Fresnel reflection loss

[nick pine]

In 984 AD, Ibn Sahl http://en.wikipedia.org/wiki/Ibn_Sahl discovered
Snell's Law (before Dutch astronomer Willebrord Snellius 1580–
1626 :-),
which says n1sin(a1) = n2sin(a2), where n1 and n2 are incident and
refracted angles media with refractive indices n1 and n2.

In 1819, Augustin Fresnel http://en.wikipedia.org/wiki/Augustin-Jean_Fresnel
wrote a book describing the reflection of light as it passes from
one medium to another.

With indices n1 and n2 and the light is normal to the interface
(with a 0 degree incidence angle) the reflection coefficient
r(0) = ((n1-n2)/(n1+n2))^2. If one medium is air and the other
has index n, r(0) = ((n-1)/(n+1))^2.

In the solar spectrum, glass has n = 1.526, so r(0) = (0.526/1.526)^2
= 0.0434, ie a single glass-air interface reflects 4.34% of the light
that passes through it. Two surfaces reflect 8.68%, even if the glass
is perfectly clean and clear and it absorbs no energy at all. So pure
glass cannot transmit more than 100-8.68 = 91.3% of the solar
spectrum.

Example 5.1.1 on page 205 of Duffie and Beckman's Solar Engineering
of Thermal Processes book shows how the reflected energy increases
with a 60 vs 0 degree incidence angle. The calculation is more
complex,
with reflection coefficients rp = sin^2(a2-a1)/sin^2(a2+a1) and
rl = tan^2(a2-a1)/tan^2(a2+a1) for the perpendicular and parallel
components of the light and an overall r = (rp+rl)/2 reflection.

Sahl's law gives a2 = sin^-1(sin(60)/1.526) = 34.58 degrees, so
r = (sin^2(-25.42)/sin^2(94.58)+tan^2(-25.42)/tan^2(94.58))/2
=(0.185+0.001)/2 = 0.093, ie a single glass surface reflects
9.3% of light with a 60 degree incidence angle.

This is more complicated for 2 surfaces... (1-rp) of the perpendicular
component of the incident light reaches the 2nd surface... rp(1-rp)
of
that is reflected back to the first, and so on. The total
transmittance
T = ((1-rp)/(1+rp)+(1-rl)/(1+rl))/2 = ((1-0.185)/(1+0.185)+(1-0.001)/
(1+0.001))/2
= 0.83 at 60 degrees.

This is even more complicated for 2 layers of non-absorbing glass,
with T = ((1-rp)/(1+3rp)+(1-rl)/(1+3rl))/2 = 0.76 at 60 degrees.

Here endeth the lesson (unless anyone would like to ARGUE with
this thousand-year-old science :-)

Nick

[Don Hull]

Any thoughts on putting attic radiant film on a roller to:

      Summer .. roll out in the day and up at night

      Winter...roll up in the day and out at night.

 

Don Hull 

--
Donald Hull
Geothermal Specialist
Ground Source HVAC
610-306-6245
groundso...@gmail.com
www.groundsourcehvac.com
Call When You're Serious About Saving Energy

[Nathan Hurst]

On Thu, May 12, 2011 at 12:50:58PM -0700, nick pine wrote:

> In the solar spectrum, glass has n = 1.526, so r(0) = (0.526/1.526)^2

(0.526/2.526)^2

> = 0.0434, ie a single glass-air interface reflects 4.34% of the light
> that passes through it. Two surfaces reflect 8.68%, even if the glass
> is perfectly clean and clear and it absorbs no energy at all. So pure
> glass cannot transmit more than 100-8.68 = 91.3% of the solar
> spectrum.

Certainly, but by the simple addition of a half wavelength layer with
the geometric mean of refractive index between air and glass you can
reduce this loss by a factor of roughly 2. More layers results in a
closer impedance match etc. I expect such coatings would be fairly
common these days?

njh

[John Canivan]

Sounds like you've done some research Nathan.
Another thing to consider is the angle of incodence of the light. A tracking
device would be needed. Without a tracking device to keep the light
perpendicular to the sun much more of the light would be reflected off the
glazing surface.
John

[Nathan Hurst]

On Sun, May 15, 2011 at 08:45:57PM -0400, John Canivan wrote:
> Sounds like you've done some research Nathan.

We did it in physics at school actually :)

> Another thing to consider is the angle of incodence of the light. A
> tracking device would be needed. Without a tracking device to keep
> the light perpendicular to the sun much more of the light would be
> reflected off the glazing surface.

For comparison, what are the theoretical limits for reflection?
we have this:
http://en.wikipedia.org/wiki/File:Image-Metal-reflectance.png

Perhaps there are better results for dielectric mirrors, or
superconductors or something. Realistically, for sunspaces 91% is
pretty good, and we're going to struggle to capture 90% of that in our
storage in any case.

If reflectors are better, it suggests that heliostats are a better
investment than glaxed designs, perhaps with a vertical collection
port that avoids convective losses. I recall Nick talking about such
a design a while back, with a mirror reflecting light upwards to a
collector in an insulated tube.

njh

[nick pine]

Nathan Hurst <n...@njhurst.com> wrote:

> ... by the simple addition of a half wavelength layer with the geometric mean of refractive index between air and glass you can reduce this loss by a factor of roughly 2.

Duffie and Beckman go on: "Surface treatment, by dipping glass in a
[hot and nasty] saturated fluosilic acid solution, can reduce the
reflection losses [from 8%] to 2% [Thomsen, 1951], and a double layer
coating [Mar, 1975] can reduce reflection losses to less than 1%."

Most of the air mass 1 solar spectrum lies between 0.2 and 2 um, with
a peak around 0.6. Mar's reflectance chart shows a quarter-wave
matching notch at 0.6, with 8% at 0.3 um and 6% at 2 um and better off-
normal incidence transmittance than unetched glass.

>I expect such coatings would be fairly common these days?

I think they are still very rare in solar collectors, vs binoculars
and camera lenses.

Nick

[nick pine]

Nathan Hurst <n...@njhurst.com> wrote:

> If reflectors are better, it suggests that heliostats are a better investment than glaxed designs, perhaps with a vertical collection port that avoids convective losses.

Like this?

The most serious mistake was making the outer container of
the receiver of plywood. We thought that the plywood would be
sufficiently insulated from the copper panel which was the receiver
proper, that it would not get too hot. The copper panel was separated
from the plywood by 4" of fiberglass insulation. Nevertheless,
the plywood caught fire and the unit was completely destroyed.
We suppose this is a success, of sorts...

from "A solar collector with no convection losses,"
(a downward-facing receiver over a 4:1 concentrating
parabolic mirror) written by H. Hinterberger and
J. O'Meara of Fermilab, ca 1976

Nick

[John Canivan]

Very interesting.

John
----- Original Message -----
From: "nick pine" <ni...@early.com>
To: "sunspace" <suns...@googlegroups.com>
Sent: Monday, May 16, 2011 6:23 AM
Subject: Re: Fresnel reflection loss

[nick pine]

The most serious mistake was making the outer container of
the receiver of plywood. We thought that the plywood would be
sufficiently insulated from the copper panel which was the receiver
proper, that it would not get too hot. The copper panel was separated
from the plywood by 4" of fiberglass insulation. Nevertheless,
the plywood caught fire and the unit was completely destroyed.
We suppose this is a success, of sorts...

from "A solar collector with no convection losses,"
(a downward-facing receiver over a 4:1 concentrating
parabolic mirror) written by H. Hinterberger and
J. O'Meara of Fermilab, ca 1976

An 8' tall reflective parabolic trough below a 4' wide east-west
shallow water tray under a sunspace roof that reflects sun from
the horizon vertically up to the edge of the tray near the glazing
would have a 45 degree slope at the bottom.

Where I live at 40 N lat, noon sun on 12/21 with a 26.5 degree
elevation would miss the south edge of the tray and get reflected
back out of the glazing.

If we make the slope at the bottom (90-26.5)/2 = 31.75 degrees,
noon sun will hit the south edge of the tray. If tan(31.75)
= dy/dx = sqrt(f)/2, the parabolic focal length f changes from 4'
to 1.53', and the bottom of the parabola is 4.95' below the tray.

I wonder how they got 4:1 concentration...

Nick