Abteilung Jülich
Contents of Lecture VIII
1 Steps to design a solar thermal plant
2 Calculation of flat-plate collectors
2.1 Effective solar power on aperture
2.2 Optical losses
2.3 Thermal losses
Lecture VIII, 22. 05. 2007
1
Steps to design a solar thermal plant (one‘s the
collector type has been choosen)
Ps
Tin
1)
Calculate the
incident power on
the collector
2)
Calculate the optical
losses (-> absorbed
power)
3)
Calculate the heat
losses
4)
Losses of the circuit
(parasitics, thermal
losses of piping,
heat exchanger eff.)
5)
Puse
Tout
POpL
PThL
Control
unit
Circulating pump
Lecture VIII, 22. 05. 2007
Tuse, Puse
Consumer
(Heat
exchanger,
Storage)
Abteilung Jülich
2
Abteilung Jülich
2. Flat-plate collector
A flat-plate collector is characterized by the fact, that its aperture area
(the solar radiation collecting area) is as large as the absorbing area
(concentration factor C = 1).
Cover
Absorber
Pipes
Lecture VIII, 22. 05. 2007
Thermal
insulation
Main components:
•The absorber with a black, absorbing
area including devices (pipes,
channels) to transmit the absorbed
energy to the heat transfer fluid
• Coverings in front of the absorber,
which are transparent for the solar
spectrum, to prevent heat losses to
the atmosphere (radiation,
convection)
•The thermal insulation at the
backside and at the sides for the
reduction of losses by thermal
conduction
3
Abteilung Jülich
General function of a flat-plate collector
ϑSky
Solar radiation (beam + diffuse)
(0.3 < λ < 2.5 µm)
Thermal radiation losses (20%)
100 %
(3 < λ < 100 µm)
Reflection
Convection losses (9%)
8%
Absorption
2%
Transmission
τ∗
G = 90 %
τ*G α*A
= 83 %
Lecture VIII, 22. 05. 2007
free
convection
Reflection 8%
Photothermal conversion
Useful heat (50 %)
Conductivity losses (4%)
4
Abteilung Jülich
Steps to design a thermal collector
(e.g.: flat plate collector)
Power balance:
PB
PD
PU = PS - PL
POpL
with
PU
PS = PB + PD ,
Incident solar radiation power
(beam + diffuse)
PL = POpL + PThL
PThL
Power losses (optical +
thermal)
PU = PB + PD – POpL – PThL
(Note: PS – POpL: Absorbed Power)
Lecture VIII, 22. 05. 2007
5
Abteilung Jülich
2.1 Calculation of the incident solar power
Power balance:
PB
PD
PU = PS - PL
with
PS = PB + PD ,
Incident solar radiation power
(beam + diffuse)
Lecture VIII, 22. 05. 2007
6
Abteilung Jülich
Solar incident radiation power on aperture (“effective power”):
PS = PB + PD
with
PB = EDNI • cos Θ • A
(effective power of beam radiation)
PD = ED • cos Θeff • A
(effective power of diffuse radiation)
(EDNI = Direct normal irradiance, i.e. beam irradiance
perpendicular to the direction of insolation [W/m²],
Θ = Angle of incidence (angle between aperture normal and
sun vector,
A = (Net) Aperture area [m²],
ED = Diffuse irradiance on horizontal surface [W/m²]
Θeff = Effective incidence angle of diffuse radiation, cos Θeff = 1
for isotropic diffuse radiation)
Lecture VIII, 22. 05. 2007
7
Abteilung Jülich
Note:
The difference between the beam radiation power perpendicular to the
direction of insulation and the effective beam power is often named „cosine
losses“ and is sometimes considered to be part of the optical losses.
Pcos = PDNI – PB
with
SUN
PDNI = EDNI•A (solar beam radiation power
normal to the direction of incidence,
ideal oriented collector)
PB = EDNI•cos Θ • A (effective solar
beam radiation power)
n
Θ
s
⇒ Pcos = EDNI • A • ( 1 - cos Θ)
Cosine-losses can be avoided by
“tracking the sun” (Θ = 0)
Lecture VIII, 22. 05. 2007
A: aperture area
8
Abteilung Jülich
2.2 Calculation of the optical power losses
(flat-plate collector)
PL = POpL + PThL
with
POpL = Optical losses: all losses which are caused on the way of
the radiation from the aperture to the surface of the
absorber due to transmission losses at covers and
absorption losses at the surface of the absorber plate
(direct losses).
PThL = Thermal losses: all losses which are caused after the
absorption of the solar radiation at the surface of the
absorber due to heat exchange between the warmed
absorber and the environment by convection, radiation
and thermal conduction (indirect losses).
Lecture VIII, 22. 05. 2007
9
Abteilung Jülich
Optical losses of a flat-plate collector
PS
Optical power losses:
POpL
Lecture VIII, 22. 05. 2007
Depend on collector‘s geometry and
material properties:
•
Number of covers
•
Transmittance of covers
•
Absorption properties of absorbing
surface
•
Reflection properties of the border
(can normaly be neglected)
10
Abteilung Jülich
2.2.1 Calculation of Transmittance
Fundamentals of reflection and refraction
If radiation passes over from a medium 1 to the
medium 2, having different refractive indices n1,
n2, reflection and refraction takes place at the
border.
I0
n1
Iρ
Energy balance:
I 0 = I ρ + Iτ
n2
with
Iτ
I0 = insolated beam radiation intensity
Iτ = transmitted beam radiation intensity
Iρ = reflected beam radiation intensity
Lecture VIII, 22. 05. 2007
11
Abteilung Jülich
Calculation of transmittance
Fundamentals of reflection and refraction
I0
n1
Division by I0 leads to
Iρ
1 = ρ +τ
τ = 1− ρ
n2
Iτ
with
reflectivity:
transmissivity:
Lecture VIII, 22. 05. 2007
ρ=
Iρ
I0
I
I0
τ= τ
12
Abteilung Jülich
Remember: Law of Fresnel
I0
n1
Θ1
ρ=
Iρ
Iρ
I0
=
1
(ρv + ρ p )
2
sin 2 (θ 2 − θ1 )
ρv = 2
sin (θ 2 + θ1 )
n2
tan 2 (θ 2 − θ1 )
ρp = 2
tan (θ 2 + θ1 )
with
ρ = reflectivity of unpolarized radiation,
Θ2
Iτ
ρv = reflectivity of vertical polarized radiation,
ρp = reflectivity of parallel polarized radiation,
Θ1 = incidence angle,
Θ2 = refractive angle.
Lecture VIII, 22. 05. 2007
13
Abteilung Jülich
Remember: Law of Snellius
I0
n1
The refractive angle Θ2 can be calculated
Θ1
from the refractive indices n1, n2 and the
Iρ
incidence angle according to Snellius‘s law:
n2 sin θ1
=
n1 sin θ 2
n2
Θ2
Iτ
Note:
nair ≈ 1
nglass ≈ 1.526
Lecture VIII, 22. 05. 2007
14
Abteilung Jülich
Calculation of reflectivity
For perpendicular incidence, Θ1, Θ2 become zero, and:
⎛ n1 − n2 ⎞
⎟⎟
ρ = ρ v =ρ p = ⎜⎜
⎝ n1 + n2 ⎠
2
If one medium is air:
⎛ n −1 ⎞
⎟
⎝ n +1⎠
2
ρ = ρv = ρ p = ⎜
Lecture VIII, 22. 05. 2007
15
Abteilung Jülich
Transmittance of covers
Absorption and multiple reflection has to be taken into account!
I0
I 0 = I ρ + Iτ + I α
Iρ
Iρ1
with
Iρ2
Iρ = Iρ1 + Iρ2 + ... + Iρn
Iα
Iτ = Iτ1 + Iτ2 + ... + Iτn
Iτ2
Iτ1
1 = ρ +τ + α
τ = 1− ρ −α
Iτ
and the absorptivity:
I
I0
α= α
Lecture VIII, 22. 05. 2007
16
Abteilung Jülich
Transmittance of covers
I
ρ•I
ρ (1 - ρ)² • I
Neglecting absorption, the
transmitted radiation
intensity is:
Iτ = Iτ 1 + Iτ 2 + ... + Iτn
= ρ 0 ⋅ (1 − ρ ) ⋅ I +
2
(1 - ρ) • I
ρ²(1 - ρ) • I
ρ (1 - ρ) • I
ρ³(1 - ρ) • I
ρ 2 ⋅ (1 − ρ )2 ⋅ I +
ρ 4 ⋅ (1 − ρ )2 ⋅ I + ...
ρ 2(n −1) ⋅ (1 − ρ )2 ⋅ I
I1 = (1 - ρ)² • I
Lecture VIII, 22. 05. 2007
I2 = ρ²(1 - ρ)² • I
17
Abteilung Jülich
Transmittance of covers
Transmittance in consideration of the reflection:
2
⎡ n 2(i −1) )
⎤
Iτ = I ⋅ ⎢ ∑ ρ
⋅ (1 − ρ ) ⎥
⎢⎣ i =1
⎥⎦
2
⎤
Iτ ⎡ n 2(i −1) )
= ⎢∑ ρ
⋅ (1 − ρ ) ⎥
I ⎢⎣ i =1
⎥⎦
1− ρ
τρ =
1+ ρ
Lecture VIII, 22. 05. 2007
18
Abteilung Jülich
Transmittance of covers
For a system of N covers of the same material the total transmittance is:
τ ρ ,N =
1− ρ
1 + (2 N − 1) ⋅ ρ
Remember:
As the reflectivity depends on the incident angle according to Fresnel‘s law, the
transmittance of glass covers depends on the incidence angle as well.
For incidence angles Θ < 40 ° the transmittance decreases smoothly, while for
incidence angles Θ > 40 ° it decreases significantly. Also the equation has to be
applied seperately for vertical and parallel poarized radiation:
Lecture VIII, 22. 05. 2007
19
Abteilung Jülich
Transmittance of covers
Transmittance calculation for incidence angles of Θ > 40 ° :
⎞
1− ρ p
1 − ρv
1 ⎛⎜
⎟
τ ρ ,N = ⎜
+
2 ⎝ 1 + (2 N − 1) ⋅ ρ v 1 + (2 N − 1) ⋅ ρ p ⎟⎠
with
ρv : reflectivity of vertical polarized radiation,
ρp : reflectivity of parallel polarized radiation.
Lecture VIII, 22. 05. 2007
20
Abteilung Jülich
Transmittance of covers
For diffuse radiation the reflectivity can be determined by calculating the
reflectivity of the cover system for an angle of incidence of ΘD = 60 ° .
For the system air/glass cover this leads to:
1 cover:
ρDiffuse,1 = 0.16 (τdiff,ρ,1 = 0.84)
2 covers:
ρDiffuse,2 = 0.24 (τdiff,ρ,1 = 0.76)
3 covers:
ρDiffuse,3 = 0.29 (τdiff,ρ,1 = 0.71)
4 covers:
ρDiffuse,4 = 0.31 (τdiff,ρ,1 = 0.69)
Lecture VIII, 22. 05. 2007
21
Abteilung Jülich
Transmittance of covers
The equation can be
plotted:
1 − ρ (Θ )
1 + (2 N − 1) ⋅ ρ (Θ )
τ(Θ=60°) for transmissivity
of diffuse radiation
0.9
0.8
Transmissivity [-]
τ ρ ,N =
1
0.7
0.6
0.5
N=1
0.4
N=2
0.3
N=3
N=4
0.2
0.1
0
0
10
20
30
40
50
60
70
80
90
Incidence Angle [deg]
Lecture VIII, 22. 05. 2007
22
Abteilung Jülich
2.2.2 Absorption calculation of covers
The absorption of radiation in a partially transparent medium is described by
Bouger`s law, which is based on the assumption, that the absorbed radiation
is proportional to the local intensity, I, in the medium and to the distance, L, the
radiation has travelled in the medium.
dI = x · I · dL ,
with
dI = locally absorbed radiation, W/m2 ,
x = extinction coefficient, assumed to be constant
in the solar spectrum, m-1,
I = local intensity of the radiation in the medium, W/m2
dL = infinitesimal path length of the radiation in the
medium, m.
Lecture VIII, 22. 05. 2007
23
Abteilung Jülich
Absorption calculation of covers
Integration along the optical path length in the medium from 0 to L, the
transmittance, τα, referred to absorption in the cover, results in:
I
θ1
air
with
θ2
L
I transmitted
τα =
= e−x ⋅ L
I incident
s
glass cover
L = s / cos Θ2 (optical path), m
x: extinction coefficient, m-1
θ1
Lecture VIII, 22. 05. 2007
air
For glass: x = 4 .... 32 m-1
24
Abteilung Jülich
Absorption calculation of covers
For a multiple cover system the transmission coefficient for absorption,
τα, for radiation under an angle of incidence, θ, yields:
τ α ,N = e
−
N ⋅ x⋅ s
cos Θ 2
with
τα= transmission coefficient for absorption,
N = Number of covers,
x = extinction coefficient,
s = thickness of cover,
θ2 = refraction angle, degree.
Lecture VIII, 22. 05. 2007
25
Abteilung Jülich
Total transmittance
The total transmittance, τ, which considers both the reflection and
the absorption, is calculated from the product:
τ = τ ρ , N ⋅τ α , N
with
τρ,Ν = transmittance due to reflection,
τα,Ν = transmittance due to absorption,
Index N: Number of covers.
Lecture VIII, 22. 05. 2007
26
Abteilung Jülich
Calculation of a flat-plate collector
Next step: Calculation of Absorption of
absorber‘s surface
PS
POpL
Lecture VIII, 22. 05. 2007
Depend on collector‘s geometry and
material properties:
>
•
Number of covers
•
Transmission properties of covers
•
Absorption properties of absorbing
surface
>
27
Abteilung Jülich
Transmittance – Absorption Product (τα)
The Transmittance-Absorption product (τα) can be calculated similar to
the calculation of the transmittance due to multiple reflection in a cover.
Incident solar radiation
System of covers
(1−α)τ
τ
Absorber plate
Lecture VIII, 22. 05. 2007
2
(1−α) τρ
(1−α)τρ
τα
τα(1− α)ρ
2
(1−α) τρ
2
2 2
τα(1− α) ρ
28
Abteilung Jülich
Transmittance – Absorption Product (τα)
i −1
⎡ n i −1
⎤
(τα ) = τ ⋅α ⋅ ⎢∑ ρ ⋅ (1 − α ) ⎥
⎢⎣ i =1
⎥⎦
τ ⋅α
=
1 − (1 − α ) ⋅ ρ
with:
τ: total transmission of cover system
α: absorption coefficient of the absorber plate
ρ: reflection coefficient of lowest cover
Lecture VIII, 22. 05. 2007
29
Abteilung Jülich
Calculation of a flat-plate collector
PB
PD
Now the absorbed power can be
calculated according to:
POpL
PU
Pabs = PS - POpL
= PB + PD – POpL
PU = PB + PD – POpL – PThL
PThL
Last step
Lecture VIII, 22. 05. 2007
30
Abteilung Jülich
Power losses:
PL = POpL + PThL
with
POpL = optical losses: all losses which are caused on the way of the
radiation from the aperture to the surface of the absorber due to
transmission losses at covers and absorption losses at the
surface of the absorber plate (optical losses).
PThL = Thermal losses: all losses which are caused after the absorption of the solar radiation at the surface of the absorber due to
heat exchange between the warmed absorber and the
environment by convection, radiation and thermal conduction
(thermal losses)
Lecture VIII, 22. 05. 2007
31
Abteilung Jülich
Thermal losses due to thermal conduction
Diffusive transport of heat through a material,
according to Fourier‘s law:
∂T ⎞
⎛
Pcond = A ⋅ ⎜ − λ
⎟
∂x ⎠
⎝
with:
A: heat exchanging surface area, m²
λ: heat conductivity coefficient, W/(m²*K
T: local temperature, K
Lecture VIII, 22. 05. 2007
32
Abteilung Jülich
Thermal losses due to thermal conduction
Integration (λ = const.) yields:
Pcond = −
λ
⋅ A ⋅ (T2 − T1 )
D
= kλ ⋅ A ⋅ (T1 − T2 )
with:
A:
heat exchanging surface area, m²
λ:
heat conductivity coefficient, W/(m²*K)
T1,T2: temperatures of surface 1 resp. 2, K
D:
thickness of material, m
and
kλ: heat loss coeff. due to conduction, W/(m²K)
Lecture VIII, 22. 05. 2007
33
Abteilung Jülich
Thermal losses due to thermal conduction
Note: In the case of cylindrical geometries, e.g. an insulated pipe, the
increasing surfaces have to be considered:
Pcond = 2πL ⋅
T0
T1
T0 − Tn
1 ⎛ d1 ⎞ 1 ⎛ d 2 ⎞
1 ⎛ dn ⎞
⎟⎟
ln⎜⎜ ⎟⎟ + ln⎜⎜ ⎟⎟ + K +
ln⎜⎜
λ1 ⎝ d 0 ⎠ λ2 ⎝ d1 ⎠
λn ⎝ d n −1 ⎠
with:
L: tube length, m
T2
Tn
λi: heat conductivity coefficient of layer i,
W/(m²*K)
di: diameter of layer i, m
Lecture VIII, 22. 05. 2007
34
Abteilung Jülich
Thermal losses due to convection
Heat transport due to movement of particels (fluids) along surfaces
y
y
Pconv
Pconv = −α ⋅ A ⋅ (Tamb − Ts )
= kα ⋅ A ⋅ (Ts − Tamb )
Tamb
T(y)
with:
α: heat transfer coefficient, W/(m²*K)
Ts
A: heat exchanging surface area, m²
T(y)
Tamb: ambient temperature (not influenced by
the surface), K
Ts: surface temperature, K
and
kα: heat loss coeff. due to convection, W/(m²K)
Lecture VIII, 22. 05. 2007
35
Abteilung Jülich
Thermal losses due to convection
The heat transfer coefficient, α, depends on the flow (material properties of
the fluid as well as flow characteristics). Models using dimensionless numbers
exist to describe the heat transfer by convection.
Nusselt number:
Nu =
Reynold‘s number,
Prandtl‘s number:
Re =
Grasshoff‘s number:
Lecture VIII, 22. 05. 2007
L ⋅α
λ
= f (Re,Gr , Pr )
ρ ⋅ d ⋅u
ν
, Pr =
η
a
Gr =
g ⋅ β ⋅ ∆T ⋅ L3
ratio of conduction /
convection resistances
forced convection
free convection
ν
36
Abteilung Jülich
Thermal losses due to convection
The meaning of the symbols are:
α
= heat transfer coefficient, W/(m2*K),
k
= heat loss coefficient, W/(m2*K),
L
= characteristic length, e.g. space between the plates, m,
λ
= thermal conductivity, W/(m*K),
g
= acceleration due to gravity, 9.81 m/s2,
β
= Volumetric coefficient of thermal expansion for air (1/T), 1/K,
∆T
= temperature difference of both plates, K,
ν
= kinematic viscosity, m2/s
ν
= η/ρ,
η
= dynamic viscosity, kg/(m*s),
ρ
= density, kg/m3.
All physical characteristics are determined for a mean temperature.
Lecture VIII, 22. 05. 2007
37
Abteilung Jülich
Thermal losses due to convection
Calculation of heat transfer coefficient resp. Nusselt number due to
free convection between two parallel plates (e.g. two covers of a flat
plate collector):
Nu = 0.152 ⋅ Gr 0.281
for horizontal plates
Nu = 0.093 ⋅ Gr 0.31
for inclined plates (45°)
(Correction factor for other inclinations follows!)
Lecture VIII, 22. 05. 2007
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Abteilung Jülich
Thermal losses due to free convection
The heat loss coefficient, k, can be calculated for a temperature of 10 °C
(mean air temperature) applying the following formulae :
Horizontal plates:
T 0.281
k10 = 1.613 ⋅ 0.157
L
45° inclined plates:
0.31
T
k10 = 1.14 ⋅ 0.07
L
with:
T = 10 (°C),
L = distance between
plates in cm,
k10:
in W/(m²K)
For the translation from 10 °C to other temperatures the following
relationship can be used:
kT = k10 ⋅ [1 − 0.0018 ⋅ (Tm − 10)]
Lecture VIII, 22. 05. 2007
39
Abteilung Jülich
Thermal losses due to convection
Calculation of heat transfer coefficient, depending on wind velocity
(e.g. first cover of a flat-plate collector):
α W = 5. 7 + 3. 8 ⋅ v w
Lecture VIII, 22. 05. 2007
vw: wind speed, m/s
40
Abteilung Jülich
Thermal losses due to radiation
Every body having a temperature T > 0 K emits thermal
radiation. The spectral intensity is:
I λ = ε (λ ) ⋅
2πhc 2
⎡
⎤
⎛ h⋅c ⎞
λ ⋅ ⎢ exp ⎜
⎟ − 1⎥
⎝ λ ⋅ k ⋅T ⎠
⎢⎣
⎥⎦
5
with:
Planck‘s constant:
h=6.62·10-34 Js
Speed of light:
c=2.99 ·108 m/s
Boltzmann’s constant: k=1.38 · 10-23 J/K
λ, m
Wavelength:
Temperature:
T, K
and the spectral emissivity coefficient ε(λ).
Lecture VIII, 22. 05. 2007
41
Abteilung Jülich
Thermal losses due to radiation
The spectral emissivity coefficient ε(λ) of a body is defined as the ratio
of its real emission at wavelength λ and temperature T, E(T,λ), to the
emission of a black body, Ebb(T), having the same temperature T:
ε (λ ) =
Lecture VIII, 22. 05. 2007
E (T , λ )
E bb (T )
42
Abteilung Jülich
Thermal losses due to radiation
Integration over the wavelength, assuming ε(λ) = ε = const.,
yields:
∞
I (T ) = ∫ ε ⋅ I λ (λ , T ) ⋅ d λ = ε ⋅σ ⋅ T 4
λ =0
With
(StefanBoltzmann)
σ : 5.67 * 10-8 W/(m²*K4)
and the following special cases:
ε = 1:
0 < ε < 1:
ε = 0:
Lecture VIII, 22. 05. 2007
“black body”
“grey body”, ε ≠ f(λ)
“ideal mirror”.
In the case of ε = ε(λ) (“coloured
body”), the integration is only
possible in sections where ε can
be assumed to be constant.
43
Abteilung Jülich
Wien‘s Law
•
The wavelength, at which the
spectral intensity reaches a
maximum shifts to shorter
wavelength with increasing
temperature
•
This wavelength can be
calculated by setting zero the
derivative of Planck‘s
formula:
∂I λ (T , λ )
=0
∂λ
λMAX ⋅ T =2897.8 [µmK ]
Lecture VIII, 22. 05. 2007
44
Abteilung Jülich
Thermal losses due to radiation
The ability of a body to emit thermal radiation, expressed by
the emissivity coefficient ε(λ), is equal to the ability to absorb
radiation, expressed by the absorptivity coefficient α(λ), at the
same wavelength:
ε (λ ) = α (λ )
Lecture VIII, 22. 05. 2007
(Kirchhoff‘s law)
45
Abteilung Jülich
Thermal losses due to radiation
Solar collectors have to absorb radiation emitted by the sun as
„good“ as possible, i.e. the absorption coefficient, α(λ), should
be close to one. On the other hand the warmed receiver shall
emit radiation as „bad“ as possible, i.e. its emissivity
coefficient, ε(λ), should be close to zero. This is possible – on
the first view in contradiction to Kirchhoff‘s law - because
absorption and emission take place at different wavelengths!
Coatings with these properties are called „(optical) selective
coatings“.
Lecture VIII, 22. 05. 2007
46
Abteilung Jülich
Thermal losses due to radiation
Selective coatings
2,0
1,5
1,5
1,5
Ideal selective coating
bodyK)
(673 K)
BlackBlack
body (673
AM
11
AM
1,0
1,0
1,0
Real selective coating
(TiNOx, Munic)
0,5
0,5
0,5
0,0
0,0
0,0
0,0
Remember:
ρ+ τ + α = 1, τ = 0
AM 1
Reflectivity
Reflectivity [-]
Spectral
irradiance/emittance
[W/m²nm]
irradiance/emittance
[W/m²nm]
Spectral
irradiance/emittance
[W/m²nm]
Spectral
irradiance [W/m²nm]
2,0
2,0
ρ=1−α
Ideal selective coating:
λ < λgr: ρ(λ)=0 (α(λ) = 1)
λ > λgr: ρ(λ)=1 (α(λ) = 0)
λgr=f(Tabsorber, c*AM)
1000
1000
1000
1000
10000
10000
10000
wavelength
[nm]
wavelength[nm]
[nm]
10000
wavelength [nm]
Lecture VIII, 22. 05. 2007
47
Abteilung Jülich
Heat exchange due to radiation
•
depends on T1, T2, ε1, ε2 and
the arrangement of the two
bodies
ϕ
ε2, T2
I (ϕ ) = Iϕ =0 ⋅ cos(ϕ )
ε1, T1
Lecture VIII, 22. 05. 2007
48
Abteilung Jülich
Heat exchange due to radiation
Special case (1): two „grey“ parallel
plates
H1
H1: all radiation emitted by surface 1
E1: directly emitted by surface 1
E1
H2: all radiation emitted by surface 2
ε1, T1
ε2, T2
E2
E2: directly emitted by surface 2
H1 = E1 + H 2 ⋅ (1 − ε 1 )
H 2 = E2 + H1 ⋅ (1 − ε 2 )
ε 2 ⋅ E1 − ε 1 ⋅ E2
H1 − H 2 =
ε1 + ε 2 − ε1 ⋅ ε 2
Lecture VIII, 22. 05. 2007
49
Abteilung Jülich
Heat exchange due to radiation
Special case (1): two parallel plates
q12, pp =
σ
1
ε1
ε1, T1
+
1
ε2
−1
(
⋅ T14 − T24
(
= ε 12, pp ⋅ σ ⋅ T14 − T24
ε2, T2
with:
ε 12, pp =
1
ε1
+
1
1
ε2
)
)
−1
„radiation exchange factor for two
parallel plates“
Lecture VIII, 22. 05. 2007
50
Abteilung Jülich
Heat exchange due to radiation
Special case (2): exchange with sky
1) All radiation is emitted to the sky.
Tsky
ε1, T1
2) As the absorber is very small in
relation to the sky, apparently no
portion of the radiation leaving the
absorber is reflected to the absorber.
Thus the sky can be regarded as a
black body (αsky = 1) with celestial
temperature Tsky.
(
4
Prad = ε 1 ⋅ σ ⋅ T14 − Tsky
Lecture VIII, 22. 05. 2007
)
51
Abteilung Jülich
Heat exchange due to radiation
For Tsky two relationships are proposed (for clear sky only):
Tsky = Tair − 6 K
or:
1.5
Tsky = 0.0552 ⋅ Tair
Lecture VIII, 22. 05. 2007
52
Abteilung Jülich
Collector overall heat losses
For the simpler calculation of the thermal losses of the collector it is
desirable to determine a heat loss coefficient, kC, which comprises all
thermal loss mechanisms of the total collector.
By that the total thermal power losses, PThL, could then simply be
calculated from the temperature difference between the absorber and
the ambient:
PThL = kC · A · (TAb - Tamb)
with:
kC:
collector overall heat loss coefficient, W/(m²K)
A:
absorber surface, m²
Tab:
temperature of absorber surface, K resp. °C
Tamb:
temperature of ambient, K resp. °C.
Lecture VIII, 22. 05. 2007
53
Abteilung Jülich
Collector overall heat losses
Note: The heat loss coefficient due to radiation depends on the
temperatures of the two heat exchanging surfaces, thus:
k rad , pp =
σ ⋅ (T1 + T2 ) ⋅ (T12 + T22 )
1
ε1
+
1
ε2
−1
≈
σ ⋅ 4T 3
1
ε1
+
1
ε2
−1
with the average temperature
1
T = (T1 + T2 )
2
Lecture VIII, 22. 05. 2007
54
Abteilung Jülich
Thermal network for a collector
•
In analogy to an electrical network, the thermal heat flows can be
treated as thermal network. The heat flow corresponds to the electrical
current, a temperature difference to the voltage and the thermal
resistance corresponds to the shunt resistance.
•
The thermal resistances are the reciprocal values of the different heat
loss coefficients:
Rcond =
Rconv =
Rrad =
Lecture VIII, 22. 05. 2007
1
kcond
1
kconv
1
k rad
=
=
L
λ
1
α
,
,
=K
55
Abteilung Jülich
Thermal network for a collector
•
Thus, calculation of a complex collector system with different heat
transfer mechanisms, appearing in parallel and/or in a row, can be
simplified.
•
Remember:
The total resistance for the sum of n resistances connected in a row is:
Rtot ,row = R1 + R2 + K + Rn
The total resistance for the sum of n resistances connected in parallel is:
1
Rtot , p
Lecture VIII, 22. 05. 2007
=
1
1
1
+
+K+
R1 R2
Rn
56
Abteilung Jülich
Thermal network for a collector
Example: Heat exchange due to radiation and free convection between
two parallel plates:
P = A ⋅ kc ⋅ (T1 − T2 )
ε2, T2
Rrad
= A⋅
Rconv
ε1, T1
Lecture VIII, 22. 05. 2007
1
⋅ (T1 − T2 )
Rtot
with:
Rtot =
1
1
1
+
Rrad Rconv
=
1
k rad + k conv
57
Abteilung Jülich
Thermal network for a collector
Example (2): Heat flow through a wall with different layers and
convective heat transfer at the surfaces:
Tin
α1
P = A ⋅ kc ⋅ (Tin − Tamb )
T0
= A⋅
T1
T2
λ1, d1 λ2, d2
α2
with:
Tamb
Rtot = Rconv ,1 + Rcond ,1 + Rcond , 2 + Rconv , 2
=
=
Lecture VIII, 22. 05. 2007
1
⋅ (Tin − Tamb )
Rtot
1
kconv ,1
1
α1
+
+
d1
λ1
1
kcond ,1
+
d2
λ2
+
+
1
kcond ,1
+
1
kconv , 2
1
α2
58
Abteilung Jülich
Thermal network for a collector
Example (3): Thermal network for a flat plate collector with three
covers:
PG
T
Losses
by
reflection
a
R6
Tp3
R2: …between backward insulation and ground,
R3: … between absorber and third cover
R5
Tp2
R4
Tp1
R3
TAb
PU
R1: Thermal resistance of backward insulation,
R1
Tb
R2
Ta
Lecture VIII, 22. 05. 2007
R4: … between third and second cover
R5: … between second and first cover
R6: … between first cover and ambient
Note: the heat flow from the absorber to the
first cover and from there to the next one
as well as from the last cover to the
ambient is constant (assuming thermal
equilibrium)!
59
Abteilung Jülich
Thermal network for a collector
The network can be simplified by summarizing the heat losses at the front side
by a front loss coefficent kf and a back loss coefficient kb:
PG
T
Losses
by
reflection
Ta
a
R6
PAb
Tp3
R5
PU
1 / kC
TAb
Tp2
kb =
R4
Tp1
R3
TAb
PU
1
kf =
R3 + R4 + R5 + R6
Ta
PAb
PU
1 / kf
Tenv
1
R1 + R2
The overall heat loss
coefficent kc yields:
R1
Tb
1 / kb
Ta
R2
Ta
Lecture VIII, 22. 05. 2007
1
kc =
R1 + R2 + R3 + R4 + R5 + R6
60
Abteilung Jülich
Empirical computation of the front loss coefficient
In the temperature range between 40 °C – 130 °C and for an inclination of 45 °C, an
empirical function was developed by Klein, 1973, with deviations for kf < 0.2 W/(m²K).
−1
⎡
⎤
⎢
⎥
2
σ ⋅ (TAb + Tamb ) ⋅ TAb2 + Tamb
N
1 ⎥
⎢
+
k f , 45° = ⎢
+
0.31
⎥
1
2 ⋅ N + f −1
αW
⎛
⎞
⎞
⎛
344
T
T
−
+
−N
Ab
amb
⎢⎜
⎥
⎟
⎟
⎜
⋅
(
)
0
.
0425
1
ε
ε
ε
+
⋅
N
⋅
−
⎟
⎜
Ab
Ab
G
⎢⎣ ⎜⎝ TAb ⎟⎠ ⎝ N + f ⎠
⎥⎦
(
N = number of glass covers,
f = (1–0.04 ⋅αw+ 5⋅10-4 ⋅αw2)⋅(d+0.058⋅N),
εG = emissivity of cover (0.88 for glass),
εAb= emissivity of the absorber,
Lecture VIII, 22. 05. 2007
)
TAb = absorber temperature, K,
Tamb = ambient temperature, K,
αW = heat transfer coefficient for
wind, W/m2K,
d = distance between covers, cm.
61
Abteilung Jülich
Empirical computation of the front loss coefficient
For other inclinations than 45°:
1.10
kf(s) / kf(45°)
1.05
kf
k f , 45°
1.00
εAb =
εAb
0.95
=
= 1 − (β − 45°) ⋅ (0.00259 − 0.00144 ⋅ ε Ab )
0 .9 5
0.
1
0.90
0.85
0
10
20
30
40
50
60
70
80
inclination angle [degree]
Lecture VIII, 22. 05. 2007
90
62