Calculating Power Loads

I am wanting to calculate power load schedule for a solar power system. I know for DC, there is no reactive power but if I take the DC power through an inverter to 110v, then the inverter to a device that supplies power to a device that is DC, how would I make a load schedule for this? Would I just take the power output of the inverter and match that with the device that has a DC load or take it with the Device power supply that plugs into the inverters AC?

There are three aspects:

1) The raw power supplied by the solar power (presumably there is a storage element connected to it, battery)
2) There is the inverter
3) There is the load

Let's say that you net 90% of the energy from the solar panels into storage, which would be W*S 0r Jouls although W*H is more common, and easier to use, it's just J*3600.

Now let's say that your inverter is 90% efficient, 1000W in, 900W out.

Now your load is typically in VA unless it's a light bulb which would be Watts. Why VA, it's because the reactive power is wasted in heat, not returned to the source or does any work on the load.

So let's say you have a load that is 900VA and you want to operate that for 10H, that is effectively 900WH (even though some of the power could be reactive, it doesn't matter because it's coming out of the battery)

So your battery/solar system needs to supply 1000W*10H or 10KWH, but the solar panel & charging system is only 90% efficient in this example so the total that the solar panels must provide is, 10KWH/0/9 = 11.11KWH of energy.

This is simplistic and assumes that the efficiency remains constant over wide ranges which usually isn't the case.

I hope I understood you question. Centrally though VA is what you should use because once it leaves the inverter to the load, it is used one way or other, call it a vacuum cleaner (VA) or a light bulb (W).

Thanks Earl,

You def gave me movement in the right direction. The panels do go into a battery bank and the bank to an invertor. So from what you said, my load requriments must be less than the output efficiency of the invertor.

Lets use the 900VA you provided from the inverter. I run that to a power supply for lets say a computer. The computer has a power supply that plugs into the AC that is ~120V*1.3A = 156VA. However, the output is at 19V*3.95A=75.05W that goes to the computer. I am assuming that I use the 156VA as my load requriment since that is the power it needs to convert to the power the computer needs. Correct? Next, is the power loss in a power supply for a laptop really that bad? I know the supply produces heat, but I expected better.

Thanks agains...

No not necessarily 156VA, this is the safety agency rating, meaning that the device will not use more that 156VA under, "Nominal," conditions. This is it's max rating to be considered safe.

Now for planning purposes, that is if you want to worse case it, then yes use 156VA as your load. You will wind up having lots of margin though given you indicated 75W, which represents 48% efficiency, not realistic if it's modern equipment and has to comply with all the agency requirements.

If you know you will never use more than 75W, then you could use 88VA as a load to the inverter without getting into trouble. Again that assume you never use more than 75W average.

I seen that efficiency, but It just confuses me when I see 100-240VAC ~1.3A. To me that says that regardless of the voltage, the current will remain the same. The only thing I can think of is at lower voltages, the efficiency is lower in terms of input power (lets say 80% of input power) which at 100V*1.3A*.8=104VA but that's still a larger efficiency gap of about 72%, but alot better then the 48%. I guess in that terms, unless the computer was running full speed, all processors and drives, it's not a that big of a deal.

So in terms of reactive loads and generators, why would there be the reason to know the wattage? If you have a 10kVA generator, what not do the loads in VA and not bother with the PF to watts?

Also, I'm sure I know this, but if you have a load that requires a higher startup current, you have to account for this and adjust your generator to accomodate this startup current, correct?

Thanks,