Friday, May 08, 2020 at 22:37
Gerry, you've almost given yourself the answer in your opening post.
Your panel is 250 watts. That is the product of its max power voltage of 34.6V and max power current of 7.23 amps. As you identified, 34.6 * 7.23 is close enough to 250 watts.
Assume everything in the system is 100% efficient - solar panel, solar regulator, no cable losses etc.
Your regulator takes the 250 watts of solar input at 34.6 volts and steps it down to the required charging voltage for the battery, in your example 14 volts. With everything at 100% efficiency your battery will get all the power your panel can deliver through the 100% efficient regulator, ie 250 watts. But now the voltage is only 14.
You can manipulate your P=VI formula to give a current if you know power and voltage. The manipulation is:
I=P divided by V (I=P/V)
so
250 watts divided by 14 volts = 17.8 amps.
Voila!
If only everything was 100% efficient. I work on about 70% for the panels and about 95% for a good MPPT, for an overall efficiency of about 66% in good conditions . HKB's 50% is not unrealistic as an average in a portable or RV system.
So using a simple 50% efficiency, the current to the battery from a 250 watt panel would be half the ideal, ie 8.9 amps.
So to get your 15 amps into the battery in real world conditions you'll need more than one panel, or more accurately, more than 250 watts of solar panels.
There are a number of variables you may be able to control to improve efficiency. Good cabling, orienting your panels toward the sun (if portable), keeping panels clean, providing ventilation to the rear of fixed panels are a few.
Cheers
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