# Ohmwork

Page 2 of 2
• posted on July 22, 2004, 7:18 pm

message wrote

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.
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
• posted on July 22, 2004, 8:06 pm
wrote

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.
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
• posted on July 22, 2004, 8:57 pm
wrote

In
Rod,
passive
electricity
energy
complain
rays
so
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
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
• posted on July 22, 2004, 10:26 pm
wrote

cells?
to
the
heat
measured,
would
to
a
energy
that
15%
first
have
to
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.

cell
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.

cell
you
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.

power
That
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.
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
• posted on July 22, 2004, 11:10 pm
wrote

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.

Nope.
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.

Nope.
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.
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
• posted on July 23, 2004, 1:03 am
wrote

surface.
30%.
12%
now
converted
to
the
the
the
compared
used
cell).
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.

No you don't have a clue.

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

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.

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.

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.

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.

get to

cooled.
discussed.
"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.

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 >>>

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 >>>

efficiency.
reaction.
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.

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.

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.

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.

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.

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

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

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.

measured.
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.
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
• posted on July 23, 2004, 2:21 am
wrote

Nothing like that bit you claimed you didnt agree with.

Nothing like that bit you claimed you didnt agree with.

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.

Duh.
Nope, he aint saying anything like that.

Having fun thrashing that straw man are you ?

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

Pathetic, really.

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

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

You were talking about a HEAT DIFFERENCE.

Nothing like your previous pig ignorant claim.

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

Corse its true.

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.

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

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

Nope, just flaunting your complete pig ignorance of 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.

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

Fraid not.

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.

And you cant with the PV SYSTEMS BEING DISCUSSED ANYWAY.

Concentrate on the word CONSUMED.

Separate issue entirely.

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.

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

No if about it.

Not even possible.

Nothing like a completely closed mind eh ?

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

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

Yes cur!!

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

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

And that is the way THE ENTIRE INDUSTRY DOES IT.

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.

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.

'consumed' isnt even relevant.

Power isnt even being created.

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

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

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

That is precisely what I have been doing, stupid.

Pathetic, really.

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

Pathetic, really.

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.

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 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.

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.

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.

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.

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.

Yep, all your crap about fuel and engines and showers is that in spades.
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
• posted on July 23, 2004, 3:32 am
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.
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
• posted on July 23, 2004, 4:05 am
message

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.

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.

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

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.

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

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.

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.

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.

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.

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

No thanks, has no relevance what so ever to 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
Keep digging, child. You'll be out in china any day now. Best take your passport.
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
• posted on July 23, 2004, 8:42 am
Timm Simpkins wrote: ...

...
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.aspx/Computers/Usenet/FAQs / http://www.usenet-replayer.com/faq/misc.consumers.frugal-living.html
Anthony
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
• posted on July 23, 2004, 10:50 am
Some fuckwit troll claiming to be
message just the puerile shit thats always pouring from the back of that fuckwit troll.
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
• posted on July 23, 2004, 10:49 am
message

<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
• posted on July 23, 2004, 3:43 pm
<snip>

more
lot
the
ratio's
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.

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.

At
watts
convert
the
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>

potential
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.

measure
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.

flawed
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.

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.

of
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?

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).

you
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
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
• posted on July 23, 2004, 9:15 pm

This
cell
only
very
Would
can
measure'.
is
restricting
(i.e.
.
efficiency.
a
the
measure
the
Measuring
do
I
air
That
PV
first
entered
is
fuel
it
create
you
if
the
efficiency
an
possibility
design
could
your
condition'.
rating
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.
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
• posted on July 23, 2004, 11:22 pm

Mindless waffle.

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
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
• posted on July 22, 2004, 9:18 pm
wrote

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.

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.

Wrong again.

Not a clue, as always.
<reams of your pig ignorant silly shit flushed where it belongs>
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
• posted on July 22, 2004, 10:32 pm
wrote

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.