THE PHYSICS OF... POWER

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FionaZuppa

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Part Time Electrician (semi retired, old) - EE retired.
Power is power. VA and VAr are not.
I don't know why I bother......

i tried to explain with my steam piston analogy a few pages back.

for all, jX is √-1X (pure reactance component of the complex impedance, or imaginary power) but it cannot dissipate/consume/convert power (watts). however, this √-1X component is partially responsible for the required amps, but not all of the energy associated with those amps is used by the load, so now you have what looks like a diff between VAgen and kWload, some of this diff is in fact wasted energy due to the real components of the gen and transmission lines, and some of that diff is real energy that is returned back to the gen. with large kVAr load the gen will see kickback which is similar to flyback from inductors. do PF correction by adding Xc near the load and the kickback and transmission losses drop due less amps needed to provide the same kW to the load.

and just for clarity, VA is associated with "power". what happens often is we forget to apply the std equations, too often just use the collapsed version for pure resistive load. one you have Φ between I and V the std I2R is no longer valid since R is really Z which carries √-1 component and not all the amps are used for load kW, etc.

I like the 'time average' bit showing what goes back and forth from poco to the inductive load has 0 average power, thus it is imaginary power, but along the way it sure makes my wires warmer - I think?
yep, the amps differential i mention is used to create that heat you speak of and needs to be accounted for in the prime mover.
 
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mike_kilroy

Senior Member
Location
United States
It isn't power. Power is expressed in Watts.
It isn't Watts.
I don't which part of that you don't get.

If there is no power in reactive va, then why does the wire it runs thru get warm due to I^2R losses from it? If it has no power, then we would not have to concern ourselves with transmission line losses, no?

I like the 'time average' bit showing what goes back and forth from poco to the inductive load has 0 average power, thus it is imaginary power, but along the way it sure makes my wires warmer - I think?
 

FionaZuppa

Senior Member
Location
AZ
Occupation
Part Time Electrician (semi retired, old) - EE retired.
if this doesn't do it I'm out of ideas lol :lol:

i read the pdf.
i think i see the issue.

"power" is a bad term to use because a characteristic of X is that it will absorb and release energy in terms of watts (J/s) but this energy is just stored and released and is not converted to work at the load. must remember, watts does not always = work. how so you ask, think of a lossless pure cap and inductor that simply move charge back and forth, during the non-zero-crossing times you will observe Ampsvector , and you can easily then describe it in terms of Watts, yet no energy is converted, it just moves back and forth, etc.
 
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Carultch

Senior Member
Location
Massachusetts
If there is no power in reactive va, then why does the wire it runs thru get warm due to I^2R losses from it? If it has no power, then we would not have to concern ourselves with transmission line losses, no?

I like the 'time average' bit showing what goes back and forth from poco to the inductive load has 0 average power, thus it is imaginary power, but along the way it sure makes my wires warmer - I think?

There is no net power due to "reactive power", after you keep track of complete cycles of AC.


However, it does directly induce additional real power consumption involved in the transmission of the intended real power, and therefore is a real issue that needs to be managed when building a power grid. There are more I^2*R losses than there otherwise would be for a real power only situation, because the current out of sync with the voltage is less effective at transmitting real power. That additional current, means more resistive power loss along the way.


But in any case, assuming that you lump the reactive power elements separate from their inherent resistance, there is no net power consumed in the ideal inductors and ideal capacitors.
 

Ingenieur

Senior Member
Location
Earth
There is no need for an analogy. Power is Watts. It really is that simple.
I don't know why such a Fandango is being made out of it.

once more
phasor notation
S = VI* = P + jQ
P = VI* cos ph ang
Q = VI* sin ph ang
note that cos and sin are unitless making S, P and Q all the same units

watt = joule/sec = v x a = joule/coulomb x coulomb/sec = joule/sec = watt = va

active power P is power
reactive power Q is power
complex power S is power
apparent power |S| is power

all the same units va = joule/sec = watts
by convention
P watts
Q var (volt-ampere reactive)
S and |S| va (volt-ampere)

as elementary as it gets :thumbsup:
 

mike_kilroy

Senior Member
Location
United States
There is no need for an analogy. Power is Watts. It really is that simple.
I don't know why such a Fandango is being made out of it.

Then how does the reactive power cause I^2R losses in the feedline?

Seems we are just hung up on reality. We ALL know reactive power has 0 avg. power as it just flows back and forth source<->load, and I think we ALL know if there is any transmission line between the source & load, it will heat at the rate of I^2R WATTS, where I is the imaginary or reactive current 90 degrees out of phase with the voltage...

This APPEARS to contradict itself but it really does not. The motor will make a magneteic field out of this reactive power with 0 average power. It makes it from the volts applied x the 90 degree shifted amps applied. But before it gets those volts from the source, the transmission line in between has I*R loss, so the voltage getting to the motor to make that reactive power is Vsource-IR! So it is slightly LESS than the source put out. So the reactive power made is made from a slightly lower voltage and THIS reactive power flows back and forth with effectively 100% efficiency. So it all works out at the end of the day, no?
 
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Ingenieur

Senior Member
Location
Earth
assume a current of 100/25 deg (90.6 + j42.3) flows to a load (pf is approximately 0.90)
assume line R is 1 ohm

loss S = I I R = 100/25 100/25 1/0 = 10000/50 = 6428 + j7666
loss P =6428 W
loss Q = 7666 var (not really lost as work)

only the active part of the current contributes to IR losses in watts, so 6428 W in this case
only 6428 of 'heating', not 10,000
but the ckt ampacity must be 100 A, not 90.6
 

FionaZuppa

Senior Member
Location
AZ
Occupation
Part Time Electrician (semi retired, old) - EE retired.
There is no need for an analogy. Power is Watts. It really is that simple.
I don't know why such a Fandango is being made out of it.

well, semantics again. the power guys will think kW, the physics guys will think "hmmmm, thats a loaded term". if you are defining it solely by power=watts then you have collapsed the last 30 pages into VavgAavg (purely resistive), which is not true in real world.

you cannot generalize "power". in physics the term "power" is defined as P=Jtransformed/t which is directly related to work. the reactive component that does no work can therefore (and should not be) marked as "power".

if you move all the math into eV units then things become a tad clearer, until then the use of "power" in the math will remain somewhat confusing if you stick within the physics of it all.

if i generalize, the input to gen does some work to increase eVtotal, some of that eVwork is used by the load, some of that eVwasted is lost due to real R of transmission, and some of that eVreturned is given back to the gen. eVwasted+eVreturned is a function of jX, which the power guys often express as PF. i can even break out eVwasted into two other terms, eVwasted(cost to do work) + eVwasted(cost to handle jX), which is nothing more than the Amps associated with the terms of Z. as we see, remove jX from Z and the amps associated with jX drop out to zero.
 
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