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RMS Power

RMS Power

RMS Power

(OP)
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.

RE: RMS Power

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

RE: RMS Power

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.

 

RE: RMS Power

"...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.
 

RE: RMS Power

(OP)
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.

RE: RMS Power

"...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.

 

RE: RMS Power

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.

RE: RMS Power

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.

 

RE: RMS Power

(OP)
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

RE: RMS Power

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 %.
 

RE: RMS Power

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

RE: RMS Power

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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RE: RMS Power

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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RE: RMS Power

"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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RE: RMS Power

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

RE: RMS Power

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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RE: RMS Power

(OP)
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."

RE: RMS Power

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

RE: RMS Power

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

RE: RMS Power

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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RE: RMS Power

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.

RE: RMS Power

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

RE: RMS Power

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

Quote:

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.


 

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RE: RMS Power

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

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RE: RMS Power

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

RE: RMS Power

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

RE: RMS Power

Gunnar,

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

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

RE: RMS Power

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

RE: RMS Power

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

RE: RMS Power

Hi

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

desertfox

RE: RMS Power

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

RE: RMS Power

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

RE: RMS Power

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

Quote:

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

Quote:

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.  

Quote:

. 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.

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RE: RMS Power

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

RE: RMS Power

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.
 

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RE: RMS Power

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.
 

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RE: RMS Power

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.
 

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RE: RMS Power

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.
 

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RE: RMS Power

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.
 

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RE: RMS Power

Five in a row. A new record.
 

RE: RMS Power

Forgot the winky smile !!

RE: RMS Power

There's a duplicate in there - that only counts once!

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RE: RMS Power

(OP)
Thank you for all your help and suggestions.  I'm going to be very conservative since while "Speed is King,"  burning up Voice Coils is verboten.  250 Watts continous gets me to an average winding temperature of 155 deg C.  This was determined by injecting a fixed DC current into the voice coil and measuring the voltage drop across the coil. At 155 deg C, the voltage increases 50% [or the resistance increases by 50%].  Roughly 6 amps RMS at 7.5 Ohms [5 Ohms cold].

But I wanted to know a "Fudge Factor" I might use based on a linear current taper from OD to ID if management wanted "Just a little more" which, of course, they always do.

RE: RMS Power

Quote:

250 Watts continuous gets me to an average winding temperature of 155 deg C.

IF   we  assume
1 - this is the temperature in a steady state temperautre
2 - your ambient temperature is 40C
THEN   we can calculate Rth = (155-40) C/ 250 watts = 0.46 degC / watt
(hey - I wasn't too far off with 0.2 degC / watt!)

But we have no idea your time constant.  If you apply DC current as a step increase from 0, how much time does it take for temperature to get 63% of the way from it's initial value to final-steady-state value?    That would be the time constant.  That info along with the ramp time, frequency, initial temperature (before the ramp-down) and ambient temperature would complete the inputs required for the spreadsheet.

To sharpen the pencil some more, one could also program the temperature coefficient of resistance into the model.  

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RE: RMS Power

Here is revised spreadsheet that includes correction of resistance for temperature and a few miscellaneous format items cleaned up as shown in the new "instructions" tab.

We start with three thermal unknowns :Rth, Cth, Tau.
We have two equations:
1 - Tau = Rth, Cth
2 – Rth = 0.46 degC

We need one more equation or unknown value to solve the thermal parameters.  Two possibilities to get there:

1 - As discussed above , Tau could be from step change in current (or as close as you can get with an inductor).
or...
2 - What might be easier is to estimate Cth.  We can guesstimate that if you tell us the mass of materials: how much mass of copper (and iron?...not sure if voice coil has a core) heat up to this temperature.
 

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RE: RMS Power

I need to clarify a few things, see posting 20 Apr 09 0:16.

I do not skim to find something to criticize. I take part in threads when I think that I have something to contribute. I think that long term members of Eng-Tips realize that.

The two "crimes" that I seem to have committed are that I used the words "RMS Power" and that I wrote "High" instead of "Low".

The first crime is an adaption to the sloppy usage of the term "RMS Power" in the thread heading and to the words used in the question I was answering. If hat is a terrible crime, I think it would have been fair to point that out long before I happened to use the words.

The second crime - well, I guess I am guilty. Slip of mind. I actually realized that just after posting, but didn't bother to post a correction. I guess I should have done that. It is a habit that I perhaps should pick up.

Gunnar Englund
www.gke.org
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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

RE: RMS Power

(OP)
The heat input is extremely complicated.  The cutting RPM may be limited by the max Velocity, the max Acceleration and the max Jerk of the fast tool [voice coil driven].  And the lenses [limits] are in random order.

There are "N" Rough Cuts [done with a fast feed rate] followed by the slow finish cut.

There ia a dead time between [load-unload] on the lenses

From a cold start with the maximum current that gets me to 155 deg C, it takes 1.5 minutes to get to 63% of the final coil voltage.  I did not test the cool down time as I already knew I was going to "Play it safe."

RE: RMS Power

I assume that your current waveform is something like an amplitude modulated waveform.  The carrier frequency is at 50hz or higher.   The envelope is slowly varying - steady in places, step change in places, ramps in places.

Since your thermal time constant (90 sec) is so much higher than the period of the carrier frequency (1/50hz   or shorter),  we can simplify the problem by ignoring the carrier frequency component and replacing the waveform with envelope (divided by sqrt2 to convert to 1-cycle-rms).   That is of course the similar to what you already did to begin with when you replaced the ramp sinusoid with a ramp to calculate the factor 1/3.   This would simplify any analytical solution and I think would also reduce the number of iterations required for numerical solution (numerical which would be my preference since it easily allows us to accmodate non-linearities such as temperature coefficient of resistance.... could also easily be adjusted to account for non-linearities in heat transfer characteristic if they are known).

If I get a chance this weekend, I will adjust the spreadsheet to add a tab where the user can easily specify an arbitrary input current waveform (which could be the full am waveform or the envelope /sqrt2).

There is now enough info to solve the thermal model (simply plug 90 sec into Tau).   If you wanted to do some rough validation of the thermal model there are some some double-checks which can be done:
1 - provide the info on mass of copper and steel and calculate thermal capacitance from that to see if it matches the thermal capacitance provided from Cth = Tau/Rth
2 - if you happen to have plot of temperature vs time during your dc test (could be constructed from resistance vs time accounting for temp coefficient), it should be roughly a straight line on a log-log scale if you have a 1-degree of freedom thermal system (one mass and one temperature).   If 2dof system, we might see two different slopes in different parts of the curve.
3 - Multiple dc tests of course provide more detailed info on the thermal model.    The form of heat transfer out of the device is most likely Ith = dT^m / Rth
where Ith is heat transfer out (watts)
dT is temperature rise above ambinet
R is thermal parameter
m is unknown coefficient. would be near 1 for conduction heat transfer but can be in the range 0.1 - 0.2 for radiation heat transfer near room temperature, approximately 0.25 for laminar convective heat transfer and 0.33 turbulent convective heat transfer.   If multiple heat transfer mechanisms exist thatn the overall effective value of m would be something like a weighted average of those numbers  Comparing the final equilibrium temperautres for different values of heat input = heat output = Ith would allow us to estimate the exponent m.

 

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RE: RMS Power

I have been assuming only I^2*R losses are relevant.  If it is an iron core device (is it?), then core losses may play a role also.

We don't know the frequency or wire size - I assume conductors are small enough and frequency low enough that skin effect / proximity effect are not significant.

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RE: RMS Power

The "max step" parameter was set to small in the last spreadsheet - makes the program do many more calculations than required.  I suggest setting it up much higher (0.02) for slowly varying envelopes like the one shown.

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