# Is DC Watts same as DC Watts?

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Okay so I know for most this is a no brainer question so yes, I AM a new to most of this. I'm trying to size my electric load requirements before going forward with panels, batteries and so on.
So on my Kill A Watt meter when I see 40 Watts can I assume this converts to 40 Watts of DC power? At 24v?
TriStar MPPT, 8 x 100w PV, MNPV6 Combiner, 4 x 12v 155ah VMAXTanks AGMs and GoPower 2000w PSW Inverter.

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edited October 2017 #2
The answer is yes and no, in a direct equation the answer is yes, indirectly converting from DC to AC involves losses, through an inverter, so the answer is no, but it is a good benchmark for calculation purposes, factoring into the equation the inverter losses, for the sake of argument let's use 15% loss, add this to the recorded watt reading should give an approximate value. If on the other hand you are using 24VDC loads, the conversion would be close to equal, but those details were not provided. I'm  assuming you are referring to AC watts, and an inverter is involved.
1500W, 6× Schutten 250W Poly panels , Schneider MPPT 60 150 CC, Schneider SW 2524 inverter, 400Ah LFP 24V nominal battery bank

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A watt is a watt.  Watts = volts x amps, so 120w is 10 amps at 12vdc, or 1 amp at 120vac.

As Mcgivor notes, more watts need to be produced than are consumed by loads.  How much more gets complicated.
Off-grid.
Main daytime system ~4kw panels into 2xMNClassic150 370ah 48v bank 2xOutback 3548 inverter 120v + 240v autotransformer
Night system ~1kw panels into 1xMNClassic150 700ah 12v bank morningstar 300w inverter
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When it comes to AC, volts x amps = VA.   Which may or may not be the same as watts.
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Also working all the way back to panels, there are additional losses across the system...
Home system 4000 watt (Evergreen) array standing, with 2 Midnite Classic Lites,  Midnite E-panel, Prosine 1800 and Exeltech 1100, ForkLift battery. Off grid for @13 of last 14 years. 1000 watts being added to current CC, @2700 watts to be added with an additional CC.
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Got it. So for my sizing exercise I will factor in a 15% converter loss. Yes, I should have said DC to AC. I have been going toward a 24v DC system but due to cost and limited end capacity at 800-1000 watts I will probably stay at 12v. Thanks for the timely replies.
TriStar MPPT, 8 x 100w PV, MNPV6 Combiner, 4 x 12v 155ah VMAXTanks AGMs and GoPower 2000w PSW Inverter.
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lzhome said:
Got it. So for my sizing exercise I will factor in a 15% converter loss. Yes, I should have said DC to AC. I have been going toward a 24v DC system but due to cost and limited end capacity at 800-1000 watts I will probably stay at 12v. Thanks for the timely replies.

If you are running a fridge you should plan on 1Kwh +, I wouldn't run one on a 12 volt system.
Home system 4000 watt (Evergreen) array standing, with 2 Midnite Classic Lites,  Midnite E-panel, Prosine 1800 and Exeltech 1100, ForkLift battery. Off grid for @13 of last 14 years. 1000 watts being added to current CC, @2700 watts to be added with an additional CC.
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
Near San Francisco California: 3.5kWatt Grid Tied Solar power system+small backup genset
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jonr said:
When it comes to AC, volts x amps = VA.   Which may or may not be the same as watts.
VA = watts needed to produce > VA work out.  VAin - VAout = losses?  WattsIN - WattsOut = losses?  A distinction without a difference?
Off-grid.
Main daytime system ~4kw panels into 2xMNClassic150 370ah 48v bank 2xOutback 3548 inverter 120v + 240v autotransformer
Night system ~1kw panels into 1xMNClassic150 700ah 12v bank morningstar 300w inverter
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
Near San Francisco California: 3.5kWatt Grid Tied Solar power system+small backup genset
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The only thing I would add is that with non-linear loads (eg, switching power supplies), it is no longer useful to talk about the phase angle between voltage and current.    They might be perfectly "in phase", but significantly different in shape.   This alone will cause < 1 power factor.