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Re: NEON- Power factor & Load Current
Tom U wrote:
>
> In my never-ending effort to better understand the functioning of
> transformers, I came up with several questions today. Any feedback would
> be great.
I will put it in a graphic, Tom Hope this helps.
First diagram gives you the uncorrected inductive load, like a
neontransformer is.
Here is the drawing:
V
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L
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____|_____
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V
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90 |
I _________|
C
___________
___________
V
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90 | 90
I _________|_________ I
C L
THE ABOVE WILL RESULT IN:
V=I (at the same time)
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Dear Tom and all,
If we add the second drawing to the first one the result we the third
drawing.
The vertical line called V is the vector for the voltage and the
horizontal
vesctor called I(L) is the inductive current.
In the first drawing you see that the current is 90 degrees behind the
volatge.
So the voltage and current are not in phase as the current is foating
90 degrees
behind the voltage.
Note: with conventional strayfield transformers it is only 60 degrees,
that the current is behind (later in time) the voltage but for
the example I give here that makes no difference, and 60 degerees I
cannot draw
in ASCCI, so the example is with 90 degerees.
Looking to the second drawing you can see that I produced a current line
I(C) being
a capacitive current. This line is as big (long) as I(L) but in the
opposite
direction. So I produced a current that is floating 90 degrees in front
of the
voltage. I did that simply by mounting (installing?) a capacitor with
the correct
value over the main terminals of the transformer.
If we add these two situations to each other like the + sign is telling
you it will
result, like the = sign tells you in a vector situation with only one
vertical line.
As you can see (there is only one vectorline) the current is now in
phase with the
voltage. In fact there are two lines, one for the voltage and another
one for the
current, but as these two line are perfectly over each other you only
can see one line.
This is the real and true situation as the current is present at the
same moment in time
as the voltage. Now there is no phase shift between current and voltage.
SO remeber Tom: inductive loads will result in a situation that there is
voltage present
but there is no current at that time. The current will be present later
in time.
Capacitive loads are acting the opposite way, so there is current
available but the
voltage is coming in later in time.
To compensate this situation complete we produce a capacitive current
(by installing a
capacitor across the mains) that has the same value as the inductive
one.
After doing that there is no (what we call) blind current, as current
and voltage are
in phase again. The above situation is the perfect one and is hard to
realise. There
always will be a slight inductive blind current, but that's not a big
deal.
When the current is only a few degrees behind the voltage we say that it
is okay.
Never overdo it, what I mean is never install a to big capacitor, so
that you produce a
capacitive blind current component as strange things can happen then. SO
we always
calculate to the inductive side of the compensation. (HELP!! TDI this
english is rubbish)
When current and voltage are in phase we say that the power factor
called cos phi is 1,
as cos zero degrees is 1 because there is no angle. If there is an angle
coming on
between the voltage and the current cos phi is not longer 1 as the
hypothenusa always is
longer then the base as there is an angle. Every compensation with a
power factor higher
then 0.9 is okay and fine as the blind current component is acceptable.
Be aware of the
fact that when cos phi is 0.9 there is still a phase shift between
voltage and current of
25 degrees.
In my example the drawing was made to a phase shift of 90 degrees and
when we compensate
and the phase shift is acceptable like the 25 degrees situation the
horizontal line is
changing in a vectorline to above coming close to the vertical voltage
line with an angle
of 25 degrees between it.
In that situation nearly no blind current is left over and all current
is used as real
(we call it the real or wattcurrent instaed of the VA current) power.
Uncompensated are using exactly two times the amount of current (and
therefore nergy)
compared to comensated ones.
I will end this mail giving you all an example: If your car is doing 40
miles a gallon
and suddenly the car is doing only 20 miles a gallon, (as there is a
little hole in your
tank) what will you do? Repair the tank or follow your natural "who
cares" instinct
thinking it is not a big deal as we can buy our petrol for so cheap
compared to the
price Europeans must pay for petrol overthere. (and what about the
environment)
I think I know the answer already so my question is: why do we think
otherwise when
we consume electricity.
Best regards from dirk a. boonstra
PS: Last remark about mercury. I visited my doctor last week asking him
if he could
test me about the merc level and he said yes I can. Here is an envelope,
come back
tomorrow with some hairs in the envelope and I will let these hairs
tested. Then I
asked my doctor, can you test my Hg level from my blood, you can have a
liter if you
need that. He said no, give me some hairs and I asked, can you do it by
operating me
looking to the inside of my body, and he said no, no, no, only a few
hairs dirk is
enough. So I will not do the test as I don't want to be complete bold
only for doing
the Hg test.
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