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Rod Speed wrote:

Problem is that the floor isnt a very good source of
lower grade heat...


Wrong again Rod. A hydronic floor with lots of surface and a low water-air
thermal resistance in an airtight house with lots of insulation works fine.
Ignoring the R1 radiation conductance, the min floor temp required to keep
a house with a 200 Btu/h-F conductance and a 2400 ft^2 floor 70 F on a 30 F
day with a U1.5 slow-moving airfilm conductance is 70+I/(2400x1.5) = 72.2 F.

Nick

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m Ransley
 
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Ken how do you insulate under the floor piping-Radiant floor heating,
here we use foamboard.
What R value are your 4 pane windows, are they sold in the US
What R value do you use in walls and ceilings

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News
 
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"Rod Speed" wrote in message
...

Some gutless ****wit troll desperately cowering behind
Gunnar wrote in message
news:ykhKc.42039$iw3.23471@clgrps13...
just what you'd expect from a desperately cowering gutless ****wit troll.


Best ignore him.



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News
 
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"Rod Speed" wrote in message
...

Some gutless ****wit troll desperately cowering behind
Gunnar wrote in message
news:ykhKc.42039$iw3.23471@clgrps13...
just what you'd expect from a desperately cowering gutless ****wit troll.


Best ignore him.




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News
 
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"Rod Speed" wrote in message
...

Some gutless ****wit troll desperately cowering behind
Gunnar wrote in message
news:ykhKc.42039$iw3.23471@clgrps13...
just what you'd expect from a desperately cowering gutless ****wit troll.


Best ignore him.






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News
 
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"Rod Speed" wrote in message
...

Some gutless ****wit troll desperately cowering behind
Gunnar wrote in message
news:ykhKc.42039$iw3.23471@clgrps13...
just what you'd expect from a desperately cowering gutless ****wit troll.


Best ignore him.




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News
 
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"Rod Speed" wrote in message
...

Some gutless ****wit troll desperately cowering behind
Gunnar wrote in message
news:ykhKc.42039$iw3.23471@clgrps13...
just what you'd expect from a desperately cowering gutless ****wit troll.


Best ignore him.




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News
 
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"Rod Speed" wrote in message
...

Your childish waffling fools absolutely no one at all cept fools
like that stupid pom desperately cowering behind 'news'


Mr Troll, you will find we are not all disciples of Nick. Do a Google and
see the conflicts. ..err is it worth saying that to him? Probably not.



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Rod Speed
 
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News wrote in message
...
Rod Speed wrote
Some gutless ****wit troll desperately cowering behind
Gunnar wrote in message
news:ykhKc.42039$iw3.23471@clgrps13...
just what you'd expect from a desperately cowering gutless ****wit troll.


Best ignore him.


I'll take a dump on him instead, thanks.


  #10   Report Post  
Rod Speed
 
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Some gutless ****wit pom desperately cowering behind
News wrote in message
...
Rod Speed wrote


Your childish waffling fools absolutely no one at all cept fools
like that stupid pom desperately cowering behind 'news'


Mr Troll, you will find we are not all disciples of Nick.


Master gutless ****wit pom, I never said you were.

Do a Google and see the conflicts.


Not interested in gutless ****wit pom **** thanks.




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News
 
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wrote in message
...
Rod Speed wrote:

...There are real downsides with superinsulation, as I said.


False.

...And there are real downsides with superinsulation too.


False.

...And there are real downsides with superinsulation, as I said.


False.

Just back from the Portland ASES workshop, where we had fun.

One workshopper had a bunch of small buildings in northern Greenland,

built
on 11,000 feet of ice above the soil. The ice increases by about 3' per

year,
so you have to raise your building or get buried. He'd like to replace his
buildings with one large one (2 stories with a 1000 ft^2 footprint) on

pylons
and add 3' to the pylons each year. He wants to solar heat it in

summertime
to reduce their $400,000 per year fuel bill (in winter it's dark, and -107

F,
with 75 knot winds for 2 weeks at a time. Drew doesn't like headwinds.)

I still like the idea of collecting heat and electricity from standard PV
panels. When I layed a 4' flat 4-year greenhouse polyethylene film duct

over
a horizontal PV panel and filled it with an inch of water, the electrical
output (Isc) only decreased by 6%. We might increase the electrical output
more than that by placing the panels next to a reflective north wall and
make hot water for showers all year, making a drain-down system in

wintertime
and keeping water between the poly films all summer to increase the PV

output
by keeping the panels cooler.


I know of a few installations of PVs where the underside of the PVs is
ducting with air florced through. A heat pump collects the heat from the
panels and house exhaust air. It wasn't that brilliant. I think it could
have been better thought out.

How much heat do PV cells emit on the back and front of the cells?



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Anthony Matonak
 
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News wrote:
....
I know of a few installations of PVs where the underside of the PVs is
ducting with air florced through. A heat pump collects the heat from the
panels and house exhaust air. It wasn't that brilliant. I think it could
have been better thought out.

How much heat do PV cells emit on the back and front of the cells?


You could think of it this way. Typical PV panels are around 12%
efficient at converting sunlight into electricity. This means
that the other 88% is converted into heat. Clearly they are much
better at making heat than electricity.

Anthony

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Fred B. McGalliard
 
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"Anthony Matonak" wrote in message
...
....
You could think of it this way. Typical PV panels are around 12%
efficient at converting sunlight into electricity. This means
that the other 88% is converted into heat. Clearly they are much
better at making heat than electricity.


I thought around 30% was reflected? This would leave about 50% as heat.


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Rod Speed
 
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Anthony Matonak wrote in
message ...
News wrote:


I know of a few installations of PVs where the underside of the
PVs is ducting with air florced through. A heat pump collects the
heat from the panels and house exhaust air. It wasn't that brilliant.
I think it could have been better thought out.


How much heat do PV cells emit on the back and front of the cells?


You could think of it this way. Typical PV panels are
around 12% efficient at converting sunlight into electricity.
This means that the other 88% is converted into heat.


Nope, that would only be true if they were a perfect
black body absorber of all energy that falls on them.

In practice quite a bit is just reflected off them.

Clearly they are much better at making heat than electricity.


But then so is say black roofing material.


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Rod Speed wrote:

Anthony Matonak wrote in


News wrote:


How much heat do PV cells emit on the back and front of the cells?


You could think of it this way. Typical PV panels are
around 12% efficient at converting sunlight into electricity.
This means that the other 88% is converted into heat.


Well, maybe 16 and 84%, eg 0.84x800 = 672 W/m^2 in AM2 sun.

...that would only be true if they were a perfect
black body absorber of all energy that falls on them.


Buth they are better than that. Rich Komp says the silicon is almost
transparent to IR, and the aluminum contact beneath is shiny, so PVs
are a selective surface.

Nick



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Fred B. McGalliard
 
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wrote in message
...
....
Buth they are better than that. Rich Komp says the silicon is almost
transparent to IR, and the aluminum contact beneath is shiny, so PVs
are a selective surface.


I think that has a lot to do with whether there is an antireflective coating
on the surface or not. Silicon is very shiny in the optical, (and I think
the reflectivity is a function of doping), where most of the energy comes
in, so, especially if the sun is at a substantial angle to the array, a lot
of the energy is reflected from the top surface. What gets in and is not
absorbed or reflected from the top surface metalization (around 10-20% I
think for typical cells) passes through the active cell and what isn't
absorbed there then passes through the rest of the bulk silicon used as a
handle. The back side metalization would be under all this heavily doped
handle. Very thin cells would be very different from the currently more
common thick cells. I am not sure about the ones that are epi deposited on
material other than silicon. If they are thrown down on a metal contact they
are probably very reflective.


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Rod Speed
 
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wrote in
message ...
Rod Speed wrote
Anthony Matonak wrote
News wrote


How much heat do PV cells emit on the back and front of the cells?


You could think of it this way. Typical PV panels are
around 12% efficient at converting sunlight into electricity.
This means that the other 88% is converted into heat.


Well, maybe 16 and 84%, eg 0.84x800 = 672 W/m^2 in AM2 sun.


...that would only be true if they were a perfect
black body absorber of all energy that falls on them.


But they are better than that.


Nope. Not even possible.

Rich Komp says the silicon is almost transparent to IR, and the
aluminum contact beneath is shiny, so PVs are a selective surface.


Waffle, the fact remains that all of what doesnt get turned
into electricity does not end up as heat. Plenty gets reflected.


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Timm Simpkins
 
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"Rod Speed" wrote in message
...

wrote in
message ...
Rod Speed wrote
Anthony Matonak wrote
News wrote


How much heat do PV cells emit on the back and front of the cells?


You could think of it this way. Typical PV panels are
around 12% efficient at converting sunlight into electricity.
This means that the other 88% is converted into heat.


Well, maybe 16 and 84%, eg 0.84x800 = 672 W/m^2 in AM2 sun.


...that would only be true if they were a perfect
black body absorber of all energy that falls on them.


But they are better than that.


Nope. Not even possible.

Rich Komp says the silicon is almost transparent to IR, and the
aluminum contact beneath is shiny, so PVs are a selective surface.


Waffle, the fact remains that all of what doesnt get turned
into electricity does not end up as heat. Plenty gets reflected.


The maximum efficiency ratings they are getting now are approaching 30%. In
normal production they are around 23%. The average cell is about 12%
efficient. Also, no matter what misguided brain fart you're having now Rod,
the rest of the sun's energy that is absorbed by the panel IS converted to
heat. In fact, the efficiency of solar cells decreases as the heat
increases so there are many people experimenting with methods, both passive
and active, to remove that heat and somehow convert that heat to electricity
as well. The measure of efficiency is not measured by the amount of energy
available, it's measured by the amount of energy produced compared to the
amount of energy used.

Here's a link discussing the heat loss:

http://www.nasatech.com/Briefs/Apr99/NPO20284.html

Now Rod, don't complain to me about how wrong you are now. You can complain
to the guys at Nasa about it.

You can measure the efficiency by taking the wattage being wasted as heat
and compare that to the wattage being output by the solar cell. If the rays
are reflected and have no impact on the process, they cannot be measured, so
the measurement is totally what is absorbed by the panel.

I don't understand why you wouldn't know how efficiency rates are
calculated. It's been common science for quite a long time now. It would
not be a true calculation of efficiency if they were to calculate the
efficiency by the total energy available. That's like calculating the
efficiency of a motor by calculating how much energy is output compared to
the amount of energy available in the entire gas tank when you only used a
quarter of a tank.


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daestrom
 
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"Timm Simpkins" wrote in message
...

"Rod Speed" wrote in message
...

wrote in
message ...
Rod Speed wrote
Anthony Matonak wrote
News wrote


How much heat do PV cells emit on the back and front of the cells?


You could think of it this way. Typical PV panels are
around 12% efficient at converting sunlight into electricity.
This means that the other 88% is converted into heat.


Well, maybe 16 and 84%, eg 0.84x800 = 672 W/m^2 in AM2 sun.


...that would only be true if they were a perfect
black body absorber of all energy that falls on them.


But they are better than that.


Nope. Not even possible.

Rich Komp says the silicon is almost transparent to IR, and the
aluminum contact beneath is shiny, so PVs are a selective surface.


Waffle, the fact remains that all of what doesnt get turned
into electricity does not end up as heat. Plenty gets reflected.


The maximum efficiency ratings they are getting now are approaching 30%.

In
normal production they are around 23%. The average cell is about 12%
efficient. Also, no matter what misguided brain fart you're having now

Rod,
the rest of the sun's energy that is absorbed by the panel IS converted to
heat. In fact, the efficiency of solar cells decreases as the heat
increases so there are many people experimenting with methods, both

passive
and active, to remove that heat and somehow convert that heat to

electricity
as well. The measure of efficiency is not measured by the amount of

energy
available, it's measured by the amount of energy produced compared to the
amount of energy used.

Here's a link discussing the heat loss:

http://www.nasatech.com/Briefs/Apr99/NPO20284.html

Now Rod, don't complain to me about how wrong you are now. You can

complain
to the guys at Nasa about it.

You can measure the efficiency by taking the wattage being wasted as heat
and compare that to the wattage being output by the solar cell. If the

rays
are reflected and have no impact on the process, they cannot be measured,

so
the measurement is totally what is absorbed by the panel.

I don't understand why you wouldn't know how efficiency rates are
calculated. It's been common science for quite a long time now. It would
not be a true calculation of efficiency if they were to calculate the
efficiency by the total energy available. That's like calculating the
efficiency of a motor by calculating how much energy is output compared to
the amount of energy available in the entire gas tank when you only used a
quarter of a tank.


That is interesting, but frankly it seems like a poor way to state
efficiency for a solar collector. How does one determine the amount of
energy absorbed and not reflected in the collector?

Since solar insolence is readily available for an area, it would seem more
proper to use useful energy out divided by solar energy insolent. This
would account for energy reflected away from the collector as well as energy
converted to non-usable forms (for example, heat in a PV cell).

Measuring efficiency like what you're saying, I could have a 1 m^2 cell that
is 80% efficient produce less electricity than a 1 m^2 cell that is only 15%
efficient. Just 'accidently' have a highly reflective coating on the first
cell. Would run much cooler, and have higher 'efficiency', but not very
useful. Makes the 'efficiency' number useless to compare cells. Would have
to revert to just electric output 'in full sun'. And *that* leaves a lot to
be desired (define 'full sun' for every location??)

Your link doesn't really explain what you are saying about calculating cell
efficiency. It discusses that cells get warm and have to be cooled. And
that heat can be used to keep batteries warm in cold weather.

But I don't see any discussion about calculating the efficiency of a solar
cell in the manner you described. I think you're wrong in this and PV cell
efficiency is calculated as (electric-energy-out) /
(solar-energy-flux*area-of-collector). A much more logical method. Do you
have another NASA link that explains your point better?

Using solar insolence in the denominator would be similar to measuring power
out of the motor divided by the fuel-flow-rate (to use your analogy). That
method *does* make a lot of sense. It even accounts for fuel losses through
evaporation or leakage (analogous to reflection from the surface of a PV
cell)

daestrom


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Rod Speed
 
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Timm Simpkins wrote in message
...
Rod Speed wrote
wrote
Rod Speed wrote
Anthony Matonak wrote
News wrote


How much heat do PV cells emit on the back and front of the cells?


You could think of it this way. Typical PV panels are
around 12% efficient at converting sunlight into electricity.
This means that the other 88% is converted into heat.


Well, maybe 16 and 84%, eg 0.84x800 = 672 W/m^2 in AM2 sun.


...that would only be true if they were a perfect
black body absorber of all energy that falls on them.


But they are better than that.


Nope. Not even possible.


Rich Komp says the silicon is almost transparent to IR, and the
aluminum contact beneath is shiny, so PVs are a selective surface.


Waffle, the fact remains that all of what doesnt get turned
into electricity does not end up as heat. Plenty gets reflected.


The maximum efficiency ratings they are getting now
are approaching 30%. In normal production they are
around 23%. The average cell is about 12% efficient.


Wasnt even commenting on that stuff that I know already.

I was JUST commenting on the original claim that what isnt
turned into electricity ALL ends up as heat in the PV. Wrong.

Like I said, quite a bit gets reflected and
isnt available as heat to heat the house.

Also, no matter what misguided brain fart you're
having now Rod, the rest of the sun's energy that
is absorbed by the panel IS converted to heat.


Wrong. What is REFLECTED isnt.

Reams of completely irrelevant crap that has nothing to do with
what was being discussed, whether what falls on the PV system
ALL ends up as electricity or heat, flushed where it belongs.

If the rays are reflected and have no impact
on the process, they cannot be measured,


Wrong again.

so the measurement is totally what is absorbed by the panel.


Not a clue, as always.

reams of your pig ignorant silly **** flushed where it belongs




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Timm Simpkins
 
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"daestrom" wrote in message
...

"Timm Simpkins" wrote in message
...

"Rod Speed" wrote in message
...

wrote in
message ...
Rod Speed wrote
Anthony Matonak wrote
News wrote

How much heat do PV cells emit on the back and front of the

cells?

You could think of it this way. Typical PV panels are
around 12% efficient at converting sunlight into electricity.
This means that the other 88% is converted into heat.

Well, maybe 16 and 84%, eg 0.84x800 = 672 W/m^2 in AM2 sun.

...that would only be true if they were a perfect
black body absorber of all energy that falls on them.

But they are better than that.

Nope. Not even possible.

Rich Komp says the silicon is almost transparent to IR, and the
aluminum contact beneath is shiny, so PVs are a selective surface.

Waffle, the fact remains that all of what doesnt get turned
into electricity does not end up as heat. Plenty gets reflected.


The maximum efficiency ratings they are getting now are approaching 30%.

In
normal production they are around 23%. The average cell is about 12%
efficient. Also, no matter what misguided brain fart you're having now

Rod,
the rest of the sun's energy that is absorbed by the panel IS converted

to
heat. In fact, the efficiency of solar cells decreases as the heat
increases so there are many people experimenting with methods, both

passive
and active, to remove that heat and somehow convert that heat to

electricity
as well. The measure of efficiency is not measured by the amount of

energy
available, it's measured by the amount of energy produced compared to

the
amount of energy used.

Here's a link discussing the heat loss:

http://www.nasatech.com/Briefs/Apr99/NPO20284.html

Now Rod, don't complain to me about how wrong you are now. You can

complain
to the guys at Nasa about it.

You can measure the efficiency by taking the wattage being wasted as

heat
and compare that to the wattage being output by the solar cell. If the

rays
are reflected and have no impact on the process, they cannot be

measured,
so
the measurement is totally what is absorbed by the panel.

I don't understand why you wouldn't know how efficiency rates are
calculated. It's been common science for quite a long time now. It

would
not be a true calculation of efficiency if they were to calculate the
efficiency by the total energy available. That's like calculating the
efficiency of a motor by calculating how much energy is output compared

to
the amount of energy available in the entire gas tank when you only used

a
quarter of a tank.


That is interesting, but frankly it seems like a poor way to state
efficiency for a solar collector. How does one determine the amount of
energy absorbed and not reflected in the collector?

Since solar insolence is readily available for an area, it would seem more
proper to use useful energy out divided by solar energy insolent. This
would account for energy reflected away from the collector as well as

energy
converted to non-usable forms (for example, heat in a PV cell).

Measuring efficiency like what you're saying, I could have a 1 m^2 cell

that
is 80% efficient produce less electricity than a 1 m^2 cell that is only

15%
efficient. Just 'accidently' have a highly reflective coating on the

first
cell. Would run much cooler, and have higher 'efficiency', but not very
useful. Makes the 'efficiency' number useless to compare cells. Would

have
to revert to just electric output 'in full sun'. And *that* leaves a lot

to
be desired (define 'full sun' for every location??)


When you buy solar cells, you don't purchase efficiency ratings, you
purchase wattage ratings. I don't agree with your assesment that you can
lower the efficiency of a solar cell by reflecting the sun's rays since the
heat increase in the solar panel should be lower as well and so the ratio's
should be similar. That reflected energy is not taken into account in
calculations of efficiency. It can't be since it's almost impossible to
measure. You can measure temperature differentials and output. A solar
cell that puts out 12 volts at 6 amps is putting out 72 watts of power. At
20% efficiency, that means for every 72 watts of power you loose 1440 watts
to heat. If you measure the heat difference in the solar panel and convert
that to watts, you can compare that to the electrical output and that's
where you get your efficiency rating. If you're inhibiting the path of the
sun, you aren't lowering the efficiency rating because you're restricting
the solar panel's fuel. Fuel that doesn't get to the panel cannot be
consumed and cannot count toward the efficiency.


Your link doesn't really explain what you are saying about calculating

cell
efficiency. It discusses that cells get warm and have to be cooled. And
that heat can be used to keep batteries warm in cold weather.


I was referring to the statement that says that the energy that doesn't
produce power produces heat. I wasn't using that to bolster efficiency
ratings arguments.


But I don't see any discussion about calculating the efficiency of a solar
cell in the manner you described. I think you're wrong in this and PV

cell
efficiency is calculated as (electric-energy-out) /
(solar-energy-flux*area-of-collector). A much more logical method. Do

you
have another NASA link that explains your point better?


I don't, but I can give you plenty of links on how to measure efficiency.
You are saying that just because there is a certain amount of energy in a
given area of light that that energy is being consumed in the reaction.
That is not how efficiency ratings are made. You only measure the potential
energy of the amount of fuel, in this case sunlight and divide that by the
amount of energy output in the reaction. Since it's impossible to measure
the exact amount of fuel being consumed by a solar cell, you need to measure
the heat.

Now, I admit that my theory that you have to measure the heat may be flawed
since I have never measured efficiency of solar panels, and have never
talked to anyone that has, but as far as the way to measure efficiency, I
know for a fact that you cannot use the amount of fuel available as the
determining factor of the efficiency since there is no possible way that
reflected energy can be consumed in the reaction. If you were to blow air
into a turbine and measure the power produced, you could not take into
account the energy in the air that was not taken in by the turbine. That
would give you a false reading of efficiency.


Using solar insolence in the denominator would be similar to measuring

power
out of the motor divided by the fuel-flow-rate (to use your analogy).

That
method *does* make a lot of sense. It even accounts for fuel losses

through
evaporation or leakage (analogous to reflection from the surface of a PV
cell)


Fuel losses are only applicable if they enter into the system in the first
place. You are talking about reflected fuel, and that cannot have entered
into the system at all.

I will admit to some lack of knowledge of solar cells, but in my studies of
fluid dynamics I have had many occasions to calculate efficiency of
different systems. We had to take special care not to calculate the fuel
not used in the reaction. Also, you cannot use the power available if you
don't have a system that is designed to consume that power. Therefore,
unused fuel must be removed from the equation. An example would be if you
were to run your system on hydrogen. If you were pumping the hydrogen
through a turbine, but not burning it, you could not take the energy
potential in the hydrogen itself into account. If you were burning the
hydrogen, but not splitting the individual atom, you could not take the
atomic energy into account. There is so much energy in everything, but if
that energy is not consumed in the reaction, you are calculating for
something that can never exist. This is the same with solar energy. If you
are calculating efficiency for a fuel that has no possibility of giving you
power because it doesn't enter into the system, you are calculating
efficiency for something that cannot exist.


  #22   Report Post  
Timm Simpkins
 
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"Rod Speed" wrote in message
...

Timm Simpkins wrote in message
...
Rod Speed wrote
wrote
Rod Speed wrote
Anthony Matonak wrote
News wrote


How much heat do PV cells emit on the back and front of the cells?


You could think of it this way. Typical PV panels are
around 12% efficient at converting sunlight into electricity.
This means that the other 88% is converted into heat.


Well, maybe 16 and 84%, eg 0.84x800 = 672 W/m^2 in AM2 sun.


...that would only be true if they were a perfect
black body absorber of all energy that falls on them.


But they are better than that.


Nope. Not even possible.


Rich Komp says the silicon is almost transparent to IR, and the
aluminum contact beneath is shiny, so PVs are a selective surface.


Waffle, the fact remains that all of what doesnt get turned
into electricity does not end up as heat. Plenty gets reflected.


The maximum efficiency ratings they are getting now
are approaching 30%. In normal production they are
around 23%. The average cell is about 12% efficient.


Wasnt even commenting on that stuff that I know already.

I was JUST commenting on the original claim that what isnt
turned into electricity ALL ends up as heat in the PV. Wrong.

Like I said, quite a bit gets reflected and
isnt available as heat to heat the house.

Also, no matter what misguided brain fart you're
having now Rod, the rest of the sun's energy that
is absorbed by the panel IS converted to heat.


Wrong. What is REFLECTED isnt.


How about reading it again. I said, "the rest of the sun's energy that IS
ABSORBED by the panel IS converted to heat." I throw out the reflected
energy since it is not used in the calculation of efficiency.


Reams of completely irrelevant crap that has nothing to do with
what was being discussed, whether what falls on the PV system
ALL ends up as electricity or heat, flushed where it belongs.

If the rays are reflected and have no impact
on the process, they cannot be measured,


Wrong again.


Show your work.


so the measurement is totally what is absorbed by the panel.


Not a clue, as always.


Show your work.

This is the only reply you'll get on this Rod. There is no point in arguing
with you because no matter how wrong you are, you intentionally ignore the
facts so that you can stir up ****. Why don't you make an intelligent
assesment? You don't seem totally ignorant of everything, but those things
you comment on that you don't have a clue about, you should be able to admit
when you're wrong. Saying I don't have a clue doesn't address your denial
of my claims, it just shows that you don't have enough information to make
an adequate rebuttal of my argument.

Do us both a favor, and unless you can back up what you say about any
statement I make don't reply to my posts. I'll do the same with yours.


  #23   Report Post  
Rod Speed
 
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wrote in
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Rod Speed wrote
Anthony Matonak wrote
News wrote

How much heat do PV cells emit on the back and front of the

cells?

You could think of it this way. Typical PV panels are
around 12% efficient at converting sunlight into electricity.
This means that the other 88% is converted into heat.

Well, maybe 16 and 84%, eg 0.84x800 = 672 W/m^2 in AM2 sun.

...that would only be true if they were a perfect
black body absorber of all energy that falls on them.

But they are better than that.

Nope. Not even possible.

Rich Komp says the silicon is almost transparent to IR, and the
aluminum contact beneath is shiny, so PVs are a selective surface.

Waffle, the fact remains that all of what doesnt get turned
into electricity does not end up as heat. Plenty gets reflected.

The maximum efficiency ratings they are getting now are approaching 30%.

In
normal production they are around 23%. The average cell is about 12%
efficient. Also, no matter what misguided brain fart you're having now

Rod,
the rest of the sun's energy that is absorbed by the panel IS converted

to
heat. In fact, the efficiency of solar cells decreases as the heat
increases so there are many people experimenting with methods, both

passive
and active, to remove that heat and somehow convert that heat to

electricity
as well. The measure of efficiency is not measured by the amount of

energy
available, it's measured by the amount of energy produced compared to

the
amount of energy used.

Here's a link discussing the heat loss:

http://www.nasatech.com/Briefs/Apr99/NPO20284.html

Now Rod, don't complain to me about how wrong you are now. You can

complain
to the guys at Nasa about it.

You can measure the efficiency by taking the wattage being wasted as

heat
and compare that to the wattage being output by the solar cell. If the

rays
are reflected and have no impact on the process, they cannot be

measured,
so
the measurement is totally what is absorbed by the panel.

I don't understand why you wouldn't know how efficiency rates are
calculated. It's been common science for quite a long time now. It

would
not be a true calculation of efficiency if they were to calculate the
efficiency by the total energy available. That's like calculating the
efficiency of a motor by calculating how much energy is output compared

to
the amount of energy available in the entire gas tank when you only used

a
quarter of a tank.


That is interesting, but frankly it seems like a poor way to state
efficiency for a solar collector. How does one determine the
amount of energy absorbed and not reflected in the collector?


Since solar insolence is readily available for an area, it would seem
more proper to use useful energy out divided by solar energy insolent.
This would account for energy reflected away from the collector as well
as energy converted to non-usable forms (for example, heat in a PV cell).


Measuring efficiency like what you're saying, I could have a
1 m^2 cell that is 80% efficient produce less electricity than
a 1 m^2 cell that is only 15% efficient. Just 'accidently' have
a highly reflective coating on the first cell. Would run much
cooler, and have higher 'efficiency', but not very useful.
Makes the 'efficiency' number useless to compare cells.


Would have to revert to just electric output 'in full sun'. And *that*
leaves a lot to be desired (define 'full sun' for every location??)


When you buy solar cells, you don't purchase
efficiency ratings, you purchase wattage ratings.


Irrelevant waffle.

I don't agree with your assesment that you can lower
the efficiency of a solar cell by reflecting the sun's rays


He didnt even say that.

since the heat increase in the solar panel should be
lower as well and so the ratio's should be similar.


Not a clue.

That reflected energy is not taken into
account in calculations of efficiency.


Irrelevant to the original claim being discussed.

It can't be since it's almost impossible to measure.


Complete and utter pig ignorant drivel. Its completely trivial
to measure the amount of energy falling on the PV system.

You can measure temperature differentials and output. A solar cell
that puts out 12 volts at 6 amps is putting out 72 watts of power.


Its actually more complicated than that too, because it isnt a
simple pair of numbers, depends on the load being driven too.

And is completely irrelevant to the original claim anyway.

At 20% efficiency, that means for every 72
watts of power you loose 1440 watts to heat.


Not a clue.

If you measure the heat difference in the solar panel


Whatever that waffle is supposed to mean.

and convert that to watts,


Not even possible.

you can compare that to the electrical output
and that's where you get your efficiency rating.


Not a clue.

If you're inhibiting the path of the sun,
you aren't lowering the efficiency rating


You can be when PV systems dont necessarily
work at the same efficiency with different light levels.

Temperature of the PV system in spades.

because you're restricting the solar panel's fuel. Fuel that doesn't get to
the panel cannot be consumed and cannot count toward the efficiency.


Pathetic, really.

Your link doesn't really explain what you are saying about calculating
cell efficiency. It discusses that cells get warm and have to be cooled.
And that heat can be used to keep batteries warm in cold weather.


I was referring to the statement that says that the
energy that doesn't produce power produces heat.


It doesnt even say that as absolutely as the original claim being discussed.

It JUST says that a lot of heat is generated as well
as the electricity. NOT that ALL the energy that falls
on the PV system ends up as electricity or heat.

What is REFLECTED clearly doesnt end up as either in the PV system.

I wasn't using that to bolster efficiency ratings arguments.


More irrelevant waffle.

But I don't see any discussion about calculating the efficiency of
a solar cell in the manner you described. I think you're wrong in
this and PV cell efficiency is calculated as (electric-energy-out) /
(solar-energy-flux*area-of-collector). A much more logical method.
Do you have another NASA link that explains your point better?


I don't, but I can give you plenty of links on how to measure efficiency.


Irrelevant to the original claim being discussed.

You are saying that just because there is a certain amount of energy in
a given area of light that that energy is being consumed in the reaction.


Nope, he aint saying anything like that either.

That is not how efficiency ratings are made.


Wrong again.

You only measure the potential energy of the
amount of fuel, in this case sunlight and divide
that by the amount of energy output in the reaction.


Wrong again. The only energy output that matters with efficiency
is the electrical energy, and that is precisely what he just said.

Since it's impossible to measure the exact
amount of fuel being consumed by a solar cell,


What matters is the TOTAL SOLAR
ENERGY FALLING ON THE SOLAR CELL.

That is completely trivial to measure.

you need to measure the heat.


Nope.

Now, I admit that my theory that you have to measure the
heat may be flawed since I have never measured efficiency
of solar panels, and have never talked to anyone that has,


And are so stupid that you cant even manage to look up
how the efficiency of solar panels is actually measured either.

but as far as the way to measure efficiency, I know for a
fact that you cannot use the amount of fuel available as the
determining factor of the efficiency since there is no possible
way that reflected energy can be consumed in the reaction.


Pathetic really.

Have a look at how solar cell efficiency is actually measured some time.

If you were to blow air into a turbine and measure the power produced,
you could not take into account the energy in the air that was not taken
in by the turbine. That would give you a false reading of efficiency.


Completely and utterly irrelevant to how solar cell efficiency is actually measured.

Using solar insolence in the denominator would be similar to
measuring power out of the motor divided by the fuel-flow-rate
(to use your analogy). That method *does* make a lot of sense.
It even accounts for fuel losses through evaporation or leakage
(analogous to reflection from the surface of a PV cell)


Fuel losses are only applicable if they
enter into the system in the first place.


Not with solar cells.

Solar cell efficiency is the percentage of the solar energy
available to the PV system that ends up in electrical energy.

You are talking about reflected fuel, and that
cannot have entered into the system at all.


Completely and utterly irrelevant to how solar cell efficiency is actually measured.

I will admit to some lack of knowledge of solar cells,


And complete pig ignorance about how solar cell efficiency is actually measured.

but in my studies of fluid dynamics I have had many
occasions to calculate efficiency of different systems.


All completely and utterly irrelevant to how
solar cell efficiency is actually measured.

We had to take special care not to
calculate the fuel not used in the reaction.


All completely and utterly irrelevant to how
solar cell efficiency is actually measured.

Also, you cannot use the power available if you don't
have a system that is designed to consume that power.


All completely and utterly irrelevant to how
solar cell efficiency is actually measured.

Therefore, unused fuel must be removed from the equation.


All completely and utterly irrelevant to how
solar cell efficiency is actually measured.

An example would be if you were to run your system on
hydrogen. If you were pumping the hydrogen through a
turbine, but not burning it, you could not take the energy
potential in the hydrogen itself into account. If you were
burning the hydrogen, but not splitting the individual atom,
you could not take the atomic energy into account.


All completely and utterly irrelevant to how
solar cell efficiency is actually measured.

There is so much energy in everything, but if that
energy is not consumed in the reaction, you are
calculating for something that can never exist.


All completely and utterly irrelevant to how
solar cell efficiency is actually measured.

This is the same with solar energy.


Nope.

If you are calculating efficiency for a fuel that has no possibility
of giving you power because it doesn't enter into the system,
you are calculating efficiency for something that cannot exist.


All completely and utterly irrelevant to how
solar cell efficiency is actually measured.


Thanks for that completely superfluous proof that you have never
had a clue about how solar cell efficiency is actually measured.



  #24   Report Post  
Rod Speed
 
Posts: n/a
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Timm Simpkins wrote in
message ...
Rod Speed wrote
Timm Simpkins wrote
Rod Speed wrote
wrote
Rod Speed wrote
Anthony Matonak wrote
News wrote


How much heat do PV cells emit on the back and front of the cells?


You could think of it this way. Typical PV panels are
around 12% efficient at converting sunlight into electricity.
This means that the other 88% is converted into heat.


Well, maybe 16 and 84%, eg 0.84x800 = 672 W/m^2 in AM2 sun.


...that would only be true if they were a perfect
black body absorber of all energy that falls on them.


But they are better than that.


Nope. Not even possible.


Rich Komp says the silicon is almost transparent to IR, and the
aluminum contact beneath is shiny, so PVs are a selective surface.


Waffle, the fact remains that all of what doesnt get turned
into electricity does not end up as heat. Plenty gets reflected.


The maximum efficiency ratings they are getting now
are approaching 30%. In normal production they are
around 23%. The average cell is about 12% efficient.


Wasnt even commenting on that stuff that I know already.


I was JUST commenting on the original claim that what isnt
turned into electricity ALL ends up as heat in the PV. Wrong.


Like I said, quite a bit gets reflected and
isnt available as heat to heat the house.


Also, no matter what misguided brain fart you're
having now Rod, the rest of the sun's energy that
is absorbed by the panel IS converted to heat.


Wrong. What is REFLECTED isnt.


How about reading it again.


No point. It was pig ignorant drivel the first time around and it
remains pig ignorant drivel no matter how many times its read.

You've proven that you havent actually got a clue about
how the efficiency of PV systems is actually measured.

I said, "the rest of the sun's energy that IS
ABSORBED by the panel IS converted to heat."


Completely and utterely irrelevant to the original claim being
discussed, whether the difference between the efficiency
percentage and 100% all ends up as heat in the PV system.

Quite a bit is as I said, reflected as well.

I throw out the reflected energy


And you cant on that question being discussed.

since it is not used in the calculation of efficiency.


It is in fact part of the solar energy that falls on the PV
system that does not get turning into electricity or heat.

Reams of completely irrelevant crap that has nothing to do with
what was being discussed, whether what falls on the PV system
ALL ends up as electricity or heat, flushed where it belongs.


If the rays are reflected and have no impact
on the process, they cannot be measured,


Wrong again.


Show your work.


Dont need to. YOU made that stupid pig ignorant claim that the
total solar energy falling on the PC system cannot be measure.

That is just plain wrong.

YOU get to sustantiate your stupid pig ignorant claim.

THATS how it works.

so the measurement is totally what is absorbed by the panel.


Not a clue, as always.


Show your work.


Dont need to. YOU made that stupid pig ignorant claim that the
total solar energy falling on the PC system cannot be measure.

That is just plain wrong.

YOU get to sustantiate your stupid pig ignorant claim.

THATS how it works.

This is the only reply you'll get on this Rod.


Great, there is only so much of your pig ignorant
**** anyone should have to put up with.

There is no point in arguing with you
because no matter how wrong you are,


Corse you are never wrong, eh ?

You are, yet again.

Even someone as stupid as you should be able to check HOW
THE EFFICIENCY OF PV SYSTEMS IS ACTUALLY MEASURED.

you intentionally ignore the facts so that you can stir up ****.


Even you should be able to do better than this puerile ****, child.

Why don't you make an intelligent assesment?


Done that, repeatedly.

Tho no real intelligence is required to work out HOW THE
EFFICIENCY OF PV SYSTEMS IS ACTUALLY MEASURED.

You don't seem totally ignorant of everything, but those
things you comment on that you don't have a clue about,


I was actually employed at one time
to implement solar systems, ****wit.

you should be able to admit when you're wrong.


I do that when I ocassionally stuff up. I often call those
brain farts so even someone as stupid as you should be
able to find plenty of examples of that using groups.google.

I did that earlier today my time in fact.

Saying I don't have a clue doesn't address your denial of my claims,


Pity I repeatedly said a hell of a lot more than just that.

it just shows that you don't have enough information
to make an adequate rebuttal of my argument.


Not a ****ing clue. As always.

Do us both a favor,


Go shove your head up a dead bear's arse.

and unless you can back up what you say about
any statement I make don't reply to my posts.


Go shove your head up a dead bear's arse.

I'll do the same with yours.


You have always been, and always will be, completely and utterly irrelevant.

What you might or might not do in spades.


  #25   Report Post  
Timm Simpkins
 
Posts: n/a
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"Rod Speed" wrote in message
...

"Timm Simpkins" wrote in message

...

"daestrom" wrote in message
...

"Timm Simpkins" wrote in message
...

"Rod Speed" wrote in message
...

wrote in
message ...
Rod Speed wrote
Anthony Matonak wrote
News wrote

How much heat do PV cells emit on the back and front of the

cells?

You could think of it this way. Typical PV panels are
around 12% efficient at converting sunlight into electricity.
This means that the other 88% is converted into heat.

Well, maybe 16 and 84%, eg 0.84x800 = 672 W/m^2 in AM2 sun.

...that would only be true if they were a perfect
black body absorber of all energy that falls on them.

But they are better than that.

Nope. Not even possible.

Rich Komp says the silicon is almost transparent to IR, and the
aluminum contact beneath is shiny, so PVs are a selective

surface.

Waffle, the fact remains that all of what doesnt get turned
into electricity does not end up as heat. Plenty gets reflected.

The maximum efficiency ratings they are getting now are approaching

30%.
In
normal production they are around 23%. The average cell is about

12%
efficient. Also, no matter what misguided brain fart you're having

now
Rod,
the rest of the sun's energy that is absorbed by the panel IS

converted
to
heat. In fact, the efficiency of solar cells decreases as the heat
increases so there are many people experimenting with methods, both
passive
and active, to remove that heat and somehow convert that heat to
electricity
as well. The measure of efficiency is not measured by the amount of
energy
available, it's measured by the amount of energy produced compared

to
the
amount of energy used.

Here's a link discussing the heat loss:

http://www.nasatech.com/Briefs/Apr99/NPO20284.html

Now Rod, don't complain to me about how wrong you are now. You can
complain
to the guys at Nasa about it.

You can measure the efficiency by taking the wattage being wasted as

heat
and compare that to the wattage being output by the solar cell. If

the
rays
are reflected and have no impact on the process, they cannot be

measured,
so
the measurement is totally what is absorbed by the panel.

I don't understand why you wouldn't know how efficiency rates are
calculated. It's been common science for quite a long time now. It

would
not be a true calculation of efficiency if they were to calculate

the
efficiency by the total energy available. That's like calculating

the
efficiency of a motor by calculating how much energy is output

compared
to
the amount of energy available in the entire gas tank when you only

used
a
quarter of a tank.


That is interesting, but frankly it seems like a poor way to state
efficiency for a solar collector. How does one determine the
amount of energy absorbed and not reflected in the collector?


Since solar insolence is readily available for an area, it would seem
more proper to use useful energy out divided by solar energy insolent.
This would account for energy reflected away from the collector as well
as energy converted to non-usable forms (for example, heat in a PV

cell).

Measuring efficiency like what you're saying, I could have a
1 m^2 cell that is 80% efficient produce less electricity than
a 1 m^2 cell that is only 15% efficient. Just 'accidently' have
a highly reflective coating on the first cell. Would run much
cooler, and have higher 'efficiency', but not very useful.
Makes the 'efficiency' number useless to compare cells.


Would have to revert to just electric output 'in full sun'. And *that*
leaves a lot to be desired (define 'full sun' for every location??)


When you buy solar cells, you don't purchase
efficiency ratings, you purchase wattage ratings.


Irrelevant waffle.

I don't agree with your assesment that you can lower
the efficiency of a solar cell by reflecting the sun's rays


He didnt even say that.


Here are a few quotes:

"How does one determine the amount of energy absorbed and not reflected in
the collector?"

"Measuring efficiency like what you're saying, I could have a 1 m^2 cell
that
is 80% efficient produce less electricity than a 1 m^2 cell that is only 15%
efficient. Just 'accidently' have a highly reflective coating on the first
cell."

He very well did say that. By adding a highly reflective surface to the
collector, you are reflecting the rays away from the collector. He claims
that the lack of those rays would lower the efficiency, and that simply is
not true.

If you took a theoretical motor that had a constant efficiency at any speed,
and put it in your car, you would get more power output if you mash down on
the skinny pedal because you are adding more fuel, but you wouldn't change
the efficiency (theoretically constant as it is). What he is saying is that
he is wanting to put a reflective surface to limit the amount of fuel. The
smaller amount of fuel doesn't change the efficiency unless there is
something in the system that causes smaller amounts of fuel to change the
efficiency, like an improperly tuned carburator.


since the heat increase in the solar panel should be
lower as well and so the ratio's should be similar.


Not a clue.


No you don't have a clue.


That reflected energy is not taken into
account in calculations of efficiency.


Irrelevant to the original claim being discussed.


It is completely relevant to the original claim I was having an issue with.


It can't be since it's almost impossible to measure.


Complete and utter pig ignorant drivel. Its completely trivial
to measure the amount of energy falling on the PV system.


That's exactly what I'm saying. If you measure that energy you're not
measuring the amount of energy being consumed, but the amount available.


You can measure temperature differentials and output. A solar cell
that puts out 12 volts at 6 amps is putting out 72 watts of power.


Its actually more complicated than that too, because it isnt a
simple pair of numbers, depends on the load being driven too.


We're talking theory here. The numbers I'm putting out are the theoretical
maximum outputs available from the solar panel, and that should be obvious.


And is completely irrelevant to the original claim anyway.

At 20% efficiency, that means for every 72
watts of power you loose 1440 watts to heat.


Not a clue.

If you measure the heat difference in the solar panel


Whatever that waffle is supposed to mean.

and convert that to watts,


Not even possible.


It's entirely possible to convert heat to watts. The measure of heat is
done in joules. You can convert joules to watts. the conversion is
k=0.0002780752 * j where w = watts and j=joules.


you can compare that to the electrical output
and that's where you get your efficiency rating.


Not a clue.

If you're inhibiting the path of the sun,
you aren't lowering the efficiency rating


You can be when PV systems dont necessarily
work at the same efficiency with different light levels.


Irrelevant when we're not talking about anything but theory. That may
likely be true, but the fact is that if you reflect that light away from the
solar panel you can't use it in your measurements.


Temperature of the PV system in spades.

because you're restricting the solar panel's fuel. Fuel that doesn't

get to
the panel cannot be consumed and cannot count toward the efficiency.


Pathetic, really.

Your link doesn't really explain what you are saying about calculating
cell efficiency. It discusses that cells get warm and have to be

cooled.
And that heat can be used to keep batteries warm in cold weather.


I was referring to the statement that says that the
energy that doesn't produce power produces heat.


It doesnt even say that as absolutely as the original claim being

discussed.

It JUST says that a lot of heat is generated as well
as the electricity. NOT that ALL the energy that falls
on the PV system ends up as electricity or heat.


"most of the solar radiation focused by solar concentrators onto solar
photovoltaic cells is converted to heat."

That is what it says, and I am giving the proof that the theory that the
efficiency rating of solar cells has anything to do with anything besides
power output and heat. FUEL NOT USED CAN'T BE USED TO CALCULATE EFFICIENCY.


What is REFLECTED clearly doesnt end up as either in the PV system.


You claimed that in a 12% efficient system that the other 88% was not
converted to heat. That is completely false.

Here's what was said:
QUOTE
You could think of it this way. Typical PV panels are
around 12% efficient at converting sunlight into electricity.
This means that the other 88% is converted into heat.


Nope, that would only be true if they were a perfect
black body absorber of all energy that falls on them.

In practice quite a bit is just reflected off them.
END QUOTE


I wasn't using that to bolster efficiency ratings arguments.


More irrelevant waffle.

But I don't see any discussion about calculating the efficiency of
a solar cell in the manner you described. I think you're wrong in
this and PV cell efficiency is calculated as (electric-energy-out) /
(solar-energy-flux*area-of-collector). A much more logical method.
Do you have another NASA link that explains your point better?


I don't, but I can give you plenty of links on how to measure

efficiency.

Irrelevant to the original claim being discussed.

You are saying that just because there is a certain amount of energy in
a given area of light that that energy is being consumed in the

reaction.

Nope, he aint saying anything like that either.


If he's not saying that, then why is he talking about the energy in a given
area of light? It has no relevance if he's not trying to take it into
account.


That is not how efficiency ratings are made.


Wrong again.


Then how are efficiency ratings made?

I'll put it simply for you. The amount of energy consumed in a system
compared to the amount of usable energy output by the system is how you get
efficiency ratings. If I am wrong, I'm waiting to be educated, even though
I know you can't.


You only measure the potential energy of the
amount of fuel, in this case sunlight and divide
that by the amount of energy output in the reaction.


Wrong again. The only energy output that matters with efficiency
is the electrical energy, and that is precisely what he just said.


That is correct, I should have said the amount of useable energy. Since
nothing is done with the heat, that figure must be thrown out. He did not
say that though.


Since it's impossible to measure the exact
amount of fuel being consumed by a solar cell,


What matters is the TOTAL SOLAR
ENERGY FALLING ON THE SOLAR CELL.


No sir! What matters is the TOTAL SOLAR ENERGY CONSUMED BY THE SOLAR CELL.
The fact that it falls on the solar cell does not guarantee that it will be
used. The fact is, you guys are arguing that reflected solar energy has
some part in the calculation of the efficiency of a solar panel in creating
power. That is simply not true.


That is completely trivial to measure.

you need to measure the heat.


Nope.


How else do you find out how much fuel is being consumed? If you can't
measure the exact amount of solar energy consumed, you have to measure the
total power created by the system. That means not only measuring the amount
of energy that was converted to electricity, but the amount of energy
converted to heat. When you add those together, you get the total amount of
energy input.

Now, in a system where there is fuel being burned, you must also take into
account fuel that does not get burned unless you are calculating how
efficient the system is at burning fuel. Fuel that does not get burned has
not been used and would skew any true efficiency figures.


Now, I admit that my theory that you have to measure the
heat may be flawed since I have never measured efficiency
of solar panels, and have never talked to anyone that has,


And are so stupid that you cant even manage to look up
how the efficiency of solar panels is actually measured either.


Apparently neither can you. If you had, and I were wrong, you would be
shoving it down my throat.


but as far as the way to measure efficiency, I know for a
fact that you cannot use the amount of fuel available as the
determining factor of the efficiency since there is no possible
way that reflected energy can be consumed in the reaction.


Pathetic really.

Have a look at how solar cell efficiency is actually measured some time.

If you were to blow air into a turbine and measure the power produced,
you could not take into account the energy in the air that was not taken
in by the turbine. That would give you a false reading of efficiency.


Completely and utterly irrelevant to how solar cell efficiency is actually

measured.

Efficiency is efficiency is efficiency. It is completely relevant. You on
the otherhand aren't.


Using solar insolence in the denominator would be similar to
measuring power out of the motor divided by the fuel-flow-rate
(to use your analogy). That method *does* make a lot of sense.
It even accounts for fuel losses through evaporation or leakage
(analogous to reflection from the surface of a PV cell)


Fuel losses are only applicable if they
enter into the system in the first place.


Not with solar cells.

Solar cell efficiency is the percentage of the solar energy
available to the PV system that ends up in electrical energy.


If the fuel isn't used, how do you count it in efficiency? It must be
introduced into the system in order for a measurement to be made.
Calculating reflected energy is as relevant to calculating actual efficiency
of a system as calculating how many hairs you loose to the sewer every time
you take a shower. No relevance whatsoever.


You are talking about reflected fuel, and that
cannot have entered into the system at all.


Completely and utterly irrelevant to how solar cell efficiency is actually

measured.

I will admit to some lack of knowledge of solar cells,


And complete pig ignorance about how solar cell efficiency is actually

measured.

but in my studies of fluid dynamics I have had many
occasions to calculate efficiency of different systems.


All completely and utterly irrelevant to how
solar cell efficiency is actually measured.

We had to take special care not to
calculate the fuel not used in the reaction.


All completely and utterly irrelevant to how
solar cell efficiency is actually measured.

Also, you cannot use the power available if you don't
have a system that is designed to consume that power.


All completely and utterly irrelevant to how
solar cell efficiency is actually measured.

Therefore, unused fuel must be removed from the equation.


All completely and utterly irrelevant to how
solar cell efficiency is actually measured.

An example would be if you were to run your system on
hydrogen. If you were pumping the hydrogen through a
turbine, but not burning it, you could not take the energy
potential in the hydrogen itself into account. If you were
burning the hydrogen, but not splitting the individual atom,
you could not take the atomic energy into account.


All completely and utterly irrelevant to how
solar cell efficiency is actually measured.

There is so much energy in everything, but if that
energy is not consumed in the reaction, you are
calculating for something that can never exist.


All completely and utterly irrelevant to how
solar cell efficiency is actually measured.

This is the same with solar energy.


Nope.

If you are calculating efficiency for a fuel that has no possibility
of giving you power because it doesn't enter into the system,
you are calculating efficiency for something that cannot exist.


All completely and utterly irrelevant to how
solar cell efficiency is actually measured.


Thanks for that completely superfluous proof that you have never
had a clue about how solar cell efficiency is actually measured.


Rod, you make a good argument for abortion, but you don't make much of an
argument against my facts. If you are talking about how efficient of a
collector of solar energy it is, that is one thing, but if you are talking
about how efficient of an electricity producer it is, that is another
entirely. A solar collector for heating water should be calculated the way
you are suggesting, because the less solar radiation it collects, the less
efficient it is. A photovoltaic cell on the other hand is measured totally
differently.

You are either arguing a point you know nothing about, or are getting
confused about the system we are talking about measuring. Measuring the
efficiency of a fuel pump is not the same thing as measuring the efficiency
of an internal combustion engine. Totally apples and triangles.




  #26   Report Post  
Timm Simpkins
 
Posts: n/a
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The more I read of your arguments Rod, the more I realize you are getting
things mixed up. You're arguing about the efficiency of a photovoltaic cell
as a collector of solar energy, and the original disagreement was about how
much solar energy was turned to electricity and how much was converted to
heat.

You are mixed up about what you are talking about. When a photovoltaic cell
collects rays from the sun, it not only absorbs the light and the radiant
heat, but it produces heat. The amount of heat collected from the sun is
not what the efficiency ratings the original post was about. Heat is
actually produced by the silicon. I'm not talking about heat retained by
the silicon that is radiated from the sun.

You need to get your systems straight, then argue with me.

"Rod Speed" wrote in message
...

Timm Simpkins wrote in
message ...
Rod Speed wrote
Timm Simpkins wrote
Rod Speed wrote
wrote
Rod Speed wrote
Anthony Matonak wrote
News wrote


How much heat do PV cells emit on the back and front of the

cells?

You could think of it this way. Typical PV panels are
around 12% efficient at converting sunlight into electricity.
This means that the other 88% is converted into heat.


Well, maybe 16 and 84%, eg 0.84x800 = 672 W/m^2 in AM2 sun.


...that would only be true if they were a perfect
black body absorber of all energy that falls on them.


But they are better than that.


Nope. Not even possible.


Rich Komp says the silicon is almost transparent to IR, and the
aluminum contact beneath is shiny, so PVs are a selective surface.


Waffle, the fact remains that all of what doesnt get turned
into electricity does not end up as heat. Plenty gets reflected.


The maximum efficiency ratings they are getting now
are approaching 30%. In normal production they are
around 23%. The average cell is about 12% efficient.


Wasnt even commenting on that stuff that I know already.


I was JUST commenting on the original claim that what isnt
turned into electricity ALL ends up as heat in the PV. Wrong.


Like I said, quite a bit gets reflected and
isnt available as heat to heat the house.


Also, no matter what misguided brain fart you're
having now Rod, the rest of the sun's energy that
is absorbed by the panel IS converted to heat.


Wrong. What is REFLECTED isnt.


How about reading it again.


No point. It was pig ignorant drivel the first time around and it
remains pig ignorant drivel no matter how many times its read.

You've proven that you havent actually got a clue about
how the efficiency of PV systems is actually measured.

I said, "the rest of the sun's energy that IS
ABSORBED by the panel IS converted to heat."


Completely and utterely irrelevant to the original claim being
discussed, whether the difference between the efficiency
percentage and 100% all ends up as heat in the PV system.

Quite a bit is as I said, reflected as well.

I throw out the reflected energy


And you cant on that question being discussed.

since it is not used in the calculation of efficiency.


It is in fact part of the solar energy that falls on the PV
system that does not get turning into electricity or heat.

Reams of completely irrelevant crap that has nothing to do with
what was being discussed, whether what falls on the PV system
ALL ends up as electricity or heat, flushed where it belongs.


If the rays are reflected and have no impact
on the process, they cannot be measured,


Wrong again.


Show your work.


Dont need to. YOU made that stupid pig ignorant claim that the
total solar energy falling on the PC system cannot be measure.

That is just plain wrong.

YOU get to sustantiate your stupid pig ignorant claim.

THATS how it works.

so the measurement is totally what is absorbed by the panel.


Not a clue, as always.


Show your work.


Dont need to. YOU made that stupid pig ignorant claim that the
total solar energy falling on the PC system cannot be measure.

That is just plain wrong.

YOU get to sustantiate your stupid pig ignorant claim.

THATS how it works.

This is the only reply you'll get on this Rod.


Great, there is only so much of your pig ignorant
**** anyone should have to put up with.

There is no point in arguing with you
because no matter how wrong you are,


Corse you are never wrong, eh ?

You are, yet again.

Even someone as stupid as you should be able to check HOW
THE EFFICIENCY OF PV SYSTEMS IS ACTUALLY MEASURED.

you intentionally ignore the facts so that you can stir up ****.


Even you should be able to do better than this puerile ****, child.

Why don't you make an intelligent assesment?


Done that, repeatedly.

Tho no real intelligence is required to work out HOW THE
EFFICIENCY OF PV SYSTEMS IS ACTUALLY MEASURED.

You don't seem totally ignorant of everything, but those
things you comment on that you don't have a clue about,


I was actually employed at one time
to implement solar systems, ****wit.

you should be able to admit when you're wrong.


I do that when I ocassionally stuff up. I often call those
brain farts so even someone as stupid as you should be
able to find plenty of examples of that using groups.google.

I did that earlier today my time in fact.

Saying I don't have a clue doesn't address your denial of my claims,


Pity I repeatedly said a hell of a lot more than just that.

it just shows that you don't have enough information
to make an adequate rebuttal of my argument.


Not a ****ing clue. As always.

Do us both a favor,


Go shove your head up a dead bear's arse.

and unless you can back up what you say about
any statement I make don't reply to my posts.


Go shove your head up a dead bear's arse.

I'll do the same with yours.


You have always been, and always will be, completely and utterly

irrelevant.

What you might or might not do in spades.




  #27   Report Post  
Rod Speed
 
Posts: n/a
Default Ohmwork


"Timm Simpkins" wrote in message ...

"Rod Speed" wrote in message
...

"Timm Simpkins" wrote in message

...

"daestrom" wrote in message
...

"Timm Simpkins" wrote in message
...

"Rod Speed" wrote in message
...

wrote in
message ...
Rod Speed wrote
Anthony Matonak wrote
News wrote

How much heat do PV cells emit on the back and front of the
cells?

You could think of it this way. Typical PV panels are
around 12% efficient at converting sunlight into electricity.
This means that the other 88% is converted into heat.

Well, maybe 16 and 84%, eg 0.84x800 = 672 W/m^2 in AM2 sun.

...that would only be true if they were a perfect
black body absorber of all energy that falls on them.

But they are better than that.

Nope. Not even possible.

Rich Komp says the silicon is almost transparent to IR, and the
aluminum contact beneath is shiny, so PVs are a selective

surface.

Waffle, the fact remains that all of what doesnt get turned
into electricity does not end up as heat. Plenty gets reflected.

The maximum efficiency ratings they are getting now are approaching

30%.
In
normal production they are around 23%. The average cell is about

12%
efficient. Also, no matter what misguided brain fart you're having

now
Rod,
the rest of the sun's energy that is absorbed by the panel IS

converted
to
heat. In fact, the efficiency of solar cells decreases as the heat
increases so there are many people experimenting with methods, both
passive
and active, to remove that heat and somehow convert that heat to
electricity
as well. The measure of efficiency is not measured by the amount of
energy
available, it's measured by the amount of energy produced compared

to
the
amount of energy used.

Here's a link discussing the heat loss:

http://www.nasatech.com/Briefs/Apr99/NPO20284.html

Now Rod, don't complain to me about how wrong you are now. You can
complain
to the guys at Nasa about it.

You can measure the efficiency by taking the wattage being wasted as
heat
and compare that to the wattage being output by the solar cell. If

the
rays
are reflected and have no impact on the process, they cannot be
measured,
so
the measurement is totally what is absorbed by the panel.

I don't understand why you wouldn't know how efficiency rates are
calculated. It's been common science for quite a long time now. It
would
not be a true calculation of efficiency if they were to calculate

the
efficiency by the total energy available. That's like calculating

the
efficiency of a motor by calculating how much energy is output

compared
to
the amount of energy available in the entire gas tank when you only

used
a
quarter of a tank.


That is interesting, but frankly it seems like a poor way to state
efficiency for a solar collector. How does one determine the
amount of energy absorbed and not reflected in the collector?


Since solar insolence is readily available for an area, it would seem
more proper to use useful energy out divided by solar energy insolent.
This would account for energy reflected away from the collector as well
as energy converted to non-usable forms (for example, heat in a PV

cell).

Measuring efficiency like what you're saying, I could have a
1 m^2 cell that is 80% efficient produce less electricity than
a 1 m^2 cell that is only 15% efficient. Just 'accidently' have
a highly reflective coating on the first cell. Would run much
cooler, and have higher 'efficiency', but not very useful.
Makes the 'efficiency' number useless to compare cells.


Would have to revert to just electric output 'in full sun'. And *that*
leaves a lot to be desired (define 'full sun' for every location??)


When you buy solar cells, you don't purchase
efficiency ratings, you purchase wattage ratings.


Irrelevant waffle.


I don't agree with your assesment that you can lower
the efficiency of a solar cell by reflecting the sun's rays


He didnt even say that.


Here are a few quotes:


"How does one determine the amount of energy
absorbed and not reflected in the collector?"


Nothing like that bit you claimed you didnt agree with.

"Measuring efficiency like what you're saying, I could have
a 1 m^2 cell that is 80% efficient produce less electricity
than a 1 m^2 cell that is only 15% efficient. Just 'accidently'
have a highly reflective coating on the first cell."


Nothing like that bit you claimed you didnt agree with.

He very well did say that.


Nope, that is saying something quite different. That isnt
even 'lower the efficiency of a solar cell by reflecting the
sun's rays' its just rubbing your nose in the fact that the
efficiency of solar cells aint measured the way you claim it is.

By adding a highly reflective surface to the collector,
you are reflecting the rays away from the collector.


Duh.

He claims that the lack of those rays would lower the efficiency,


Nope, he aint saying anything like that.

and that simply is not true.


Having fun thrashing that straw man are you ?

If you took a theoretical motor


Hasnt got any relevance what so ever to HOW THE EFFICIENCY
OF SOLAR CELLS IS ACTUALLY MEASURED.

since the heat increase in the solar panel should be
lower as well and so the ratio's should be similar.


Not a clue.


No you don't have a clue.


Pathetic, really.

That reflected energy is not taken into
account in calculations of efficiency.


Irrelevant to the original claim being discussed.


It is completely relevant to the original claim I was having an issue with.


Nope.

It can't be since it's almost impossible to measure.


Complete and utter pig ignorant drivel. Its completely trivial
to measure the amount of energy falling on the PV system.


That's exactly what I'm saying. If you measure that energy you're not
measuring the amount of energy being consumed, but the amount available.


Pity about how the efficiency of solar cells is actually measured.

You can measure temperature differentials and output. A solar cell
that puts out 12 volts at 6 amps is putting out 72 watts of power.


Its actually more complicated than that too, because it isnt a
simple pair of numbers, depends on the load being driven too.


We're talking theory here.


Nope.

The numbers I'm putting out are the theoretical maximum
outputs available from the solar panel, and that should be obvious.


Even you should be able to bull**** your way out of
your predicament better than that pathetic effort.

And is completely irrelevant to the original claim anyway.


At 20% efficiency, that means for every 72
watts of power you loose 1440 watts to heat.


Not a clue.


If you measure the heat difference in the solar panel


Whatever that waffle is supposed to mean.


and convert that to watts,


Not even possible.


It's entirely possible to convert heat to watts.


You were talking about a HEAT DIFFERENCE.

The measure of heat is done in joules. You can convert joules to watts.
the conversion is k=0.0002780752 * j where w = watts and j=joules.


Nothing like your previous pig ignorant claim.

you can compare that to the electrical output
and that's where you get your efficiency rating.


Not a clue.


If you're inhibiting the path of the sun,
you aren't lowering the efficiency rating


You can be when PV systems dont necessarily
work at the same efficiency with different light levels.


Irrelevant when we're not talking about anything but theory.


Even you should be able to bull**** your way out of
your predicament better than that pathetic effort.

That may likely be true,


Corse its true.

but the fact is that if you reflect that light away from
the solar panel you can't use it in your measurements.


Pity about how the efficiency of solar cells is actually measured.

Try telling the industry that they havent got a clue about how to
measure it. Dont be too surprised when they laugh in your face.

Temperature of the PV system in spades.


because you're restricting the solar panel's fuel. Fuel that doesn't get to
the panel cannot be consumed and cannot count toward the efficiency.


Pathetic, really.


Your link doesn't really explain what you are saying about calculating
cell efficiency. It discusses that cells get warm and have to be cooled.
And that heat can be used to keep batteries warm in cold weather.


I was referring to the statement that says that the
energy that doesn't produce power produces heat.


It doesnt even say that as absolutely as the original claim being discussed.


It JUST says that a lot of heat is generated as well
as the electricity. NOT that ALL the energy that falls
on the PV system ends up as electricity or heat.


"most of the solar radiation focused by solar concentrators
onto solar photovoltaic cells is converted to heat."


Nothing like the claim that all the energy falling
on the PV system ends up as heat or electricity.

That is what it says,


Pity its nothing like the claim that all the energy falling
on the PV system ends up as heat or electricity.

and I am giving the proof that


Nope, just flaunting your complete pig ignorance of
how the efficiency of solar cells is actually measured.

the theory that the efficiency rating of solar cells has anything
to do with anything besides power output and heat. FUEL
NOT USED CAN'T BE USED TO CALCULATE EFFICIENCY.


Try telling the industry that they havent got a clue about how to
measure it. Dont be too surprised when they laugh in your face.

What is REFLECTED clearly doesnt end up as either in the PV system.


You claimed that in a 12% efficient system
that the other 88% was not converted to heat.


I ACTUALLY said that the other 88% wasnt ALL converted
to heat IN THE PV system, a different matter entirely.

That is completely false.


Fraid not.

Here's what was said:
QUOTE
You could think of it this way. Typical PV panels are
around 12% efficient at converting sunlight into electricity.
This means that the other 88% is converted into heat.


Nope, that would only be true if they were a perfect
black body absorber of all energy that falls on them.


In practice quite a bit is just reflected off them.
END QUOTE


Which was correct when I said it and is still correct.

All you have ever done is flaunt your complete pig ignorance
of how the efficiency of PV systems is actually measured.

I wasn't using that to bolster efficiency ratings arguments.


More irrelevant waffle.


But I don't see any discussion about calculating the efficiency of
a solar cell in the manner you described. I think you're wrong in
this and PV cell efficiency is calculated as (electric-energy-out) /
(solar-energy-flux*area-of-collector). A much more logical method.
Do you have another NASA link that explains your point better?


I don't, but I can give you plenty of links on how to measure efficiency.


Irrelevant to the original claim being discussed.


And you cant with the PV SYSTEMS BEING DISCUSSED ANYWAY.

You are saying that just because there is a certain amount of energy in
a given area of light that that energy is being consumed in the reaction.


Nope, he aint saying anything like that either.


If he's not saying that, then why is he talking
about the energy in a given area of light?


Concentrate on the word CONSUMED.

It has no relevance if he's not trying to take it into account.


Separate issue entirely.

That is not how efficiency ratings are made.


Wrong again.


Then how are efficiency ratings made?


Basically be MEASURING the amount of solar energy that falls
on the PC system and measuring the amount of electrical energy
it produces, and that is the efficiency, at the particular set of
conditions on solar energy level and temperature and load etc.

You dont approve ? Your problem.

Try telling the industry that they havent got a clue about how to
measure it. Dont be too surprised WHEN they laugh in your face.

I'll put it simply for you. The amount of energy consumed
in a system compared to the amount of usable energy
output by the system is how you get efficiency ratings.


Pity about how the efficiency of PV systems is actually measured.

If I am wrong,


No if about it.

I'm waiting to be educated,


Not even possible.

even though I know you can't.


Nothing like a completely closed mind eh ?

You only measure the potential energy of the
amount of fuel, in this case sunlight and divide
that by the amount of energy output in the reaction.


Wrong again. The only energy output that matters with efficiency
is the electrical energy, and that is precisely what he just said.


That is correct, I should have said the amount of useable energy.


Doesnt help.

Pity about how the efficiency of PV systems is actually measured.

Since nothing is done with the heat, that figure
must be thrown out. He did not say that though.


Yep, he's pointing out to you that you havent got a clue
about how the efficiency of PV systems is actually measured.

Since it's impossible to measure the exact
amount of fuel being consumed by a solar cell,


What matters is the TOTAL SOLAR
ENERGY FALLING ON THE SOLAR CELL.


No sir!


Yes cur!!

What matters is the TOTAL SOLAR ENERGY CONSUMED BY THE SOLAR CELL.


Pity about how the efficiency of PV systems is actually measured.

The fact that it falls on the solar cell does not guarantee that it will be used.


Pity about how the efficiency of PV systems is actually measured.

The fact is, you guys are arguing that reflected
solar energy has some part in the calculation
of the efficiency of a solar panel in creating power.


And that is the way THE ENTIRE INDUSTRY DOES IT.

That is simply not true.


Fraid so.

Try telling the industry that they havent got a clue about how to
measure it. Dont be too surprised WHEN they laugh in your face.

That is completely trivial to measure.


you need to measure the heat.


Nope.


How else do you find out how much fuel is being consumed?


There is no 'fuel'.

The industry measures the amount of solar energy falling on the PV system.

Try telling the industry that they havent got a clue about how to
measure it. Dont be too surprised WHEN they laugh in your face.

If you can't measure the exact amount of solar energy consumed,


'consumed' isnt even relevant.

you have to measure the total power created by the system.


Power isnt even being created.

That means not only measuring the amount of energy
that was converted to electricity, but the amount of
energy converted to heat. When you add those
together, you get the total amount of energy input.


Pity about how the efficiency of PV systems is actually measured.

Now, in a system where there is fuel being burned,


Completely and utterly irrelevant to how the
efficiency of PV systems is actually measured.

you must also take into account fuel that does not get
burned unless you are calculating how efficient the system
is at burning fuel. Fuel that does not get burned has not
been used and would skew any true efficiency figures.


Completely and utterly irrelevant to how the
efficiency of PV systems is actually measured.

Now, I admit that my theory that you have to measure the
heat may be flawed since I have never measured efficiency
of solar panels, and have never talked to anyone that has,


And are so stupid that you cant even manage to look up
how the efficiency of solar panels is actually measured either.


Apparently neither can you. If you had, and I were
wrong, you would be shoving it down my throat.


That is precisely what I have been doing, stupid.

but as far as the way to measure efficiency, I know for a
fact that you cannot use the amount of fuel available as the
determining factor of the efficiency since there is no possible
way that reflected energy can be consumed in the reaction.


Pathetic really.


Have a look at how solar cell efficiency is actually measured some time.


If you were to blow air into a turbine and measure the power produced,
you could not take into account the energy in the air that was not taken
in by the turbine. That would give you a false reading of efficiency.


Completely and utterly irrelevant to how
solar cell efficiency is actually measured.


Efficiency is efficiency is efficiency.


Pathetic, really.

It is completely relevant.


Completely and utterly irrelevant TO HOW SOLAR
CELL EFFICIENCY IS ACTUALLY MEASURED.

You on the otherhand aren't.


Pathetic, really.

Using solar insolence in the denominator would be similar to
measuring power out of the motor divided by the fuel-flow-rate
(to use your analogy). That method *does* make a lot of sense.
It even accounts for fuel losses through evaporation or leakage
(analogous to reflection from the surface of a PV cell)


Fuel losses are only applicable if they
enter into the system in the first place.


Not with solar cells.


Solar cell efficiency is the percentage of the solar energy
available to the PV system that ends up in electrical energy.


If the fuel isn't used, how do you count it in efficiency?


Pity about how the efficiency of solar cells is actually measured.

It must be introduced into the system in order for a measurement to be made.


Try telling the industry that they havent got a clue about how to
measure it. Dont be too surprised when they laugh in your face.

Calculating reflected energy is as relevant to calculating actual
efficiency of a system as calculating how many hairs you loose to
the sewer every time you take a shower. No relevance whatsoever.


Pity about how the efficiency of solar cells is actually measured.

Try telling the industry that they havent got a clue about how to
measure it. Dont be too surprised when they laugh in your face.

You are talking about reflected fuel, and that
cannot have entered into the system at all.


Completely and utterly irrelevant to how
solar cell efficiency is actually measured.


I will admit to some lack of knowledge of solar cells,


And complete pig ignorance about how
solar cell efficiency is actually measured.


but in my studies of fluid dynamics I have had many
occasions to calculate efficiency of different systems.


All completely and utterly irrelevant to how
solar cell efficiency is actually measured.


We had to take special care not to
calculate the fuel not used in the reaction.


All completely and utterly irrelevant to how
solar cell efficiency is actually measured.


Also, you cannot use the power available if you don't
have a system that is designed to consume that power.


All completely and utterly irrelevant to how
solar cell efficiency is actually measured.


Therefore, unused fuel must be removed from the equation.


All completely and utterly irrelevant to how
solar cell efficiency is actually measured.


An example would be if you were to run your system on
hydrogen. If you were pumping the hydrogen through a
turbine, but not burning it, you could not take the energy
potential in the hydrogen itself into account. If you were
burning the hydrogen, but not splitting the individual atom,
you could not take the atomic energy into account.


All completely and utterly irrelevant to how
solar cell efficiency is actually measured.


There is so much energy in everything, but if that
energy is not consumed in the reaction, you are
calculating for something that can never exist.


All completely and utterly irrelevant to how
solar cell efficiency is actually measured.


This is the same with solar energy.


Nope.


If you are calculating efficiency for a fuel that has no possibility
of giving you power because it doesn't enter into the system,
you are calculating efficiency for something that cannot exist.


All completely and utterly irrelevant to how
solar cell efficiency is actually measured.


Thanks for that completely superfluous proof that you have never
had a clue about how solar cell efficiency is actually measured.


Rod, you make a good argument for abortion, but
you don't make much of an argument against my facts.


You havent even presented a SINGLE fact relevant to HOW
THE EFFICIENCY OF PV SYSTEMS IS ACTUALLY MEASURE.

Try telling the industry that they havent got a clue about how to
measure it. Dont be too surprised when they laugh in your face.

If you are talking about how efficient of a collector of solar
energy it is, that is one thing, but if you are talking about how
efficient of an electricity producer it is, that is another entirely.


Pity about how the efficiency of solar cells is actually measured.

Try telling the industry that they havent got a clue about how to
measure it. Dont be too surprised when they laugh in your face.

A solar collector for heating water should be calculated
the way you are suggesting, because the less solar
radiation it collects, the less efficient it is. A photovoltaic
cell on the other hand is measured totally differently.


Pity about how the efficiency of solar cells is actually measured.

Try telling the industry that they havent got a clue about how to
measure it. Dont be too surprised when they laugh in your face.

You are either arguing a point you know nothing about, or are
getting confused about the system we are talking about measuring.


Or I actually understand how the efficiency
of solar cells is actually measured.

Try telling the industry that they havent got a clue about how to
measure it. Dont be too surprised when they laugh in your face.


Measuring the efficiency of a fuel pump
is not the same thing as measuring the
efficiency of an internal combustion engine.


Pity about how the efficiency of solar cells is actually measured.

Try telling the industry that they havent got a clue about how to
measure it. Dont be too surprised when they laugh in your face.

Totally apples and triangles.


Yep, all your crap about fuel and engines and showers is that in spades.


  #28   Report Post  
Rod Speed
 
Posts: n/a
Default Ohmwork


Timm Simpkins wrote in
message ...

The more I read of your arguments Rod, the
more I realize you are getting things mixed up.


You're the one that keeps flauting your complete pig ignorance
about how the industry measures the efficiency of PV systems.

Try telling the industry that they havent got a clue about how to
measure it. Dont be too surprised WHEN they laugh in your face.

You're arguing about the efficiency of a photovoltaic
cell as a collector of solar energy,


Nope. I JUST pointed out that what doesnt get converted
into electricity USING THE EFFICIENCY NUMBER THAT
THE INDUSTRY USES TO STATE THE EFFICIENCY OF
PV SYSTEMS, doesnt all end up as heat in the PC system.
Quite a bit gets reflect from the PV system as well.

and the original disagreement was about how much solar energy
was turned to electricity and how much was converted to heat.


Like hell it was. The original claim was that ALL of the difference
between 100% and the THE EFFICIENCY NUMBER THAT THE
INDUSTRY USES TO STATE THE EFFICIENCY OF PV SYSTEMS,
ends up as heat in the PV system. Wrong.

You are mixed up about what you are talking about.


Nope, you keep flauting your complete pig ignorance about
how the industry measures the efficiency of PV systems.

Try telling the industry that they havent got a clue about how to
measure it. Dont be too surprised WHEN they laugh in your face.

When a photovoltaic cell collects rays from the sun, it not only
absorbs the light and the radiant heat, but it produces heat.


Duh.

The amount of heat collected from the sun is not
what the efficiency ratings the original post was about.


The original post used THE EFFICIENCY NUMBER THAT THE
INDUSTRY USES TO STATE THE EFFICIENCY OF PV SYSTEMS,

Heat is actually produced by the silicon.


Utterly mangled all over again. Quite a bit of the heat that
ends up in the PV system isnt produced by the silicon at all.

I'm not talking about heat retained by
the silicon that is radiated from the sun.


You have always been, and always will be, completely and
utterly irrelevant. In spades when you cant even manage to
work out how the industry measures the efficiency of PV systems.

You need to get your systems straight,


You need to work out how the industry
measures the efficiency of PV systems.

then argue with me.


Pointless, you've never had a ****ing clue and couldnt
even bull**** your way out of a wet paper bag even if
your pathetic excuse for an existence depended on that.


"Rod Speed" wrote in message
...

Timm Simpkins wrote in
message ...
Rod Speed wrote
Timm Simpkins wrote
Rod Speed wrote
wrote
Rod Speed wrote
Anthony Matonak wrote
News wrote


How much heat do PV cells emit on the back and front of the

cells?

You could think of it this way. Typical PV panels are
around 12% efficient at converting sunlight into electricity.
This means that the other 88% is converted into heat.


Well, maybe 16 and 84%, eg 0.84x800 = 672 W/m^2 in AM2 sun.


...that would only be true if they were a perfect
black body absorber of all energy that falls on them.


But they are better than that.


Nope. Not even possible.


Rich Komp says the silicon is almost transparent to IR, and the
aluminum contact beneath is shiny, so PVs are a selective surface.


Waffle, the fact remains that all of what doesnt get turned
into electricity does not end up as heat. Plenty gets reflected.


The maximum efficiency ratings they are getting now
are approaching 30%. In normal production they are
around 23%. The average cell is about 12% efficient.


Wasnt even commenting on that stuff that I know already.


I was JUST commenting on the original claim that what isnt
turned into electricity ALL ends up as heat in the PV. Wrong.


Like I said, quite a bit gets reflected and
isnt available as heat to heat the house.


Also, no matter what misguided brain fart you're
having now Rod, the rest of the sun's energy that
is absorbed by the panel IS converted to heat.


Wrong. What is REFLECTED isnt.


How about reading it again.


No point. It was pig ignorant drivel the first time around and it
remains pig ignorant drivel no matter how many times its read.

You've proven that you havent actually got a clue about
how the efficiency of PV systems is actually measured.

I said, "the rest of the sun's energy that IS
ABSORBED by the panel IS converted to heat."


Completely and utterely irrelevant to the original claim being
discussed, whether the difference between the efficiency
percentage and 100% all ends up as heat in the PV system.

Quite a bit is as I said, reflected as well.

I throw out the reflected energy


And you cant on that question being discussed.

since it is not used in the calculation of efficiency.


It is in fact part of the solar energy that falls on the PV
system that does not get turning into electricity or heat.

Reams of completely irrelevant crap that has nothing to do with
what was being discussed, whether what falls on the PV system
ALL ends up as electricity or heat, flushed where it belongs.


If the rays are reflected and have no impact
on the process, they cannot be measured,


Wrong again.


Show your work.


Dont need to. YOU made that stupid pig ignorant claim that the
total solar energy falling on the PC system cannot be measure.

That is just plain wrong.

YOU get to sustantiate your stupid pig ignorant claim.

THATS how it works.

so the measurement is totally what is absorbed by the panel.


Not a clue, as always.


Show your work.


Dont need to. YOU made that stupid pig ignorant claim that the
total solar energy falling on the PC system cannot be measure.

That is just plain wrong.

YOU get to sustantiate your stupid pig ignorant claim.

THATS how it works.

This is the only reply you'll get on this Rod.


Great, there is only so much of your pig ignorant
**** anyone should have to put up with.

There is no point in arguing with you
because no matter how wrong you are,


Corse you are never wrong, eh ?

You are, yet again.

Even someone as stupid as you should be able to check HOW
THE EFFICIENCY OF PV SYSTEMS IS ACTUALLY MEASURED.

you intentionally ignore the facts so that you can stir up ****.


Even you should be able to do better than this puerile ****, child.

Why don't you make an intelligent assesment?


Done that, repeatedly.

Tho no real intelligence is required to work out HOW THE
EFFICIENCY OF PV SYSTEMS IS ACTUALLY MEASURED.

You don't seem totally ignorant of everything, but those
things you comment on that you don't have a clue about,


I was actually employed at one time
to implement solar systems, ****wit.

you should be able to admit when you're wrong.


I do that when I ocassionally stuff up. I often call those
brain farts so even someone as stupid as you should be
able to find plenty of examples of that using groups.google.

I did that earlier today my time in fact.

Saying I don't have a clue doesn't address your denial of my claims,


Pity I repeatedly said a hell of a lot more than just that.

it just shows that you don't have enough information
to make an adequate rebuttal of my argument.


Not a ****ing clue. As always.

Do us both a favor,


Go shove your head up a dead bear's arse.

and unless you can back up what you say about
any statement I make don't reply to my posts.


Go shove your head up a dead bear's arse.

I'll do the same with yours.


You have always been, and always will be, completely and utterly

irrelevant.

What you might or might not do in spades.






  #29   Report Post  
Timm Simpkins
 
Posts: n/a
Default Ohmwork

Where do you get your credentials Rod? You seem to think you know how "THE
INDUSTRY" does anything. If "THE INDUSTRY" takes into account solar energy
that is not used, it is doing it wrong. Calculations of efficiency don't
change just because of the industry or the system being calculated for. You
can calculate efficiency all day long, but if you are taking in the wrong
numbers, you are doing it wrong.

I on the other hand have a background in engineering. I have a degree in
electrical engineering, and I took many other classes to learn things like
fluid dynamics, structural engineering, and chemistry. I have worked little
with solar cells, but I have worked extensively with electronics and
calculating efficiency. I worked for motorola for 5 years, and for IBM for
2 years before I found that I can make much more money doing other things.

I can tell you with great certainty that if they are calculating the
efficiency of a solar cell in producing electricity from solar energy in the
way you describe, they are doing it wrong.

Let's compare something a little more relevant. If I take a circuit that is
used to convert electricity to rotational inertia, otherwise calculated in
horsepower, I can take the voltage of the circuit and multiply that by the
amp draw in order to calculate the wattage that the motor is using. I can
then convert the horsepower created in the motor and convert that to watts.
If I make sure that there are no other sources for the current draw, I can
subtract the wattage output at the shaft of the motor from the wattage taken
in by the motor and easily calculate the efficiency. Now, if I add another
circuit to that motor, such as a control relay, and the electricity that
runs through that relay comes from the same power source, I can no longer
measure the amp draw at the power source because some of that power is being
used in the relay circuitry. That power never gets to the motor and
therefore would invalidate my results. That is the same as trying to
calculate the efficiency of a photovoltaic cell in converting solar energy
to electricity by including the energy diverted elsewhere in your
calculations. If what you are saying is true, the true efficiency of a PV
cell would be significantly higher than the stated efficiency.

Let's be conservative and say that up to 60% of the sun's rays never reach a
point where they can be converted, and that would mean that a 30% efficient
solar cell would actually be 75% efficient with the energy that comes into
it. That can't be true. In reality, I would have to say that less than 10%
of the sun's rays reach a point where they can be converted to electricity.
That would make a 30% efficient solar cell, if measured by your method, 300%
efficient. Totally illogical. If I'm not correct on the amount of solar
energy that is able to be converted then figure it out for yourself with
your figures, but please make the source of your figures available. Let's
say that 30% of the sun's rays are processed through the PV cell, that would
mean that a 30% efficient solar cell is actually 100% efficient. I
guarantee you that the amount of rays that actually get processed are much
less than 50% of the total rays. Even at 50%, the cell would be 60%
efficient, and that's not likely.

You need to do some more math on the matter and figure out that what you're
saying is totally illogical.


  #30   Report Post  
Rod Speed
 
Posts: n/a
Default Ohmwork


Timm Simpkins wrote in
message ...

Where do you get your credentials Rod?


Dont need any of those to work out how the industry
measures the efficiency of PV systems, child.

Even you should be able to manage it if someone was
actually stupid enough to lend you a seeing eye dog and a
white cane and pointed you in the general direction of google.

You seem to think you know how "THE INDUSTRY" does anything.


More of you childish lying. I JUST said that I do know
how the industry measures the efficiency of PV systems.

And even you should be able to work it out if someone was
actually stupid enough to lend you a seeing eye dog and a
white cane and pointed you in the general direction of google.

If "THE INDUSTRY" takes into account solar
energy that is not used, it is doing it wrong.


Try telling the industry that they havent got a clue about how to
measure it. Dont be too surprised WHEN they laugh in your face

Calculations of efficiency don't change just because
of the industry or the system being calculated for.


Wrong again. How efficiency is measure with say engines is
NOTHING like how the efficiency of a PV system is measured.

You havent managed to work that out yet ? Your problem, child.

You can calculate efficiency all day long, but if you are
taking in the wrong numbers, you are doing it wrong.


Try telling the industry that they havent got a clue about how to
measure it. Dont be too surprised WHEN they laugh in your face

I on the other hand have a background in engineering.


Fat lot of good that has ever done you when you
clearly havent managed to work out how the industry
measures the efficiency of PV systems, child.

I have a degree in electrical engineering, and I took many other classes
to learn things like fluid dynamics, structural engineering, and chemistry.


Fat lot of good that has ever done you when you
clearly havent managed to work out how the industry
measures the efficiency of PV systems, child.

I have worked little with solar cells, but I have worked
extensively with electronics and calculating efficiency.


Fat lot of good that has ever done you when you
clearly havent managed to work out how the industry
measures the efficiency of PV systems, child.

I worked for motorola for 5 years, and for IBM for 2 years before
I found that I can make much more money doing other things.


Fat lot of good that has ever done you when you
clearly havent managed to work out how the industry
measures the efficiency of PV systems, child.

I can tell you with great certainty that if they are calculating
the efficiency of a solar cell in producing electricity from
solar energy in the way you describe, they are doing it wrong.


Try telling the industry that they havent got a clue about how to
measure it. Dont be too surprised WHEN they laugh in your face

Let's compare something a little more relevant.


No thanks, has no relevance what so ever to how
the industry measures the efficiency of PV systems.

You need to do some more math on the matter and
figure out that what you're saying is totally illogical.


Try telling the industry that they havent got a clue about how to
measure it. Dont be too surprised WHEN they laugh in your face

Keep digging, child. You'll be out in china
any day now. Best take your passport.




  #31   Report Post  
Timm Simpkins
 
Posts: n/a
Default Ohmwork

It appears that you are lost for words again. You can't face the facts,
that's why you deleted them. You can't back up your argument with any facts
because there are none to support you. You can't even make a decent logical
argument. You sit there ****ing your pants, showing that you know nothing.
You have offered no proof that you are correct about how photovoltaic cells
efficiency calculations are made. You claim knowledge but can't back up
that putative knowledge with any actual fact or meaningful discussion.

You are wasting bandwidth with your total lack of intelligence. With your
measurement of efficiency, it is easily possible to get a theoretical
efficiency of greater than 100% at the actual system, which is impossible,
so your method is flawed. You don't back that up either. No matter what,
there would be no way for your measurement of efficiency to reach 100% even
if none of the energy were reflected, so again your method is flawed.

Now go suck your thumb after another major mistake.


"Rod Speed" wrote in message
...

Timm Simpkins wrote in
message ...

Where do you get your credentials Rod?


Dont need any of those to work out how the industry
measures the efficiency of PV systems, child.

Even you should be able to manage it if someone was
actually stupid enough to lend you a seeing eye dog and a
white cane and pointed you in the general direction of google.

You seem to think you know how "THE INDUSTRY" does anything.


More of you childish lying. I JUST said that I do know
how the industry measures the efficiency of PV systems.

And even you should be able to work it out if someone was
actually stupid enough to lend you a seeing eye dog and a
white cane and pointed you in the general direction of google.

If "THE INDUSTRY" takes into account solar
energy that is not used, it is doing it wrong.


Try telling the industry that they havent got a clue about how to
measure it. Dont be too surprised WHEN they laugh in your face

Calculations of efficiency don't change just because
of the industry or the system being calculated for.


Wrong again. How efficiency is measure with say engines is
NOTHING like how the efficiency of a PV system is measured.

You havent managed to work that out yet ? Your problem, child.

You can calculate efficiency all day long, but if you are
taking in the wrong numbers, you are doing it wrong.


Try telling the industry that they havent got a clue about how to
measure it. Dont be too surprised WHEN they laugh in your face

I on the other hand have a background in engineering.


Fat lot of good that has ever done you when you
clearly havent managed to work out how the industry
measures the efficiency of PV systems, child.

I have a degree in electrical engineering, and I took many other classes
to learn things like fluid dynamics, structural engineering, and

chemistry.

Fat lot of good that has ever done you when you
clearly havent managed to work out how the industry
measures the efficiency of PV systems, child.

I have worked little with solar cells, but I have worked
extensively with electronics and calculating efficiency.


Fat lot of good that has ever done you when you
clearly havent managed to work out how the industry
measures the efficiency of PV systems, child.

I worked for motorola for 5 years, and for IBM for 2 years before
I found that I can make much more money doing other things.


Fat lot of good that has ever done you when you
clearly havent managed to work out how the industry
measures the efficiency of PV systems, child.

I can tell you with great certainty that if they are calculating
the efficiency of a solar cell in producing electricity from
solar energy in the way you describe, they are doing it wrong.


Try telling the industry that they havent got a clue about how to
measure it. Dont be too surprised WHEN they laugh in your face

Let's compare something a little more relevant.


No thanks, has no relevance what so ever to how
the industry measures the efficiency of PV systems.

You need to do some more math on the matter and
figure out that what you're saying is totally illogical.


Try telling the industry that they havent got a clue about how to
measure it. Dont be too surprised WHEN they laugh in your face

Keep digging, child. You'll be out in china
any day now. Best take your passport.




  #32   Report Post  
Anthony Matonak
 
Posts: n/a
Default Ohmwork

Timm Simpkins wrote:
....
You are wasting bandwidth with your total lack of intelligence.

....

It's called Trolling and the main goal is to elicit responses.
You will notice, by the number of your responses, that it sometimes
works.

http://www.hyphenologist.co.uk/killfile/
http://www.searchbug.com/directory.a...s/Usenet/FAQs/
http://www.usenet-replayer.com/faq/m...al-living.html

Anthony

  #33   Report Post  
Rod Speed
 
Posts: n/a
Default Ohmwork


Timm Simpkins wrote in
message ...

It appears that you are lost for words again.


Even you should be able to bull**** your way out of
your predicament better than that pathetic effort, child.

You can't face the facts, that's why you deleted them.


None of your **** I deleted had any relevance what so ever TO HOW
THE INDUSTRY MEASURES THE EFFICIENCY OF PV SYSTEMS.

reams of your puerile **** any 3 year old
could leave for dead flushed where it belongs

Try harder, little boy. You might actually manage to fool someone, sometime.


"Rod Speed" wrote in message
...

Timm Simpkins wrote in
message ...

Where do you get your credentials Rod?


Dont need any of those to work out how the industry
measures the efficiency of PV systems, child.

Even you should be able to manage it if someone was
actually stupid enough to lend you a seeing eye dog and a
white cane and pointed you in the general direction of google.

You seem to think you know how "THE INDUSTRY" does anything.


More of you childish lying. I JUST said that I do know
how the industry measures the efficiency of PV systems.

And even you should be able to work it out if someone was
actually stupid enough to lend you a seeing eye dog and a
white cane and pointed you in the general direction of google.

If "THE INDUSTRY" takes into account solar
energy that is not used, it is doing it wrong.


Try telling the industry that they havent got a clue about how to
measure it. Dont be too surprised WHEN they laugh in your face

Calculations of efficiency don't change just because
of the industry or the system being calculated for.


Wrong again. How efficiency is measure with say engines is
NOTHING like how the efficiency of a PV system is measured.

You havent managed to work that out yet ? Your problem, child.

You can calculate efficiency all day long, but if you are
taking in the wrong numbers, you are doing it wrong.


Try telling the industry that they havent got a clue about how to
measure it. Dont be too surprised WHEN they laugh in your face

I on the other hand have a background in engineering.


Fat lot of good that has ever done you when you
clearly havent managed to work out how the industry
measures the efficiency of PV systems, child.

I have a degree in electrical engineering, and I took many other classes
to learn things like fluid dynamics, structural engineering, and

chemistry.

Fat lot of good that has ever done you when you
clearly havent managed to work out how the industry
measures the efficiency of PV systems, child.

I have worked little with solar cells, but I have worked
extensively with electronics and calculating efficiency.


Fat lot of good that has ever done you when you
clearly havent managed to work out how the industry
measures the efficiency of PV systems, child.

I worked for motorola for 5 years, and for IBM for 2 years before
I found that I can make much more money doing other things.


Fat lot of good that has ever done you when you
clearly havent managed to work out how the industry
measures the efficiency of PV systems, child.

I can tell you with great certainty that if they are calculating
the efficiency of a solar cell in producing electricity from
solar energy in the way you describe, they are doing it wrong.


Try telling the industry that they havent got a clue about how to
measure it. Dont be too surprised WHEN they laugh in your face

Let's compare something a little more relevant.


No thanks, has no relevance what so ever to how
the industry measures the efficiency of PV systems.

You need to do some more math on the matter and
figure out that what you're saying is totally illogical.


Try telling the industry that they havent got a clue about how to
measure it. Dont be too surprised WHEN they laugh in your face

Keep digging, child. You'll be out in china
any day now. Best take your passport.






  #34   Report Post  
Rod Speed
 
Posts: n/a
Default Ohmwork

Some ****wit troll claiming to be
Anthony Matonak wrote in
message ...
just the puerile **** thats always pouring
from the back of that ****wit troll.


  #35   Report Post  
daestrom
 
Posts: n/a
Default Ohmwork


"Timm Simpkins" wrote in message
...

"daestrom" wrote in message
...

snip

That is interesting, but frankly it seems like a poor way to state
efficiency for a solar collector. How does one determine the amount of
energy absorbed and not reflected in the collector?

Since solar insolence is readily available for an area, it would seem

more
proper to use useful energy out divided by solar energy insolent. This
would account for energy reflected away from the collector as well as

energy
converted to non-usable forms (for example, heat in a PV cell).

Measuring efficiency like what you're saying, I could have a 1 m^2 cell

that
is 80% efficient produce less electricity than a 1 m^2 cell that is only

15%
efficient. Just 'accidently' have a highly reflective coating on the

first
cell. Would run much cooler, and have higher 'efficiency', but not very
useful. Makes the 'efficiency' number useless to compare cells. Would

have
to revert to just electric output 'in full sun'. And *that* leaves a

lot
to
be desired (define 'full sun' for every location??)


When you buy solar cells, you don't purchase efficiency ratings, you
purchase wattage ratings. I don't agree with your assesment that you can
lower the efficiency of a solar cell by reflecting the sun's rays since

the
heat increase in the solar panel should be lower as well and so the

ratio's
should be similar.


Only if the reflector is equally reflective for all bands of the spectrum.
Clearly, that is not the case. Just changing from a glass covering to one
of some plastic could change this.

That reflected energy is not taken into account in
calculations of efficiency. It can't be since it's almost impossible to
measure.


The reflected energy *should* be. It is not 'almost impossible to measure'.
If you know the *total* insolent light energy, then efficiency is simply
(useful energy out) / (total energy in). And total insolent light energy is
measurable with a black-body instrument, or simply looking it up from data
tables for your location.

You can measure temperature differentials and output. A solar
cell that puts out 12 volts at 6 amps is putting out 72 watts of power.

At
20% efficiency, that means for every 72 watts of power you loose 1440

watts
to heat. If you measure the heat difference in the solar panel and

convert
that to watts, you can compare that to the electrical output and that's
where you get your efficiency rating. If you're inhibiting the path of

the
sun, you aren't lowering the efficiency rating because you're restricting
the solar panel's fuel. Fuel that doesn't get to the panel cannot be
consumed and cannot count toward the efficiency.


Only if you draw the system boundary as just under the reflective surface.
Much more practical to consider the entire cell, including the surface as
the system boundary. In that case, some energy enters the system and is
reflected back out again, while other energy is converted to heat and
finally some energy is converted to useful electric output. Since the
surface of the cell is certainly a *part* of the cell, its performance (i.e.
its reflectivity) is a valid part of the cell's overall performance.

To use your previous analogy, if the fuel gets to the carbuerator, but
doesn't go into the cylinder, you have a less efficient engine. Claiming
that such a loss doesn't enter into the efficiency of the engine is silly.
Claiming that it 'is almost impossible to measure' is not a valid argument.

snip

I don't, but I can give you plenty of links on how to measure efficiency.
You are saying that just because there is a certain amount of energy in a
given area of light that that energy is being consumed in the reaction.
That is not how efficiency ratings are made. You only measure the

potential
energy of the amount of fuel, in this case sunlight and divide that by the
amount of energy output in the reaction.


Exactly. If the potential energy is the energy of the incoming sunlight,
and the output is the electricity produced, then you have a complete measure
of the cell's ability to convert sunlight to electricity. This includes the
losses due to reflection, heat production, and any others.

Since it's impossible to measure
the exact amount of fuel being consumed by a solar cell, you need to

measure
the heat.


It is *not* impossible to measure the energy flux from sunlight. Measuring
the total energy flux in the same location can be done quite easily. How do
you think those tables of solar radiation are derived? Measure the energy
flux at the site, multiply by the total area of the cell that is normal to
the flux and you have 'the exact amount of fuel being consumed'. Measure
the output of the cell and the efficiency is trivially calculated.

Now, I admit that my theory that you have to measure the heat may be

flawed
since I have never measured efficiency of solar panels, and have never
talked to anyone that has, but as far as the way to measure efficiency, I
know for a fact that you cannot use the amount of fuel available as the
determining factor of the efficiency since there is no possible way that
reflected energy can be consumed in the reaction. If you were to blow air
into a turbine and measure the power produced, you could not take into
account the energy in the air that was not taken in by the turbine. That
would give you a false reading of efficiency.


Wind turbine efficiency is based on the total energy available in the wind
per unit area and the area swept through by the turbine. This is very
similar to the method I suggest for measuring PV input.


Using solar insolence in the denominator would be similar to measuring

power
out of the motor divided by the fuel-flow-rate (to use your analogy).

That
method *does* make a lot of sense. It even accounts for fuel losses

through
evaporation or leakage (analogous to reflection from the surface of a PV
cell)


Fuel losses are only applicable if they enter into the system in the first
place. You are talking about reflected fuel, and that cannot have entered
into the system at all.


You are defining the 'system' differently than I am. You define the
'system' as the cell just below the reflective surface. So by your
definition, reflected energy does not enter the system. But as you say,
that is very hard to calculate. I define the 'system' as the cell
*including* the surface. So by my definition, reflected energy is energy
that enters the system, then is reflected back out without much change
(except direction). Measuring the total energy that flows into my system is
simply the solar flux times the area normal to said flux. Very
straight-forward measurement.

I will admit to some lack of knowledge of solar cells, but in my studies

of
fluid dynamics I have had many occasions to calculate efficiency of
different systems. We had to take special care not to calculate the fuel
not used in the reaction.


I work in thermodynamics everyday too. If you have a situation where some
fuel is lost without entering the system, it is still a loss. From the
thermodynamics of the boiler's furnace, no it doesn't enter the burner so it
isn't a loss there. But from the standpoint of total fuel needed to create
a given output, unburned fuel losses are a factor. Why else do people get
excited about stopping fuel leaks?

Also, you cannot use the power available if you
don't have a system that is designed to consume that power. Therefore,
unused fuel must be removed from the equation. An example would be if you
were to run your system on hydrogen. If you were pumping the hydrogen
through a turbine, but not burning it, you could not take the energy
potential in the hydrogen itself into account. If you were burning the
hydrogen, but not splitting the individual atom, you could not take the
atomic energy into account. There is so much energy in everything, but if
that energy is not consumed in the reaction, you are calculating for
something that can never exist.


Not always as simple as that. If you use hydrogen to power the fuel pump
itself (such as in a space shuttle engine), then the total hydrogen
delivered to the main engine is *not* the same as what was delivered to the
fuel pump. Some of it was burned in the fuel pump to power it so it could
deliver the rest to the main engine. When calculating the total efficiency
of the system, one would look at total hydrogen consumed (in both the fuel
pump and main engine) and the total output of the main engine. The fuel
that doesn't make it to the main engine (consumed in the fuel pump) *is* an
energy loss that is relevent and *lowers* the *efficiency* of the system.
But at the same time, since it greatly increases the fuel flow rate, it
greatly *raises* the power level (despite the drop in efficiency).

This is the same with solar energy. If you
are calculating efficiency for a fuel that has no possibility of giving

you
power because it doesn't enter into the system, you are calculating
efficiency for something that cannot exist.


But we are *not* 'calculating efficiency for a fuel that has no possibility
of giving you power'. Reflective losses *are* under the control of the
designer. Energy that is reflected off the surface in one design, needn't
necessarily be reflected in another design. The theoretically perfectly
efficient panel would have no reflected energy at all. Second law doesn't
require that this energy *must* be unrecoverable, only limitations in design
technology.

If you do not consider the reflective losses, an unscrupulous designer could
coat his entire system in silver foil. Measuring the efficiency using your
methods, it would be a very efficient PV cell. He/she could advertise PV
cell with efficiency in the high 80's perhaps 90's. But no one would be
happy with such a design.

The cell must be rated in electric-watts/area in some 'standard condition'.
And the 'standard condition' is some value of solar flux. Any other rating
is useless.

daestrom




  #36   Report Post  
Timm Simpkins
 
Posts: n/a
Default Ohmwork


"daestrom" wrote in message
...

"Timm Simpkins" wrote in message
...

"daestrom" wrote in message
...

snip

That is interesting, but frankly it seems like a poor way to state
efficiency for a solar collector. How does one determine the amount of
energy absorbed and not reflected in the collector?

Since solar insolence is readily available for an area, it would seem

more
proper to use useful energy out divided by solar energy insolent.

This
would account for energy reflected away from the collector as well as

energy
converted to non-usable forms (for example, heat in a PV cell).

Measuring efficiency like what you're saying, I could have a 1 m^2

cell
that
is 80% efficient produce less electricity than a 1 m^2 cell that is

only
15%
efficient. Just 'accidently' have a highly reflective coating on the

first
cell. Would run much cooler, and have higher 'efficiency', but not

very
useful. Makes the 'efficiency' number useless to compare cells.

Would
have
to revert to just electric output 'in full sun'. And *that* leaves a

lot
to
be desired (define 'full sun' for every location??)


When you buy solar cells, you don't purchase efficiency ratings, you
purchase wattage ratings. I don't agree with your assesment that you

can
lower the efficiency of a solar cell by reflecting the sun's rays since

the
heat increase in the solar panel should be lower as well and so the

ratio's
should be similar.


Only if the reflector is equally reflective for all bands of the spectrum.
Clearly, that is not the case. Just changing from a glass covering to one
of some plastic could change this.

That reflected energy is not taken into account in
calculations of efficiency. It can't be since it's almost impossible to
measure.


The reflected energy *should* be. It is not 'almost impossible to

measure'.
If you know the *total* insolent light energy, then efficiency is simply
(useful energy out) / (total energy in). And total insolent light energy

is
measurable with a black-body instrument, or simply looking it up from data
tables for your location.

You can measure temperature differentials and output. A solar
cell that puts out 12 volts at 6 amps is putting out 72 watts of power.

At
20% efficiency, that means for every 72 watts of power you loose 1440

watts
to heat. If you measure the heat difference in the solar panel and

convert
that to watts, you can compare that to the electrical output and that's
where you get your efficiency rating. If you're inhibiting the path of

the
sun, you aren't lowering the efficiency rating because you're

restricting
the solar panel's fuel. Fuel that doesn't get to the panel cannot be
consumed and cannot count toward the efficiency.


Only if you draw the system boundary as just under the reflective surface.
Much more practical to consider the entire cell, including the surface as
the system boundary. In that case, some energy enters the system and is
reflected back out again, while other energy is converted to heat and
finally some energy is converted to useful electric output. Since the
surface of the cell is certainly a *part* of the cell, its performance

(i.e.
its reflectivity) is a valid part of the cell's overall performance.

To use your previous analogy, if the fuel gets to the carbuerator, but
doesn't go into the cylinder, you have a less efficient engine. Claiming
that such a loss doesn't enter into the efficiency of the engine is silly.
Claiming that it 'is almost impossible to measure' is not a valid argument

..

snip

I don't, but I can give you plenty of links on how to measure

efficiency.
You are saying that just because there is a certain amount of energy in

a
given area of light that that energy is being consumed in the reaction.
That is not how efficiency ratings are made. You only measure the

potential
energy of the amount of fuel, in this case sunlight and divide that by

the
amount of energy output in the reaction.


Exactly. If the potential energy is the energy of the incoming sunlight,
and the output is the electricity produced, then you have a complete

measure
of the cell's ability to convert sunlight to electricity. This includes

the
losses due to reflection, heat production, and any others.

Since it's impossible to measure
the exact amount of fuel being consumed by a solar cell, you need to

measure
the heat.


It is *not* impossible to measure the energy flux from sunlight.

Measuring
the total energy flux in the same location can be done quite easily. How

do
you think those tables of solar radiation are derived? Measure the energy
flux at the site, multiply by the total area of the cell that is normal to
the flux and you have 'the exact amount of fuel being consumed'. Measure
the output of the cell and the efficiency is trivially calculated.

Now, I admit that my theory that you have to measure the heat may be

flawed
since I have never measured efficiency of solar panels, and have never
talked to anyone that has, but as far as the way to measure efficiency,

I
know for a fact that you cannot use the amount of fuel available as the
determining factor of the efficiency since there is no possible way that
reflected energy can be consumed in the reaction. If you were to blow

air
into a turbine and measure the power produced, you could not take into
account the energy in the air that was not taken in by the turbine.

That
would give you a false reading of efficiency.


Wind turbine efficiency is based on the total energy available in the wind
per unit area and the area swept through by the turbine. This is very
similar to the method I suggest for measuring PV input.


Using solar insolence in the denominator would be similar to measuring

power
out of the motor divided by the fuel-flow-rate (to use your analogy).

That
method *does* make a lot of sense. It even accounts for fuel losses

through
evaporation or leakage (analogous to reflection from the surface of a

PV
cell)


Fuel losses are only applicable if they enter into the system in the

first
place. You are talking about reflected fuel, and that cannot have

entered
into the system at all.


You are defining the 'system' differently than I am. You define the
'system' as the cell just below the reflective surface. So by your
definition, reflected energy does not enter the system. But as you say,
that is very hard to calculate. I define the 'system' as the cell
*including* the surface. So by my definition, reflected energy is energy
that enters the system, then is reflected back out without much change
(except direction). Measuring the total energy that flows into my system

is
simply the solar flux times the area normal to said flux. Very
straight-forward measurement.

I will admit to some lack of knowledge of solar cells, but in my studies

of
fluid dynamics I have had many occasions to calculate efficiency of
different systems. We had to take special care not to calculate the

fuel
not used in the reaction.


I work in thermodynamics everyday too. If you have a situation where some
fuel is lost without entering the system, it is still a loss. From the
thermodynamics of the boiler's furnace, no it doesn't enter the burner so

it
isn't a loss there. But from the standpoint of total fuel needed to

create
a given output, unburned fuel losses are a factor. Why else do people get
excited about stopping fuel leaks?

Also, you cannot use the power available if you
don't have a system that is designed to consume that power. Therefore,
unused fuel must be removed from the equation. An example would be if

you
were to run your system on hydrogen. If you were pumping the hydrogen
through a turbine, but not burning it, you could not take the energy
potential in the hydrogen itself into account. If you were burning the
hydrogen, but not splitting the individual atom, you could not take the
atomic energy into account. There is so much energy in everything, but

if
that energy is not consumed in the reaction, you are calculating for
something that can never exist.


Not always as simple as that. If you use hydrogen to power the fuel pump
itself (such as in a space shuttle engine), then the total hydrogen
delivered to the main engine is *not* the same as what was delivered to

the
fuel pump. Some of it was burned in the fuel pump to power it so it could
deliver the rest to the main engine. When calculating the total

efficiency
of the system, one would look at total hydrogen consumed (in both the fuel
pump and main engine) and the total output of the main engine. The fuel
that doesn't make it to the main engine (consumed in the fuel pump) *is*

an
energy loss that is relevent and *lowers* the *efficiency* of the system.
But at the same time, since it greatly increases the fuel flow rate, it
greatly *raises* the power level (despite the drop in efficiency).

This is the same with solar energy. If you
are calculating efficiency for a fuel that has no possibility of giving

you
power because it doesn't enter into the system, you are calculating
efficiency for something that cannot exist.


But we are *not* 'calculating efficiency for a fuel that has no

possibility
of giving you power'. Reflective losses *are* under the control of the
designer. Energy that is reflected off the surface in one design, needn't
necessarily be reflected in another design. The theoretically perfectly
efficient panel would have no reflected energy at all. Second law doesn't
require that this energy *must* be unrecoverable, only limitations in

design
technology.

If you do not consider the reflective losses, an unscrupulous designer

could
coat his entire system in silver foil. Measuring the efficiency using

your
methods, it would be a very efficient PV cell. He/she could advertise PV
cell with efficiency in the high 80's perhaps 90's. But no one would be
happy with such a design.

The cell must be rated in electric-watts/area in some 'standard

condition'.
And the 'standard condition' is some value of solar flux. Any other

rating
is useless.


So, you're saying that the rating of efficiency is based on the efficiency
of the cell as a collector and as a converter. Consequently it is not a
true efficiency rating of the cell as a converter only. That leaves the
question of how efficient of a collector is a given cell.

On the other hand, I am still correct in saying that the actual fuel
consumed in the cell is converted 100% into heat and electricity.


  #37   Report Post  
Rod Speed
 
Posts: n/a
Default Ohmwork


"Timm Simpkins" wrote in message ...

"daestrom" wrote in message
...

"Timm Simpkins" wrote in message
...

"daestrom" wrote in message
...

snip

That is interesting, but frankly it seems like a poor way to state
efficiency for a solar collector. How does one determine the amount of
energy absorbed and not reflected in the collector?

Since solar insolence is readily available for an area, it would seem

more
proper to use useful energy out divided by solar energy insolent.

This
would account for energy reflected away from the collector as well as
energy
converted to non-usable forms (for example, heat in a PV cell).

Measuring efficiency like what you're saying, I could have a 1 m^2

cell
that
is 80% efficient produce less electricity than a 1 m^2 cell that is

only
15%
efficient. Just 'accidently' have a highly reflective coating on the
first
cell. Would run much cooler, and have higher 'efficiency', but not

very
useful. Makes the 'efficiency' number useless to compare cells.

Would
have
to revert to just electric output 'in full sun'. And *that* leaves a

lot
to
be desired (define 'full sun' for every location??)

When you buy solar cells, you don't purchase efficiency ratings, you
purchase wattage ratings. I don't agree with your assesment that you

can
lower the efficiency of a solar cell by reflecting the sun's rays since

the
heat increase in the solar panel should be lower as well and so the

ratio's
should be similar.


Only if the reflector is equally reflective for all bands of the spectrum.
Clearly, that is not the case. Just changing from a glass covering to one
of some plastic could change this.

That reflected energy is not taken into account in
calculations of efficiency. It can't be since it's almost impossible to
measure.


The reflected energy *should* be. It is not 'almost impossible to

measure'.
If you know the *total* insolent light energy, then efficiency is simply
(useful energy out) / (total energy in). And total insolent light energy

is
measurable with a black-body instrument, or simply looking it up from data
tables for your location.

You can measure temperature differentials and output. A solar
cell that puts out 12 volts at 6 amps is putting out 72 watts of power.

At
20% efficiency, that means for every 72 watts of power you loose 1440

watts
to heat. If you measure the heat difference in the solar panel and

convert
that to watts, you can compare that to the electrical output and that's
where you get your efficiency rating. If you're inhibiting the path of

the
sun, you aren't lowering the efficiency rating because you're

restricting
the solar panel's fuel. Fuel that doesn't get to the panel cannot be
consumed and cannot count toward the efficiency.


Only if you draw the system boundary as just under the reflective surface.
Much more practical to consider the entire cell, including the surface as
the system boundary. In that case, some energy enters the system and is
reflected back out again, while other energy is converted to heat and
finally some energy is converted to useful electric output. Since the
surface of the cell is certainly a *part* of the cell, its performance

(i.e.
its reflectivity) is a valid part of the cell's overall performance.

To use your previous analogy, if the fuel gets to the carbuerator, but
doesn't go into the cylinder, you have a less efficient engine. Claiming
that such a loss doesn't enter into the efficiency of the engine is silly.
Claiming that it 'is almost impossible to measure' is not a valid argument

.

snip

I don't, but I can give you plenty of links on how to measure

efficiency.
You are saying that just because there is a certain amount of energy in

a
given area of light that that energy is being consumed in the reaction.
That is not how efficiency ratings are made. You only measure the

potential
energy of the amount of fuel, in this case sunlight and divide that by

the
amount of energy output in the reaction.


Exactly. If the potential energy is the energy of the incoming sunlight,
and the output is the electricity produced, then you have a complete

measure
of the cell's ability to convert sunlight to electricity. This includes

the
losses due to reflection, heat production, and any others.

Since it's impossible to measure
the exact amount of fuel being consumed by a solar cell, you need to

measure
the heat.


It is *not* impossible to measure the energy flux from sunlight.

Measuring
the total energy flux in the same location can be done quite easily. How

do
you think those tables of solar radiation are derived? Measure the energy
flux at the site, multiply by the total area of the cell that is normal to
the flux and you have 'the exact amount of fuel being consumed'. Measure
the output of the cell and the efficiency is trivially calculated.

Now, I admit that my theory that you have to measure the heat may be

flawed
since I have never measured efficiency of solar panels, and have never
talked to anyone that has, but as far as the way to measure efficiency,

I
know for a fact that you cannot use the amount of fuel available as the
determining factor of the efficiency since there is no possible way that
reflected energy can be consumed in the reaction. If you were to blow

air
into a turbine and measure the power produced, you could not take into
account the energy in the air that was not taken in by the turbine.

That
would give you a false reading of efficiency.


Wind turbine efficiency is based on the total energy available in the wind
per unit area and the area swept through by the turbine. This is very
similar to the method I suggest for measuring PV input.


Using solar insolence in the denominator would be similar to measuring
power
out of the motor divided by the fuel-flow-rate (to use your analogy).
That
method *does* make a lot of sense. It even accounts for fuel losses
through
evaporation or leakage (analogous to reflection from the surface of a

PV
cell)

Fuel losses are only applicable if they enter into the system in the

first
place. You are talking about reflected fuel, and that cannot have

entered
into the system at all.


You are defining the 'system' differently than I am. You define the
'system' as the cell just below the reflective surface. So by your
definition, reflected energy does not enter the system. But as you say,
that is very hard to calculate. I define the 'system' as the cell
*including* the surface. So by my definition, reflected energy is energy
that enters the system, then is reflected back out without much change
(except direction). Measuring the total energy that flows into my system

is
simply the solar flux times the area normal to said flux. Very
straight-forward measurement.

I will admit to some lack of knowledge of solar cells, but in my studies

of
fluid dynamics I have had many occasions to calculate efficiency of
different systems. We had to take special care not to calculate the

fuel
not used in the reaction.


I work in thermodynamics everyday too. If you have a situation where some
fuel is lost without entering the system, it is still a loss. From the
thermodynamics of the boiler's furnace, no it doesn't enter the burner so

it
isn't a loss there. But from the standpoint of total fuel needed to

create
a given output, unburned fuel losses are a factor. Why else do people get
excited about stopping fuel leaks?

Also, you cannot use the power available if you
don't have a system that is designed to consume that power. Therefore,
unused fuel must be removed from the equation. An example would be if

you
were to run your system on hydrogen. If you were pumping the hydrogen
through a turbine, but not burning it, you could not take the energy
potential in the hydrogen itself into account. If you were burning the
hydrogen, but not splitting the individual atom, you could not take the
atomic energy into account. There is so much energy in everything, but

if
that energy is not consumed in the reaction, you are calculating for
something that can never exist.


Not always as simple as that. If you use hydrogen to power the fuel pump
itself (such as in a space shuttle engine), then the total hydrogen
delivered to the main engine is *not* the same as what was delivered to

the
fuel pump. Some of it was burned in the fuel pump to power it so it could
deliver the rest to the main engine. When calculating the total

efficiency
of the system, one would look at total hydrogen consumed (in both the fuel
pump and main engine) and the total output of the main engine. The fuel
that doesn't make it to the main engine (consumed in the fuel pump) *is*

an
energy loss that is relevent and *lowers* the *efficiency* of the system.
But at the same time, since it greatly increases the fuel flow rate, it
greatly *raises* the power level (despite the drop in efficiency).

This is the same with solar energy. If you
are calculating efficiency for a fuel that has no possibility of giving

you
power because it doesn't enter into the system, you are calculating
efficiency for something that cannot exist.


But we are *not* 'calculating efficiency for a fuel that has no

possibility
of giving you power'. Reflective losses *are* under the control of the
designer. Energy that is reflected off the surface in one design, needn't
necessarily be reflected in another design. The theoretically perfectly
efficient panel would have no reflected energy at all. Second law doesn't
require that this energy *must* be unrecoverable, only limitations in

design
technology.

If you do not consider the reflective losses, an unscrupulous designer

could
coat his entire system in silver foil. Measuring the efficiency using

your
methods, it would be a very efficient PV cell. He/she could advertise PV
cell with efficiency in the high 80's perhaps 90's. But no one would be
happy with such a design.

The cell must be rated in electric-watts/area in some 'standard

condition'.
And the 'standard condition' is some value of solar flux. Any other

rating
is useless.


So, you're saying that the rating of efficiency is based on the efficiency
of the cell as a collector and as a converter. Consequently it is not a
true efficiency rating of the cell as a converter only. That leaves the
question of how efficient of a collector is a given cell.


Mindless waffle.

On the other hand, I am still correct in saying that the actual fuel
consumed in the cell is converted 100% into heat and electricity.


Pity about how the industry actually measures the efficiency of PV systems.

Try telling the industry that they havent got a clue about how to
measure it. Dont be too surprised WHEN they laugh in your face


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