What is the r.m.s. value of a current? 10

[davrom]

I responded to a question of this forum (cable selection based on duty cycle) where I afirmed that the r.m.s. value is similar to the thermal equivalent current. I realized this may be a stupidity (in fact, I am pretty sure).

I analyzed the problem and now I believe that the r.m.s. value of a current is the so called "effective" value of the current (for ac currents).

I.e. for an ac current i(t) = I*sin(omega*t - fi), the effective value is sqrt(2)*I, where I is the amplitude.

Can anyone explain me what the r.m.s. value is?

 

[jbartos]

Comment on the three previous postings: It appears that there is a lack of understanding of Iav and Irms with respect to physical phenomena, e.g. meters principles of operation, heating. Visit

http://cybertron.vlsi.uwindsor.ca/85-124/Docs/85-124-Lab-2.pdf

for:
""Average Responding Type RMS Meter
An average responding type RMS analog meter is designed for sinewave use only and rectifies (full-wave rectification) the input sinewave to form a DC waveform with an average value of 0.636 VMAX. The meter movement responds to the average DC value. An internal multiplying scale factor of 1.11 (=0.707/0.636) is used to obtain the correct RMS value.""

Where the meter without adjustment to rms would be indicating Iav, therefore, there would be smaller amount of heat calculated or measured over voltage and current. This is a basic principle between sinusoidal rms and average.
I am surprised that such basics of electrical engineering are posing so big problems to so many. Perhaps, an additional coursework at University, e.g. Windsor (originator of the above link content), would be beneficial.

 

[jbartos]

Suggestion to stevenal (Electrical) Dec 12, 2003 marked ///\\Jbart,
Try the square root of the area under the square of the sinusoid.
///O.K. This yields Irms=0.707xImax, and Vrms=0.707xVmax
Prms=Vrms x Irms=0.707xImax x 0.707xVmax=(1/2) x Imax x Vmax
\\ Wutz a wormer heater?
///RMS figures or values of the sinusoidal wave are bigger than Average values of rectified sinusoid. The most practical way to obtain the dc equivalent of the sinusoidal waveform is over the sinusoid full wave rectification, i.e. by absolute value of |sinwt|.\\ Sounds nasty.
///Not necessarily, if a proper education is in the place.\\\

 

[jbartos]

Suggestion to skogsgurra (Electrical) Dec 13, 2003 marked ///\\\
jb,
I can see that you are busy convincing the engineering community that they have been using the wrong definition of RMS for about one hundred years.
///To the contrary, the rms has been used very shrewdly, especially by the Utility, since the ac values for power show 19% higher values than DC values for power. What this means is that the Utility would lose 19% of profits by supplying customers DC electricity. I would love to get 19% more heat energy for the same money; especially in cold weather.\\ Would it not be worth while to sit down and think.
///About what?\\ You might find that you and Don Quixote are in the same business.
///I have clearly been seeing what is going on for quite a some time, have you? If not, try to figure out why the meter has to be adjusted in the following link:

http://cybertron.vlsi.uwindsor.ca/85-124/Docs/85-124-Lab-2.pdf

\\\

 

[jbartos]

Suggestion to jghrist (Electrical) Dec 14, 2003 marked ////\\\jb wrote:
"(quoting me)'In the case of current through the heater, W = I²•R.'
///Not clear what you mean by I. Do you mean Idc=Iav or Irms?\\\"
I mean that the instantaneous power is equal to the instantaneous current squared time the resistance. It doesn't matter if the current is dc, sinusoidal, exponential, square wave, or worm shaped.
////True.\\\ To get the energy, you integrate the instantaneous power over whatever time period you want.
////True.\\\ The result is the integral of I²·dt over that time period.
////True. However, for sinusoidal waveforms:
Prms x t = kWhr = (1/2) Vmax x Imax x t = (0.707 x Vmax) x (0.707 x Imax) x t = (0.5) x Vmax x Imax x t = Vrms x Irms x t

For absolute value of sinusoid, |sinwt| averaged to dc:

Pav x t = kWhr = (2/pi) x Vmax x (2/pi) x Imax x t = 0.64 x Vmax x 0.64 x Imax x t = (0.405) x Vmax x Imax x t = Vav x Iav x t

The 19% is coming from (1-0.405/0.5)x100%=19%.\\\

 

[jghrist]

jbartos wrote:

Where the meter without adjustment to rms would be indicating Iav, therefore, there would be smaller amount of heat calculated or measured over voltage and current.

Exactly the point. The meter without adjustment would indicate a value that would calculate the wrong amount of heat.

jbartos writes a lot about rectified sinewaves. The rms value of a full-wave rectified sinewave is the same as the rms value of unrectified ac. Since power equals I²·R, to get the average power, you have to average I², not I. The average of unrectified sinusoidal current squared equals the average of full-wave rectified current squared. This comes from the well-known fact that

(+N)² = (-N)² = |N|² where N is any number.

 

[jbartos]

Suggestions to the previous posting marked ///\\jbartos wrote:
Where the meter without adjustment to rms would be indicating Iav, therefore, there would be smaller amount of heat calculated or measured over voltage and current.

Exactly the point. The meter without adjustment would indicate a value that would calculate the wrong amount of heat.
///Finally, got it.\\jbartos writes a lot about rectified sinewaves. The rms value of a full-wave rectified sinewave is the same as the rms value of unrectified ac.
///Exactly true, never denied.\\ Since power equals I²•R, to get the average power, you have to average I², not I.
///Yes, you can average I^2, however, the result is Irms^2 as I showed in my previous posting, not Iav^2. Therefore, there is a DC average and AC average (often expressed in in rms values. See the reference below.\\ The average of unrectified sinusoidal current squared equals the average of full-wave rectified current squared. This comes from the well-known fact that

(+N)² = (-N)² = |N|² where N is any number.
///True, never denied. However, the average of the |Imax x sinwt| is Iav=0.64 x Imax, which is different from Irms=0.707 x Imax
There have been several postings on this issue within past four years in this Forum.

The following reference is available:
Ned Mohan, Tore M. Undeland, William P. Robbins, “Power Electronics, Converters, Applications, and Design,” Third Edition, John Wiley & Sons, Inc., 2003.

Page 14 Section 1-7
The uppercase symbols V and I refer to their values computed from their instantaneous waveforms. They generally refer to an average value in dc quantities and root-mean-square (rms) value in ac quantities.

Page 382 Section 13-5-1 Form Factor = Ia(rms)/Ia(average)..Equation (13-15)
The form factor will be unity only if Ia is a pure dc. The more Ia deviates from a pure dc, the higher will be the value of the form factor. The power input to the motor (and hence the power output) varies proportionally with the average value of Ia whereas the losses in the resistance of the armature winding depend on Ia(rms)^2. Therefore, the higher the form factor of the armature current, the higher the losses in the motor (i.e. higher heating) and, hence, the lower the motor efficiency.

The above paragraph implies that the dc value derived from sinusoidal waveform will produce less heat. This means that the dc value has to be adjusted to match the ac rms value to produce the same value of heat. Therefore, the rms value is somewhat inflated value in comparison with the dc value. I.e. the Utility is selling 19% more energy over the rms than it would be sold over the dc values.

Please, would you kindly support your statements with references, e.g. textbooks, papers, websites, handbooks, etc. to add more credibility to your statements.\\


\\

 

[jghrist]

jbartos' reference page 382 is discussing dc motors. I will grant him that if he is paying for dc service from his utility, he is getting cheated if the utility is providing a lot of ripple and charging him for the rms value of the current. Note that jbartos' reference states the losses in the motor (i.e. heating) depend on Ia(rms)^2. This is just what everyone else has been trying to convince him of; the rms value of current is the correct value to use when calculating heat produced.

 

[Skogsgurra]

Do not expect that some of the participant(s) in this discussion will ever accept facts. The only way of ending this thread is probably not to comment on his postings any more. It has once again gone much too far.

 

[jbartos]

Comment: I learned what I posted more than 40 years ago. It was included in texbook and labs demonstrated it. Some readily available references on web have been posted above. I know what I know, do you?

Comment to jghrist (Electrical) Dec 20, 2003 marked ///\\jbartos' reference page 382 is discussing dc motors.
///Actually, the paragraph is discussing application of the Form Factor that has been downplayed in the Standard Handbook for Electrical Engineers, and some other electrical engineering publications. By looking at the above posting, it appears that the Irms=10A would produce the same heat, electrical energy, electrical power as Idc=10A. This is essentially what has been discussed as incorrect, at least in my posting.\\\
I will grant him that if he is paying for dc service from his utility, he is getting cheated if the utility is providing a lot of ripple and charging him for the rms value of the current. Note that jbartos' reference states the losses in the motor (i.e. heating) depend on Ia(rms)^2. This is just what everyone else has been trying to convince him of; the rms value of current is the correct value to use when calculating heat produced.
///Yes, the rms value is correct for rms power, i.e. Prms=Irms x Vrms. This one is different from Pav=Pdc=Idc x Vdc.
There is also a statement:
""The power input to the motor (and hence the power output) varies proportionally with the average value of Ia ...""
Now, the dc motor is supplied with Vdc and Idc leading to power consumption Pdc=Vdc x Idc in watts. Corresponding HPs are indicated on the motor nameplate. Now, ac motor is supplied with Irms and Vrms leading to power VArms=Vrms x Irms. Corresponding HPs are posted on the motor nameplate. The input power is then Prms=Vrms x Irms x PF in watts. Considering relationships between Irms=.9 x Idc and Vrms = .9 x Vdc, there are different powers around. I do not see this addressed in many textbooks on AC machinery. If anyone happens to come across this in any literature, please, would you kindly post the reference and the Author. I see that some contributors have access to EPRI literature or industry standards worth of hundreds of thousands of dollars, then perhaps there might be some content dealing with this subject to clarify the differences.\\

 

[Skogsgurra]

jb,

I also learnt what I know about Effective Value, RMS, Average Value and DC value about 40 years ago, but I obviously did not learn the same thing as you did.

I have been applying this knowledge and I have also been doing Fourier analysis (manually in the beginning and using Excel, MatCad and LabView later), working with ABB, Siemens, as a consultant and lecturer and in my own business. My field of work is drive systems, availability and power quality and I can assure you that none of my reports was ever questioned. At least not with regard to the RMS/Average/DC issues.

I also developped DSP based thermal protection relays for Adtranz (now Bombardier) and that is really where you have to consider current waveforms (traction motors with inverter and/or SCR control) and heating. And do you know what? My definitions (and most other engineers' definitions) worked flawlessly and were approved by the customer and the authorities.

So, I urge you to sit down and think again. And I wish you a Merry Christmas (and some insight).

 

[pennsy]

Yeah, this thread has taken on a life of its own, rms and effective values have been defined, restated and re-defined again.
Maybe it could migrate to eigen values. ?

WOC

 

[allmend]

Hi to all!

I believe that everybody knows the definition of the “effective value of a current”, but the discussion is about the measurement and the calculation of this “effective value”.

Mr Ohm discovered his law before the invention of a AC-Generator, but AS-Current can also be used for heating (there were no non-linear loads then). So you have to define a current, which produces the same amount of heat in a resistor.

A true responding Rms-Meter measures the heat dissipated in a reference resistor.

You can also calculate the effective value of a current using the root-mean-squaring, but with less accuracy, because you do not know what frequencies and curves you are measuring.

That’s just a problem of definition!

In a 50-60-Hz-Grind with sinusoidal current there is no difference between Effective-Value and RMS-Value.

I wish you a Merry Christmas and a Happy New Year!