320 watts

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Re: 320 watts

You are driving this conversation to a level of math that I don?t think you want to get anywhere near. But you are also using invalid simplifications. So I?ll try to clarify a few things. I might not succeed in making it more clear, but I?ll try.

What I said is that you can take any signal, be it a power or a voltage or a financial or a set of sports scores, and calculate the RMS value of that set of numbers. I also said that if the original values followed a standard sine wave, then the RMS value is going to be 70.71 % of the peak value.

That 70.71% rule does not apply, if the original signal is not a sine wave. In a single phase AC circuit, even one with a purely resistive load, ?power? is not a sine wave. Therefore, you can?t use the 0.7071 factor. To get the RMS value of power, you have to start with the numbers that represent the moment-by-moment values of power. You get those by taking the moment-by-moment values of voltage and of current, and multiplying them. This is the set of numbers with which you do the ?square, then average, then square root? process.
Originally posted by coulter: Well, one approach is: The power delivered to the load is .707Vpeak X .707Ipeak = (1/2)(Vpeak)(Ipeak)
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That part is right, but only because you are talking about resistive loads.
(1/2)(Vpeak)(Ipeak) = (1/2)(Peak Power)
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That part is also right, but again only because you are talking about resistive loads.

If there are inductive or capacitive loads (i.e., if the power factor is not 1.0), then the voltage and current do not reach their peaks at the same time. In that case, multiplying (Vpeak) by (Ipeak) will not give you anything closely related to (Ppeak)
Originally posted by coulter: The power delivered to the load is . . . (not) .707 (Ppeak)
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You are right. It?s not. Nor is there any reason to expect that it would be. My integral calculus is a bit rusty, but I calculated that the RMS value of power delivered to a single phase resistive the load was about 61.24% of the peak value of power. I had to find a table that gave me an expression for the ?indefinite integral of cosine raised to the power of 4.? Wanta see it? :D
 
Re: 320 watts

Originally posted by physis:
My integral calculus is a bit rusty
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I'm gonna have to learn that stuff. Looks tedious. I'm not particularly exitied about the prospect. :(
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Only a complete MEW would be interested in this. :p

Yeah, I like it :eek:

carl
 
Re: 320 watts

I've been playing with physics again. That's fun for me, the conceptual part anyway. To actually understand it though you gotta do the yukkie part too.
 
Re: 320 watts

Sam try to locate a copy of "The PA Bible" It was produced by Electro Voice several years ago. It has all the formulas in it as well as everything you need to know about audio systems. It's out of print but maybe you could snag a copy on E-Bay of other web auction site. I'll see if I can locate one from this end. :D
 
Re: 320 watts

Sam I found all 18 chapters on the web for free:
It's in PDF format down load every one. Check out chapter 17. there are photos of the MT4's that have 4 18" sub's in one cabnet with the magnets back to back. They thump! Here's The link:
EV's PA Bible Ch's 1-18

Good reading :D

[ May 13, 2005, 03:49 AM: Message edited by: hurk27 ]
 
Re: 320 watts

What is suprising to me is the claim of the power at 100 Hz. Usually, low-powered amplifiers are rated at 1 KHz (1000 Hz).
 
Re: 320 watts

I bookmarked it Wayne. I haven't looked at anything yet but when I do I'll let you know if I think it's cool stuff. :cool:
 
Re: 320 watts

Comedy and engieering, far too often considered to be mutually exclusive.

Edit: Most of the engineers I've worked with have had an above average sense of humor. It might be a degradation of sanity from all the numbers. :D

[ May 15, 2005, 12:29 PM: Message edited by: physis ]
 
Re: 320 watts

Originally posted by charlie b:
My integral calculus is a bit rusty, but I calculated that the RMS value of power delivered to a single phase resistive the load was about 61.24% of the peak value of power. I had to find a table that gave me an expression for the ?indefinite integral of cosine raised to the power of 4.? Wanta see it? :D
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Charlie B., I would argue that the term "RMS power" is undefined and meaningless. RMS currents and/or voltages are used to compute average power. Furthermore, I am unaware of any application of the RMS formula to anything other than voltage or current even if it is mathematically possible.

Let me say also that RMS values can only be computed for periodic waveforms in which case definite integrals are used.

Rattus
 
Re: 320 watts

(RMS V) X (RMS I)

I think I see your point Rattus.

Edit: They don't use fourier transforms for amplifier specs. :D So I guess your stuck with sine waves. :p

[ May 15, 2005, 08:25 PM: Message edited by: physis ]
 
Re: 320 watts

Originally posted by rattus: Charlie B., I would argue that the term "RMS power" is undefined and meaningless. . . . Let me say also that RMS values can only be computed for periodic waveforms in which case definite integrals are used.
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You just disagreed with yourself. Since I can compute the RMS for any periodic waveform, and since power as a function of time is a periodic waveform, then the RMS value of power does have a definition and a meaning.

I agree that you can calculate the RMS value of any periodic waveform, and I described the process for doing so. I used non-mathematical terms; I didn?t say for example that I used a definite integral, but I did use one. I will add that you can take any non-periodic waveform, look at one period of time, and compute an RMS value for that time period by assuming that its wave shape would be repeated thereafter.
 
Re: 320 watts

Charlie -

I apologize for not getting back sooner. I actually had to like "work" for a living this week. Sheesh, the gall of some employers :) You are right, it doesn't go equally above and below zero which is why the .707 doesn't fit. The RMS rating for ampliers doesn't multiply the delivered power by 141%, rather 122%. Still looks pretty arbitrary to me. I'm still not seeing the connection between the RMS power calculation and the actual power delivered to the load.

The RMS/Power connection gets even fuzzier if one looks at other than resistive loads. I had used the resistive load case because I had recalled that is what is normally used for amplifier ratings, since the amp mfg would not know the inductive impedance of the intended speakers. However, no sweat. For the case where V and I are not in phase, then:

P = V Sin(t) I Sin(t + a) where "a" is the phase angle between the two functions.

the result is P = VI(1/2 Cos(a) - 1/2 Cos(2t+a))

As my math instructor said some years ago, "I'll leave this simple derivation for you." ;)

For V and I in phase, then a = 0, then Cos(a) = 1, and the power function is the same as for a resistive load (okay, no surprise)
P = VI(1/2 - 1/2 Cos(2t))

We already know the RMS value, .61 VI. So now calculate the Average Power:

From P = VI(1/2 - 1/2Cos(2t)) integrate it, evaluate 0 to pi, multiply by the period average factor 1/(pi) and we get:
P(ave) = 1/2VI hummm interesting

For 0 less than "a" less than 90 deg,
Cos(a) gets smaller until at a = 90, Cos(a) = 0. So, as the phase angle gets larger, the DC bias gets smaller until at a = 90 deg, the bias = 0. Here, half the Cos function is greater than 0 and half the Cos function is below 0. This is consistent with power transferred to inductive loads. For pure inductive loads, there is no power transfer, rather power is transferred to the load on one part cycle and transferred back on the other.

Lets try this with the RMS calculation where a = 90 deg:

P = -1/2 VI Cos(t - pi/2)
= 1/2VI Sin(2t)

So, calculate Power(rms): square it, integrate, evaluate between 0 and pi, (don't forget 1/(pi) factor to average over one period), extract square root.

And we get .707VI. Oh - oh. What happened, the power should be zero. Well, when we squared the sine function, we got rid of any negative (out of the load) power flow.

So calculate Average Power, same conditions:
P = 1/2 VI Sin(2t), integrate it, evaluate 0 to pi, whoops, stop right here. P(ave) = 0. As it should.

There is a simple explanation. RMS power has nothing to do with the real world. It is a made up concept from Madison Avenue. The power transferred should be calculated as Average Power.


carl
 
Re: 320 watts

Something you might appreciate Carl. If you're not already aware of it.

Provided you use windows, from the start menu:

All programs/Accessories/System Tools/Character Map

You probably know it's a program, (and what it does) open it. You can copy character codes from it with the copy function and paste them in the forum's post editor.

You'll only be able to use the characters that the forum font supports, which, unfortunately, isn't enough of the Greek alphabet but is certainly still useful.

It's a bit of a hassle but a lot of fun, at least for a while.

Edit: And, I didn't read your thesis, but I'd guess you're asking for it giving Charlie that much material. :D

[ May 18, 2005, 02:43 AM: Message edited by: physis ]
 
Re: 320 watts

That would make 'Θ' an upper-case Theta, and 'θ' a lower-case.

Thanx for the tip!
 
Re: 320 watts

Originally posted by coulter: For a resistive load:
Phase angle = 0, Cos(0) = 1, P(ave) = Vrms x Irms = .707V x .707I = .5VI Not .61 RMS shown in your calculation.
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I won?t comment on your assessment of Madison Avenue?s impact on electrical theory. But you are getting ?Average Power? confused with ?RMS Power,? at least as it pertains to my calculations.
</font>
  • <font size="2" face="Verdana, Helvetica, sans-serif">Let V(t) = Vm cos ( wt)</font>
<font size="2" face="Verdana, Helvetica, sans-serif"></font>
  • <font size="2" face="Verdana, Helvetica, sans-serif">Let I(t) = Im cos ( wt - a)</font>
<font size="2" face="Verdana, Helvetica, sans-serif"></font>
  • <font size="2" face="Verdana, Helvetica, sans-serif">Then P(t) = V(t) I(t) = Vm Im cos ( wt) cos ( wt - a)</font>
<font size="2" face="Verdana, Helvetica, sans-serif">For a purely resistive load, ?a? equals zero.
</font>
  • <font size="2" face="Verdana, Helvetica, sans-serif">So P(t) = Vm Im cos ^2(wt)</font>
<font size="2" face="Verdana, Helvetica, sans-serif">Taking the integral of this function over the period of ?0? to ?w,? and then dividing by ?w,? shows us that the average value of the ?power as a function of time? is indeed, as you calculated, half the product of Vm and Im.

But to calculate the RMS, you first square the function P(t)

</font>
  • <font size="2" face="Verdana, Helvetica, sans-serif">P^2(t) = (Vm Im)^2 cos ^4(wt)</font>
<font size="2" face="Verdana, Helvetica, sans-serif">This is the formula that required me to look up the indefinite integral of cosine raised to the fourth power. Taking the integral of this function over the period of ?0? to ?w,? and then dividing by ?w,? and finally extracting the square root, shows us that the RMS value of power is indeed, as I had calculated, about 61% of the product of Vm and Im.
 
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