PV Gap

Lance Said:

What type of solar power system would I need?

We Answered:

2500 kwh/month is an average consumption of 3.5 kilowatts.
(2500 divided by the number of hours in a month (24*30))
3.5 kilowatts is 3500 watts.
If you only run 9 out of 24 hours then you would need (24/9)*3500 watts.
Which is 9333 watts.
This site lists solar panels:
http://www.wholesalesolar.com/solar-pane…
As you can see a 210 watt panel sells for $970.
So you would need 9333/210 of these panels.
Or 44 of these panels at a total cost of $970 * 44.
or $42,680.00
You might get the panels cheaper somewhere.
This wattage data on the panels is also probably peak output.
You would probably need as many as 88 panels to provide the power you want.
On top of all this you have to add battery and controller costs.

Amber Said:

How much computational power is required to simulate the Solar System?

We Answered:

It depends on how accurate you want to be. I can simulate every atom in the solar system, with zero accuracy, with hardly any computational power at all. If you want to predict their position to within a few million kilometers for the next few seconds, that takes a bit more power.

There are about 10^57 atoms in the solar system, practically all of them in the sun. The world has about 10^27 flops of computing power (my guesstimate). I takes about 10^6 flops (at least) to simulate each atom (with reasonable error), so we could handle around 10^21 atoms. Which means it would take all the computer power on Earth times 10^36, to simulate the solar system at the atomic level, in real time.

Allen Said:

Inverter losses for solar power system?

We Answered:

Inverter losses vary from design to design, but typically, losses are pretty close to a fixed amount for a particular inverter. It's the power the inverter uses just by being turned on. As a general rule you can figure 15% of maximum. If your inverter is 2000 watts max, a typical loss might be 300 watts. It doesn't matter how much or even if you are using equipment that's plugged in. If the inverter is turned on, it's using 300 watts and is "lost".

Inverter loss is usually only important when you're running off a battery and want to calculate how long a battery or battery bank with a certain amp hour rating will last. If you're on a solar system, it doesn't really matter, until you swiitch to battery backup.

Personally, I'd try to work from the equipment to be powered back to the solar panel. Let's say for example you need 2000 watts of power for a maximum. At 120v that's just under 17 amps. Not wanting to run at absolute maximum anyway, I'd ignore inverter loss for now and opt for something about 25% more than I needed just to be easy on the electronics in the inverter. I'd be looking for an inverter or combination of inverters that could provide 22 amps or more... , or 2500 watts or more.

Assuming 12 v, you're talking 12 volts at 170 amps from your solar system. (2000 watts / 12v)

So... now it's dark and you need to run on battery...

Lets suppose your battery bank is 12v and 1,000 amp hours, and your more typical usage is 1200 watts. NOW add the 300 watts the inverter uses, for a total of 1500 watts. 12v * 1000 Ah / 1500 watts = 8 hours on battery. Only have a few light's on? Three 100 watt bulbs = 300 watts
300 w for lights + 300 w for the inverter = 600 watts. 12 * 1000 / 600 = 20 hours.
Using those energy efficient bulbs?
3 * 26w for the lights + 300w inverter loss = 380
12 * 1000 / 380 = 31.5 hours

Clear as mud?
Hope that helped.

Nicholas Said:

4kw solar power system produce how much kw output per day?

We Answered:

In the US, you can use this map http://rredc.nrel.gov/solar/old_data/nsr… to find out how many peak equivalent sun-hours your area gets per day. On that page, select average, annual, flat plate tilted at latitude, for the most common type of system.

Suppose the number you get for your area is 5 hours per day. Then that's 5 hours x 4 kW = 20 kWh/day on average. Mind you, some days could be zero, some days could be 30, depending on weather and season.