I suggest starting with a very simple look at the continuous output
rating of the battery and match that to the inverter continuous rating.
This ends up with quite a bit more battery than you might think at
first. However, if the customer paid the big bucks for a battery backup
system, they should at least get enough battery to operate the inverter
at its rated power. Anything less, I would consider poor design. The
same goes for lead acid batteries, they just don't work well, even on
grid, when the battery is undersized. Even lead acid batteries have
maximum current output ratings that should at least be matched to the
inverter rating.
If you WERE going to under size the battery, then you should keep it
safe by reducing the size of the inverter battery breaker as well. I
know this "micro battery for GTB" is becoming a trend, but it doesn't
mean its right.
Ray Walters
Remote Solar
303 505-8760
On 5/23/20 5:48 PM, Mick Abraham wrote:
Greetings, All~ Mark Frye had asked: "...with a 4000w Outback Radian,
AC coupled with 2000w of PV, what is the smallest Li battery I would
want for stable operation?"
Mick's reply: In the lithium arena, one often finds the ratings in
kilowatt-hours rather than in amp-hours. This is an easy conversion
for us energy nerds to make, so below I'm mainly using watts &
watt-hours with occasional conversions to amp-hours.
I suggest that you start with the spec sheet (of whichever brand
lithium battery is on your radar), & see what the manufacturer says
about "allowable rate of recharge". You may find that C/2 is an
acceptable rate with several manufacturers, Mark--much faster than
what we're accustomed to with typical lead-acid batteries. Let's assume:
* ...that your AC coupled PV array crests around 1800 watts to the
battery when there are no loads
* ...that the battery is below 100% state of charge, and
* ...that the solar conditions are optimal.
Let's further assume that those conditions represent your battery's
highest energy rate in either direction.
1800w X two hours = 3,600 watt-hours dividing by 48v nominal = 75
amp-hours in the above hypothetical situation.
++++++++++++++++++++++++++++++
The rate at which a battery can receive energy is (usually) also the
rate at which it can comfortably deplete, so let's briefly look at the
flip side: Assuming the battery mfr. wants the watt-hours to be no
less than two times the maximum wattage, a 3600 watt-hour (C/2)
battery should not be depleted faster than 1800 watts. Your 4kW
Radian, Mark, could exceed that if the loads demand it--& maybe your
loads don't. Motor starting surges should probably be part of the
thinking & it wouldn't hurt to ask the mfr.'s opinion about short term
surges which briefly deplete the battery faster than the hypothetical
two hour rate.
By approaching this question based on the battery manufacturer do's &
don'ts, one improves the chance of getting warranty coverage should it
later be needed. "What the large print giveth, the fine print taketh
away."
The Wrench List is the Bomb!
Mick Abraham, Proprietor
www.abrahamsolar.com <http://www.abrahamsolar.com>
Landline: 970-731-4675
Cell phone or for text messaging: 970-946-6584
ᐧ
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