THE PHYSICS OF... POWER

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winnie

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Location
Springfield, MA, USA
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Electric motor research
va = volt x amp = joule/coloumb x coulomb/sec = joule/sec = watt

show me using unit analysis what is not correct
very simple, except you can't because it is valid

I believe that I answered the above point in post 465.

You have two different measurements that have the same units. This does not mean you are measuring the same thing. Dimensional analysis is great; if after manipulation you get different units, then you _know_ you have two different things. However getting the same units doesn't mean you have the same thing.

The unit for V(rms) is Volts, but V(rms) is a time average, and a related but different thing than V(instantaneous).

If I have a supply which produces 10KV pulses for 100uS every second (100V RMS) would you be comfortable connecting 600V romex to the output of the supply?

-Jon
 

Ingenieur

Senior Member
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Earth
I believe that I answered the above point in post 465.

You have two different measurements that have the same units. This does not mean you are measuring the same thing. Dimensional analysis is great; if after manipulation you get different units, then you _know_ you have two different things. However getting the same units doesn't mean you have the same thing.

The unit for V(rms) is Volts, but V(rms) is a time average, and a related but different thing than V(instantaneous).

If I have a supply which produces 10KV pulses for 100uS every second (100V RMS) would you be comfortable connecting 600V romex to the output of the supply?

-Jon

volts are volts
Vrms = Vinst/sqrt 2
units are the same since sqrt 2 is unitless

I have to disagree
the units are the same
the convention is different
P watts
Q va (reactive)
S va

S = P + jQ
for those to add up they MUST be the same
the following text states that S, P and Q area all the same units
no ifs, ands, or buts
can't be any other way
 

Carultch

Senior Member
Location
Massachusetts
If I have a supply which produces 10KV pulses for 100uS every second (100V RMS) would you be comfortable connecting 600V romex to the output of the supply?


I think the rating of most cables is based on the expectation that the RMS nominal voltage of the circuit will relate to the peak voltage of the circuit by a factor of sqrt(2), assuming that it is either a continuous DC waveform, or a standard AC waveform. There might be some safety margin on that, such as 1.5 rather than sqrt(2), to account for small amounts of irregularities.

If the waveform isn't standard, you'd have to confirm with the manufacturer on whether or not you could use your cables and equipment.
 

Ingenieur

Senior Member
Location
Earth
Vpeak or max is the same exact units as Vrms or effective, only differing by the sqrt 2 factor.

reference the attached
some excerpts

reactive power Q has the same units as real P, but practice is to use 'var'

View attachment 15665
I did not write the text, pretty much the standard for power system analysis and design (used for seniors and first year grad students)
 

Ingenieur

Senior Member
Location
Earth
That's false as written. Perhaps you mean that for a sine wave, Vrms = Vmax / sqrt(2).

Cheers, Wayne
correct
or Vrms = Vpeak/sqrt2

what's true is P, Q and S all are the same units
but by practice or convention
P watts
Q var
S va

that is difficult for some to grasp, even though textbook excerpts show this
as does the unit analysis
they are using the physics definition of as W/t not the electrical engineering definitions
that power has to perform mechanical work...that is not true
 
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ggunn

PE (Electrical), NABCEP certified
Location
Austin, TX, USA
Occupation
Consulting Electrical Engineer - Photovoltaic Systems
I am an EE, but my educational experience was a long time ago and my experience since then has been in other areas, so I freely admit my ignorance of detailed power factor analysis. I have been slogging through this thread and reading some of the contributions in more detail than others, as well as trying to ignore the little putdowns and snide remarks.

Anyway, enough disclaimers. Here's what I see: When I consider a pair of AC waveforms, one for voltage and one for current which are not in phase, as would be the case for a load with a reactive component, if I multiply the RMS values of voltage and current I get one number for "power". If I multiply all the instantaneous values of V and I, I get something different, and the waveform I would see if I plot the result over time has two differences from what I would get if the V and I waveforms were in phase i.e., when it and the RMS product would be the same. One is that there are four zero crossings instead of two and the other is that twice per cycle the VI product is negative. If I were to integrate that waveform over a full cycle I would get something less than I would if the load were purely resistive, where I understand the result would be equal to the product of RMS V and I. The ratio of the integrated V*I waveform to the RMS product is the power factor, correct?
 

mbrooke

Batteries Included
Location
United States
Occupation
Technician
Perhaps i can slip a Q in here , one that was forwarded to me by an apprentice that had me stumped?

'Why does Hot to Neutral trip a breaker w/ a load , as opposed to a simple load" ? (think light fixture)

As a career Master , i'm embarrassed to say i was at a loss to respond...:(

~R(egg on face)J~

Maybe I read it to quick, but I don't understand the question. :?
 

wwhitney

Senior Member
Location
Berkeley, CA
Occupation
Retired
but the units are the same
the statement was made that they are not
Not sure what you're referring to, but my response in post 486 was only to the "differing by sqrt(2)" part, not the units part. But I can see how it could be read differently, sorry for the ambiguity.

Cheers, Wayne
 

Ingenieur

Senior Member
Location
Earth
Not sure what you're referring to, but my response in post 486 was only to the "differing by sqrt(2)" part, not the units part. But I can see how it could be read differently, sorry for the ambiguity.

Cheers, Wayne

correct
the OP was about power generation, as in 3 phase power, which afaik is usually a sinusoidal wave form
so Vrms = Vpeak/sqrt2
both units are in volts, the 'volt' unit definition is the same for both
 

winnie

Senior Member
Location
Springfield, MA, USA
Occupation
Electric motor research
but the units are the same
the statement was made that they are not
they are

What I have been saying is that the _same_ units represent _different_ quantities with different meanings.

The product of V(rms) times A(rms) has the same units as power.

But it _is not_ the rate that work is being delivered to the load, except in the common special case of a pure resistive load.

So even though it has the same units as power, it is _not_ power.

That is why it is called 'apparent power'. From a simple measure of V(rms) and A(rms) we get something that 'looks like' power but is measuring something else.

-Jon
 

Ingenieur

Senior Member
Location
Earth
What I have been saying is that the _same_ units represent _different_ quantities with different meanings.

The product of V(rms) times A(rms) has the same units as power.

But it _is not_ the rate that work is being delivered to the load, except in the common special case of a pure resistive load.

So even though it has the same units as power, it is _not_ power.

That is why it is called 'apparent power'. From a simple measure of V(rms) and A(rms) we get something that 'looks like' power but is measuring something else.

-Jon

I disagree
mixing physics/mechanics with electrical engineering muddies the water
all 3 are power, P, S and Q
all 3 are the same units va (if physics and the textbooks are to be trusted, lol)
they are the same 'thing' manifested differently (for P and Q, S is a composite of the 2)
that is why by convention only (not physical fact) that
P watts
Q va reactive
S va

if 100 VAC is impressed across a R, L and C in parallel a multimeter will read the same across them, phasing is another topic
 

FionaZuppa

Senior Member
Location
AZ
Occupation
Part Time Electrician (semi retired, old) - EE retired.
Responding to my statement that 1000VA and 1000W are similar in terms of losses in the wires feeding the load


I agree with your example above, with different load currents being served, you will see different losses.

What I was trying to say was that for the _same_ load current but different power factor, the losses in the wires will be roughly the same.

This is different from saying that kVAr directy means power 'lost'. In principal, the kVArs mean energy that is flowing back and forth without being consumed. kVArs are associated with _increased_ losses, but are not losses themselves.

I am uncomfortable with your use of 'impedance mismatch', but don't have a good enough understanding of the implications to say that your use is wrong.

In RF work, the goal is to match impedance because impedance mismatch results either in reduced power output or standing waves on transmission lines. But in 60 Hz power distribution you have different goals. You want a 'stiff' low impedance supply, you don't worry about the 'characteristic impedance' of the wires, and you don't match the load to the supply, and you don't match the impedance of the wires to the load. The closest thing you do is to match capacitive and inductive loading, without matching the resistive part of the impedance.

-Jon


there will always be a src Z and a load Z, for everything. post #1 describes classic impedance mismatch.
the kVAr's are indeed just lost power. kVAr has amps associated with it, you think those amps are created and flow free of penalty? its the exact reason why gen side needs to account for it, its capacity headroom based on the general understanding of what the load will be, but yet load characteristics will change, load is very dynamic in many terms (real power, Q, etc etc).

i build a gen station so i can bill for 20MW of power per month, great, thats all the $$ you get, but it will cost some $$ to deliver it, which is partially dictated by the very dynamic customer load.

here's another simple analogy. 1950's milk truck takes cold milk by truck, they create ice blocks at the milk plant, by the time delivery is done for the day the ice block has melted and is gone, not usable any longer, its the cost (penalty) for delivering the milk. milk co only charges for the milk though consumed by Mom, Pop, Kids, etc. in the end, both models account for the extra costs, built into kW and milk cost rate, both also adjust for equipment, bigger truck to accommodate ice blocks, bigger electrical gear to supply those extra amps, etc etc etc. then came along R22, so the ice blocks are no longer needed, so you can either downsize the truck and carry same milk, or keep bigger truck and carry more milk. R22 added some efficiency thus less waste, while at the same time the milk cost rate remains the same.
 
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