Is DC Watts same as DC Watts?

Generally, the end to end system loses (sun to dc, dc through charge controller, to battery charging, discharging battery, dc to a.c. inverter) are around 52% efficiency. For example:

1,000 watt array * 0.52 system eff * 4.0 hours of sun = 2,080 watt*hours per day

-Bill

VA is different than Watts. But, it depends. They differ by the "power factor" or PF.

Power Factor is sort of like efficiency--But instead of being a pure loss, it is, sort of, how efficiently the AC current and AC voltage is being used.

An analogy--Peddling a bicycle. If you are peddling smoothly and apply force "in phase" with the peddles, that is the most efficient. Your PF=1.0

If, however, you sometimes peddle backwards for part of the cycle, or you "jump" on the peddles for part of the cycle, your PF~0.5 ... Your peddles+chain have to be 2x stronger to manage the irregular and extra force for part of the cycle. But you are not loosing any "efficiency" because, eventually, all of the force+RPM is being used to move the bicycle forward

Another analogy, pulling on a car with a rope. Standing in front of the car, 100% of your force is pulling the car forward. (Cosine 0 degrees = 1.0 = "PF"). Stand 60 degrees to the side, then only (COS 60 degrees=) 0.5 of the fource is moving the car forward, the rest is trying to pull the car sideways--But no work is being done, so no "loss of energy". The rope has to 2x stronger to pull the same car forward when standing to the side.

Similar with pulling for 1 second and not pulling for 1 second. 1/2 the cycle you are pulling forward, 1/2 the cycle you are not--The rope has to be twice a strong to do the same amount of work as standing in front of the car.

In the AC power world, some items have a natural near 1.0 PF... Heating elements and filament bulbs are example. Others have a vary poor PF--For example Cheap Twisty Florescent Bulbs may have a 0.5-0.6 PF. Electric induction motors may have a PF of ~0.6 to 0.8

Computer power supplies may have a PF~0.6-0.7, but "power factor corrected" power supplies generally have a PF~0.95 (close enough to perfect 1.0).

When you measure AC voltage and AC current--Then the proper equation is VA=V*I ... Your "simple meter" cannot measure the "phase between" voltage and current (that P=Watts=Volts*Amps*Cos phase angle between current and voltage). You need a special "Power Meter" to measure both at the same time to give you an accurate Watts calculation.

Why does it matter? Wiring, transformer, AC inverter outputs, Generators, and even utility lines, need to be design heavier to support the possible out of phase current (and non-sine wave current draw for computer/electronic power supplies). So, V*A is good for the wiring/system design. AC motor example:

• 10 amps AC * 1/0.60 "bad PF" = 16.7 Amps
• 120 VAC * 16.7 amps = 2,004 VA
• Watts = V*A*PF = 120 VA * 16.7 amps * 0.60 PF = 1,202 Watts

Now, for the DC input to your AC inverter--More or less, the Inverter only converts "power" (or Watts), bad VA on the output does not show up on the DC input as power usage--So the above DC calculation would be:

• 1,202 Watts AC * 1/0.85 typical AC inverter efficiency =  1,414 Watts DC battery bus power
• 1,414 Watts DC power * 1/12 volt AC inverter input voltage = 118 Amps DC input current

There are more details, but that is the basics.

The summary... Use V*A to calculate wiring AWG, transformers, AC inverter and Genset VA size (if no VA given, assume VA output = Watts name plate rating). To calculate the size of the battery bank, solar panels, and generator fuel usage, use Watt*Hours (power*hours of energy usage).

-Bill