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.

 

[desertfox]

hi sreid

Have a look at this site it might help:-

http://www.allaboutcircuits.com/vol_2/chpt_11/1.html

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

If your feeding current into a voice coil would it be purely resistive? I assume a coil would have reactance as well.

desertfox

 

[VEBill]

I'm sure you already know most of this, but to start at the begining...

First, don't assume that the voice coil is a resistor. Depending on the inductance and frequency, the reactance might define most of the impedance which defines the ratio of voltage and current.

If you multiply the RMS voltage by the RMS current then you have calculated the power. In your case, the power is a linear function of time.

It can get a bit 'sticky' if the period of the sinwave is not small in comparison to the duration of the linear drop. In that case if might be better to do the calculation in MS-Excel on an instantaneous basis and average it over various time frames.

If the sinwave is high frequency, then you can easily calculate the average power within the triangle shape. Don't forget it's a square (^2) function of RMS current.

If the overall cycle repeats, then you can then calculate the average power over the period by taking into account the relative duty cycle.

If the cycle never repeats, then the power starts at some peak value and then averages towards zero as time passes.

 

[VEBill]

"...the power is a linear function of time."

Doh! The current is a linear function of time, so the power would be a curve based on the square of I.

 

[sreid]

1. The sine drive is current so the per cycle power is Irms^2 x R.

There are at least 100 sine cycles from the start of the linear ramp to zero amplitude.

The reason I ask this question is that I am often guilty of coming up with Average Power rather than RMS Power. And my math skills are lacking these days [old age and non-use of math].

By bench experiment I know the DC Current I can drive through the coil and keep the winding temperature below 155 deg C.

 

[VEBill]

"...Average Power rather than RMS Power..."

Once you correctly calculate the power using RMS I & V, then it is just 'power' (over some period, depending on how you've calculated it). The result can be a function of time. And with 100 cycles, you can skip the sub-cycle instantaneous calculations.

Then you can average the power over whatever duration you wish using normal (linear) averaging techniques. So the end result can be 'average power' (over some stated duration) correctly calculated using RMS I & V.

You may have to reassure the audience that the 'average power' doesn't mean that you've made the noobie mistake of calculating the power using the average I and the average V.

So, 'average power' can be a correct phrase. But you should state the duration (or it's 'steady state') to make it clear.

'RMS power' isn't really an accurate phrase. You might say 'power calculated using the correct RMS I & V', but it should go without saying.

 

[stevenal]

RMS power (the root of the average value of the squared P(t) waveform) is not a useful concept. No benefit gained, and directional information is lost.

 

[Noway2]

The sinewave would be the function y = sin(2*pi*f*t/Ts), where TS is the sample rate, t is the time count (0,1,2,..), f is the frequency. The linearly decaying amplitude would be a line y = mX+b where m is the slope and b is the initial amplitude (x would also be t).

You should be able to multiply the two functions, point by point to get the combined waveform. Then you could compute the RMS of that waveform to get the effective power for that period of time.

It wouldn't be a closed form solution, but it would give you an answer. This should be relatively easy to perform with a basic math package such as matlab or one of the free clones.

 

[sreid]

OK. I believe that the power curve can be approximated by saying the power curve is [for A=1]

y=x^2

Integrating gives the area area from x = 0 to 1 as

x^3/3 = 1/3

 

[zekeman]

RMS power is meaningless in your case unless you tell us how repetitive it is. For example, if you have one burst of power of 1 second duration as you mention in an hour I would say your RMS power is nearly zero. However , if this is continuously repeating, say 1 burst followed by another then it has some meaning.
I did this problem on that basis and instead of assuming a linear decay I used for the sake of mathematcal convenience the classical damping exponential decay e^-1.5t which has the sine function decaying at approximately the rate you mention.
The RMS power over T the time of interest is
SQRT( 1/T*integral A^2*sin^2(wt)*e^-1.5t*dt limits of t are 0,T.
I get
A*sqrt(1/6T)
For 1 second this is .408A
For 1.5 seconds .33A where decay is 90 %.

 

[zekeman]

correction RMS equation should read
SQRT( 1/T*integral A^2*sin^2(wt)*e^-3t*dt limits of t are 0,T.

 

[electricpete]

I agree RMS power is a dicey concept, not very useful and better to avoid. I assume you want to bound the temperature under a conservative adiabatic assumption (no heat transferred out) and so you really want the total power input or the average power which is total power over the length of time.

IF we are truly assuming the coil is a resistor (I leave it to others to decide if that is good enough for a voice coil), and if we can neglect resistance changes with temperature, than rms current is good enough to get there. The average power is the resistance times the rms current.

Let's assume the envelope starts at magnitude I0pk (peak) and drops linearly to 0 over a time T
I(t) = I0pk *cos(w*t) * (1 - t/T)

The rms current of your waveform will change depending on the interval.

Since it ramps down over at least 100 cycles, T >> 1/w I think it's reasonable to consider the rms current constant within a cycle. For example rms of the first cycle is I0/sqrt(2).... rms half way through is 0.5*I0/sqrt(2). That assumption allows us to simplify the problem by replacing with a simpler current I2 which starts at I0pk/sqrt(2) and decreases linearly to 0 over time T.

I2 = I0pk/sqrt(2) * (1-t/T)
I2^2 = I0pk^2 / 2 * (1 - 2*t/T + t^2/T^2)
<I2^2> = I0pk^2 / 2 * [<1 - 2*t/T> + <t^2/T^2>]

Solve the pieces of stuff in square brackets [ ]
<1 - 2*t/T> over the interval (0,T) starts at 1 and decreases linearly to -1. The average value is 0.

<t^2/T^2> = (1/T) * int(<t^2/T^2> = (1/T) * (t^3/3)/T^2 at t = T
<t^2/T^2> = (1/T) * (T^3/3)/T^2 = 1/3

Plug these items < > back into the expression for <I2^2>
<I2^2> = I0pk^2 / 2 * [0+ 1/3]

I2rms = sqrt(<I2^2>) = I0pk / sqrt(2*3).

So the rms value of the waveform during the period of linear decrease is a factor of sqrt(6) less than the initial peak value and a factor of sqrt(3) less than the initial rms value.

Again we assumed that we ramped down over many cycles T >> 1/w to allow using the simpler waveform I2(t) and we assumed that the circuit was resistive and the resistance was constant.


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[electricpete]

And just to clarify, that value of rms current (I2rms) solved above applies only for purposes of calculating the total energy dissipated over that entire period of linear decrease to 0 (the interval from t=0 to T)

W = T * Pavg = T * R * I2rms^2
W = T * R * {I0pk / sqrt(2*3)}^2
W = T * R * I0pk^2 / 6

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[electricpete]

"The average power is the resistance times the rms current."
should have obviously been:
"The average power is the resistance times the rms current squared"

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[waross]

A little off topic, but perhaps a valid observation:
RMS horse power may be calculated for a motor with a varying duty cycle. A motor may be loaded above its rated HP if an RMS horse power calculation shows that there is sufficient time at light loading to allow the temperature to stay below the rated temperature.
See the Cowern papers on the Baldor Motor site.

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

 

[electricpete]

The assumption inherent in the Cowern Papers' rms horsepower calculations are:
1 - Motor current is proportional to motor horsepower
2 - Motor heating comes from I^2*R losses.

Accordingly, we can manage the total I^2*R heating over an interval below that which would occur at rated power over the same interval by limiting this rms horsepower calculated over the interval below rated.

From assumption 1 you can see that horsepower is used in this approach simply as a surrogate for current. In contrast for assumed resistor exposed to non-sinusoidal current, there is no power which we can consider roughly proportional to current.


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[sreid]

Electricpete,

Thanks for your analysis, it was once again complete and elegant. And it confirms my primative analysis.

My use of the term "RMS Power" was inappropiate. I'll fall back on my standard excuse, "I'm Old and I forget things."

"Speed is King" in the eyeglass making business. Forced Air Cooling is going to let me go from 12 g to 18 g acceleration on the Fast Tool Servo. And this is based on continous current, not the ramped current you just analyzed. However, I may be limited by the amplifier driving the Voice Coil. But I've got another amplifier "Warming up in the Bullpen."

 

[waross]

Well Pete has proved conclusively that you, I and Mr. Cowern are wrong to use the terms "RMS power or RMS horsepower".
If we can only come up with an acceptable name for Mr. Cowern's method it may be an appropriate method to calculate how far you can push your equipment.

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

 

[Skogsgurra]

Bill,

You couldn't say that it is wrong to use RMS. The concept is valid all the way from DC to microwaves, and beyond. It is mostly a question of time scales, especially thermal time constant of an object in comparison to the frequencies contained in the signal that heats that object. 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 there must be an exact expression for the integral f(x)*sin(x), where f(x) is a linearly increasing (or decreasing, in this case) function of x, in the CRC handbook. I shall look it up sometime when I think it is worthwhile.

Gunnar Englund

http://www.gke.org

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

 

[electricpete]

Bill - Sorry if I am about to beat a dead horse. But when you lump yourself together with Cowern and say I am contradicting both of you, then I take that as a disagreement. I don't believe anything I said was incorrect. And I don't think I said there was anything wrong with Cowern papers' terminology or approach.

The concept of rms power applies directly in the case of Cowern papers but has absolutely no application within this thread. Let us examine why.

The quantity of interest for resistive heating is generally average resisitive loss power <Presistive(t)>


Presistive(t) = I(t)^2 * R
<Presistive(t)> = <I(t)^2> * R
add {sqrt( )} ^2
<Presistive(t)> = {SQRT<I(t)^2>} ^2 * R
recognize sqrt(<I(t)^2> = Irms
<Presistive(t)> = Irms^2 * R
where < > denotes average value

Now lets look how it applies to this thread. We want average power <Presistive>, so we determine RMS current and plug it into the above equation. RMS power has got nothing to do with it. Does anyone disagree with the contention that rms power has zero application in this thread?

Now back to the Cowern papers already discussed. There we have two different "powers" that we could talking about for the motor and we cannot interchange them. One is the resistive heating <Presistive> where the ultimate objective is to control heating which is assumed resistive. The other is the power transmitted out of the motor Pmotor. The power which we apply RMS concept in that situation is the power output by the motor Pmotor based on the ASSUMPTION that motor power is proportional to current: Pmotor = K*Imotor where K is proportionality constant.

Then we have:
Presistive(t) = Imotor(t)^2 * R = {Pmotor(t)/K}^2 * R
<Presistive(t)> = <Pmotor(t)^2> * R/K^2
add {sqrt( )} ^2
<Presistive(t)> = {sqrt(<Pmotor(t)^2>)}^2 * R/K^2
recognize sqrt(<Pmotor(t)^2> = Pmotor_rms
<Presistive(t)> = Pmotor_rms^2 * R/K^2


If we have two loading patterns which have the same Pmotor_rms, they will give the same resistive heating. So we can limit resistive heating below that at nameplate by limiting the Pmotor_rms below nameplate. There are a few more simplifying assumptions inherent in this approach (for example we neglect heat from starting and assume motor heating is entirely resistive), but that is the basic approach, it is useful, and it is correct within the assumptions.


But nowhere in that entire motor rms power approach did we ever consider rms{Presistive|}. We only looked at rms{Pmotor}. But there is no power anywhere in the current thread that is analogous to rms{Pmotor}.

Once again, my apologies if I have misinterpretted the comment or over-beat a dead horse.


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