nearly spherical body that has a radius Rsun = 7 108 m and is located at a mean distance from the Earth of d = 1.51 1011 m. it can be approximated as a black body emi`ng at Tsun = 5800 K. The solar heat flux is distributed uniformly over the sphere centered on the sun of radius d. Thus, the rate at which energy arrives at the top of the atmosphere is:

S0 is called solar constant and it represents the rate at which solar energy is incident on a surface normal to the sun’s rays at the outer edge of the atmosphere when the Earth is at its mean distance from the sun and its value is S0≅1380W/m2.

S0 represents the maximum possible power that the Sun can deliver to the Earth. However, basicgeometrylimitstheactualsolarenergyinterceptedbyEarth. Since the Earth is approximately a sphere, only half of the sphere is ever lit by the sun at one 4me, which halves the total solar irradiance. Moreover, only areas near the equator at midday come close to being perpendicular to sun’s rays; everywhere else the light comes in at an angle. The progressive decrease of the angle of solar illumina4on with increasing la4tude reduces the solar constant by an addi4onal one-half. Therefore, averageovertheen+replanet, therateatwhichenergyperunitareaarrivesatthesurfaceisonlyS0/4.

Exercise:thegreenhouseeffect

S0 =Esun

4⇡d2=

�T 4sun4⇡R

2sun

4⇡d2


Exercise1). In the first idealized model we assume the atmospherecompletelytransparenttobothincomingandoutgoingradia+on. A frac4on ρ (which is called planetary albedo) of the intercepted incident radia4on is reflected back to the space by clouds, ice, snow, lands… (since the atmosphere is assumed totally tranparent to the incident radia4on, it does not ma8er whether this albedo is caused by reflec4on at the surface or at the top of the atmosphere or a mixture). On the average, theEarth’salbedoisabout0.3. Calculatetheeffec+veemissiontemperatureoftheEarth assuming the Earth to be a perfect black body emi`ng at temperature Te and that, in the long-term, thermodynamicequilibriumis reached (all the energy absorbed is re-emi8ed at different wavelengths).


S0/4 (100%)

⇢S0/4 (30%)

�T 4e (70%)


Exercise2). In the second idealized model we assume the atmospherecompletelytransparenttotheincomingradia+onandpar+allyopaquetotheoutgoingradia+on. Assume the Sun and the Earth to be perfect black bodies emi`ng at Tsun and Ts, respec4vely. Note that, since Tsun >> Ts, the Sun mostly emits in the visible part of the spectrum, short wavelengths, while the Earth emits radia4on at much longer wavelengths, in the infrared. Assume the atmosphere to be a “one-layer atmosphere” at temperature Ta that can be approximated as a diffuse gray body with emissivity ε. Because of Kirchhoff’s law, at any given wavelength the emissivity of the atmosphere will be equal to the absorp4vity. Calculate the average surface temperature (Ts) for ε = 0.78.


S0/4 (100%)

⇢S0/4 (30%)

�T 4s (118%)

✏�T 4a (48%)

✏�T 4a (48%)

(1� ✏)�T 4s (22%)

isothermalatmosphereatTaandε=α=0.78


Exercise3)The idealized greenhouse model is based on the fact that certain gases in the Earth’s atmosphere, including CO2, H2O and CH4, are almost transparent to the short-wavelengths solar radia4on, but are much more opaque to the longer-wavelengths infrared radia4on leaving the surface of the Earth. Thus heatarrivingfromthesuncaneasilypassthroughtheatmosphere,butitispar+allytrappedasittriestoleaveit.Thus greenhouse gases cause the atmosphere (and it turn the Earth’s surface) to be warmer that it would be without them.

The balance between absorbed and radiated energy determines the average surface temperature of our planet. The radia4on balance may be altered by many factors: the intensity of solar radia4on, reflec4vity of clouds and gases and also by the amount of greenhouse gases in the atmosphere. Any such altera4on is called radia4ve forcing, and changes the balance. The es4mated radia+veforcingfor doubling the amount of CO2 in the atmosphere is dQ = 3.71 W/m2. Using the second idealized greenhouse model, calculate the global warming for a doubling of carbon dioxide.



Viewfactorforblackbodies. We have seen that when two black bodies (α=1,ε=1), say surface 1 and surface 2, exchange radia4on to each other thenetradia+ondependsontheviewfactor. In general, the view factor F1-2 is the frac4on of field of view of (1) occupied by (2). When the surfaces are each isothermal and diffuse, this correspond to:

F1-2=frac+onofenergyleaving(1)thatreaches(2)

Since all the energy leaving (1) must reach “something else”, it follows from energy conserva4on that:

where F1-1 accounts for the fact that surface (1) may also views itself (concave surface).

We have also seen that the net radia4on from (1) to (2) can be wri8en as Qnet1-2 = Q1-2 - Q2-1. Similarly, Qnet2-1 = Q2-1 - Q1-2.

Radiantheatexchangebetweentwofiniteblackbodies

F1�1 + F1�2 + . . .+ F1�n = 1

Qnet1�2 = A1F1�2��T 41 � T 4

2

�

Qnet2�1 = A2F2�1��T 42 � T 4

1

�

(10.12)

(10.13)

(10.14)


Viewfactorreciprocity. For energy conserva4on we have, Qnet1-2 = - Qnet2-1, or

Eq. (10.12) and (10.15) allow to reduce the number of view factor that need to be calculated for N surfaces, which would be N2. It can be shown that the number of view factors that need to be calculated directly is N(N-1)/2 (for example, for N=2 only 1 view factor must be calculated directly instead of 22 = 4).


A1F1�2��T 41 � T 4

2

�= �A2F2�1�

�T 42 � T 4

1

�

A1F1�2 = A2F2�1

§10.3 Radiant heat exchange between two finite black bodies 537

Figure 10.6 Some configurations for which the value of theview factor is immediately apparent.

of heat exchange among diffuse, nonblack bodies, it is the only correctionof the Stefan-Boltzmann law that we need for black bodies.

Some evident results. Figure 10.6 shows three elementary situations inwhich the value of F1–2 is evident using just the definition:

F1–2 ≡ fraction of field of view of (1) occupied by (2).

When the surfaces are each isothermal and diffuse, this corresponds to

F1–2 = fraction of energy leaving (1) that reaches (2)

A second apparent result in regard to the view factor is that all theenergy leaving a body (1) reaches something else. Thus, conservation ofenergy requires

1 = F1–1 + F1–2 + F1–3 + · · · + F1–n (10.12)

where (2), (3),…,(n) are all of the bodies in the neighborhood of (1).Figure 10.7 shows a representative situation in which a body (1) is sur-rounded by three other bodies. It sees all three bodies, but it also views


Figure 10.6 Some configurations for which the value of theview factor is immediately apparent.

of heat exchange among diffuse, nonblack bodies, it is the only correctionof the Stefan-Boltzmann law that we need for black bodies.

Some evident results. Figure 10.6 shows three elementary situations inwhich the value of F1–2 is evident using just the definition:

F1–2 ≡ fraction of field of view of (1) occupied by (2).

When the surfaces are each isothermal and diffuse, this corresponds to

F1–2 = fraction of energy leaving (1) that reaches (2)

A second apparent result in regard to the view factor is that all theenergy leaving a body (1) reaches something else. Thus, conservation ofenergy requires

1 = F1–1 + F1–2 + F1–3 + · · · + F1–n (10.12)

where (2), (3),…,(n) are all of the bodies in the neighborhood of (1).Figure 10.7 shows a representative situation in which a body (1) is sur-rounded by three other bodies. It sees all three bodies, but it also views

(1)innersphere

(2)outersphere

F1-2=1;F2-1=A1/A2;F1-1=0;F2-2=1-A1/A2someevidentviewfactors



Figure 10.8 Radiant exchange between two black elementsthat are part of the bodies (1) and (2).

where s is the distance from (1) to (2) and cosβ2 enters because dA2 isnot necessarily normal to s. Thus,

dQ1 to 2 =σT 4

1

π

!cosβ1 cosβ2 dA1dA2

s2

"

By the same token,

dQ2 to 1 =σT 4

2

π

!cosβ2 cosβ1 dA2dA1

s2

"

Then

Qnet1–2 = σ#T 4

1 − T 42

$%

A1

%

A2

cosβ1 cosβ2

πs2 dA1dA2 (10.16)

The view factors F1–2 and F2–1 are immediately obtainable from eqn.(10.16). If we compare this result with Qnet1–2 = A1F1–2σ(T 4

1 − T 42 ), we

get

F1–2 =1A1

%

A1

%

A2

cosβ1 cosβ2

πs2 dA1dA2 (10.17a)

• 

•  Since body (1) is a black-body, the intensityofradia+on is (see Eq. 10.7b):

•  The element ofsolidangleis given by (where s is the distance between (1) and (2) and cosβ2 enters because we are interested in the normal component of the incident radia4on and dA2 is not necessarily normal to s):

Calcula+onoftheblack-bodyviewfactorF1-2. In presence of two surfaces with general shape and orienta4on, the view factor must be calculated explicitly. Consider two elements dA1 and dA2 of larger black-bodies (1) and (2) both isothermal. The radiant heat from (1) to (2)


dQ1�2 = (I1d!1)(dA1cos�1)

I1 =�T 4

1

⇡

d!1 =

cos�2dA2

s2


Calcula+onoftheblack-bodyviewfactorF1-2. Thus we have (similarly for Q2-1):

The net heat radia4on from (1) to (2) is dQnet = dQ1-2 – dQ2-1. Thus:

If we compare with Eq. (10.13) and (10.14), it follows that:

The evalua4on of the above integrals becomes rather complicated, even for the simplest configura4ons. Usetable10.2(2D)and10.3(3D).


dQ1�2 =

�T 41

⇡

✓cos�1 cos�2dA1dA2

s2

◆

dQ2�1 =

�T 42

⇡

✓cos�1 cos�2dA1dA2

s2

◆

dQnet1�2 = �(T

41 � T

42 )

Z

A1

Z

A2

cos�1 cos�2

⇡s

2dA1dA2

F1�2 =

1

A1

Z

A1

Z

A2

cos�1 cos�2

⇡s

2dA1dA2

(10.16)

F2�1 =

1

A2

Z

A2

Z

A1

cos�1 cos�2

⇡s

2dA1dA2



Table 10.2 View factors for a variety of two-dimensional con-figurations (infinite in extent normal to the paper)

Configuration Equation

1.F1–2 = F2–1 =

√

1+( hw

)2

−( hw

)

2.

F1–2 = F2–1 = 1− sin(α/2)

3.F1–2 =

12

⎡⎣1+ h

w−√

1+( hw

)2⎤⎦

4.

F1–2 = (A1 +A2 −A3)/2A1

5.

F1–2 =r

b − a

[tan−1 b

c− tan−1 a

c

]

6.Let X = 1+ s/D. Then:

F1–2 = F2–1 =1π

[√X2 − 1+ sin−1 1

X−X

]

7.

F1–2 = 1, F2–1 =r1

r2, and

F2–2 = 1− F2–1 = 1− r1

r2

543

Table 10.3 View factors for some three-dimensional configurations


1. Let X = a/c and Y = b/c. Then:

F1–2 =2

πXY

⎧⎨⎩ln

[(1+X2)(1+ Y 2)

1+X2 + Y 2

]1/2

−X tan−1X − Y tan−1 Y

+X√

1+ Y 2 tan−1 X√1+ Y 2

+ Y√

1+X2 tan−1 Y√1+X2

⎫⎬⎭

2. Let H = h/ℓ and W = w/ℓ. Then:

F1–2 =1πW

⎧⎪⎨⎪⎩W tan−1 1

W−√H2 +W 2 tan−1

(H2 +W 2

)−1/2

+H tan−1 1H+ 1

4ln

⎧⎨⎩

[(1+W 2)(1+H2)

1+W 2 +H2

]

×[W 2(1+W 2 +H2)(1+W 2)(W 2 +H2)

]W2 [H2(1+H2 +W 2)(1+H2)(H2 +W 2)

]H2⎫⎬⎭

⎫⎪⎬⎪⎭

3.

Let R1 = r1/h, R2 = r2/h, and X = 1+(1+ R2

2

)/R2

1 . Then:

F1–2 =12

[X −

√X2 − 4(R2/R1)2

]

4.

Concentric spheres:

F1–2 = 1, F2–1 = (r1/r2)2, F2–2 = 1− (r1/r2)2

544


A heater (h) as shown in the figure radiates to the par4ally conical shield (S) open at the top (T) that surrounds it. If the heater and the shield are black, calculate the net heat transfer from the heater to the shield.

Example10.2


Figure 10.11 Heat transfer from a disc heater to its radiation shield.

for R1 = r1/h = 5/20 = 0.25 and R2 = r2/h = 10/20 = 0.5. Theresult is Fh−i = 0.192. Then

Fh−s = 1− 0.192 = 0.808

Thus,

Qneth−s = AhFh−s σ!T 4h − T 4

s

"

= π4(0.1)2(0.808)(5.67× 10−8)

#(1200+ 273)4 − 3734

$

= 1687 W

Example 10.3

Suppose that the shield in Example 10.2 were heating the region wherethe heater is presently located. What would Fs−h be?

Solution. From eqn. (10.15) we have

AsFs−h = AhFh−sBut the frustrum-shaped shield has an area of

As = π(r1 + r2)%h2 + (r2 − r1)2

= π(0.05+ 0.1)&

0.22 + 0.052 = 0.09715 m2

Table 10.3 View factors for some three-dimensional configurations


1. Let X = a/c and Y = b/c. Then:

F1–2 =2

πXY

⎧⎨⎩ln

[(1+X2)(1+ Y 2)

1+X2 + Y 2

]1/2

−X tan−1X − Y tan−1 Y

+X√

1+ Y 2 tan−1 X√1+ Y 2

+ Y√

1+X2 tan−1 Y√1+X2

⎫⎬⎭

2. Let H = h/ℓ and W = w/ℓ. Then:

F1–2 =1πW

⎧⎪⎨⎪⎩W tan−1 1

W−√H2 +W 2 tan−1

(H2 +W 2

)−1/2

+H tan−1 1H+ 1

4ln

⎧⎨⎩

[(1+W 2)(1+H2)

1+W 2 +H2

]

×[W 2(1+W 2 +H2)(1+W 2)(W 2 +H2)

]W2 [H2(1+H2 +W 2)(1+H2)(H2 +W 2)

]H2⎫⎬⎭

⎫⎪⎬⎪⎭

3.

Let R1 = r1/h, R2 = r2/h, and X = 1+(1+ R2

2

)/R2

1 . Then:

F1–2 =12

[X −

√X2 − 4(R2/R1)2

]

4.

Concentric spheres:

F1–2 = 1, F2–1 = (r1/r2)2, F2–2 = 1− (r1/r2)2

544

T

S

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