# Re: Ohmwork - Page 6

Posted by Rod Speed on July 22, 2004, 7:18 pm

Nope. Not even possible.

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

Posted by Timm Simpkins on July 22, 2004, 8:06 pm

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.

Posted by daestrom on July 22, 2004, 8:57 pm

passive

electricity

complain

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??)

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

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

Posted by Timm Simpkins on July 22, 2004, 10:26 pm

measured,

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.

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.

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.

through

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.

Posted by Rod Speed on July 22, 2004, 11:10 pm

Irrelevant waffle.

He didnt even say that.

Not a clue.

Irrelevant to the original claim being discussed.

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

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.

Not a clue.

Whatever that waffle is supposed to mean.

Not even possible.

Not a clue.

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

Temperature of the PV system in spades.

Pathetic, really.

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.

More irrelevant waffle.

Irrelevant to the original claim being discussed.

Nope, he aint saying anything like that either.

Wrong again.

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

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

That is completely trivial to measure.

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

Pathetic really.

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

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

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.

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

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

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

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

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

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

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

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

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

Thanks for that completely superfluous proof that you have never

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 Re: Ohmwork Rod Speed 07-22-2004
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 Re: Ohmwork Rod Speed 07-22-2004
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