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Re: Help with inverter size
Well, I tried. In my earlier post:If you want to run 5 hours of AC power per eveningThat would be around: 392 Watt*Hours / 5 hours a night = 78 Watt average AC load
 90
AH * 24 VDC * 0.85 AC inverter eff * 1/8 hour discharge rate = ~230
Watt AC typical maximum AC inverter (continuous discharge rate)
 90 AH * 24 VDC * 0.85 AC inverter eff * 1/5 hour discharge rate = 367 Watt AC maximum supportable (reliably) AC inverter
I am not going to sticky this threadEverybody has bad days and we don't need to make a scarlet letter out of it when it happens.
The account has been "deleted".
Bill
1 
Re: Help with inverter size
To add to Photowhit's information... I like to call the battery bank the "heart" of your system. Normally, one defines the loads, then defines the battery bank.. 24 volts * 90 Amp*Hours * 0.85 AC inverter eff * 1/2 days of storage * 0.50 maximum discharge (longer battery life) = 459 WH per day for 2 days AC loads
 90 AH * 29.0 volts charging * 1/0.77 solar panel+controller deratings * 0.05 rate of charge = 169 Watt array minimum
 90 AH * 29.0 volts charging * 1/0.77 solar panel+controller deratings * 0.10 rate of charge = 339 Watt array nominal
 90 AH * 29.0 volts charging * 1/0.77 solar panel+controller deratings * 0.13 rate of charge = 441 Watt array cost effective maximum
Next, how much energy per day would you get from solar. Assuming a fixed array somewhere around Philadelphia PA:
http://www.solarelectricityhandbook.com/solarirradiance.htmlPhiladelphia
Measured in kWh/m2/day onto a solar panel set at a 50° angle:
Average Solar Insolation figures
(For best yearround performance)Jan Feb Mar Apr May Jun 3.07
3.77
4.23
4.46
4.74
4.84
Jul Aug Sep Oct Nov Dec 4.95
4.79
4.55
4.21
3.12
2.77
 200 Watts panels * 0.52 off grid solar AC system end to end efficiency * 3.77 hours of sun (Feb break even month) = 392 WH per day average February sun
If you want to run 5 hours of AC power per eveningThat would be around: 392 Watt*Hours / 5 hours a night = 78 Watt average AC load
 90 AH * 24 VDC * 0.85 AC inverter eff * 1/8 hour discharge rate = ~230 Watt AC typical maximum AC inverter (continuous discharge rate)
 90 AH * 24 VDC * 0.85 AC inverter eff * 1/5 hour discharge rate = 367 Watt AC maximum supportable (reliably) AC inverter
If you expect/hope to get "useful" amount of energy from the wind turbine portionYou probably want to look around for how to site and mount a wind turbine:
Wind Power Links
www.otherpower.com (good forum for DIY Wind Power)
Hugh Piggott  Scoraig Wind Electric site for tons of info (from mike90045)
Scoraig Wind "Recipe Book" for DYI Turbines (from Chris Olson... From his 4/11/2013 post)
www.greenpowertalk.org (added from "russ"Like here but more wind/less solar)
Small windpower a scam ? Survey says SO
Truth About Skystream & SWWP
Windmax HY2000 2kW Wind Turbine
Note, I did not predict how much power your turbine will produceAt best (good siting, good winds, higher tower), you might get 1015% of nameplate capacity. A 400 Watt turbine with 10% capacity: 400 Watts * 10% of capacity * 24 hours = 960 WH per day
Bill
1  24 volts * 90 Amp*Hours * 0.85 AC inverter eff * 1/2 days of storage * 0.50 maximum discharge (longer battery life) = 459 WH per day for 2 days AC loads

Re: Is DC Watts same as DC Watts?
VA is different than Watts. But, it depends. They differ by the "power factor" or PF.
Power Factor is sort of like efficiencyBut instead of being a pure loss, it is, sort of, how efficiently the AC current and AC voltage is being used.
An analogyPeddling 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 sidewaysBut 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 notThe 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 PFFor example Cheap Twisty Florescent Bulbs may have a 0.50.6 PF. Electric induction motors may have a PF of ~0.6 to 0.8
Computer power supplies may have a PF~0.60.7, 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 currentThen 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 nonsine 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
 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
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
1 
Re: New build offgrid
Inverters tend to be a key component in the system design. 2,400 watt 24 volt DC inverters are typically easier to find vs 2,400 watt 48 vdc inverters.Also, if you want to run some DC loads, 24 vdc appliances are easier to find. Also 24 vdc fuses, breakers, fans are a bit easier to find too.Roughly, if you want/need a 2,400 to 4,000 watt or larger inverter, then 48vdc battery bus would usually be better.Do several paper designs and see what works best for you. Inverters that are higher wattage than you need, tend to waste more power too.Bill1 
Re: charging battery bank with a mains power charger
What voltage is your battery bank?
Connect the charger to your battery bank (less than 80% SOC) and see how hot the battery charger gets after charging for 1560 minutes of charging.
If the a.c. battery charger does not get hot, you probably are fine. And yes, the battery bank will take longer to charge.
 Bill1