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Re: Ohmwork - Page 8

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Posted by Anthony Matonak 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


Posted by Rod Speed on July 23, 2004, 10:50 am
 
Some fuckwit troll claiming to be
just the puerile shit thats always pouring
from the back of that fuckwit troll.



Posted by Rod Speed on July 23, 2004, 10:49 am
 


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


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.




Posted by daestrom on July 23, 2004, 3:43 pm
 
<snip>

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.


convert

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.


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.


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).


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



Posted by Timm Simpkins on July 23, 2004, 9:15 pm
 

measure'.

restricting

efficiency.

measure

Measuring

entered

efficiency

possibility

condition'.

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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| |--> Re: Ohmwork Fred B. McGalli...07-06-2004
|   `--> Re: Ohmwork Fred B. McGalli...07-07-2004
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| ---> Re: Ohmwork Anthony Matonak07-21-2004
| | |--> Re: Ohmwork Fred B. McGalli...07-21-2004
| ---> Re: Ohmwork nicksanspam07-22-2004
| | |--> Re: Ohmwork Fred B. McGalli...07-22-2004
| ---> Re: Ohmwork Timm Simpkins07-22-2004
| |   ---> Re: Ohmwork Timm Simpkins07-22-2004
| |     | ---> Re: Ohmwork Timm Simpkins07-23-2004
| |     |     ---> Re: Ohmwork Timm Simpkins07-23-2004
| |     |         ---> Re: Ohmwork Timm Simpkins07-23-2004
| |     |           ---> Re: Ohmwork Anthony Matonak07-23-2004
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