Solar Panel System Help

  • Thread starter jeffc0987
  • Start date
  • #1
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Main Question or Discussion Point

Right now I am working on a small pump that is capable of running on solar power for 24 hours a day.

So far, I have solar panels with specs at typical irradiance/temp of 1000Wm^-2 and 25C:
Pmax = 135 W
Vpmax = 17.7 V
Ipmax = 7.63 A
Voc = 22.1 V
Isc = 8.37 A
Area = 1m^2

and batter(ies) with specs:
12V, 100 Ah
@20 hour rate to 1.75 VPC (No idea what this is useful for)

For my area I have found that during the worst month (December) we receive 2.52 kWh/m^2*day (I believe this is sun hours per day) so I am trying to model a pump around this scenario to be autonomous for 3 days with a safety factor of 1.2.

I am able to measure the Ah draw of the pump I am running through a meter and I am wondering how I can figure out the amount of solar panels / batteries I will need with this number. If needed I am also able to find out the Wh, A(peak), W(peak), V(max) of the pump during its operation.

Right now the formula I am using to calculate the amount of solar panels is:

[tex]\frac{[(Ah)*(24)*(1.2)]/(Sun Hours Per Day)}{8}[/tex]

The 24 is for a continuous runtime throughout the day and the 1.2 represents the safety factor I would like. The formula was suggested to me, but it doesn't seem to add up in units to me. I have no idea where the 8 came from in this equation so it is hard for me to trust. I think it may have to do with the Ipmax value of 7.63A.

I also have an equation for a smaller 85W solar panel which is

[tex]\frac{[(Ah)*(24)*(1.2)]/(Sun Hours Per Day)}{5}[/tex]

Which may somewhat help with understanding what the 8 is for in the first equation.

The battery equation I am using right now is:

[tex]\frac{(Ah)*(24)*(1.2)*(3)}{100}[/tex]

Where the 100 is the total Ah capacity of the battery, and the 3 is the days of autonomy. This seems to make more sense to me (although I think I should reduce the batteries Ah by ~30% to account for efficiency).

Throughout my time using these equations, the numbers for panels / batteries required has always seemed extremely high and I know next to nothing about energy / electricity / solar panels so I look forward to any feedback and criticism.
 
Last edited:

Answers and Replies

  • #2
vk6kro
Science Advisor
4,081
40
You need to measure the current that the pump draws and the voltage supplied to the pump.

Then you can work out the number of amp hours the pump uses (ie multiply the current in amps by 24 hours) and your solar panels have to supply at least this much power at the same voltage to the batteries while the sun is shining.

Then you need to worry about what would happen if there was no sunlight for 5 days, or maybe 10 days. Could you supplement the solar panel with a small wind generator?
If the pump was supplying drinking water for animals, this may be important.
 
  • #3
8
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I have measured the Ah draw that the pump needs (Assume in this case to be simple it draws 2Ah).

The problem is figuring out how many solar panels I would need to supply this current. I will be using a voltage regulator to supply 12V to the motor. I have also stated the amount the sun shining will be 2.52 hours/day and the solar panels will need to charge the battery enough to last for 3 days without sunlight.
 
  • #4
vk6kro
Science Advisor
4,081
40
Motor current is measured in amps, so assume this motor draws 2 amps.

If the battery was fully charged, the pump could run for 100 AH / 2 amps or 50 hours.

These amp-hour figures are fairly unreliable, so you might need two batteries.


The solar panel figure that is most useful is the max current of 7.63 amps. If your panel is not going to track the sun, then this figure can be halved. Call it 4 amps.
You would have to check the output under test with your level of sunlight, or consult the manufacturer.

4 amps for 2.52 hours per day is 10 amp hours per day. So, it would take 20 days to charge the two 100 amp hour batteries with one panel. (200 amp hours / 10 amp hours per day = 20 days)

Actually it is worse than that, because 2 amps will go to running the pump, so it would take 40 days to charge the battery.

I think the economics of this are becoming clear. This is going to be very costly and possibly unsuccessful.
Is it windy where you live? Have you considered getting a wind generator?
 
  • #5
uart
Science Advisor
2,776
9
Actually it is worse than that, because 2 amps will go to running the pump, so it would take 40 days to charge the battery.
Actually it's MUCH worse, as he said he wants to run the pump 24 hours per day. So that's 48AH per day to run the pump. He'll need at least 5 panels.

Of course it could be that when he said 2Ah for the pump he meant 2AH per day and not 2 AH per hr. In any case the OP needs to figure out how to give information unambiguously and in correct units before anyone can really help properly.
 

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