Radiation characteristics of opaque
materials
•  Selective surfaces
•  98% of the solar radiation energy is from 0-3 µm
•  99,9% of the energy emitted by a black or a grey body
at 100ºC is from 3 µm - ∞
Radiation characteristics of opaque
materials
•  Selective surfaces
•  Coatings transparent for long wavelengths and with high
absorptance for short wavelengths;
•  On substrates of low emittance for long wavelengths
•  (metal oxides on metal)
Radiation characteristics of opaque
materials
•  Absorption enhancement
•  Antireflective coatings (like in camera lenses);
•  Cavities to induce multiple reflections
Radiation characteristics of opaque
materials
•  Selective surfaces
•  Usually there is a interdependence of α and ε. The
performance of the solar collector is more sensitive to α.
Radiation characteristics of opaque
materials
Problem 3.2: Calculate the absorptance for the terrestrial
solar spectrum and the emittance at 100ºC of a dull nickel
coated with black chrome surface (after some humidity
exposure).
R: ε=0.09 ; α=0.95
Radiation characteristics of opaque
materials
Problem 3.2: Calculate the absorptance for the terrestrial
solar spectrum and the emittance at 100ºC of a dull nickel
coated with black chrome surface (after some humidity
exposure).
R: ε=0.09 ; α=0.95
Problem 3.2: Calculate the absorptance for the terrestrial
solar spectrum and the emittance at 100ºC of a dull nickel
coated with black chrome surface (after some humidity
exposure).
R: ε=0.09 ; α=0.95
Radiation characteristics of opaque
materials
•  Angular dependence of solar absorptance
•  Data is scarce. The available data suggests:
Radiation characteristics of opaque
materials
•  Specularly reflecting surfaces
Typical values 0.85-0.94
ρλ, s =
∫
ρ=
∫
I λ, s
I λ, i
∞
0
∞
0
I λ, s d λ
I λ, i d λ
0.85 ! back-aluminized acrylic
0.94 ! back-silvered low-reflectance glass
4. RADIATION TRANSMISSION THROUGH
GLAZING
Radiation transmission through glazing
•  Reflection
•  Snell law
n1 sin θ1 = n2 sin θ 2
Radiation transmission through glazing
•  Reflection
•  Fresnel reflection coefficients
tan 2 (θ 2 − θ1 )
r|| =
=
I i|| tan 2 (θ 2 + θ1 )
I r||
r⊥ =
Ir ⊥
Ii ⊥
=
sin 2 (θ 2 − θ1 )
sin 2 (θ 2 + θ1 )
Radiation transmission through glazing
•  Reflection
•  For unpolarized radiation, the reflectance is
I r r⊥ + r||
r= =
Ii
2
Radiation transmission through glazing
•  Reflection
•  For normal incidence
" n2 − n1 %
rn = $
'
# n2 + n1 &
2
•  And if n1=1 (air) and n2=n (the glass)
" n −1 %
rn = $
'
# n +1 &
2
Radiation transmission through glazing
Problem 4.1: Solar radiation has an incident angle of 45º
on a glass that has a mean index of refraction of 1.5. Find
the reflectance of the glass and compare it with the
reflectance for normal incidence.
R: r45º=0.051 ; rn=0.040 ; rn/r45º=0.79
Radiation transmission through glazing
•  Transmission through a slab without absorption
•  For each component of the incident radiation (if the
incidence is normal there is only one)
1− r
τr =
1+ r
1− r
τr =
1+ r
Radiation transmission through glazing
•  Transmission through a slab without absorption
•  If the radiation is unpolarized
1 # 1− r|| 1− r⊥ &
τr = %
+
(
2 $ 1+ r|| 1+ r⊥ '
•  For normal incidence
1− r
τ r, n =
1+ r
Radiation transmission through glazing
Problem 4.2: Solar radiation has an incident angle of 45º
on a glass that has a mean index of refraction of 1.5. Find
the transmittance of the glass and compare it with the
transmittance for normal incidence.
R: τr,45º=0.906 ; τr,n=0.923 ; τr,45º/τr,n=0.98