[Date Prev][Date Next][Thread Prev][Thread Next][Date Index][Thread Index]
Re: Average, RMS and Power Factor made easy!
Original poster: "Albert Hassick by way of Terry Fritz <twftesla-at-uswest-dot-net>" <uncadoc-at-juno-dot-com>
Hi members. But does not R.M.S. power have significance when used as a
means of rating various electrical components such as loudspeakers and
power amplifiers? And would not these same ideas apply to a Tesla coil?
After all, a loudspeaker is a coil within a magnetic field that becomes
excited when a voltage is applied to it, just like a Tesla coil. And we
all know how terrible our sound equipment begins to sound once we have
exceeded the R.M.S. value of the rated components. AL.
On Mon, 08 Jan 2001 22:53:14 -0700 "Tesla list" <tesla-at-pupman-dot-com>
writes:
> Original poster: "by way of Terry Fritz <twftesla-at-uswest-dot-net>"
> <free0076-at-flinders.edu.au>
>
>
> 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..
>
>
>
________________________________________________________________
GET INTERNET ACCESS FROM JUNO!
Juno offers FREE or PREMIUM Internet access for less!
Join Juno today! For your FREE software, visit:
http://dl.www.juno-dot-com/get/tagj.