how many volt amps is 4.0 kwa equal to?
Im sorry, that was a 4.0kw whats that equal to va?
Depends:
KW= 1000Watts
Watts= (I)X(E)
I=current
E=Voltage
So 4kw=4kva, but only in simple terms, it gets more involved when you factor in 3ph or PF.
Dnk....
[This message has been edited by Dnkldorf (edited 10-21-2005).]
[This message has been edited by Dnkldorf (edited 10-21-2005).]
Asking how many kVA = 4kW is almost meaningless.
It is like saying 'I just drove 4 miles. How many miles west did I go?' The question has meaning if you supply additional information.
In an AC circuit, both the voltage and the current continuously cycle from 0 to + to 0 to - and back again. The current will change at the same frequency as the voltage, but does not need to be perfectly in step with the voltage. However the _real power_ delivered to the load depends upon that portion of the current flowing that is exactly in step with the applied voltage. kVA is the raw product of voltage and current; kW is the real power delivered to the load. kW and kVA are equal if the voltage and current are perfectly in step.
kVA and kW are related by the 'power factor' of the load. kVA = kW/pf
When pf=1.0, then kVA = kW.
pf can never be greater than 1.0 .
Most (but not all) loads have pf > 0.7. So as a reasonable _guess_ you can say 'kVA will certainly be greater than or equal to kW, and probably be less than kW * 1.4 '
Some inductive loads can have pf that approaches 0, so without knowing the load pf, you can never be sure.
-Jon
If power factor is unknown or cannot be determined, it is common to assume pf=0.8.
So, for your case
kw = 4.0
kva = kw/pf = 4.0/0.8 = 5.0
Note, this assumes the circuit your dealing with is predominatly resistive & inductive. For a resistive & inductive circuit the power factor is termed a "lagging" power factor. On the other hand, if the circuit is predominately capacitive, power factor is termed a "leading" power factor.
Hope this sheds some electrons on the situation for you ;-)
Rich
Maybe if you were a little more specific as to your project and what it involves, the responses here could be more meaningfull.
If we were using DC, or AC with onlt resistance loads (IE: simple light bulbs), a watt and a volt-amp would be the same thing.
Unfortunately, we don't live in that kind of world. Large motors, lots of motors, variable spped controls, electronics, all sorts of things we use distort the wave form...and can fool our meters into thinking we're using more electricity than we actually are.
For the most part, this difference is of interest only to engineers. Most of us can get by equating one watt to one volt-amp.
Reno, so is it safe to say then that volt-amps be considered only when computing loads for fluorescent lighting?
Now there's a trick question if there ever was one!
For fluorescent lighting calculations, you use the figure on the ballasts, rather than adding up light bulbs.
For traditional ballasts and when using T-12 lamps, yes, volt-amps can be treated as watts.
The same applies for any fluorescents on single-phase systems.
However- on some fixtures you will see a sticker warning you about sharing the neutral wire between different circuits. This sticker is a major clue that not only are the ballasts electronic- but that "harmonics" may be an issue. And, since with three phase systems no single leg can completely "cancel out" another, you will want to up-size the neutral, or run separate neutrals.
All fixtures using the narrower T-8 and T-5 lamps use electronic ballasts.
The explanation here as to "why" volt-amps and watts are not the same is that the distorted waveform can fool an amp meter into reading lower than the actual current being used.
I didn't mean for it to be a trick question, but your explanation was spot-on. I understand the harmonics part of the circuit and the effect a nonlinear load has on a grounded conductor, but what I don't understand is what causes the waveform to become distorted. I'm probably making this more difficult to comprehend then I need to. lol.
I'm a little confused here. Are you asking about Power Factor or harmonic distortion?
VA (volts x amps) are what you are supplying. The watts (V x A x cosine angle) is how efficiently your load works. The VAR's (volts X amps x sine angle) are how much reactive energy the load supplies.
In resistive loads, the voltage and current peak simultaneously (as Winnie said).
In inductive loads (motors, ballasts, etc.) the counter electromotive force (emf) created as the magnetic field builds and collapses across the winding opposes the current that is creating it, causing the rise in current to occur behind (lag) the voltage. So the farther apart the two peak, the less efficient the use of the VA. When they get to 90 degrees apart (cos = zero, a perfect inductor), the watts are zero regardless of the VA. The sine of the angle would be 1, so VAR's would equal the VA.
In a capacitive circuit, the opposite occurs, but from electrostatic forces rather than magnetic ones. In any event, the current peaks AHEAD of the volt, with the same effect of being less efficient.
Since the two are opposite effects, they cancel each other out. That is why you see capacitor banks installed next to large motor loads. A motor (inductive) load running at 100 VARs with a 100 VAR capacitor bank connected to it will bring the source VA back to unity (100% PF).
Say you have a motor pulling 10 amps at 100 volts with the angle between the current and voltage being 30 degrees.
100 volts X 10 amps = 1000 VA
cosine of 30 degrees = .83
sine of 30 degrees = .5
100 volts x 10 amps x cos of .83 = 830 watts
100 volts x 10 amps x sine of .5 = 500 vars
Now, if you put a 500 var capacitor in parrallel with the motor, the vars cancel each other out. Since the source now only has to supply the VA to match the watts, the current from the source can drop to 8.3 amps, saving transformer capacity, line losses, voltage drop, etc.
Harmonic distortion. If you are familiar with SCR's, you know that they turn on and off almost instantly. Picture a voltage sine wave feeding a resistive load in your mind. It is smoothly building and dropping to zero at 60 cycles. So if the SCR turned on halfway through this wave, you would see the current wave form go from zero to maximum almost instantly, creating a wave form that appears chopped. If the SCR did this several times in the cycle, the current waveform would have a very sawtooth appearance.
Now, you know from ohms law that whenever you apply a load to a voltage source, you will get a voltage drop across that source. So in the scenario above, each time the SCR fires, there will be a corresponding distortion of the voltage sinewave as the sudden current flow causes a small drop at the source. As long as this drop is insignificant, nobody sees it. But when it becomes larger, you start seeing the same distortion on the voltage at the source, and anyone connected to that source starts having that distortion applied across their loads. And if that particular distortion happens to create a third harmonic, then you can start getting neutral problems in 3 phase systems.
Did that make any sense?