RMS Power

[sreid]

I need some conformtion of a power calculation.

If a resistor [R] is droven by a sinusoidal current source that at time zero is

i = Asin[wt]

and "A" decreases to zero linearly, what is the RMS Power.

This is not a home work assignment, this is an approximation of the current waveform I'm driving into a Voice Coil.

 

[N1755L]

Just to clarify things for desertfox,

in a helpful way, I hope it would be taken, from which statement:

"To my knowledge the RMS power is the equivalent power to that of a DC circuit."

I would say that DC power is equal to "RMS power x 0.9".

Is this correct folks, is that all there is to it, is DC power 90% of RMS power, or is that an over-simplification?

Just curious.

 

[dpc]

Watts are watts. There is no difference in 100 watts used in an ac circuit versus 100 watts used in a dc circuit. Or any other 100 watts.

The only hard part is computing the power used in an ac circuit. Using rms values of voltage and current is generally used to find average watts dissipated over number of cycles.

"The more the universe seems comprehensible, the more it also seems pointless." -- Steven Weinberg

 

[electricpete]

http://en.wikipedia.org/wiki/Audio_power

In common use, the terms "RMS power" or "watts RMS" are erroneously used to describe average power

Yup. In a resistive circuit we use rms current to compute the average power. I thought the horse was killed but he just keeps coming back.




=====================================
Eng-tips forums: The best place on the web for engineering discussions.

 

[electricpete]

By the way I also agree wtih dpc's comments.

=====================================
Eng-tips forums: The best place on the web for engineering discussions.

 

[waross]

I misunderstood that the application was determining how far above nominal current a device could be pushed if there was a cool down time in the cycle. For such a duty cycle, where the temperature developed by the resistive heating may be the limiting factor and the resistive heating may proportional to the square of the current, I mistakenly felt that RMS analysis was appropriate.
Sorry.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[Skogsgurra]

N1755L (Computer)

You are correct in your assumption that "DC power is equal to RMS power x 0.9" is an oversimplification.

It is a gross oversimplification that is correct only in one very special case, where you compare rectified sine AC without any harmonics to TRMS. The exact factor there is 2*sqrt(2)/PI, which is 0.900316316...

I get the impresson that many of you have the impression that there is a factor to convert AC RMS power to DC power. There is no such factor.

Pete is correct when he says "In a resistive circuit we use rms current to compute the average power" but there are many ways to misinterpret that statement and only one correct way. The correct way involves integrating instantaneous power in a component over the time span in interest and then divide the energy obtained with that time span.

There is a very simple practical way of determining TRMS of a voltage and that is to light an incandescent lamp with the AC voltage and then apply a DC voltage that makes the lamp shine equally bright. The DC voltage is then the TRMS voltage. Using a relay that switches at a 1 - 2 Hz ratio between the two voltages makes the comparison quite exact.

Gunnar Englund

http://www.gke.org

--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

 

[ScottyUK]

Gunnar,

I hope you have a traceable calibration certificate for your eyes.



----------------------------------

If we learn from our mistakes I'm getting a great education!

 

[Skogsgurra]

Scotty,

It is a comparative measurement. No calibration necessary.

Actually, that is quite true. When you cannot see any change in light as the relay switches, then you have the TRMS value on the DC meter (which has to be calibrated to the precision and accuracy needed).

To make the light variation independent of the transfer time between the two voltage sources, a closed or overlapping transition can be used.

Gunnar Englund

http://www.gke.org

--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

 

[waross]

Are we talking about what I know as the form factor? The ratio between the average value of a wave form and the RMS value of the same wave form?
I was taught that the deflection of a D'Arsonval meter movement was proportional to the current. As a result, when a D'Arsonval meter (most analogue multi meters use a D'Arsonval meter movement) is used with a full wave rectifier to measure an AC voltage (inferred from the current through a series resistor) the form factor of 1.11 or .9 must be used to convert the average to RMS.
By the way, Gunnar's factor is valid only for a sine wave. If the sine wave is distorted the indication on the meter is no longer accurate.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[desertfox]

Hi

I see the error of my statement now, thanks for pointing it out N1755L and others.

desertfox

 

[Skogsgurra]

Yes Bill. The PI/(2*sqrt(2))is what we once called the Form Factor of a waveshape. The expression "Form Factor" now seems to be used in many different meanings. Form Factor is related to, but not identical to, Crest Factor.

Gunnar Englund

http://www.gke.org

--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

 

[waross]

I have had a couple of adventures over the years with D'Arsonval meters and distorted wave forms. (No longer a sine wave)

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[electricpete]

Gunnar, thanks very much for your ostensiblely genuine concern about how others might be confused by misinterpetting my comments. However I would say my comments stand on their own quite well thank you since my assumptions have been clearly stated and the interval over which I computed rms was clearly stated. I think rather than skimming to search for something to criticize, you should read with a view toward understanding (which clearly you did not based on your continuing use of the term rms power) and perhaps you should pay some attention to your own comments

You couldn't say that it is wrong to use RMS. The concept is valid all the way from DC to microwaves, and beyond

I get the impresson that many of you have the impression that there is a factor to convert AC RMS power to DC power. There is no such factor.

Do you not understand the fact that rms power has no significance in this thread? RMS current has relevance and average power has relevance. We can use rms current to compute average power or we can compute average power directly. In either the case of an rms and an average we need to define an interval over which those statistics are relevant as you say. But rms power (with the exception rms horsepower for motors based on special unique circumstances not relevant to this thread) is just an irrelevant and misleading term. A horse which I really thought had been beaten to death.

. If ratio between thermal time constant and period time of the highest frequency component is large, then RMS is OK to use. If not, then an analytical expression is needed

I think everyone will agree the system thermal time constant needs to be considered in deciding whether an average power can be used and over what interval. However it is the lowest (not highest) frequency component of the signal that is relevant for this determination.

=====================================
Eng-tips forums: The best place on the web for engineering discussions.

 

[waross]

Still searching for a word. How about RMS losses? If the losses are approximately proportional to the square of the current and the time interval is less than a couple of thermal time constants can we use this concept to predict how much varying current may be fed to a device without over heating it?
That was my understanding of the original question.
I suspect that average power may not accurately describe the heating of a device. Power factor may be important.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[electricpete]

Hi Bill,

In my use of the term "average power", power referred to losses in the resistor. There was no component other than a resistor mentioned in the original post, so I thought that was clear the power referred to the resistor. Further it was stated "The average power is the resistance times the rms current squared" which defines exactly which average power we are talking about. But if you wanted to use the phrase "average resistive power", or "average power dissipated in resistor R1 during the interval (a,b)", those would certainly be more descriptive.

The use of the term average power is certainly correct (when the averaging interval is small compared to time constant as you say) although one could argue about how many qualifiers are required. In contrast, RMS losses and RMS power are simply not correct for this problem (heating of a resistor by a specified current). They do not mathematically describe anything related to this problem. There is a perfectly logical reason to compute rms of a current (to determine average power). There is no logical reason to compute rms of a power (rms horsepower of a motor is a special case where horsepower is used as a surrogate for current and so rms horsepower becomes relevant in that one special case).

=================

Maybe a math example will help to illustrate that rms power is a completely different thing than average power:

Look at a current pulse train which is 1 for 25% of the cycle and and 0 for 75% of the cycle.
This pulse train is applied through a one-ohm resistor.

i.e.
i(t) = 1 for 0 < t < 0.25*T, i(t) = 0 for 0.25 < t < T, repeating at period T
R=1

i(t)^2 = 1 for 0 < t < 0.25*T, 0 for 0.25 < t < T
<i(t)^2> = 1/4
Irms = sqrt<i^2(t)> = 1/2
<p(t)> = <I^2*R> = Irms^2 * R = 1/4 [AVERAGE POWER IS 1/4]

p(t)= I(t)^2 * R
p(t)^2 = I(t)^4 * R^2
p(t)^2 = 1 for 0 < t < 0.25*T, i(t) = 0 for 0.25 < t < T
<p(t)^2> = 0.25
Prms = sqrt<p^2(t)> = 1/2 [RMS POWER IS 1/2]

So we see from this example Prms = 1/2 and Paverage = 1/4. They are two different things. Paverage is relevant since we can multiply it by a time to determine the total energy added over the time interval. Prms cannot be used in this manner and has very little practical application. That is why I object to the use of Prms. It is a different quantity from Paverage.


=====================================
Eng-tips forums: The best place on the web for engineering discussions.

 

[electricpete]

sreid - One comment to add - if the system time constant is not much larger than the interval over which time was ramped, then a dynamic analsyis is required as skogsgurra stated.

I have previously posted a spreadsheet that solves an initial value problem using Runge-Kutta method. That would solve a resistor heating problem very well.

Inputs: i(t) waveform, Re = electrical Resistance, Rth = "Thermal Resistance" = degrees C rise per watt which could be computed from your DC test, Tau = time constant measured in response to a step change during dc test, Cth = thermal capacity can be computed as Cth = Tau/Rth.

Assume for simplicity electrical resistance does not change with temperature and Rth does not change with temperature.

Then
Heat balance: watts In - watts Out = Cth * d/dt (T)
i(t)^2 *R - (T-Tambient)/Rth = Cth * d/dt(T)

Solve for d/dt(T):
d/dt(T) = [i(t)^2 *R - (T-Tambient)/Rth]/Cth

This is the form required for a numerical solution since d/dt(T) is given in terms of known variables and T. Given initial value of T at t=0, we can "integrate" to find T for all other times.


=====================================
Eng-tips forums: The best place on the web for engineering discussions.

 

[electricpete]

Attached I have done an analysis using the spreadsheet as I described above. I pulled the values out of my hat for the green input cells. My input values are probably orders of magnitude away from your reality, but you can adjust the green input cells any way you'd like.

In order to get the "wiggle" from the 60hz sin wave to even show up in the temperature profile I had to adjust the time constant way down to 1 sec. I'll bet for your oil bath system the time constant is much higher. If so, that means for all practical purposes you'd get the same result if you simply used a ramp function (without the sinusoid) starting at Ipk/sqrt(2) and ramping down to 0.

It would also be not too much trouble to change the current to whatever waveform you want - just go into the vba code and change the function named "current1"

I should mention another assumption beyond those mentioned above - the whole system is assumed represented by one temperature and one time constant - assumes uniform temperature.


=====================================
Eng-tips forums: The best place on the web for engineering discussions.

 

[electricpete]

I forgot to mention:
Your thermal resistance would be calculated from your steady DC test result.
Rth = (Tdcss-Tambient) / (Idc^2*R)
where Tdcss is steady state temperature reached during DC test.

Your time constant would be meausured based on response to a step change in input power.


=====================================
Eng-tips forums: The best place on the web for engineering discussions.

 

[electricpete]

I forgot to mention:
Your thermal resistance Rth would be calculated from your steady DC test result.
Rth = (Tdcss-Tambient) / (Idc^2*R)
where Tdcss is steady state temperature reached during DC test.

Your thermal time constant tau would be meausured based on response to a step change in input power.


=====================================
Eng-tips forums: The best place on the web for engineering discussions.

 

[VEBill]

Five in a row. A new record.