## Free Energy - from Tom Napier, Mike C and Boyd.

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• People, sorry if people are tired of hearing about this; but the following post has some worthy science. Dear Boyd, Your amended Thermo-dyne write-up, and your
Message 1 of 1 , Aug 2, 1999
People,

post has some worthy science.

Dear Boyd,

eventually worked their way down the pipeline to me.

As I suspect others have already pointed out, the drinking bird works
only so long as the humidity of the surrounding air is less than 100%. In
this case evaporating water produces enough local cooling to drive the
bird. There is energy available because the water is not in equilibrium
with its surroundings. If you put the bird in a sealed container it will
stop working as soon as the air can no longer accept more water vapor.
The Second Law is still intact.

I earlier sent you an analysis of your Thermo-dyne based on
thermodynamic theory and also a practical analysis of its heating and
cooling cycles based on the temperatures you quote. Both analyses show
that the Thermo-dyne will not and cannot work.

One advantage of the college courses which you malign is that after
taking one you know where the numbers in the tables of refrigerant
properties come from and what they mean.

Let me explain things for the third time. The theoretical maximum
thermodynamic efficiency of a heat engine and the theoretical maximum CoP
of a heat pump are statements about input and output energies. They are
not functions of which gas you use. (They are also pure numbers so your
reference to "CoPhp" is meaningless.)

The CoP is the reciprocal of the efficiency only in this ideal case.
Only an ideal, and physically unrealizable, device can achieve these
maximum efficiencies and CoPs. Losses reduce both the CoP and the
efficiency, making their product always less than unity. Since the
theoretical CoP is not a function of the gas, only of the input and output
temperatures, there's no way you can gain net energy by using different
gases in the heat pump and the heat engine.

However, in a practical device how close you can come to the
theoretical figure does depend on how efficiently you can move heat energy
around the system. This is easier when you use the latent heat of
vaporization and condensation to move heat energy from place to place.
Thus the CoP attainable in practice is a function of where in the
temperature and pressure range a gas liquefies. An ideal gas system,
although just as efficient in theory, would be less efficient in practice.

People use different gases to match the properties of the gas to the
application. For example, a compressed gas won't liquefy if the condenser
is hotter than its critical temperature. On the other hand, a gas with a
low boiling point will generate too high a pressure if used in a heat
engine. And so on.

The different CoP figures in your tables probably refer to different
temperature ranges. From the calculation you do it looks as if the CoPs
quoted are the theoretical ones. I would have thought that quoting the
CoPs attainable in practice would have been more useful since any
scientist knows how to calculate the theoretical figures. (Though I grant
you that an engineer might find it easier to look up a table.) Where a
gas is quoted as having a low CoP you will probably find that it is being
specified over a higher temperature range.

Of course you can always reduce the CoP of a heat pump by increasing
the losses. By your reasoning this would increase the efficiency of a
heat engine using the same lossy system. I hope it is obvious that this
is utter nonsense.

I'm sorry but there really is no Santa Claus. No matter how often you
ask the question or how you phrase it the answer remains the same. You
can't get around the Second Law.

Tom Napier August 1, 1999

--

Eric Krieg eric@...

http://www.phact.org/e/more.htm

PS: for anyone still with us - the following are some thoughts from
Mike Carpenter:

----------------------------------------------------------------------------
-----------------------------------------------

BTW, I'd like to point out that the reciprocal of 1 is 1, and of course,
100% is the same thing as 1. Although that wont help you any. Examining
Bob's math....I find no flaws.

I found Bob's point about where do you draw the line for a gas of low COP
to be a very good one. Obviously a vacuum wont do in a refrigerator coil.
His point is that your theory is flawed.

I'm sure you know that division by 0 is undefined. Therefore I find it
absolutely necessary to consider your even an accidental mention of Celcius
scale to be a gross conceptual error, not mearly a typo or slip up.

No matter what gas(s) you decide to use, I think you will find that it will
take more energy to compress the gas(s) than any energy you can get out of
the gas(s). No matter what efficiencies are delivered, if you hook up a
heat engine to a heat pump, the net will be a loss. If you insist

Did you notice that when you joined this list it was specificly noted that
it would be a low traffic list? Why does everybody complain when the
moderator (Eric in this case) points out that your post frequency is too high?

I'm posting this in the background via private email to avoid excessive
list traffic. This post only concerns you, Bob, Eric and me now since I'm
butting in. It's not meant to be behind anybody's back. I give anybody
permission to post anything I write publicly.

I think you will find that most people are not purposely posting behind
other people's back. Most of the offended members are just WAITING to find
something to be mad about. I dont know Eric from Hitler, but he does make
very sharply and accurately pointed points.

I post to people I want to talk to. You are one of them. I think almost
every one does the same. I believe your anger is unfounded, or your anger
is pointed in the wrong direction. You may be upset because you have
struggled to find the answer to a puzzle that 1000s of people have tried to
solve. Many have claimed to have the answer to this puzzle. I believe
the problem is the puzzle itself, and not the answer to the puzzle. When
you find that the real answer is that the puzzle is flawed, you will no
longer seek the answer to the puzzle.

There is no free energy.

When you can define the source of the energy, you will find that it is not
free.

Mike Carpenter

========== and lastly, some more from Boyd:

PS: I close with the following from Boyd:

7-31-99
Eric,
.
>From Boyd to all members:
There are some things that I want to straighten out.

#1
Eric finds a great difference in being asked to stop posting so often and
being told to stop posting. I don't want it to go down as an untruth. He
did not tell me to actually stop posting forever. He asked me to not post
so often. It was only after that last letter that I recieved the
unsubscribe notice.

#2
About the movie film analogy that I wrote. It was more for Bobs benefit
than anyone else so I wrote it in such a way that he could use his quanitys
version of the Carnot equations. ( his Q2-Q1 type stuff ). But I forgot
to say it. So don't try to use the Temperature version of the Carnot
equations. It would result in the same answer for all gasses wheather they
be real or ideal. It's only the quanity equations that go along with each
different gas.

#3
Yesterday I got a letter from a Mike Carpender telling me that Bobs math
was correct in that letter where he gave a vacuum to use as a refrigerant
so this morning I went back and read it. Well you are right Bob. You got
the correct answer. Here is what happened, When I first read it and saw
that you gave me a vacuum for a refrigerant I didn't read closley anymore.
I just instantly jumped to the conclusion that (we ain't going to get
nothing out of a vacuum ) so I began telling you about the work of
compression.

Well today I went back and saw that you did indeed compress a vacuum and
convert your work into heat. I don't know how you did that but your math
is indeed correct. Does it not prove my point? You came out with an
answer of 1 and The reciprocal of that is unity. But then you ask ( where
did it break down ? ) I don't know of a break down. If you run the film
backwards you see that all of the heat was converted back into work.

This morning I got your letter saying that someone wrote to you and told
you of the difference between the COP of a refrigerator and the COP hp of a
heat pump. I swear I thought I posted that. Wait let me look.

Okay yes it's in digest 200 of the free_energy list. In fact it's the
letter where I forgot to say that you must use your Carnot Quanity