Whats difference betw. KVA and vars?

watts is real power. volt and amps in phase. power dissipated in heat or that did work.

vars is volts amps reactive??

KVA the same thing as vars?

Reply to
Loading thread data ...


Yes. Volts and amps 90 degrees out of phase.

VA is volt-amperes. (k = kilo or 1000) What you get if you multiply volts by amperes, ignoring phase angle. Useful for sizing transformers. It is the "sum" of watts and vars. I put sum in quotes because they don't add directly but as vectors, and the vectors are at right angles. If you have a circuit that draws 4 watts and 3 vars it will be 5 VA, not 7. (sqrt(3^2 + 4^2) = 5)

Reply to
Michael Moroney

dissipated in heat

K simply means multiply by 1000. VA means volts x amps... thats 'watts' and that is absorbed cleanly by pure resistance loads such as lighting and electric resistance heating. KW = KVA

Motors are different in that regard they draw amps (billions of electons) after the voltage peak,,,in other words they draw peak amps out of phase with the voltage peaks ...so the meter gives a false low reading... volts x amps but out of sycn dont meter properly.. the correct power is stated in VARS vars takes this shift into account by multiplying the VA read by the cosine of the shift angle (you can see that on an occiliscope).... usually in the 15 to 25 degree range.... so its cosine 15 or 25 or whatever....then the utility company charges a fee over and above what the meter reads in V xA because of this out of phase condition they call that a power factor charge.

(which can be corrected at the motor with the addition of capacitors) this reduces the amp/voltage phase angle...and the power factor improves, and you can have the utility company come out and test that..and reduce yer power factor related charges..thats how power factor correction capacitors pay for themselves... an issue the electric utilities sometimes like to talk about, sometimes not.

If your peak demand is largely not reactive..say from heavy lighting and electric restance loads.. then your 15 minute peaks define the 'demand charge' you will be billed for...if you have heavy surges that last for more than 15 minutes at one shot per month... the utility company might multiply yer bill by 20 to 50% or more to pay for the capability they are providing you.

In addition to this, if they notice that they are sending you a lot more VA than your meter is reading, from that they can compute the reactance load VARS ... thats the actual power that you are using....or use an occiloscope to notice the voltage/ amperage peak angle deviation...(voltage peaking before amperage for instance) then multiply that angle in degrees by cos of the amperage voltage phase angle shift. this results in a power factor charge. You can look the cosine of any angle up in yer geometry book.

Beyond that there are harmonic distortions that any large facility can feed back into the power grid, screwing things up... the utility companies just hate that....motors and transformers can both create unwanted harmonic effects.

This in the extended grid screws up the true power available to anyone on the grid.

Paragraph 2 of the following article addresses these harmonics and power grid issues

formatting link
"However, just providing the raw power isn't enough these days. Motor drives present a nonlinear load to the power grid feeding them. Characteristically the diodes in an inverter's input stage create harmonic distortion as they switch on and off. Reflected back into the line, these harmonics tend to sap usable power and cause overload in transformer neutrals, circuit breakers, and motors. They can even damage other sensitive equipment connected to the same source. Some industry estimates predict that harmonics will eventually consume up to 50% of the energy on the nationwide power grid-clearly anunacceptable situation.

The solution, of course, is to suppress harmonics."

long article balance at URL above

Phil Scott

Reply to
Phil Scott

That's correct. Spelling and grammar aside, watts IS the difference between kVA and VARs.

Reply to
Paul Hovnanian P.E.


formatting link

Reply to
Gerald Newton

What a moron.

Volts * Amps * cos (phase angle) is *NOT* vars. It is Watts. And 'the correct power is stated in VARS' is another idiotic statement. Real power is measured in Watts, only reactive power is measured in VARS.

VARS = Volts * Amps * sin(phase angle) VA = Volts * Amps. (VA)^2 = (VARS)^2 + (Watts)^2

When voltage and current are out of phase, a kwh meter does *not* give a false low reading. It gives a *true* reading of the watt-hours consumed because it is designed to properly take the cosine of the phase angle between them. The utility does not "charge a fee over and above what the meter reads in V xA". The utility charges are based on what the meter reads in kwh (since the standard meter reads kwh, automatically correcting for any phase displacement). Only in very rare situations does the utility add a second meter to measure VARS and charge a second fee for supplying an inordinately large amount of VARS to a customer (in addition to the kwh charges).

Your inability to even explain the difference between reactive power (VARS) and real power (watts) shows that any/all of your explanations are suspect.

Another elementary school level mistake. The reactive load VARS is *not* 'the actual power that you are using...'

'multiply that angle in degrees by the cos of the amperate voltage phase angle shift'????? What nonsense.

So if voltage is 480, current is 10 and the phase angle between them is 30 degrees, by your explanation above, "...multiply that angle in degrees by cos of the amperage voltage phase angle shift...", we would have...

30 *cos(30) = 30* 0.5 = 15 ????

And it sure is a good thing we don't measure the angle in radians instead of degrees. That would probably blow your calculations all to h___!!!

What you *might* have meant was 480 * 10 * cos(30), but that isn't what you wrote. And if you *did* mean that, what you describe is the exact calculation performed *inside the meter* to determine *watts* not VARS. So the meter correctly would register 2400 watt-hours for every hour this power draw exists.

Anyone that follows your calculations and advice is getting complete garbage.


Reply to











Watts. And 'the

statement. Real power


You are correct of course in your last sentence...and spinning with your trash job... having been totally out to lunch on your assertion along with charles and his PhD waving fiend and now proven utterly wrong across the boards you choose to do more trashing... perhaps I could criticize you for not giving the VARS factor on a Sears chains saw model 361 or something, then assert that you are an idiot... but that would be bogus... Your attempt to mix harmonic issues with vars is likewise bogus as you continue now to ignore the world class article from top industry sources on the issue,,,, and expect that my one sentence remark should be all inclusive.


the brain you are screwing up is your own... such upset and stress and attempts as you and Charlie and the PhD waving associate just generate cortisols in the brain and while the surge is beneficial, beyond that brain matter is destroyed... for examples of that end game perhaps Charles and whomever can look around thier cubicle cages at thier place of containment and take notice.

Regards, Phil Scott....who is not proven entirely correct on the harmonics feed back into the grid issue.

sorry son...thats just how it is....and its not a secret.. books are written on the subject and you have now this article for refrence and google to find a hundred more.

*not* give a

watt-hours consumed

phase angle

above what the

the meter reads

correcting for any

utility add a

supplying an

to the kwh

reactive power (VARS)

explanations are suspect.




VARS is *not*

voltage phase

between them is 30

in degrees by


radians instead of


isn't what you


*watts* not VARS. So

hour this power


Reply to
Phil Scott

Nevertheless, your formula for VARS that you posted, "vars takes this shift into account by multiplying the VA read by the cosine of the shift angle (you can see that on an occiliscope).... ", is still wrong. VARS = VA * sin(phase angle), not cosine. Watts = VA*cos(phase angle)

And you also posted, "...the correct power is stated in VARS." Another

*wrong* statement. Real power is measured in watts, not VARS.

Your silliest statement (so far), "...then multiply that angle in degrees by cos of the amperage voltage phase angle shift." Multiplying the angle in degrees times the cosine of the angle, now *that* will give you interesting results. Totally meaningless, but 'interesting'.

Why don't you address these three mistakes of yours that you have made in this thread??? Admit you made a mistake, you will find it quite cathartic.

"not proven entirely correct..." ??? You mean you admit that you may have made some error?? Perhaps there is hope for you after all.

And strangely, *all* of the articles that I've read support Charles and my positions and *none* of them claim that induction motors create significant harmonics. Shows that you have severe reading comprehension problems if you think they support your claims. The 'world class' article you've cited clearly states that electronic motor drive systems are a source of harmonics, not induction motors. You have admitted that you have extrapolated from that article that induction motors and transformers are also a source of harmonics. Your extrapolation is wrong.

"sorry son.. thats just how it is..." You have erred and you have been caught making incorrect statements.


Reply to

















takes this shift

shift angle



Phil Scott

VARS." Another


angle in degrees by

the angle in

you interesting

have made in

quite cathartic.

that you may have


Charles and my

create significant

problems if you

you've cited

source of


transformers are

you have been

Reply to
Phil Scott

No. "Not in phase" (not zero degrees) is not the same as "90 degrees out of phase".

Reply to

------ He is correct with regard to vars as being related to the component of current that is 90 degrees out of phase with the voltage - that is the non-power portion of the current (single frequency). Example: load voltage =100V at phase angle 0 (reference) current =10A at a phase angle lagging voltage by 30 degrees. VA =100*10 =1000 Volt-amps Associated angle of 30 degrees Power =100*10* cos(30)=866 watts due to component of current in phase with V Volt-amps reactive =100*10*sin(30)=500 vars due to component of current 90 degrees out of phase with V Power factor =power/VA =0.866

Reply to
Don Kelly

PolyTech Forum website is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners.