VA vs. Watts

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I have heard that for all intents and purposes, VA and watts are the same.

Is this true? Does anyone have a good explination of the difference?
 

infinity

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New Jersey
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Technically they're not the same. I'll defer the textbook explanation to someone better equipped to give a concise explanation.
 

bphgravity

Senior Member
Location
Florida
VA is the apparent power.

Watts is the true power.

To get the Watts from the VA, you must include the power factor (pF) of the circuit.
 

charlie b

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I wrote this up some time ago, and post it from time to time, as the question of power factor is raised again.
There are three different kinds of electrical power. There is ?real power,? and there is ?reactive power,? and the two add up (in a trigonometric sort of way) to give you ?apparent power.? What causes ?reactive power? to come into the picture are things that involve magnetic fields (such as the windings of motors, ballasts of fluorescent lights, and welding machines) and things that involve electric fields (such as capacitors and synchronous generators).

I don?t really like the ?beer and foam? analogy for power factor. But here it is anyway. Real power is like the liquid that fills most of a glass of beer. Reactive power is like the foam at the top of the glass. What you really want is the liquid, but the foam is there and you can?t avoid it being there and it doesn?t change the amount of liquid at the bottom. Suppose you start with a glass with some amount of liquid, and a little foam at the top. In order to contain both, the glass has to be a bit larger than necessary for the liquid alone. Suppose you can generate some extra foam, and suppose you pour it into the glass. You haven?t changed the amount of liquid, but you need a bigger glass to contain it all.

Now back to electrical stuff, and a bit of trigonometry. Take a sheet of paper, and draw a horizontal line about one inch long. Label the left hand end of the line as point ?1? and the right hand end as point ?2.? From point 2, draw a vertical line upwards one inch. Label the top of that line as point ?3.? Finally connect points 1 and 3 to form a triangle. Line 1-2 represents the real power, and it is a measure of the ?real work? being done by the load equipment. Line 2-3 represents reactive power, and is a measure of the energy being exchanged (you might even say ?wasted?) between the magnetic field of the utility generator and the magnetic field of the customer?s motors and other inductive loads. Line 1-3 represents apparent power, and is a measure of the power that the utility has to provide, in order for the customer to run the motors.

Power factor is a measure of the angle located at point 1. In terms of trigonometry, PF is defined as the cosine of that angle. If that angle is zero degrees (which happens if line 2-3 has no length at all, so that line 1-2 and line 1-3 lie one on top of the other), then PF is unity. If that angle is large (which happens if line 2-3 is long), the PF is considered ?poor.?

What would happen to that triangle, if you drew the vertical line as 2 inches long, instead of 1 inch? The real work, line 1-2, (like the liquid portion of the glass of beer) does not change. But the apparent power provided by the utility (line 1-3) is now much longer. Thus, in order to allow the customer?s motors to do the same amount of work, the utility has to provide more power (i.e., you need a larger glass to hold the same amount of liquid with a larger amount of foam).

What would happen to that triangle, if you drew the vertical line as 1/2 inches long, instead of 1 inch? Here again, the real work, line 1-2, does not change. But the apparent power provided by the utility (line 1-3) is now much shorter. Thus, in order to allow the customer?s motors to do the same amount of work, the utility has to provide less power (i.e., you can get by with a smaller glass, since there is less foam to contain). You can see why the utility does not like the customer to have a lot of reactive power in their loads. You can see why the utility will charge some customers (large industrial plants, for example) a penalty for having a poor power factor.

What a power factor correction device does is to reduce the length of line 2-3. It does that by adding a reactive power load of a type that is opposite to that in the customer?s loads. Usually, a customer?s loads are motors and other inductive loads. To that, you add a capacitive load. What that does to the triangle is like starting at point 3, and drawing a line downwards towards (but not as far as) line 2. Call the end of this new line point ?4.? Here?s what you get: Line 2-3 is the customer?s inductive loads. Line 3-4 is the power factor correction capacitors. Line 2-4 is the new value of total reactive power. Line 1-4 (I didn?t tell you to draw that yet, did I?) is the new value of total apparent power. You will note that line 1-4 is shorter than line 1-3. So the amount of power supplied by the utility has gone down.

What is physically happening in the power factor correction device is that, where you used to have an exchange of energy between the between the magnetic field of the utility generator and the magnetic field of the customer?s motors and other inductive loads, you now have an exchange of energy between the magnetic field of the customer?s motors and the electric field of the power factor correction capacitors. The utility company meter does not record that energy exchange, because it takes place entirely within the customer?s facility.

One final note. The units of measure for the three lines described above are fundamentally the same. They all relate to the rate of use of energy, and can all be expressed in terms of ?joules of energy per second of time.? But to keep track of them separately, they are given separate names. The ?real power? unit (line 1-2) is expressed in ?watts (W)? or ?kilowatts (KW).? The ?reactive power? unit (line 2-3) is expressed in ?volt-amperes reactive (VAR)? or kilo-volt-amperes reactive (KVAR).? The ?apparent power? unit (line 1-3) is expressed in ?volt-amperes (VA),? or kilo-volt-amperes (KVA).?
 

laketime

Senior Member
been in the electrical industry for 20 years and have never had it explained to me so well. Thanks!
 

Jraef

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I use the Beer and Foam analogy when I have to explain why the utility wants bigger customers to correct power factor. The PoCo is the barkeep, he buys beer and resells it, foam or no foam. The user is the pub patron. He buys beer, not foam, so he expects the bartender to pour off the foam. That pour off is a loss of revenue to the barkeep, but for the occasional tourist who comes in for a beer once in a while and doesn't ask for a discount, it doesn't matter much. But he insists that his biggest drinker local patrons who expect a better price bring their own beer mugs that reduce foaming, or pay a premium for the beer to make up for the expected loss.

For the end user's perspective, I like to use the analogy of horses pulling a rail car. If the horses are walking right on the track in front of the rail car, all of their power is being used to pull the car along the track, so that is like having a PF of 1.0 or Unity. But if the horses move away from the rail, an amount of the energy they exert is wasted on pulling the rope sideways, not in advancing the car down the track. The difference between the total exertion by the horses vs the force used in moving the car along the track is the Power Factor. If the PF = .80, that means that 80% of the horses' strength is used to pull the car along, 20% is just keeping the rope tight. The extreme is if the horses move out to where the rope is 90degrees from the rail car; all of the energy is pulling sideways, the car cannot move. That's like having a PF of zero. PFC capacitors are basically like a teamster that guides the horses back to being closer to the track again so that more of the force they exert is used to do useful work.

Then when the "energy saver" guys try to tell me that their teamster (capacitor) saves energy, I can point out that the number of horses hasn't changed, nor has the amount of hay they expect to eat.:grin:
 

Besoeker

Senior Member
Location
UK

From me, you may or may not get "interesting," but I hope you at least get "clear." However, "concise" is not in my nature. :wink:

It's a concept I have understood for many years so it is difficult for me to judge the clarity from the perspective of someone who is not entirely familiar with it but, given the positive feedback from others, it has obviously been received well. Kudos for that.:smile:

However, I do think you should correct the last paragraph, Specifically:

The units of measure for the three lines described above are fundamentally the same. They all relate to the rate of use of energy, and can all be expressed in terms of ?joules of energy per second of time.?
As I'm sure you know, Joules/second is power in Watts. There is no power associated with the vertical line. VArs are Wattless, thus can't be correctly expressed in terms of "joules of energy per second of time?.
 

drbond24

Senior Member
Actually I believe charlie b is right.

V = (kg*m^2) / (C*s^2)
A = C / s

V*A = (kg*m^2) / (s^3) = (kg*m^2) / (s^2) * (1/s) = J / s

Broken down into it's most fundamental pieces, a Volt-Amp is still Joules per second.

:)
 

charlie b

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However, I do think you should correct the last paragraph
No need. It is correct as written.
As I'm sure you know, Joules/second is power in Watts.
Not quite true. It is like the difference between saying “a dog is an animal” and “an animal is a dog.”

A “Watt” is a “Joule per second,” but a “Joule per second” can be a “Watt” or it can be one of at least two other things. Specifically, a “Joule per second” could be a “Volt-Ampere Reactive,” which is a measure of reactive power, and a “Joule per second” could also be a “Volt-Ampere,” which is a measure of apparent power.
There is no power associated with the vertical line. VArs are Wattless, thus can't be correctly expressed in terms of "joules of energy per second of time”.
If you take each type of power down to the very fundamental units (i.e., length, mass, time, and charge), then,

? “Real Power” is measured in units of “Kilograms x (meters squared) / (seconds cubed)"
? “Reactive Power” is measured in units of “Kilograms x (meters squared) / (seconds cubed)"
? “Apparent Power” is measured in units of “Kilograms x (meters squared) / (seconds cubed)"
 

Besoeker

Senior Member
Location
UK
?Joule per second? could also be a ?Volt-Ampere,? which is a measure of apparent power.
So you want to redefine the Joule?
A Joule per second is a Watt which power, not apparent power.
Take a reactive load like a capacitor.
The current it takes times the applied voltage gives it's VA.
Ignoring losses (a close approximation at power frequencies), there will be no power and thus no Watts. It will consume no Joules of energy in any one second period.
In short, in an AC circuit, Joules per second is not VA.
 

drbond24

Senior Member
It will consume no Joules of energy in any one second period.

You missed the point. No one is saying that a Watt is the same as a Volt-Amp. When boiled down to their fundamental units of measure, both are quantified the same way. They are applied very differently, yes, but both have the same building blocks.

The sentence you questioned in charlie b's dialog is correct as written. You are reading too much into it. Your points are valid, but you are correcting something that he didn't say. :)
 

Besoeker

Senior Member
Location
UK
The sentence you questioned in charlie b's dialog is correct as written)
Since he is dealing with AC circuits, it is incorrect.
VA and Watts are not the same thing except for the specific case of unity power factor.
Specifically, a “Joule per second” could be a “Volt-Ampere Reactive,”
That's just wrong.
 

drbond24

Senior Member
VA and Watts are not the same thing except for the specific case of unity power factor.

You're going to have to show me where he said "a VA and a Watt are the same thing." That is what you are arguing against, but that isn't what he said. Of course they are not the same thing.

Besoeker said:
Specifically, a ?Joule per second? could be a ?Volt-Ampere Reactive?

Break it down to fundamental units. Watts, VA, and VAr all break down to (kg*m^2) / (s^3).

I think I would be correct in saying that the point here is that Watts, VA, and VAr all describe the same quantity of energy. They all describe different kinds of energy, of course, but not different amounts.

Imagine you're on the USS Enterprise and you can convert mass and energy back and forth at your whim. You tell the computer to convert one Watt of energy into a mass. Then you tell the computer to convert one VA of energy into a mass. The computer will create you two identical masses.
 
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