Power Factor on inverter output

alkit

Hi,

Generally, to calculate how many Wh I can get from a battery, I do the following over simplified calculation (let's use a 100ah 12v battery as an example):
100x12 = 1200Wh
50% DOD = 600Wh
80% for losses = 480Wh

As an example (forget inverter sizing for now):

I have an item that uses 50W/h, but has a power factor of 0.5:
Do I now need to say that the item is ultimately pulling 100W/h now because of power losses (in which case my battery will last 4.8 hours)
Or do I look at real wattage - 50W/h (in which case the battery will last 9.6 hours)?

Thanks in advance for the help

SolarPowered

You have to pay attention to inverter voltage operation. The "good" efficient inverters will not operate under 11V or above 15V which will only allow a 85% D.O.D of the battery. So you can't ignore the inverter you would be using, a good inverter in standby will draw .02kWh. Bad inneficient inverters can draw .10kWh, that is a factor that has to be considered.

Typically the battery shouldn't be depleted more than 40% DOD for AGM/GEL/FLA if you want the battery to last longer than a few years.
If the battery is used frequently and DOD drops frequently it becomes more difficult for the CC to charge the battery in bulk charge.

Power factoring is not consistent to actual (PTC) as the voltage drops with DOD. 50% DOD is roughly 12.1V with AGM, and 11.8V with SLA/GEL. Its important to know the battery chemistry you are using.

Knowing if your battery is a C8, C10, or C20 has alot to do with power factoring.

SolarPowered

This is a good read.
http://solarhomestead.com/battery-amp-hour-ratings/

inetdog

alkit wrote: »

Hi,

Generally, to calculate how many Wh I can get from a battery, I do the following over simplified calculation (let's use a 100ah 12v battery as an example):
100x12 = 1200Wh
50% DOD = 600Wh
80% for losses = 480Wh

As an example (forget inverter sizing for now):

I have an item that uses 50W/h, but has a power factor of 0.5:
Do I now need to say that the item is ultimately pulling 100W/h now because of power losses (in which case my battery will last 4.8 hours)
Or do I look at real wattage - 50W/h (in which case the battery will last 9.6 hours)?

Thanks in advance for the help

Concentrating on the power factor question alone, if the actual power consumption is 50w (not w/H!) then the voltage times the current will be 100VA.
If the VA draw is 50, then the actual power consumption will be only 25W.

As the inverter sees it, the current it has to provide will correspond to the VA number, while the power it actually draws from the battery will be closer to the watts number. Most inverters will just recycle the reactive power and not drain the battery to supply it.

SolarPowered

inetdog wrote: »

. Most inverters will just recycle the reactive power and not drain the battery to supply it.

When you state "most" you mean the differences between PSW and MSW?
MSW isn't capabe of "recycling power", unlike PSW. Most MSW's will operate at an efficiency of 90% and lower, PSW's 92% and better.
MSW's tend to deplete batteries faster than PSW's.