Re: Free energy machine? I have such a headache.

Well you basily settled my argument.
My Boss who is well educated and smart as a whip in some way's is arguing
that the power losses are not acumlative becuase it's a parelell circut.

That is obviusly false. So he is correct that like you are that the
percentage is not and I cleary see that now.
But my main point being you can't get back the loss
of enegery through the volatge drop across each branch.
That must be added up.

He is also wrong as he stated
that there would be a 50% current measurment error
at some points becuase of the chopped waveform.

This would be true if my NON True RMS current sensors were installed in
line with the controllers and load but they are not. They take the
average. . They are at the power source. The voltage is not copped there ,
it's constant.

All indications are that the possible error is 10% across the board at any
load level. Subtract the 2% total loss and we are down to 8% error in
curent rerading. That's a simple correction in math to remove the 8% error
that the system is reporting. Of course it's yet to be determined if the 8%
is real or not. As Gregg stated we can take voltage measurement across the
load with a True RMS Volt meter. We know the resistance of each bank so we
can get close. For our purposes 5% accuracy would be just dandy.


"Androcles" <Headmaster@xxxxxxxxxxxxxxxxxx> wrote in message
news:YdMYm.106120$lP6.103448@xxxxxxxxxxxxxxxx

You can lose power, you cannot "lose" current.
Only resistive loads require power, reactive loads
(capacitors and inductors) involve no power loss
yet pass current.
The ideal motor, generator or transformer will be
superconducting rather than just copper, and copper
is used in preference to other metals because its
resistance is low. The lower the better, you don't
want to waste heat.
So how do you have current without power?
That's what puzzles most people, but it's really quite
simple.
http://www.androcles01.pwp.blueyonder.co.uk/AC/AC.htm









"Tom" <txgiorgi@xxxxxxxxxxxx> wrote in message
news:hh00n3$uul$1@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx
So 1 amp loss each leg does = 5 amps but percentage is not.
I think I get it. Nice way of explaining it.


"Androcles" <Headmaster@xxxxxxxxxxxxxxxxxx> wrote in message
news:LwLYm.67282$IZ1.20243@xxxxxxxxxxxxxxxx

"Nospam" <Nospam@xxxxxxxxxx> wrote in message
news:hgvtia$45p$1@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx
How simple is This? Or is it no so simple? I have 5 phase angle fired
Controller in line with 5 heating coils. 3 are on one side of a 240V
line and 2 are the other half so we have have 5 load circuts total
carrying 120 AC. Each controller has a max efficency of 98%. I am
being told by that the loss being 2% on each device is not acumaltive.
So here I am scratching my head.
How was each device suddenly turned into a more effiecent one? How was
energy created? I am just a high school drop out, so I guess I know
nothing.

Exuse me but 2% loss X 5 Devices and 5 loads = 10% loss. I know this
is to simple for you guys but for love of god can some please tell me I
am not crazy? Much appreacited. If I am wrong that;s fine, I would
just like to understand why I am wrong? If I am correct you don't want
to know who is telling me I am wrong.

Excuse me but by your illogic 50% loss x 5 devices and 5 loads = 250%
loss,
more than 100%.
Or if you prefer, 2% loss * 100 devices = 200% loss, again more than
100%.

If you had £100 and lost £2 your loss would be 2%.
If you lost £2 every week of the year your loss would be £104
or 104% (of £100), but if your income was £100 a week your
loss would only be 2% of £5200.

Cost of running 1 device = $1
2% loss if $1 is 2 cents.
Cost of running 5 devices = $5
10% loss of $1 is 10 cents.
2% loss of $5 is 10 cents.

So you lose 10 cents, but you operated 5 devices, not 1.

The devices are cumulative, the losses are cumulative,
the percentage is not.









.

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