THE PHYSICS OF... POWER

[GoldDigger]

then no current flows
it's counted once for each pass thru the plane regardless of direction
the net is not 0, it is 2 times the 1/2 cycle

get your $$$ back lol
you should know the basics

Your definition of net current flow as counting the number of times an electron crosses a plane, regardless of direction is at odds with most of the world's used of that word. (And that is based on both formal academic and practical use of the word "net", so I am not arguing a popularity contest.)
If I pay someone $10 and he later gives me $10 in return, there has not been a net money transfer between us of $20.

FWIW, even if I accepted your definition of net, that would mean that there is also, in the exact same way, net current flow in a purely reactive circuit, so your distinction of a net flow indicating real power is wrong.

 

[Ingenieur]

Your definition of net current flow as counting the number of times an electron crosses a plane, regardless of direction is at odds with most of the world's used of that word. (And that is based on both formal academic and practical use of the word "net", so I am not arguing a popularity contest.)
If I pay someone $10 and he later gives me $10 in return, there has not been a net money transfer between us of $20.

FWIW, even if I accepted your definition of net, that would mean that there is also, in the exact same way, net current flow in a purely reactive circuit, so your distinction of a net flow indicating real power is wrong.

it better not be
if we consider it to be opposite signs no current flows, sums to zero lol

if you toss a ball over a net back and forth an even amount of times some amount work is done, not 0, lol
in your example there has been a cash 'flow' of $20

now I got to get some work done
they don't pay me $120/hr (gross, not net, lol) for my good looks
just think, with only a 6th grade education to boot lol

 

[GoldDigger]

in your example there has been a cash 'flow' of $20

But not a net cash flow!

Net is not always a synonym for total.


Fortunately your job involves much more sophisticated calculations, so this disagreement about the fundamentals does not appear to be adversely affecting your results. :angel:

 

[cuba_pete]

When I try to explain how low pf makes equipment sizes bigger with no increase in ability to do work I show them the beer mug analogy. pf is the ratio of foam to beer. Mug sized to handle both but a pf closer to unity maximizes your drinkable beer for the same mug size.

So, my kid poured me a beer the other day and I ended up with too much foam. She said to drink some of the foam off of the top so she could fit the rest of the bottle.

First off, she should have used the chilled glass and poured correctly, which she now knows.

I told her it was a waste of the reactive beer...and everybody just stared at me.

I said thank you and took the glass and the bottle to the living room.

 

[Besoeker]

only if non-oscillatory

It it moves further in one direction than the other, that's DC.

 

[wwhitney]

if we consider it to be opposite signs no current flows, sums to zero

Precisely right! But the voltage is also changing sign. So, e.g., in a purely resistive circuit where the voltage and current are in phase, the power is always nonnegative, P=IV. The electrons are carrying energy in the same direction both when the current is positive and the current is negative. Net energy flow, no net electron flow.

Cheers, Wayne

 

[Ingenieur]

Precisely right! But the voltage is also changing sign. So, e.g., in a purely resistive circuit where the voltage and current are in phase, the power is always nonnegative, P=IV. The electrons are carrying energy in the same direction both when the current is positive and the current is negative. Net energy flow, no net electron flow.

Cheers, Wayne

but there is electron flow
or P = V I, and if I = 0, so therefore P = 0

 

[wwhitney]

P = V I, and if I = 0, so therefore P = 0

This would be true if I, V, and P were scalars, e.g. for DC.

With AC, each of I, V, and P are functions of time.

So all of the following can be true simultaneously, when integrating over some finite interval of time:

Integral (I) = 0.
Integral (V) = 0
Integral (I * V) > 0.

Cheers, Wayne

 

[jtinge]

So, my kid poured me a beer the other day and I ended up with too much foam. She said to drink some of the foam off of the top so she could fit the rest of the bottle.

First off, she should have used the chilled glass and poured correctly, which she now knows.

I told her it was a waste of the reactive beer...and everybody just stared at me.

I said thank you and took the glass and the bottle to the living room.

You needed a mug with a larger rating to contain the beer with a low beer factor.

 

[K8MHZ]

Can someone explain this to me?

If reactive power isn't power, why does it take more power at the generating end to deal with it? To me, that seems like a violation of the law of conservation of energy.

If reactive power just oscillated to and fro, wouldn't the result be 'nothing' to the 'eye' of the generating end?

I am probably overlooking something that to the rest of you is obvious.....

 

[GoldDigger]

Can someone explain this to me?

If reactive power isn't power, why does it take more power at the generating end to deal with it? To me, that seems like a violation of the law of conservation of energy.

If reactive power just oscillated to and fro, wouldn't the result be 'nothing' to the 'eye' of the generating end?

I am probably overlooking something that to the rest of you is obvious.....

I would not say that it is obvious, but I will try to explain it simply:

1. The prime mover of the generator (steam or gas turbine, diesel, etc.) needs to be able to deliver enough power to keep the generator shaft turning at the regulated design speed. To do this it ONLY has to be able to produce power to match the "real" power consumed by the loads and the sum of the losses in the distribution network. (If the generator is three phase, then for balanced loads of matching PF the prime mover will see a constant torque load, regardless of whether there is reactive power flowing out and back at different times in the cycle.)
2. Adding in loads with low PF increases the current that the generator must deliver, and so increases the distribution network losses in the same proportion. To that extent the prime mover must be able to deliver more power, but this is not nearly in direct proportion to the apparent power.
3. The generator itself (windings, etc) must be able to deliver the current to match the apparent power, even if this does not show up as a load on the prime mover. To deliver this additional current the generator needs to be more heavily built than it would be if it only had to deal with the real power.
4. An additional consideration is that the generator output must be regulated to match the needs of the grid, including maintaining output voltage or even increasing it to deal with generator and distribution network losses. This regulator is a feedback loop and under some conditions of out of phase current and voltage the regulator may not be stable. To that extent delivering a large amount of reactive power will stress the design and operating parameters of the generator itself, again independent of the prime mover.

If that does not answer your question, please let me know what parts are still giving you problems.

 

[Besoeker]

Can someone explain this to me?

If reactive power isn't power, why does it take more power at the generating end to deal with it? To me, that seems like a violation of the law of conservation of energy.

If reactive power just oscillated to and fro, wouldn't the result be 'nothing' to the 'eye' of the generating end?

Pretty much so apart from losses in conductors etc.

 

[dionysius]

If reactive power just oscillated to and fro, wouldn't the result be 'nothing' to the 'eye' of the generating end?

I am probably overlooking something that to the rest of you is obvious.....

you are correct when you state that reactive power is not real power

 

[GoldDigger]

you are correct when you state that reactive power is not real power

But regardless of how you stand on the power issue, reactive current is perfectly real current.

 

[ATSman]

A delicious analogy!

I agree................ I'm just here for the beer!!!!!!!!!!! :thumbsup::thumbsup:

 

[Smart $]

Can someone explain this to me?...

... I will try to explain it simply...

It's all real power, but when the load(s) is reactive and assuming the power generation is remote, some real power is realized locally and the balance is not.

:angel:

 

[Fulthrotl]

I agree................ I'm just here for the beer!!!!!!!!!!! :thumbsup::thumbsup:

yup. and two pages after the first beer, they are all lining up to see
who can pee the farthest......

but, we shall need a mechanical engineer to properly calculate
the total developed length of the streams, as elevation of the
streams might be considered (by some) as "reactive distance".

this isn't going to turn out well......

 

[mivey]

pf is the ratio of foam to beer.

pf is ratio of beer to mug size.

 

[mivey]

Power is power.
Apparent power isn't power.
It's as simple as that.

Active power is active power. It's as simple as that.

There are also other terms that include the word "power" that are not active power. Any of these terms use the short-hand "power" when the clarifier is dropped. Arguing that one term is or isn't "power" hinges on the definition used and is kinda pointless. But carry on if you must.

 

[mivey]

I'm with Golddigger, it's 3 to 1 against you.

I'll join in as well. Otherwise you have a DC component as was said earlier by others.