VA vs. Watts

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ok ok, so i didnt read all 13 pages of posts, by like 8 or 9 i knew i know nothing......

i absolutly love the knowledge and debate of this place, it helps us to keep learning, growing, and last but not least, humble.

thanks.
 
What is all the fuss about?

What is all the fuss about?

Besoeker said:
OK.
Fine.
Interchange 55kVA to the equivalent kW.
What is your answer?
Click to expand...

My Freshman physics book states, "power is the time rate of doing work"--real power that is. Energy flows in one direction, and we measure this flow in watts.

Then, by this definition, "reactive power" is really a misnomer because no work is done. Energy is merely moved back and forth, and we measure this flow in VARs.

Real and reactive power combine in the power triangle to obtain apparent power which we measure in VAs.

Clearly, these units, although similar in nature, are not interchangeable.
 
Besoeker said:
Interchange 55kVA to the equivalent kW.
What is your answer?
Click to expand...
55KVA = 55 KW. You didn't read anything I wrote, did you?

If you asked me to convert the "Total Power" of 55KVA to the "Real Power" given a specific powerfactor, that is a legitimate question with a different answer. But KVA and KW are the same, so you cannot convert one from the other. You convert the "Function"!!
rattus said:
Then, by this definition, "reactive power" is really a misnomer because no work is done. Energy is merely moved back and forth, and we measure this flow in VARs.
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No, you also didn't read what I wrote. Electrons don't just go bouncing around for no apparent reason, and they don't move back and forth between the load and the source doing nothing. These electrons are flowing through a voltage differential, and that by definition, is expending or producing work. Work is expended in one portion of the cycle and given back in the other portion. Just because the net/average work is canceling out doesn't mean that no work is being done in the process.

If you climbed up a big hill and then rolled down the other side, you have not completed any net work, but does that mean that you won't be out of breath from your efforts? No! That is exactly how reactive power works. It is both Give and Take!

As I have already stated, you are misusing average power concepts by assumptively applying it to all concepts.
 
If I may add my take on the debate...

The argument seems to be whether real power watts and reactive power watts are the same thing. In my opinion, it depends on what you're really discussing.

No, the reactive power gets nothing productive done, but yes, it heats up the wire it flows in just the same. Watts is watts, they're just not all productive.

The reactive power has volts and amps, just as the real power does. However, the reactive power represents energy that flows through the system, but does not reflect a transmission of energy from the power source to the load.

How'd I do? :)
 
Cold Fusion said:
Rick -
Still waiting on an answer (RE: VARS)
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I am sorry, I could have sworn I posted a response to this last night, but maybe I am just getting senile. VARs is an unusual case because we have taken a basic unit of measure and added a secondary condition to it. The unit of measure is the same, which is VA or Watts or j/s, but we have added a condition to this indicating that the units are limited to being "reactive".

This is comparable to using units of Feet-Right or Feet-Left when dealing with distance. The core units are still Feet, but we have added the exterior condition of Right or Left. If you stepped 5 feet to your left, but were instructed to use the dimension of Ft-R, then you would have to say that you moved (-5) Ft-R. The core units are still feet, but you have moved a portion of the function into the units.

Edited to add, that this does not apply to VA, Watt, or j/s. This is ONLY the special case of VAR, and it is the "R" in VAR that compels this.
Cold Fusion said:
Perhaps you could take time to read some of the posts.
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I have read the posts and I fully understand the topic. Your VARs example is the only situation that comes close to fully understanding the situation. But note that VARs is the ONLY unit throughout this discussion that imposes this additional condition.
LarryFine said:
No, the reactive power gets nothing productive done, but yes, it heats up the wire it flows in just the same. Watts is watts, they're just not all productive.

The reactive power has volts and amps, just as the real power does. However, the reactive power represents energy that flows through the system, but does not reflect a transmission of energy from the power source to the load.

How'd I do? :)
Click to expand...
Sorry Larry, but this is the same thing that Rattus stated. Reactive power does do work, but like Rattus and others, you are confusing the average work with the belief that no actual work is being performed.

Reactive elements, by their very definition, store energy for a short period of time. They expend energy as they are charged and expel energy as they discharge. For example, it takes energy to build the magnetic field around a coil of wire, and energy is given off as this magnetic field collapses. This is real and actual electrical work, but when you examine the power over a complete cycle, this "giving" of energy and "taking" of energy cancel out, which makes it seem as though the current (electron movement through a voltage) does no work.

I modified the original drawing from my previous posting, but this time I added some color. The Red portion of the graph is where the reactive element (capacitor/inductor) puts power back into the supply, and the Green portion of the graph is the equivalent amount of this power that gets cancelled out (just a photoshop estimate, but these two areas should be mathematically equal.)

When you look at the time period of the cycle, the time beginning at the start of the red through the end of the green actually amount to zero average power. The cancelation of these two very real and exiting powers is what we now classify as "reactive power". One is positive, and the other is negative, and the total of the two is zero.

These Red and Green areas are what we "ignore" or "remove" from your normal equation of power. But, just because we remove them (by cancellation) does not mean that they do not exist by themselves.

When you remove these two portions of the instantaneous power, you get what we call the "real power". It is not that these two portions do not have actual power, but that they continually cancel out with every cycle of the graph.

When people say that the current just bounces back and forth doing nothing, it is the cancellation of the red and green areas that make it seem this way. But in reality, one is giving work (red) and the other is expending work (green).

As I said earlier, the Red and Green areas should be mathematically equal, but I simply used photoshop to add these colors, so I think the green area is too big.

InstantaneousPower2.jpg
 
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Oh, one more follow-up comment to the above.

VARs is not the Red area. VARs is not the Green area. VARs is the sum of both Red and Green, and this is why VARs "appears" to provide no actual work.

"Appears" is the key word here, because it only "Appears" that no work is completed. It "Appears" that no work is completed because the positive work and the negative work cancel out. Work was done, and it was real.
 
Besoeker said:
Interchange 55kVA to the equivalent kW.
Click to expand...
Besoeker said:
What is your answer?
Click to expand...

I don't quite know what you are trying to do here. KVA and KW are not equivalent. I, for one, never said they were, and I am not sure that anyone else did either. Real power, reactive power, and apparent power are different physical phenomena, and we use different names for their respective units of measure. But their units of measure break down to the same fundamental unit: kilogram-meter squared per second cubed. Do you dispute that statement?
 
Rick Christopherson said:
This is comparable to using units of Feet-Right or Feet-Left when dealing with distance. The core units are still Feet, but we have added the exterior condition of Right or Left. If you stepped 5 feet to your left, but were instructed to use the dimension of Ft-R, then you would have to say that you moved (-5) Ft-R. The core units are still feet, but you have moved a portion of the function into the units.
Click to expand...
What if you used feet up for real power, and feet left and right for reactive power? Would that make for a better analogy?



Sorry Larry, but this is the same thing that Rattus stated. Reactive power does do work, but like Rattus and others, you are confusing the average work with the belief that no actual work is being performed.
Click to expand...
What I meant is that the reactive power does not get taken from the motor shaft and do that kind of real work.

I imagine the fact that the reactive power is capable of contributing to conductor heat makes it real power (and real watts.)
 
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Rick Christopherson said:
--- VARs is an unusual case because we have taken a basic unit of measure and added a secondary condition to it. The unit of measure is the same, which is VA or Watts or j/s, but we have added a condition to this indicating that the units are limited to being "reactive".---
Click to expand...
Rick Christopherson said:

---Edited to add, that this does not apply to VA, Watt, or j/s. This is ONLY the special case of VAR, and it is the "R" in VAR that compels this.---

--- But note that VARs is the ONLY unit throughout this discussion that imposes this additional condition.---
Click to expand...

Rick -
I'm trying not to selectively quote. I'm trying to pick out the pieces of your discussion centered on this one topic, because what you are saying just doesn't make any sense.

You think vars are a special case because they are reactive. Well, watts are a special case because they are real. VA is pointless (that would be a pun)

Watts have a direction
Vars have a direction
Watts + Vars = Complex power (not apparent power)
Complex power has a direction

VA is apparent power.
Apparent power = magnitude [complex power]
Apparent power does not have direction

You know all of this. charlie b knows all of this. I don't understand why there is this insistance to dump the vector information.

Interestingly, your position is very similar to where charlie b was by post 35.

Repeat after me:
Watts + Vars = Complex power.
Complex power is a vector

Unless we are sizing wire, (which includes generators, transformers, switchgear) apparent power is pointless (not a vector)


cf
 
charlie b said:

I don't quite know what you are trying to do here. KVA and KW are not equivalent. I, for one, never said they were, and I am not sure that anyone else did either.
Click to expand...

OK, I know they are not the same, and you know they are not the same. But, this from post #125 claims not just that they are equivalent, but identical.
The units are Identical. They are Equal. They are Interchangeable!!!!!
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Do you agree with that statement?

charlie b said:
Real power, reactive power, and apparent power are different physical phenomena, and we use different names for their respective units of measure. But their units of measure break down to the same fundamental unit: kilogram-meter squared per second cubed. Do you dispute that statement?
Click to expand...
I already answered that in post #44.

I do.
"The dimensional analysis is not disputed.
One Volt times one Amp is one Watt. Provided they occur at the same time*.
It works for instantaneous and steady state values. For time varying or periodic waveforms the instantaneous values of current and voltage give Watts.
For an alternating circuit, displace the current and voltage by a quarter of a cycle and you still have VA but no W.
They are not the same units."

I still think it is quite a lucid explanation of the quantities being discussed.
*The need for VA is when they don't occur at the same time.
 
rattus said:
My Freshman physics book states, "power is the time rate of doing work"--real power that is. Energy flows in one direction, and we measure this flow in watts.

Then, by this definition, "reactive power" is really a misnomer because no work is done. Energy is merely moved back and forth, and we measure this flow in VARs.

Real and reactive power combine in the power triangle to obtain apparent power which we measure in VAs.

Clearly, these units, although similar in nature, are not interchangeable.
Click to expand...
Thank you.
You made the point with the calrity and brevity that we should all seek to emulate.
 
Looks like I missed out on some fun. Well, no time like the present to dive in.

It is true that VA, W, and var really have the unit of watts. By convention, we use descriptive units (VA, W, var) to reduce confusion.

Once we accept this convention and use the descriptive units, we have essentially limited W to designating the average (real) power and really should not go back to using W in the general sense. Also, once we use this convention, it would be wrong to say that VA=W=var, even though they all have the same pre-convention unit of W.

The instantaneous power is composed of an average component and a sinusoidal component of frequency 2ω. When we use P, we are normally talking about the average, or real component of power and use the descriptive unit W. When we use Q, we are normally talking about the imaginary component of power and use the descriptive unit var.

When we use S, we are normally talking about the complex power which is the combination of P & Q (W & var) and use the descriptive unit VA. The apparent power is |S| and also uses the unit VA.

It seems a lot of the discussions here are mixing pre-convention and post-convention units, which can cause confusion.

Yet, strictly speaking, the units are all watts and convert to the base units of kg*m^2/s^3.

I'm trying to think of another analogy: it might be similar to kinetic energy and potential energy. A roller coaster car running down the track has KE=1/2*m*V^2. When it climbs the hill, the energy is converted to PE=m*g*h. Both are energy, and will convert to the same base units, but KE & PE describe two different things.

Besoeker said:
Only for the very specific case of unity power factor.

55kW tells you precisely the rate of doing work.

55kVA tells you the product of current and voltage.

You simply cannot infer or equate it to the rate of doing work. It could be some. Or none.
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VA is a measurement of the rate of work or energy flow. For our circuit, two things happen:
1) W (the real component of VA) is used as a measure over a specific time to indicate the heat dissipated by a resistor, or energy converted to mechanical energy, etc
2) var (the imaginary component of VA) is used as a measure over a specific time to indicate how much work was done to move the charges to store energy in the capacitor's electric field, or the work done to store energy in the inductor's magnetic field.

The energy stored still has the capacity to do other work.

One other problem may be that some are using the term power as a general term when they are really referring to average power (as most of us probably do).
 
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