Ok. Gotcha. Thanks folks.
So is Vars basically equal to watts?
No... not exactly. VAR = Volt Ampere Reactive which is an expression about the relationship between the voltage waveform and the current/ampere waveform. So if we look at Watts vs. VAR's:
When you have a perfectly resistive load, the voltage and current waveforms are in-phase:
Resistive loads are things like incandescent lights, or a heating element, etc...
Reactive loads are either inductive or capacitive in nature. Things like motors with big windings and magnetics are inductive, whereas things like computers and other electronics can be more capacitive. Inductive loads pull current up to 90 degrees out of phase in one direction (+), and capacitive loads pull current 90 degrees out of phase with voltage in the opposite direction. (-)
A good way to remember this is: ELI the ICE man, where:
E = Electromotive Force, or Voltage
I = Current or Amperes
L = Inductive
C = Capacitive
So when E is in front of I, it's an inductive load. (Voltage leads current.) When I is in front of E, it is a capacitive load. (Voltage lags current.)
Inductive Load:
Capacitive Load:
So when we look at the loads, we will have a combination of these items typically. The inductive and capacitive loads will actually cancel each other out and it will look more like a resistive load in nature. So if we had an inductive load that was pulling the waveforms 30 degrees out of phase, and a capacitive load pulling it 10 degrees out of phase, the net result is 20 degrees out of phase.
So now that we know that we're talking about voltage and amperage, when we combine the two we have Wattage/Power. (Volts * Amps = Watts) When we have 1,000 watts, we call it 1 kilowatt, or 1kW. When we start accounting for the leading and lagging of the current waveforms, we start to see that we aren't really getting all of the power we are supposed to be getting. Here's why:
When volts and amps are in phase, they reach their highest peak at the same time. So if you have 20 volts and 10 amps, you have 200 watts.
But what happens when they are out of phase? We have to account for the phase angle between them, so when voltage is at it's peak, current isn't... so if we hit our 20 volt peak, we might only be at 7 amperes... and that's only 140 watts. As we drop down to 14 volts, we hit the peak of 10 amps... and again, that is only 140 watts.
So to express these values, we talk about Kilowatts (kW) Kilo Volt Amperes (kVA) and Kilo Volt Ampere Reactive. (kVAR) KW is 'real' power, KVA is apparent power, and KVAR is reactive power. The easiest way to understand this is a big, cold glass of beer.
kVARs are just foam... it's technically beer, but it's not worth anything. Our glass is kVA and is 'apparently' full, but the only thing we can really drink is the liquid beer.
When we are talking about power factor what exactly does this mean? Does it mean the percentage of good power? I know you want to be on lag side of 1 but if you are on the positive side of 1 then what is exactly taking place? Does this mean I’m not selling all of the power?
Yes, that's pretty close. Power Factor is nothing more than a representation of the relationship between the phase angles between voltage and amperage, expressed as the kVA = kW vs. kVAR. When voltage and amperage are in-phase, Power Factor is 1. If we go either direction (inductive or capacitive) then that number is less than 1. So the more kVAR's we have (+ or -) then the more the number drops away from 1:
So if you had a +0.80 power factor, you would have a slightly inductive load. Your kW would be 80% of your kVA rating. If we use the example above, 125kVA * 0.80 = 100kW. If the number were negative, it would be a capacitive load.
Hope this helps.
Mark
