[Prev][Next][Index][Thread]
Re: NEON- Power factor & Load Current
dirk wrote:
>
> 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
> |
> |
> |
> |
> |
> |
> |
> | 90
> |_________ I
> L
>
> |
> |
> ____|_____
> |
> |
> |
>
> V
> |
> |
> |
> |
> |
> |
> |
> 90 |
> I _________|
> C
>
> ___________
> ___________
>
> V
> |
> |
> |
> |
> |
> |
> |
> 90 | 90
> I _________|_________ I
> C L
>
> THE ABOVE WILL RESULT IN:
>
> V=I (at the same time)
> |
> |
> |
> |
> |
> |
> |
> |
> |
>
> 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.
References: