Heating Effect - DC vs AC 5

[Lakey]

Hi,

Although it is known that the heating effect of DC is equal to AC at the same RMS, can anyone point me to any publications or references that substantiate this?

Thanks in advance.

 

[electricpete]

P(t) = I^2(t)*R
<P(t)> = <I^2(t)>*R where < > denotes average over time.

For dc current I(t)=I0, <I^2(t)> = I0.
For ac sinusoidal, <I^2(t)> = Irms = Ipeak/sqrt(2)
For any general signal nonsinudsoial or otherwise, <I^2(t)> = Irms (definition of rms).
Irms determines heating and plays the same role as I0 does for dc

Aside - If the frequency becomes high and/or conductor is large, skin effect may play a role to increase the effective resistance.

 

[CarlPugh]

Try electronics dictionary.
1974-75 Radio Shack dictionary of electronics
&quot;rms amplitude&quot;
&quot;The value assigned to an alternating current or voltage that results in the same power dissipation in a given resistance as dc current or voltage of the same numerical value.&quot;

 

[electricpete]

I left out the square roots. Should have been:

For dc current I(t)=I0, sqrt(<I^2(t)>) = I0.
For ac sinusoidal, sqrt(<I^2(t)>) = Irms = Ipeak/sqrt(2)
For any general signal nonsinudsoial or otherwise, sqrt( <I^2(t)>) = Irms (definition of rms).

Irms determines heating and plays the same role as I0 does for dc. It comes in straightfoward manner from the definition of rms.

 

[buzzp]

Any basic electronics text book will state this. It can also be shown easily by an experiment.

 

[buzzp]

I think it is also worthwhile to state 'True RMS' although technically, RMS is correct. There are too many devices that state RMS but they use an averaging technique for measuring that is only applicable for a pure sine wave.

 

[jbartos]

Suggestion: 10A of battery DC current or rectified and smoothed full wave sinusoidal current will produce more heat than 10A RMS sinusoidal current since there is a relationship between the average value of rectified sinusoid and rms sinusoidal current. Irms = 1.1 x Iaverage. The Iaverage is true DC, for example battery DC. The heat effect of positive half-wave of sinusoid adds to the heat effect by the negative half-wave of sinusoid resulting in a cumulative heat effect corresponding to the average value of current Iaverage=Idc of the sinusoidal current. In the past, the Form factor between Irms and Iaverage been used. Lately, this concept has been downplayed by IEEE and the Standard Handbook for Electrical Engineers without any proofs or scientific explanations. I have the old publications and other books e.g. electronics handbooks that indicate the relationships. The RMS is pure mathematical expression Irms=(1/T) integral from 0 to T of (i(t) x dt) which was developed for sinusoidal electrical variables way at the beginning of AC power generation and transmission.

 

[buzzp]

It is not clear what your stating Jbartos. I do not agree with your first sentence.
I think what your trying to say is if we have a signal that contains DC and AC then the current that will cause heating is the TRMS of the AC portion plus the DC portion. If we simply have an AC waveform with no DC (average voltage = 0) then the TRMS value of this AC waveform by itself, has the same heating effect of just a DC signal with the DC=AC (TRMS).
If there is some new revelation into this age old information then I would like to know some sources of your information.

 

[electricpete]

I agree with buzz that the following statement is incorrect:

&quot;10A of battery DC current or rectified and smoothed full wave sinusoidal current will produce more heat than 10A RMS sinusoidal current&quot;

 

[Lakey]

Thank you all for your input. I’ve been running some experiments to prove that the maths is correct. Theory is OK, but I need prove. A publication made by a recognised test house would satisfy my requirements.

FYI
The test:
Thermocouples were inserted into a device at selected points. The device passed 200A AC (60Hz) for 2 hours until thermal stabilisation was achieved. Then the supply was switched to batteries (DC 500 Ah).

Result:
Alas, it was observed that the temperatures initially rose by several degrees C before the battery current began to drop off.

I had two significant problems with this test; firstly, maintaining a constant battery current, and secondly controlling the airflow and ambient temp around the device.

 

[electricpete]

One more time on the theory:

If you believe conservation of energy and ohm's law, that leads to p(t)=i(t)^2*R

<p(t)>= <i(t)^2>*R

Definition of Irms is sqrt(<i(t)^2>)
therefore substitute Irms^2 = <i(t)^2>

<p(t)>= Irms^2*R

Assuming that R is constant there is nothing left to prove. What is the purpose of your test?

 

[Lakey]

Thx Pete, and I’m certainly not going to dispute Ohms Law!!

You must understand that some people/customers are not always convinced with the math, they like to see the physical evidence - that's all!

 

[buzzp]

You might try your test on a smaller scale to eliminate your battery power problem (or build a quick DC supply). Try using 10A or even smaller on a simple resistor network or something. Or maybe your customer wants it proved on his equipment.

 

[RadarRay]

You never said what the &quot;Device&quot; was that you were connecting to your AC and DC supply, so this may be a mute point. However, the power in an AC circuit is determined by the Impedance of the load. So unless you have a purely resistive load, your measurements may be off.

 

[electricpete]

Good point by radarray: if the device is not a resistor then all bets are off.

Also, how/why do you believe that the dc current created by the battery was equal to the rms current during the ac part of the test... what measurements did you take? And as buzz says when you measured ac quantities did you really measure true rms?

 

[electricpete]

you mention the airflow so you are apparently aware temperature may affect the resistance.

 

[jbartos]

Clarification/correction (I beg your pardon):
Well known Form Factor F is:
F=Irms/Iav=Irms/Idc=(1/0.9)=1.1111...,
for the sinusoidal wave.

 

[Lakey]

Thank you all for your help.

Cheers.