> It appears that you are lost for words again.
Even you should be able to bullshit 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 shit I deleted had any relevance what so ever TO HOW
THE INDUSTRY MEASURES THE EFFICIENCY OF PV SYSTEMS.
<reams of your puerile shit any 3 year old
could leave for dead flushed where it belongs>
Try harder, little boy. You might actually manage to fool someone, sometime.
> >
> >
> > > 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.
> >
> >
<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
> >
> <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.
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