How do you measure insolation?

[mike]

I'd like to do some experiments with solar collectors, both PV and non-PV.
So how do I measure the energy coming out of the sky to calibrate
the measurements?
The obvious solution is to point a photographic light meter at the sky.
Where do I point it? Does pointing directly at the sun overstate
the average energy? And what about frequency response? Does the
measurement intended for photography correlate with the energy spectrum
that concerns absorbing light/heat? Don't clouds distort the energy
spectrum? In the northwest, overcast is the predominant
weather pattern.

The objective is to make measurements under different conditions and
correlate the results to assess relative performance of collection
systems.

 

[Al Forster]

I'd like to do some experiments with solar collectors, both PV and
non-PV.
So how do I measure the energy coming out of the sky to calibrate
the measurements?
The obvious solution is to point a photographic light meter at the
sky.
Where do I point it? Does pointing directly at the sun overstate
the average energy? And what about frequency response? Does the
measurement intended for photography correlate with the energy
spectrum
that concerns absorbing light/heat? Don't clouds distort the energy
spectrum? In the northwest, overcast is the predominant
weather pattern.

The objective is to make measurements under different conditions and
correlate the results to assess relative performance of collection
systems.
--

Usually with a pyranometer (http://en.wikipedia.org/wiki/Pyranometer)
or a pyrheliometer (http://en.wikipedia.org/wiki/Actinometer)

For frequency distributions, you could measure with a referenced solar
cell and a selection of appropriate colour filters and normalize to
your pyranometer readings.

Al

 

[Bill Kaszeta / Photovoltaic Resources]

http://www.licor.com/env/Products/Sensors/200/li200_description.jsp
about $195

I'd like to do some experiments with solar collectors, both PV and non-PV.
So how do I measure the energy coming out of the sky to calibrate
the measurements?
The obvious solution is to point a photographic light meter at the sky.
Where do I point it? Does pointing directly at the sun overstate
the average energy? And what about frequency response? Does the
measurement intended for photography correlate with the energy spectrum
that concerns absorbing light/heat? Don't clouds distort the energy
spectrum? In the northwest, overcast is the predominant
weather pattern.

The objective is to make measurements under different conditions and
correlate the results to assess relative performance of collection
systems.

Bill Kaszeta
Photovoltaic Resources Int'l
Tempe Arizona USA
[email protected]

 

[mike]

http://www.licor.com/env/Products/Sensors/200/li200_description.jsp
about $195




Bill Kaszeta
Photovoltaic Resources Int'l
Tempe Arizona USA
[email protected]

Quoting the site

The spectral response of the LI-200 does not include the entire solar
spectrum, so it must be used in the same lighting conditions as those
under which it was calibrated. Therefore, the LI-200 should only be used
to measure unobstructed daylight. It should not be used under
vegetation, artificial lights, in a greenhouse, or for reflected solar
radiation.

end quote

This is the whole reason I asked the question. I'd like to make
comparative measurements of different configurations in different
lighting situations and get meaningful comparisons of which is better.
The first experiments are scheduled to be hot-box heat collectors.
Visible light calibration may or may not bear any relationship to
wavelengths used for heating air/liquid.

And $195 is out of the question. Cheap/free/homebrew is what I desire.

Painting a thermocouple black is the current front runner.
Thanks, mike

 

[[email protected]]

... $195 is out of the question.

Omigod, serious money

Cheap/free/homebrew is what I desire.

Steve Baer paints an aluminum plate black and measures the temperature rise
over time with an IR thermometer. In 250 Btu/h-ft^2 full sun, 1 ft^2 would
absorb 250x0.5/60 = 2.08 Btu in 30 seconds. A 1/8" 1.76 pound plate with
1.76x0.214 = 0.377 Btu/F would warm 2.08/0.377 = 5.5 F without much heat
loss to the surroundings, if it starts at the same temperature.

Nick

 

[mike]

Omigod, serious money


Steve Baer paints an aluminum plate black and measures the temperature rise
over time with an IR thermometer. In 250 Btu/h-ft^2 full sun, 1 ft^2 would
absorb 250x0.5/60 = 2.08 Btu in 30 seconds. A 1/8" 1.76 pound plate with
1.76x0.214 = 0.377 Btu/F would warm 2.08/0.377 = 5.5 F without much heat
loss to the surroundings, if it starts at the same temperature.

Nick

NOw, that's what I'm talkin' about. cheap, simple, results supported by
math. I like it.
Thanks, mike