Power factor

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wwhitney

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if by sinusoidal you mean square, sawtooth, triangle, etc yes
perhaps periodic is a better term
although the sqrt term varies
The formula as posted is only valid for sinusoidal waveforms with 0 average. As you mention, there are similar formulas for other periodic waveforms with 0 average.

Cheers, Wayne
 

wwhitney

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Total watts through out the cycle.
Not sure what your comment means, but let me point out that RMS is just one way to average a time varying waveform. It is the choice made so that the equation P = I * V will work when the I and V waveforms have the same shape and are in phase (their ratio is constant).

Cheers, Wayne
 

mbrooke

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Not sure what your comment means, but let me point out that RMS is just one way to average a time varying waveform. It is the choice made so that the equation P = I * V will work when the I and V waveforms have the same shape and are in phase (their ratio is constant).

Cheers, Wayne

I mean average. Ingenieur's equations are average power, not DC offset or the like. At least my take...
 

mivey

Senior Member
Yes, it has a constant component, the disagreement is in terminology, should that constant component be called a "DC" component, when it arises strictly from AC? I would just call it a constant component.

Cheers, Wayne
It is common industry and educational terminology to refer to the sinusoidal component as "AC" and the constant (or decaying or rising) component as "DC".

Don't have time at the moment to read the thread in detail but from scanning the latest it seems for the others to deny that common terminology is just them being argumentative. It is a very, very long-standing use of terminology.
 

Smart $

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It is common industry and educational terminology to refer to the sinusoidal component as "AC" and the constant (or decaying or rising) component as "DC".

Don't have time at the moment to read the thread in detail but from scanning the latest it seems for the others to deny that common terminology is just them being argumentative. It is a very, very long-standing use of terminology.
It's not a problem when referring to a voltage or current plot.... or even a power plot when there truly is a DC component.
 

Ingenieur

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Earth
It is common industry and educational terminology to refer to the sinusoidal component as "AC" and the constant (or decaying or rising) component as "DC".

Don't have time at the moment to read the thread in detail but from scanning the latest it seems for the others to deny that common terminology is just them being argumentative. It is a very, very long-standing use of terminology.

yepper
if pf=1 and 100% offset is power alternating?
does it reverse polarity ie drop below zero?
a full wave rectifier without filter caps
is its output ac or dc? Basically a sine wave with <0 cycle inverted
it has a freq component so how can it be 'DC'?

this may help:
36" dia pipe
100' long
1% grade
postive reciprocating piston pump at load side
pump rate set so pipe is 2' full at bottom and 1 at top
Pumps 60 strokes/min of a fixed volume

we have steady state or 'dc' flow on which 'waves' from the slugs form on top
so we have a dc level + ac level
due to the slope the ac basically establishes a standing wave on top but does not contribute much if any to the discharge

Clear as mud
j/k
lol
 

wwhitney

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Location
Berkeley, CA
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It is common industry and educational terminology to refer to the sinusoidal component as "AC" and the constant (or decaying or rising) component as "DC".
I certainly agree with that with respect to a voltage or current waveform. Does the same apply to a power waveform? That would mean that every power waveform in an AC system has a DC component (unless the power factor is zero).

I don't really know what is standard, that just sounds a little confusing to me. So in the case of a power waveform, I think it would be clearer to call it a constant component. The phrase "DC component of the power wave form" suggests to me the constant power attributable to the DC components of the current and voltage waveforms.

Cheers, Wayne
 

mbrooke

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It is common industry and educational terminology to refer to the sinusoidal component as "AC" and the constant (or decaying or rising) component as "DC".

Don't have time at the moment to read the thread in detail but from scanning the latest it seems for the others to deny that common terminology is just them being argumentative. It is a very, very long-standing use of terminology.

No one is denying that it exists, just the way its being used.
 

wwhitney

Senior Member
Location
Berkeley, CA
Occupation
Retired
Brain fart. What is RMS vs the total power in a wave forum?
Not sure I follow your question, but Smart$ points out an imprecision in my earlier comments. So let me expand on my earlier comments:

Say you have a function of time over some interval [f(t) defined on the interval (0,T)], such as one cycle of a periodic function, and you'd like to know the average value of the function. One obvious way to do this is to take the actual average, i.e. pick some (equally) distributed points in time, find the function values at those points in time, and average the function values [more precisely, average f = 1/T * integral(f(t)dt)].

So you can do this for a current waveform I, and for a voltage waveform V, and for the power waveform P = I * V. In DC, where I and V are contstant functions of time, then you get (average P) = (average I) * (average V). But in AC, that won't work, as average I = average V = 0 (assuming no DC component). In general, the average of the product is not the product of the averages.

In the case of the power waveform, the average we want really is the usual average described above. We just need to a new notion of average for current and voltage so that (average P) = (new_average I) * (new_average V). When the I and V waveforms have the same shape and phase (i.e. V = alpha * I for some constant alpha), then the notion of average for I and V that we need is root mean square, RMS. So we have (average P) = (RMS I) * (RMS V).

Cheers, Wayne
 

mbrooke

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Technician
Not sure I follow your question, but Smart$ points out an imprecision in my earlier comments. So let me expand on my earlier comments:

Say you have a function of time over some interval [f(t) defined on the interval (0,T)], such as one cycle of a periodic function, and you'd like to know the average value of the function. One obvious way to do this is to take the actual average, i.e. pick some (equally) distributed points in time, find the function values at those points in time, and average the function values [more precisely, average f = 1/T * integral(f(t)dt)].

So you can do this for a current waveform I, and for a voltage waveform V, and for the power waveform P = I * V. In DC, where I and V are contstant functions of time, then you get (average P) = (average I) * (average V). But in AC, that won't work, as average I = average V = 0 (assuming no DC component). In general, the average of the product is not the product of the averages.

In the case of the power waveform, the average we want really is the usual average described above. We just need to a new notion of average for current and voltage so that (average P) = (new_average I) * (new_average V). When the I and V waveforms have the same shape and phase (i.e. V = alpha * I for some constant alpha), then the notion of average for I and V that we need is root mean square, RMS. So we have (average P) = (RMS I) * (RMS V).

Cheers, Wayne

So RMS I times RMS V = RMS watts (actually work exerted)? I say this because most EE books have actual work as a complex equation, at least for 3 phase power.
 
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