killawatt reading

mshen11 · November 2009

there is a reading for "watts" and "VA" (volt amps)... sometimes they are off. i thought i understood the difference but since im asking - i obviously dont.

i know what voltage and amperage is. i thought voltage times amperage = wattage? if so then what is VA?

mikeo · November 2009

Re: killawatt reading

i thought voltage times amperage = wattage? if so then what is VA?

If the phase of your AC voltage and current is close to 1, then the above is true. If you have inductive or capacitive loads, the voltage and current are out of phase and the VA reading reflects this difference.

BB. · November 2009

Re: killawatt reading

Try reading this link...

The issue is that Volts and Amps are "Vector" quantities... They not only have a value, they have a direction.

If you stand in front of the car and pull, all of your strength goes into pulling the car forward. If you stand off to one side and pull, a lot of your force goes into pulling the car sideways and less goes into pulling the car forward. The actual amount of force that goes into pulling the car forward over a distance is represented by:

Work=Force * Distance * Sine Angle

Where the Angle is 90 degrees if you are standing in front of the car and Angle is 0 degrees if you are standing off to the side of the car.

Notice that if you are standing in front of the car, Sine of 90 degrees = 1.0 and all of your energy goes into pulling the car forward. Notice that Sine A is Power Factor and PF=1 in this case.

If you are standing of to the side of the car, the angle is 0 degrees, and Sine of 0 is equal to 0.0. PF=0 and no matter how hard you pull, the car will not move forward.

Notice that the farther off to the side you stand, the "stronger" the rope needs to be... At Sine(90)=1.0=PF the rope just needs to be strong enough to take all of your strength.

If you are standing at Sine(45)=0.707=PF, the rope needs to be 1/PF=1/0.707=1.41 time stronger (41% stronger) to do the job.

Even though you are pulling harder, the actual useful work you are doing is no more (because the motion forward and fraction of the force pulling forward is still the same).

Does that make sense (it is hard to explain without the drawings.

-Bill

PS: You will notice that sometimes we use Sine and other times we use the COS function... The two are identical except they are offset by 90 degrees. It depends on how you measure the angles and exactly what you are doing...

In the case of mechanical equations, we typically use Sine and the angle of pulling/pushing at 90 degrees is 1.00 of the force is used for motion.

For electricity, we use Cosine because we say that when the voltage and current is "in phase"--they are working together to provide useful work. So in that case Cos(0 degrees)=1.0=Power-Factor.

Either Sine or Cosine will give the identical results--with just using the appropriate angle reference/measurement.

mshen11 · November 2009

Re: killawatt reading

got it... gives me warm fuzzy feeling of high school physics. makes more than perfect sense since i have a pretty good science background... you couldve stopped at "vector" but the detail is appreciated

so normally we use watts but VA is a "better" measurement?

and also... the utiltiies companies use - watts?

BB. · November 2009

Re: killawatt reading

I guess the simple answer is you cannot use just one--You need both to characterize the system for operation...

You need VA to size the wiring, transformers, inverters and such. And you need Watts to calculate fuel/sun/solar panels/inverters/etc...

Regarding Utilities, if you are a home owner, they charge for Watts (KWatt*Hours).

If you happen to own an oil refinery (or other large concern)--they do effectively charge you for kVAR. I have not seen one of those types of bills, but it appears that they charge you for (Watts * 1/PF) where PF factor is your worse peak PF for a billing period which works out a version of KVAR.

I have read of talk with the new "Smart Meters" that they want to charge by KVAR instead to force Power Factor Correction onto homes.

And remember, when we talk about, for example, a CFL (compact florescent lamp) having a PF of 0.5 to 0.6 -- this is a "conversion/corruption" of a mathematical/linear transformation (cosine of the angle between voltage and amperage as sine waves) to a non-linear relationship between Voltage (a sine wave) and Current (short shots of current at peak voltage--i.e., for example a diode connected capacitor that only takes current from peaks of the voltage wave form).

Non-Linear current profiles cannot be "corrected" with motor-run capacitors and when CFLs (and other electronic devices) are connected.

The PF number for a CFL is "real" in the sense that it represents how the wiring and (to a degree) the inverter will respond to the non-linear load.

-Bill

mshen11 · November 2009

Re: killawatt reading

woah wait. can you talk more about CFLs.

are you implying saying CFLs (havent actually measured one) "save" money in the sense that we are billed by the utilties w/out the PF rating? if they were to bill us for the PF rating, it would be alot more (equivalent to incandescent?)

so in a sense, we arent truely saving w/ CFL, especially not w/ solar?

RCinFLA · November 2009

Re: killawatt reading

Your electric company charges for watts not VA in your home. Wire losses will be higher as they are based on current running in wire.

Watts is true power, also called real power.

VA (just magnitude, absolute value, multiplication of volts times amps) is also called apparent power.

The imaginary part is called reactive power which can be positive (inductive) or negative (capacitive) value.

The difference between the two is called power factor. Motors, being inductive and drawing significant power, are the primary significant cause of poor power factor. The power company often puts oil filled capacitors (stacked square cans with two terminals on top) on intermediate distribution poles to correct out the inductive power factor. Power factor matters to power company as it effects how much loss in their distribution system (cables and transformer losses).

BB. · November 2009

Re: killawatt reading

mshen,

As usual, the answer is sort of yes and no...

Yes, you are saving real power with a CFL... A 23 watt bulb can be used where a 100 watt filament bulb was used before (I think that is a bit optimistic--but that is what the package says). So, you are saving 75% of the electricity required for light.

On the negative side, if the CFL has a PF=0.50 -- then, for the utility, for wiring, transformers, generator stations, etc. -- the utility sees this as a 23w/0.50=46VAR load... So, your ~75% savings in power is a ~50% savings in equipment charges/requirements.

Now--because of the 2x amount of current due to the PF=0.50 -- They have losses of the extra current. One place is in resistive losses:

Power=I^2 * R

For every 2x increase in current, that value is squared--so the waste heat is 4x worst because of the heavier current (due to power PF in this case).

When you talk about a huge network of CF bulbs (such as the entire US)--it starts making sense to either charge for poor power power factor and/or make a new law that requires CFL bulbs to be made with a PF correction circuit and get them to >0.95 PF. The additional costs may add $0.50 or so to the cost of each bulb (just a guess).

For the normal person--they will see the extra costs for the lamp and/or changing out to new meters that charge for KVARs (more government rules driving up costs for the little guy).

But, the CFL requirements (outlawing filament lamps) is what cut the billing rate by the utility by 75% but only cost their costs to provide the power (with poor PF) by 50%... Which was a real, net negative hit for the utility bottom line.

-Bill

AntronX · November 2009

Re: killawatt reading

mshen11 wrote: »

woah wait. can you talk more about CFLs.

are you implying saying CFLs (havent actually measured one) "save" money in the sense that we are billed by the utilties w/out the PF rating? if they were to bill us for the PF rating, it would be alot more (equivalent to incandescent?)

so in a sense, we arent truely saving w/ CFL, especially not w/ solar?

You are still saving with CFLs. 12 watt CFL measures 20 VA. Still 3 times less than 60W incandescent bulb equivalent. This is not a problem in residential installation, since you will not have more than 10 - 15 of them on at the same time. If it were a large commercial install needing thousands of them, then it would make a much better sense to use T8 fluorescent tubes with solid state 0.99 PF ballasts. Notice that office buildings, libraries and supermarkets do use T12 or T8 lights and not CFLs.

Cariboocoot · November 2009

Re: killawatt reading

RCinFLA wrote: »

The 'imaginary' part is called reactive power which can be positive (inductive) or negative (capacitive) value.

And this is an excellent explanation of the whole 'start/run capacitors on motors' thing: the 'negative' aspect of the caps to try and nullify the 'positive' aspect of the induction motors.

killawatt reading

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