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Re: Ohmwork - Page 7

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Posted by Timm Simpkins on July 23, 2004, 1:03 am
 


surface.

converted

compared

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.


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.



Posted by Rod Speed on July 23, 2004, 2:21 am
 


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.


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.


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


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.


used.

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.



Posted by Timm Simpkins 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.



Posted by Rod Speed on July 23, 2004, 4:05 am
 


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.



Posted by Timm Simpkins on July 23, 2004, 7:32 am
 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 pissing 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.



chemistry.


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