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Re: Average, RMS and Power Factor made easy!



Original poster: "Terry Fritz" <twftesla-at-uswest-dot-net>

Hi Darren,

Many thanks for this fine explanation!  I think the area many coiler's have
trouble with is that Tesla coil's produce some pretty nasty power problems
for the "student".  I whipped up a MicroSim analysis of power to an output
streamer over a single cycle at:

http://hot-streamer-dot-com/temp/PowerCalc.gif

To the layman, trying to find the delivered power to the streamer from
these waveforms and the Radio Shack booklet that came with the meter (sorry
Ted ;-)), you are simply doomed...  Toto, you aren't in Kansas anymore...

The RMS voltage to the streamer is 12772 volts while the peak is ~300kV.
The RMS current is 28.973 mA RMS while the peak is ~500mA.  Simply
multiplying the RMS voltage and current gives a streamer power of 370
watts.  However, the true delivered average power is actually 237.75 watts.
 The problem is the voltage and current are not in sync and the calculated
resistance (v/i) is varying wildly at each instant of time.  At the voltage
peak, the current is negative so at that point the streamer is actually
sending power back to the coil!!  (Yes! that does happen since it is
reactive!!)

So, to make a long story short, this is really darn complex stuff!!  Those
P = V x I  DC equations and  P = V x I x COS(theta)  AC steady state sine
equations are just simplified forms of the nasty  P = integral (0,T) (v(t)
x i(t) dt)  power equation needed here.  In the case of Tesla coils, the
v(t) and I(t) are nasty pulse functions that barely have closed forms
(needs tricks).  Computers grind the numbers with ease which is the ONLY
way to figure this stuff out...

Even us with the "EE" brand need some refreshing when it gets to this
level.  So don't anyone feel bad if they don't get this stuff.  It is nasty
business and trying to do this stuff by actual measurements gets even worse
unless you have some big $$$ "stuff" to do it all for you.

Rest assured that no one needs to know any of this to build a fine Tesla
coil.  Us "big doggy" theoretical folks really get into this stuff because
that's the way we are (math teacher hit us in head with ruler too many
times :-)).  Occasionally, we find some wonderful thing that ends up
helping everybody.  However, 99% of it is math test with little practical
use.  However, when the going gets rough, we do seem to shine :-)))

One really needs to crack open those college level electrical engineering
books with the calculus in them to "get" this stuff or have a good teacher
to explain it.  My sister has a Ph.d in math so I was set (although, I
taught her everything she knows, so how did she end up smarter than me.  :-)).

Cheers,

	Terry (who hasn't rambled in long time :-))

PS - Some people don't like the "over unity" or "free energy" threads.
However, after the initial "jolt", they really seem to bring out some darn
good posts... 



At 12:53 PM 1/9/2001 +1030, you wrote:
>
>Many people seem to have trouble with the ideas of RMS current, RMS
>voltage, average power and Power Factor. I will take some time here to
>explain what it is all about, and hopefully any remaining doubts will be
>dispelled forever and we can get back to coiling (one of the finer points
>of life! =)
>
>
>RMS stands for Root Mean Squared, and it can be
>applied to anything at all. It will only be useful in electronics to talk
>about RMS current and voltages however and average power, however. RMS
>energy doesn't make much sense but there would be applications where you
>might want the average energy per pulse, for example, and you might choose
>RMS rather than average. But I very much doubt that anybody on this list
>has a use for talking about any RMS measurements that aren't voltage or
>current.
>
>The method of finding RMS is as follows. Say I have five numbers, I can
>find the mean quite easily. Just add them and divide by five. Now lets say
>I want to find the mean squared value. I first square all 5 numbers, then
>I find out what the mean of the new list is. So I have the Mean Squared
>value of all 5 numbers. Now lets say that I want to square-root the mean
>squared value, I end up with the Root Mean Squared value, or RMS value (of
>the 5 numbers). If the mean of the values is equal to 0, ie some numbers
>are positive, some negative, and they add up to 0, then the RMS value
>equals the standard deviation as found in statistics.
>
>Talking about a list of numbers is fine, but doesn't really apply to a
>continuous function of time, such as voltage or current. I will get to
>that as soon as I explain why we even use RMS.
>
>Lets consider an AC waveform. Assume that it is periodic, like the mains
>voltage. If I connect a heater and turn it on, a certain power is
>dissipated. If I then use a DC supply and find the voltage that produces
>exactly the same average heating (and at 50/60Hz you wouldn't notice that
>the power is fluctuating) then I can say that the mains waveform has the
>same heating effect as that voltage, and that's the basis of why we use
>RMS. If we consider only the _average_ power dissipation, we can quote it
>as say 2400W.
>
>So how do we find the DC equivalent voltage? Lets say that I have a
>resistance hooked up to a voltage or a current source (could be AC or DC).
>Then the equation for instantaneous power is (i.e. at any instant in the
>cycle):
>
>P = i^2 x R   or   P = v^2 / R
>
>Note that the instantaneous power depends on the current or voltage
>squared. If we took an average of the power, since we want average power
>of course, then we would get either the Mean Squared value of current,
>multiplied by the resistance, or the Mean Squared value of voltage, then
>divide by the resistance. Either way we get the average power. In a DC
>system the Mean Squared value is just the DC value squared, since it never
>changes. So we have a way of finding the DC equivalent. If instead of
>finding the Mean Squared value of voltage or current and comparing it to
>the DC value squared, we can just square root the whole thing. So we see
>that the DC equivalent of either a voltage or current waveform is just the
>RMS value of that waveform.
>
>Now it should be clear why we talk about RMS voltage or current, we can
>use it in Ohm's law, we can use it to find power, we can use it where we
>like as if it were a DC circuit. But only if we are talking about
>resistors. The danger is that many people don't know that when you
>multiply RMS voltage by RMS current you _only_ get the average power _if_
>the voltage and current waveforms are proportional to each other at each
>instant in time. In other words, the two waveforms must be exactly in
>phase, like in the case of a resistor where V = I x R shows that at every
>instant in time V and I must be proportional by the constant R.
>
>If you have inductors and capacitors the current and voltage are not
>necessarily in phase, one signal lags behind the other. Then the actual
>average power will be lower than the value you get by multiplying RMS
>current by RMS voltage. It turns out that the formula for real power in an
>AC circuit is:
>
>P = V I cos(theta)                 <--- theta is often replaced with phi
>
>The extra cos(theta) bit is called the power factor, or PF. It can vary
>between 1 and -1, where 1 is called unity power factor and corresponds to
>a purely resistive load. Why -1? If we were talking about a source of
>energy like the wall socket, we might like to talk about a negative power
>like -2400W in the heater example, because it is delivering power. The
>cos(theta) term can handle that too, since a negative answer comes out if
>you use an angle between 90 and 270 degrees. By the way, the angle
>represents the phase difference. Just imagine the whole cycle being 360
>degrees, then a quarter of a cycle out of phase (like a capacitor or
>inductor) would give 90 degrees, or a PF of 0. Note that in that case NO
>power flows (if the capacitor or inductor were perfect with no
>resistance). Even though we have our regular voltage present across the
>thing and a measurable current through it, NO POWER FLOWS!!
>
>Note in the above example you might still get charged for the power you
>aren't using... The metering systems that I have actually seen in my own
>area won't compensate for the power factor, and I doubt that yours does
>either, they just measure the current and charge you for it. So if you
>draw 10A and your capacitor isn't getting warm you probably are still
>paying for it.
>
>That's what PF correction is all about. If you run your coil at a PF
>that's less than unity, you draw more current than you need to. In the
>case of a coil, it's usually an inductive load so the answer is to place a
>capacitor in parallel with the circuit and keep changing the capacitance
>until the capacitance best cancels the inductance in your coil (mainly
>transformer inductance I suspect). The result is less current drawn from
>the mains and maybe a little lower bill. It may be that you can actually
>get more power into the coil if you were close to blowing a fuse, since
>the current drawn has gone down. It's as if the capacitor supplies the
>extra current needed by the inductance of the coil circuit. Remember that
>the power factor of the current delivered by the capacitor to the coil is
>close to zero and it doesn't actually supply any power, just the current
>that the inductance was planning to drain from the wall.
>
>
>I hope that many people had a good read and maybe some fun too =)
>
>
>Darren Freeman
>
>
>PS I'm studying Electrical/Electronic Engineering and everything I've said
>I'm 100% confident of..
>