Process EngineeringThermodynamicsusers.abo.fi/rzevenho/PET15-2a-RAD.pdf · ÅA 424304 Åbo Akademi...

Thermal radiationrevisited

Ron ZevenhovenÅbo Akademi University

Thermal and Flow Engineering Laboratory / Värme- och strömningstekniktel. 3223 ; [email protected]

Process EngineeringThermodynamicscourse # 424304.0 v. 2015

ÅA 424304

25.1.2015Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, 20500 Turku 2/64

2a.1 Heat radiation; black bodies


Thermal radiation A body radiates heat if its temperature is > 0 K. Radiative heat transfer involves the transfer of

heat between surfaces of different temperature separated by a transparent (”diathermal”) medium, by electromagnetic waves

Radiant energy can be exchanged withoutany intervening medium and across(very) long distances.

For heat transfer, most important is the thermal radiation wavelength range0.1 ≤ λ≤ 1000 µm (100 ≤ λ≤ 1×106 nm), in the infrared (IR) region)

A complicating factor is that radiant energy and also radiant properties of materials are dependent on wavelength λ Pic: T06

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25.1.2015 Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, 20500 Turku

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Thermal radiation, heat radiation

Pics: T06

.

.

The radiation QR (W) from a surfacewith emissivity ε (-) (sv: emissivitet), surfaceA (m2) and temperature T(K) equals

QR = ε· σ·A·T4

with Stefan-Boltzmann coefficientσ = 5.67×10-8 W/m2K4

For a blackbody surface– ε =1 in the Stefan-Boltzmann Law

– all incident radiation is absorbed– radiation is maximum for its temperature at

all wavelengths (Planck, Wien)– the intensity of emitted radiation is

independent of direction diffuse emitter

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Tσ

λd)λ,T(E)T(Ebλb

ħ = h/2π, h = Planck’s constant 6.626×10-34 J.s c0= vacuum speed of light 2.998×108 m/skB = Boltzmann’s constant 1.381 ×10-23 J/KT = temperature K, λ = wavelength m

Blackbody radiation

Planck’sradiation Lawgives the spectraldistribution of the radiationemitted by a blackbody

The area under the curve equalsthe radiation:

Pic: T06

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Wien’s displacement law

Note:radiation heat transfer can be important also at low temperatures such as room temperature !

For a blackbody, the wavelength λmax for whichthe intensity is maximal is related to temperature T by

λmax· T = 2898 µm· K

Wien’s displacement law

λmax = 10 µm @ T = 300 Kλmax = 0.5 µm @ T = 6000 K

(Tsun ~ 5800 K) Pic: T06

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Bandwidths: blackbody radiationfunctions

Radiant energy emitted by a blackbody per unit area (i.e. radiant energy flux) for a wavelength band (0, λ) is equal to

where f0-λ represents the fraction of the total emission, with Planck’s radiation law

Values for f0-λ are given in the so-called Blackbody radiation functions tables – see next page

For a finite bandwidth (λ1, λ2) the fraction of the total emission is

Tσ)T(fλd)λ,T(E)T(fλd)λ,T(Eλλbλ

λ

λb

)T(f)T(f)T(fλλλλ

)Tλ

cexp(λ

c)λ,T(E

λbc1 = 3.743×108 W∙µm4/m2

c2 = 1.439×104 µm∙K


Tab: BÖ88Blackbody radiation functions


Two examples 1. What is the temperature of a blackbody that gives an emissive

power of 1000 W/m2 at wavelength 4 µm. Planck’s radiation law gives T = 609 K.

2. For a material at 2500 K, what fraction of the heat radiation is in the visible range 0.4 - 0.7 µm?

Use the blackbody radiation functions: for λ1·T = 0.4·2500 = 1000 µmK this gives f0-λ1 = 0.00032, similarly for λ2·T = 0.7·2500 = 1750 µmK4 this gives f0-λ2 = 0.03392. Thus for 0.4 -0.7 µm, fλ1-λ2 = 0.03392 -0.00032 = 0.03360 = 3.36%. Source: BÖ88

Pic: H89 λ∙T (nm.K)

f 0-λ

Blackbody radiation functions The blackbody radiation functions can be

approximated by - here symbol F instead of f :

with x = c2/(λ·T), λ in m, c2 = c0· h/kB = 14388,5 µm· K

for which n = 4 or 5 summations gives < 0.2% error

Chang, S.L. and Rhee, K.T., 1984, ‘’Blackbody Radiation Functions,” Int. Comm. Heat and Mass Transfer, Vol. 11, pp. 451-455

See alsohttps://www.thermalfluidscentral.org/encyclopedia/index.php/Blackbody_fraction_for_radiation


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2a.2 Radiation properties of materials


Emissivity, non-black surfaces Real emission depends on λ, T,

angle θ (and time...) Emissivity of a diffuse surface is

defined by

for so-called hemi-sphericalemissivity ε, or ε(T).

For a specific wavelength, the spectral hemi-sphericalemissivity is ελ or ελ(T).

For a graybody, no effect of wavelength, or direction :ε = ελ ≠ f(λ), but ε = ε(T)

)T,λ(E

)T,λ(E)T(ε

Tσ)T(ε

λd)T,λ(E)T,λ(ελd)T,λ(E

λb

λ

λ

λbλ

Pic: T06

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Tables and pictures: KJ05

Radiation interaction with (a) a general surface; (b) a black surface.

Real bodies - emissivity data PTG


Emissivity, radiation from real bodies

Spectral emissivity of a few materials

Radiation from a blackbody, a real body, a graybody at 1922 K

Pics: BÖ88, H89


Emission, reflection, transmission An energy balance for a

surface depends on reflected, absorbed, transmitted and of courseemitted energy

Reflectivity ρ (-), absorptivity α (-) and transmittivity τ (-) are defined as

Energy balance → ρ + α + τ = 1For an opaque material τ = 0 → ρ + α = 1More exact, for incident angle θ, temperature T and wavelength λ

ρ = ρ(λ,θ,T), α = α(λ,θ,T), τ = τ(λ,θ,T), as also ε = ε(λ,θ,T)

Pic: KJ05

energy incident

energy dtransmitte ;

energy incident

energy absorbed ;

energy incident

energy reflected

Incident angle θ with respect to normal

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Absorptivity A blackbody absorbs all

incident radiation: absorptivity α = α(θ,λ,T) = 1 ≠ function of θ, λ, T

A real surface absorbsonly partly; hemispherical(spectral) absorptivity α(T) or αλ(λ,T) is 0 < .. < 1

Large temperaturedifferences between the absorbing material and the emitting radiation sourceaffect the value for the absorptivity – see Fig.

Pic: BÖ88

Variation of absorptivity ofseveral materials at roomtemperature versus incident radiation source temperature.


Ideal and real surfaces

Emissive power as a function of direction: (a) black surface; (b) real surface; (c) diffuse surface.

Reflected energy as a function of direction: (a) diffuse reflection; (b) specular reflection; (c) reflection from a real surface

Pictures: KJ05


Kirchhoff’s Law The surface emissivity ελ(T) by a body at temperature T is equal

to the surface absorptivity αλ(T) (for a given angle θ) of the bodyfor radiation originating from a blackbody at temperature T.

For gray surfaces, or other situations where radiationproperties are independent of wavelength: α(T) = ε(T)

Note: an exception to this law would mean a violation of the Second Law of Thermodynamics, since it would allow heat transfer from cold to hot surfaces

According to Petela [P10]: can be used for ε >0.5

Consider the situation in the Figure: a small, gray, diffuse body with surface A at temperature T1 in an evacuatedoven with black walls at temperature T2, and T2 > T1. The object receives radiation from the walls: Q = α∙σ∙T2

4 (W) and after some time its temperature has risen to T2. It willemit radiation which is then at a rate Q = ε∙σ∙T2

4 (W) Thus α = ε, but strictly speaking only if the two surfacesthat ”exchange” radiation are at the same temperature ! In practice OK for ΔT up to a few 100 K.

Picture: KJ05

.

.

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2a.3 Radiosity, radiation resistance


Radiosity: opaque, gray surfaces

For non-black surfaces the radiation flux Q”R,out leaving a surface equals the sum of emittedown radiationε· Eb = ε· σ·T4 plus (partially) reflected incoming radiation flux ρ· Q”R,in = ρ· G

Thus

Q”R,out = Q”R,own,out + ρ· Q”R,in or J = ε· Eb + (1-ε)·G

for an opaque gray medium: τ = 0, α = ε = 1- ρ

For this, the term radiosity, J, is defined as all the radiationleaving a surface including emitted and reflected radiation

.

.

.

Pic: KJ05

. .

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Radiosity, radiation resistance

The energy balance for the surfacestates that the net leavingradiation flux Q”R,out,net equals the own emission minus the absorbed incoming radiation:

Q”R,out, net = Q”R, out - α· Q”R,in = J – G = ε·Eb + (1-ε)·G - G = ε·Eb – ε·G which gives after elimination of G: Q”R,out, net = ε· (Eb - J)/(1 - ε)

The net heat flow leaving the surface A (m2) is thenQR,out,net = A· ε· (Eb - J)/(1 - ε) = (Eb - J)/RR, with surfaceresistance to radiation RR = (1-ε)/(A·ε)

For a blackbody surface, RR = 0.

.

Pic: KJ05

...

.

.

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2a.4 View factors


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View factors

Pics: KJ05

Fi→j = 0, Fj→i =0 Fi→j = 1, Fj→i = ½Ai = πR2, Aj = 2πR2

ij

Ai Aj

ji

iji dAdA

Sπ

θcosθcos

AF

:geometry any for ,Definition

N

jjiF

The view factors Fi→j (also referred to as configuration factors, shape factors, angle factors)quantify how much (i.e. what fraction) of the radiation from surface ”i” reaches othersurface ”j”, (by a straight-line route) as determined by Lambert’s cosine law, QR(θ) = QR(θ)·cos(θ), , and vice versa.

For any two surfaces Ai and Aj in anygeometrical arrangement: Fi→j·Ai = Fj→i·Aj

0 ≤ Fi→j ≤ 1, and for an enclosure with many surfaces:

PTG


View factor diagrams /1

Pics: KJ05

Example: Radiation occurs betweenfloor, walls and ceiling of a room as shown in the Figure. Calculate the view factors from the end wall (1) to the other five surfaces (2...6) usingthe diagram given.

Answer:

For F1→2 : X=Y=8 and L=12, gives X/L = Y/L = 0.667. From diagram: F1→2 ≈ 0.11

For the other surfaces: F1→3 = F1→4 = F1→5 = F1→6 and ΣF1→j = 1

This gives F1→2 + 4· F1→3 = 1

F1→3 = (1 - 0.11)/4 = 0.2225

= F1→4 = F1→5 = F1→6

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View factor diagrams /2

Pics: KJ05

ÅA 424304


2a.5 Radiation heat exchange, network models


Radiation between black surfaces

For two black surfaces, some of the radiation form surface ”1” strikes surface ”2” and vice versa

For two black surface with different temperatures, the net radiative heat transfer between them equals

Q12 = radiation leaving ”1” and arriving at ”2” minus radiation leaving ”2” and arriving at ”1”= Eb1A1F1→2 - Eb2A2F2→1 ,

and with A1F1→2 = A2F2→1 this gives

Q12 = A1F1→2 (Eb1 – Eb2) = A1F1→2 σ(T14 – T2

4)

Pic: KJ05.

.

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Radiation between gray surfaces

This can be written as Q12 = (J1 – J2) / R1→2 , with radiation resistance RR= 1/(A1F1→2).

Thus, there are three resistances: 1-ε/A1·ε1, 1-ε2/A2·ε2, and 1/(A1F1→2) = 1/(A2F2→1)

In analogy with radiation betweenblackbodies, with J instead of Eb :Q12 = radiation leaving ”1” and arriving at ”2” minus radiationleaving ”2” and arriving at ”1”

= J1A1F1→2 – J2A2F2→1 ,

and with A1F1→2 = A2F2→1 this gives

Q12 = A1F1→2 (J1 – J2)

Pic: KJ05

.

.Two isothermal, gray, diffuse surfaces exchanging heat by radiation..

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Heat radiation: two surfacesSurface A1

at temperature T1

with emissivity ε1

Surface A2

at temperature T2

with emissivity ε2

F1→2 = view factor A1 → A2

Radiative heat transferSurface A1 → surface A2

εε

AA

Fεε

)TT(Aσ

εAε

FAεAε

)TT(σQ

and similar for Q2→1

PTG


Radiation in two-surface enclosuresPicture: KJ05

Two gray surfaces radiation PTG


Example: radiation between gray surfaces

A tube that transports steam at 150°C runs along a long corridor as illustrated in the Figure. The temperature of all walls is 10°C and wall emissivity is 0.85.The temperature of the tube on the outside of the isolation at diameter 0.15 m is measured to be 25°C, at an emissivity of 0.85.

a. Calculate the heat losses per meter tube length as a result of heat radiation, and

b. By how much would the heat losses decrease if the isolation is painted with Al-paint with emissivity 0.42

Source: ÖS96-7.4

PTG


Example: radiation between gray surfaces

Answer: Radiation losses Q’ (W/m) are calculated using for a surface surrounded by another surface:

with A1’= A1 / length, A2’= A2 / length The view factor F1→2 = 1 for this situation,

A1’ = π·dtubeoutside = 0.47 m2/m, and A2’ = 3m + 2.5m + 3m + 2.5 m = 11 m

a: Q’12 with A1’, A2’, ε1 = ε2 = 0.85, T1 = 298 K, T2 = 283 K gives

Q’12 = 33.1 W/m (assuming A2 » A1 gives 33.5 W/m) b: changing ε1 = 0.85 → ε1 = 0.42 gives Q’12 = 16.0 W/m which means

a reduction of ((33-16)/33)×100% = 51.6%

2

1

21

42

411

12

111

A

A

TTAQ

.

.

.

.

Source: ÖS96-7.4

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Two-, three-surface enclosures, and equivalent networks

Resistance analogy for two or three-surface enclosures

Pics: KJ0532

Q and Q for similar

AFR

AFR

Aε

εR

R

JTσQ


Example: three-surface enclosures /1

Source: BÖ88

p.24p.25


Example: three-surface enclosures /2

Source: BÖ88

= 1 / 5.4


Radiation shields Radiation heat transfer

between (two) surfaces can be significantly reduced by putting a radiation shield made of low-emissivity material between them.

Using, for surfaces 1 and 2 without a shield

gives after inserting a material with emissivities ε3,1 and ε3,2, that

1

11

21

42

41

12

TTAQ

,,εεεε

TTσAQ

Example: T1, T2, ε1 = 0.8, ε2 = 0.4, inserting shield ε3,1 = ε3,2 = 0.05

QR drops to 6.6 %


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Radiative heat transfer coefficient

_

_ _

If the temperature difference ΔT = T1 – T2 of a radiative heat transfer process is small compared to T1 and T2, a radiativeheat transfer coefficient hR can be defined by (for the radiationfrom surface ”1”)

Q”R = hR· ΔT = QR / A1 = ε1· σ· (T14-T2

4) (W/m2)

→ hR = ε1· σ· (T14-T2

4) / (T1-T2) = ε1· σ· (T12 + T2

2)· (T1+T2)

using (x4 - y4) = (x2 - y2)·(x2 + y2) = (x - y)·(x + y)·(x2 + y2)

This can also be simplified to hR = ε1· σ· 4·T3 where

T = ½· (T1+T2) → 4T3 ≈ (T12 + T2

2)· (T1+T2)

PTG

Double glass window set-up and radiation network

25.1.2015Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, 20500 Turku 38

EbA EbBJ1 J2

J3 J4

EbG

R1

R2 R3R4 R5

R6 R8 R9 R7

R10

RA RB

EbA EbBJ1 J2

J3 J4

EbG

R1

R2 R3R4 R5

R6 R8 R9 R7

R10

RA RB

TA

εA

TεG

τG

TB

εB

TG2

εG2

A2

τG2

TG1

εG1

A1

τG1

L

”room” ”window” ”sky”

TA

εA

TεG

τG

TB

εB

TG2

εG2

A2

τG2

TG1

εG1

A1

τG1

L

”room” ”window” ”sky”

After calculating the values for network nodesJ1,J2, J3, J4 and EbG the temperatures of the windows and the enclosed gas can be calculated.

Ref:ZF10

The model equations


Gas temperature TG& Heat flux QA-B

Ref:ZF10

5 equations5 variables J1 J2 J3 J4 & EbG

Matrix inversion

ÅA 424304


2a.6 Gas radiation


Radiation characteristics of gases

Pic: M02

Emission/absorption wave-lenghts (in µm) for CO2, H2O, CH4, CO between 0 and 100 µm. (data M02)

Gases can absorb and emit thermal radiation in different wavelengths (bands), depending on their possible transitionsin vibrational and rotational energy levels.

Important for engineering and heat transfer applications are gases like CO2 and H2O, but also CO, CH4, SO2, NH3, O3, and others. But – note also soot particles, etc.!


Gas radiation: emissivity /1

Data by Hottel (1954) and Leckner (1972) give values for emissivity of CO2 and H2O and correction factors for mixtures of these (taking into account spectral overlap).

__ Hottel

---Leckner

Pics: M02

CO2

H2O



For CO2 or H2O at total gas pressure 1 bar and low partial pressures × thickness up to (a few) 100 bar·cm, graphs aregiven on the previous slide; for other cases see proceduresgiven in SH92, or (as used here) M02.

For gas thickness L (cm!), absorbing gas partial pressure pa(bar), gas temperature Tg (K) the emissivity ε0 (-) for lowpartial pressures (→ 0), is calculated using M02, as given in the graphs on the previous slides

with parameters M,N, cji,(paL)0 as given in Tables -see next slides



For higher pressures a corrections factor is used (M02):

were pE is an effective pressure (bar) and a,b,c,d and (paL)mare correlation parameters – see Tables below.

H2O CO2



For mixtures containing both CO2 and H2O, bands overlap must be accounted for (M02):

Thus the procedure can be summarised as (M02):

Note for α:Ts of the external source


Example: gas emissivity Calculate the emissivity of a 1 m thick gas layer at 1000

K, 5 bar, consisting of 10% CO2, 20% H2O, 70% N2. pco2·L = 50 bar·cm, at 1000 K gives ε0,CO2 = 0.157

(from diagrams or (10.144)); with pE = 5.14, a = 1.1, b = 0.23, c = 1.47 and (paL)m = 0.225 bar·cm, (10.145) gives (ε/ε0)CO2 ≈ 1 εCO2 = 0.157.

pH2O·L = 100 bar·cm, at 1000 K gives ε0,H2O = 0.359 (from diagrams or (10.144)); with pE = 7.56, a = 1.88, b = 1.1, c = 0.5 and (paL)m = 13.2 bar·cm, (10.145) gives (ε/ε0)H2O ≈ 1.414 εH2O = 0.508.

for the band overlap, (10.147) and (10.146) gives ζ = 2/3, Δε = 0.072

Thus, the final value is ε = 0.157+0.508-0.072 = 0.593.

Emissivities of CO2 and CO2-containing gases


0

0.2

0.4

0.6

0.8

1

0 2 4 6 8 10 12 14 16 18 20 22

ab

so

rpti

vit

y,

em

iss

ivit

y (

-)

wavelength (µm)

_____ CO2 (p,xCO2,L,T)------------ CO2 atmosphere

0

0.2

0.4

0.6

0.8

1

0 2 4 6 8 10 12 14 16 18 20 22

ab

so

rpti

vit

y,

em

iss

ivit

y (

-)

wavelength (µm)

_____ CO2 (p,xCO2,L,T)------------ CO2 atmosphere

pressure

CO2

average value (range)Air

average value (range)

p = 0.1 bar 0.045 (0.038 – 0.049) 0.023 (0.018 – 0.027)

p = 1 bar 0.106 (0.103 – 0.109) 0.092 (0.089 – 0.095)

p = 10 bar 0.190 (0.186 – 0.193) 0.175 (0.171 – 0.179)

Emissivities of CO2 and air at different pressures,path-length L = 0.1 m, for the temperature range250 – 350 K

Absorptivity/emissivity for TIR for Atmospheric CO2 compared with an enclosure of 100% CO2 at p = 5 bar, path length L = 0.1 m, for T = 300 K.

Ref:ZF10

ÅA 424304


2a.7 Solar radiation

Picture: T06


Radiation from the sun /1

Consider the sun as a blackbody radiator at 5800 K, with diameter ~ 1.39×109 m. The diameter of earth is ~ 1.29×107 m; the distance between earth and sun is ~ 1.5×1011 m

The solar radiation output equals

Qsun = π· d2sun· σ·T4

sun = 3.89×1026 W

The fraction of Qsun that is inter-cepted by earth equals Qsun-earth

= Qsun × (π/4)· d2earth/(4π·x2

earth-sun)

= Qsun× d2earth/(16·x2

earth-sun) = 1.80×1017 W

heat flux to spherical earth equalsQsun-earth = Qsun/(4π·x2

earth-sun) = 1376 W/m2

Pic: T06

Pic: KJ05”Solar constant”

PTGSe

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soco

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Radiation from the sun /2

Spectral distribution of solar radiation

Solar irradiation on a horizontal surface under average atmospheric conditions Pic: DB74

Table: H89

1 Ly = 1 Langley = 1 cal/cm2


Radiation from the sun /3 Albedo

The albedo of a surface, defined as the ratio reflected energy / incident energy, depends on material and angle:

Pic + Table: H89

ÅA 424304


2a.8 Environmental radiation

ÅA 424304

Earth-Atmosphere Energy Balance


Gho

niem

, Ahm

ed F

.,“N

eeds

, res

ourc

es a

nd c

limat

e ch

ange

: Cle

an a

nd

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ient

con

vers

ion

tech

nolo

gies

” P

rog.

Ene

rgy

Com

bust

. Sci

. 37

((20

09)

15-5

1

Tsky

Tair

Tamb

31.2% 22.5% 68.8%

< 4 µm > 4 µm


Incoming/outgoing atmospheric IR /2

Approximate spectra for solar and earth radiation (note different vertical axes – intensity difference by a factor 106 !)

10-8 10-2


55/64

IR to/from earth’s surface /1

Radiation heat transfer QR(W) in an enclosure between two surfaces A1and A2, emissivities ε1 and ε2 and temperatures T1 and T2, with surface A2completely surrounding surface A1 is given by (M02, SH92):

T = Tsur

Ground levelsurface Asur

Surroundingatmospheresurface Asky

Surroundinguniversesurface Auni

T = Tsky

T = Tuni = uni

= sky

= surT = Tsur




T = Tsky

T = Tuni

T = Tsur




T = Tsky

T = Tuni = uni

= sky

= sur

Pic: Z08

2

1

21

42

411

111

A

A

TTAQR


56/64


Discarding incoming solar irradiation (night-time), and with Tuni = 3 - 4 K << Tsky ≈ Tsur , Auni >> Asky ≈ Asur

gives the following heat balance for the sky, for a steady-state situation:

T = Tsur




T = Tsky

T = Tuni = uni

= sky

= surT = Tsur




T = Tsky

T = Tuni

T = Tsur




T = Tsky

T = Tuni = uni

= sky

= sur

Pic: Z08,ZF10

skysur

skysur

skysky

sur

sky

skysur

skysursur

uni

sky

unisky

uniskysky

εε

T½TT½ε

A

A

εε

)T½T(σA

A

A

εε

)TT(½σA


57/64


This gives

where the (view) factor ½ accounts for the fact that half the radiation from the sky is away from earth, into space, and the other half is towards earth.

With typical ground level emissivity εsur = 0.8 - 0.9 and emissivity of the sky ε sky = 0.6 - 0.9, temperature differ-ence Tsur - Tsky ≈ 5 - 10 K(IdWBL06, MM77)

T = Tsur




T = Tsky

T = Tuni = uni

= sky

= surT = Tsur




T = Tsky

T = Tuni

T = Tsur




T = Tsky

T = Tuni = uni

= sky

= sur

sursky

sky

sur

ε½ε

T

T

εsur = 0.6 ... 0.9 (Tsur / Tsky)4 = 1.06 ... 1.09Pic: Z08,ZF10


Earth-to-space IR: passive cooling

see ongoingresearch at ÅA Thermal and Flow Engineering

Pics: FZ11

Naturalconvectiondriven by participatinggases

Input data:Tambient≠ Tsky !!

Results:(see alsonext page)

Temperature (°C, left) and velocity (cm/s, right) profiles for a skylight (width 0.5 m, height 0.1 m, not to scale) containing CO2 with absorptivity = emissivity = 0.19, cooling during summer (Helsinki, Finland). Heat flux with CO2 117 W/m2, with air 15 W/m2. Simulated with Comsol 4.1


cm/s°C

0.5 m

0.1 m

Earth-to-space IR: passive cooling

Pics: FZ11

ÅA 424304


2a.9 Wavelength dependence(for passive cooling modelling)

ÅA 424304

Four-band model: SW + 3x LW


Radiation throughthe atmosphereat 4 wavelengthbands.

Short wave< 4 µm

Long wave range 1:4 – 8 µm

Long wave range 2:8 – 14 µmthe atmospheric window

Long wave range 3:> 14 µm

Pic: ZFG14

ÅA 424304

Four-band model: SW + 3x LW


Thermal radiationfrom a buildingor room to the atmosphere /skyusing the four-band model and an equivalent networkfor a doubleglasswindow

Pic: ZFG14


Sources 2a /1

B97: A. Bejan “Advanced engineering thermodynamics” 2nd ed. Wiley (1997) BS06: H.D. Baehr, K. Stephan ”Wärme- und Stoffübertragung”, 5. ed., Springer (2006)

Chapter 5 BÖ88: Y. Bayazitoglu, M.N. Özisik “Elements of heat transfer” McGraw-Hill (1988)

Chapter 11 DB74: Duffie, J.A., Beckman, W.A. Solar energy thermal Processes Wiley & Sons (1974) FZ11: M. Fält, R. Zevenhoven “Combining the radiative, conductive and convective heat

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the enhanced greenhouse effect” ECOS’2008, Krakow-Gliwice, Poland, June 24-27, 2008

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DI Martin Fält is acknowledged forcourse material update support (Jan. 2013)

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