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#129813 - 09/01/05 08:50 AM RMS  
PEdoubleNIZZLE  Offline
Joined: Jun 2004
Posts: 176
McKeesport, PA, USA
Ok, here's what I know about RMS voltage...
The RMS voltage is Vpk/sqrt2
my question is, where do you come up with the square root of 2 (im just beginning calculus, and haven't touched it ever in mathematics, so I'm assuming it's a complex thing.)

Also, let's say you have 3 phase, and put it thru a rectifier so you have pulsating DC, with 3 sine waves, 120 degrees between phases. How do I find RMS voltage with the 3 waves combined like that?

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#129814 - 09/01/05 11:09 AM Re: RMS  
winnie  Offline
Joined: Sep 2003
Posts: 649
boston, ma
RMS is short for _R_oot _M_ean _S_quare. It is a form of averaging.

I am sure that you are familiar with the ordinary 'average'; you take all of the values, add them up, and divide by the number of values. Say you have 10 people in a room. You take their ages, add them up, and divide by 10 to get the average age in the room.

You are just starting calculus, and one of the things that you will learn is the concept of a _continuous_ sum, and _continuous_ averages. An example would be the average temperature over the course of a day. You could go to your thermometer every hour, and make 24 temperature measurements, and then average these 24 values. This gives you an idea of what the average temperature is. But wait: the temperature is changing between each of these measurements...wouldn't it be better to take a measurement every half hour, add them up, and then divide by 48...and if that is better, then why not take a measurement every minute??? Every second.... You se where this is going: in calculus, you will learn techniques that let you take the _continuous_ average of a changing value.

So now we get to a sine function. What is the _average_ value of a sine function? Remember that it spends equal time + and -, so it turns out that the _continuous_ _average_ value of a sine wave is _zero_, nada, zilch. But those wiley mathematicians come up with a cute little trick: the root mean square average.

To find an RMS average, you take you list of values, _square_ each of the values, find the average of these squares, and then take the square root of the average. This is a sort of weighted average, were negative and positive values count the same, and larger values count more than smaller values. You can take the RMS average of a list of numbers, and you can also take the continuous RMS average of a function.

As you get deeper into calculus, you will take the RMS average of a sine function; it isn't too difficult, but there are a bunch of details, so I'll refer you back to your textbook for that detail. But I'll give you the answer for now: If you have a sine function with a peak value (or _amplitude_) of 1, then the RMS average value is 1/sqrt(2). This is where the sqrt(2) RMS to peak relationship comes from.

This relationship is specific to pure sine functions. For any other function, you get a different RMS to peak relation. This is why you need 'true RMS' meters for dealing with harmonics; the harmonic waveform is not a pure sine wave, so the sqrt(2) relation doesn't hold. A 'true RMS' meter actually does the Root Mean Square averaging to find the RMS voltage/current; cheaper meters my measure the peak voltage and then assume that the sqrt(2) relation holds.

To find the RMS voltage of rectified 3 phase, you 'simply' figure out what the voltage function looks like after rectification, and then use an integral to calculate the RMS average. You will be learning all of these techniques, and it really isn't too hard. The pulse DC after 3 phase rectification is a series of humps, each the _60_ degree band around the peak of the sine curve, at 6x the original AC frequency.

Finally, RMS measurements are particularly useful in electrical work because the power delivered to a resistive load scales as the square of the voltage. Thus the average of the square of the voltage tells you the _average_ voltage delivered to a resistive load. 100V DC and 100V AC RMS will both deliver the same number of watts to a resistive load.


#129815 - 09/01/05 11:30 AM Re: RMS  
Alan Belson  Offline
Joined: Mar 2005
Posts: 1,803
Mayenne N. France
A brilliant explanation. Enough to answer the question, and yet to whet the appetite to learn more.


edit for grammar.

[This message has been edited by Alan Belson (edited 09-01-2005).]

Wood work but can't!

#129816 - 09/01/05 02:16 PM Re: RMS  
John Crighton  Offline
Joined: May 2005
Posts: 177
Southern California
Jon, +1 for the great explanation.

If I may emphasize your last point: The whole reason for measuring AC voltage as an RMS value is to provide the simplest and best indication of the power it will deliver to a resistive load. The reason why the RMS calculation squares the voltage is that the power is equal to E^2/R. (See the square there?)

During the AC cycle, as the voltage rises from zero, the instantaneous power delivered to the load increases as the square of the voltage. The power falls to zero again at the next crossover point.

So, what happens during the negative half-cycle? The power rises again to the same peak value, because the square of a negative number is a positive number.

In fact, if you plot the square of a sine-wave voltage, you get another sine wave of twice the frequency! The lower peaks of the squared waveform "sit" on the zero line, and the upper peaks are at Epk^2, where Epk is the peak voltage.

Since the new double-frequency sine wave represents power delivered to the load, the average power is at the midpoint, which is (Epk^2)/2.

The voltage that represents that average power level is the square root of that level, which is sqrt((Epk^2)/2), which reduces to Epk/sqrt(2). Look familiar?

#129817 - 09/22/05 12:16 PM Re: RMS  
PEdoubleNIZZLE  Offline
Joined: Jun 2004
Posts: 176
McKeesport, PA, USA
Thanks Jon and John. I could never connect RMS with sqaring the wave before, but it seems simple enough.

Also, I think I was able to answer my own question about the rectified 3 phase. Since the waves would be superimposed on eachother, I took the RMS between the lowest voltage and peak voltage and added it to the lowest voltage . It took me a while to figure out, but on the bus ride home from school, I was messing with it a bit, trying it in a graphing calculator, and I came up with this:

Vrms = Vpk * {sin(π/Φ) + [1-sin(π/Φ)]/√2}

Where Vrms is the RMS DC voltage of a fully rectified symmetrical polyphase AC voltage, Vpk is the peak DC voltage, and phi is the number of symetrical phases.

Also, we shall call this equation f(Φ) so I don't have to write it all over again. (Just the part inside the curly braces)

lim f(Φ)=1

That means the more and more phases you have, the closer you will get the RMS voltage to the peak voltage, but you will never get it to be the exactly same.

However, my current need for any of this has kinda gone out the window, since single phase to 3 phase transformers don't exist (another one of my posts). It would help for my windmills tho (I think I just heard a bunch of electricians sighing in disgust :p )

[This message has been edited by PEdoubleNIZZLE (edited 09-22-2005).]Edited because my equation automatically turned into smilies

[This message has been edited by PEdoubleNIZZLE (edited 09-22-2005).]

#129818 - 10/06/05 12:31 AM Re: RMS  
PEdoubleNIZZLE  Offline
Joined: Jun 2004
Posts: 176
McKeesport, PA, USA
scratch that, the correct calculation should be:

Vrms = Vpk * {sin[(Φ+1)π/2Φ] + [1-sin[(Φ+1)π/2Φ]]/√2}

The first equation didn't work, i made a dummy error (made lots of them, I'm finally passing Calc 1 on my 4th try!)

#129819 - 02/27/06 02:53 PM Re: RMS  
RODALCO  Offline
Joined: Dec 2005
Posts: 854
Titirangi, Akld, New Zealand
a RMS voltage will give the same amount of heating in a resistor as a DC voltage.

Regards Ray

The product of rotation, excitation and flux produces electricty.

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