Hello Chris,
Thank you for the example. It helped .. but not completely.
See if this is what you have in mind:
PV -> Inverter ---V
^ V
V ATS -> Grid
Battery Loads
PV = 2 kW STC rating.
Inverter has an internal MPPT charge controller.
Automatic Transfer Switch "ATS" disconnects the system in the event of a
utility failure.
Battery provides storage in the event of a utility failure. Chemistry not
identified.
Presuming the above is correct:
There isn't a clear-cut answer to your question due to the number of
uncontrolled variables involved.
However, some variables can be generally addressed:
1. Battery chemistry. Lead-acid? Li-Ion?
2. Site location (for average temperature, annual insolation, etc.)
3. PV mechanical considerations: Azimuth? Tilt? Mounting method?
It appears you're seeking an estimated value for total annual energy
production. Subsequently, I'll proceed on that basis:
For very generalized analysis, actual overall PV output is 85-88% of their
label ratings. Much of the decrease is due to cells operating above STC
temperature of 25C. Haze, pollution, weather, dirt on the PV, and other
factors are also considerations. Use 86%.
There are two key aspects to charge controllers in this discussion: 1. How well
do they track the actual maximum power point? 2. Conversion efficiency? For
your purposes, estimate an overall efficiency factor of 95%.
Whether battery efficiency plays into this or not depends on how often the system operates on
battery power only, depth of discharge, age of the batteries, and battery chemistry .. for openers.
If lead-acid, there's another aspect called the "Peukert's Law" (or the "Peukert
formula") that has to be factored in. Peukert says the higher the rate you withdraw stored
energy from a L-A battery, the less you get. It's akin to saying, the faster you drive, the worse
your miles per gallon.
Inverter efficiency varies depending on the point of their curve on which it's operating.
The "sweet spot" tends to be mid-range. For a 3.6 kW, and only 2 kW of PV,
you'll never exceed the inverter's mid-range. For a typical grid-interactive inverter
today, use 97%.
Batteries have a round-trip efficiency of approximately 90% for new lead-acid, and
slightly better than 90% for Li-Ion. As L-A batteries age, efficiency decreases. Near
end of life, L-A batteries are in the 80-85% range. This also varies with the method of
charging. L-A also require periodic conditioning charges (also known as an
"equalizing" charge). Li-Ion are better in that respect, losing approximately
0.5% per year, and don't require conditioning. However, efficiency in both topologies
depends on how often the battery is discharged, how deeply it's discharged, and for L-A,
how long the battery sits in a less than 100% state of charge condition before being
fully recharged. Heat also takes its toll on L-A batteries in life and efficiency. When
used in repeated discharge/recharge cycles, for L-A, use 88%. For Li-Ion, use 92%.
Battery float current also varies with chemistry. L-A batteries consume
nominally 0.3 to 0.5% of the rated amp-hour value. Thus for a 48 volt, 300
amp-hour battery .. the float current would be 1 to 1.5 amps. This value will
vary with battery temperature. It's higher in L-A batteries than Li-Ion.
Compromise and use 1.25% for L-A, and 1% for Li-Ion.
Wiring loss in a well-designed system is recommended to be 2.5% for best
practices. This is inclusive of all BOS hardware, connectors, and so forth.
Use 2.5%.
Finally, on reading your reply below .. I infer from the information the site gets 4.5
hours of "good" sunlight per day.
I'm unable to proceed beyond this point due to lack of information related to
the PV azimuth, tilt, mounting method, and other variables. Hopefully you can
plug those in .. and with the information above .. derive a better estimate of
annual AC energy from the system.
Regards,
Dan
P.S. You may not get admonished by Michael W for inadvertently including my identifying
information in your re-post to the Wrenches. I'm not an installer, and no longer with
any manufacturer. For that reason, Mr. Welch has deemed me "unqualified" to
post to the Wrench BBS, and persona non grata where the the BBS is concerned. One of
your colleagues on the board posted one of my comments a year or so back .. and was
contacted by Michael for doing so. Michael told him if he ever posted information to the
BBS that originated from me, he too would be blocked from making any future posts.
You're welcome to post anything I send to you. I would ask for your own sake, and
continued participation on the Wrench board, that in re-posting information .. you omit
anything that would identify me as the source. I have a somewhat unique style of
writing, so it's also suggested you change the ".." and other characteristics
to resemble what you'd type. Plagiarize away. I'm OK with it.
Dan Lepinski, P.E.
Professional Solar Consultant & Electrical Design Engineer
Serving the Solar Industry Since 1972
817.884.6081 (cell)
* 2010 State of Texas Renewable Energy Industries Association Honoree
and Award Recipient for Meritorious Achievement in Renewable Energy.
* 2016 Memnosyne Institute Sustainable Leadership Award
* 2017 United States Department of Energy National BCAP Project - 2017 & Ongoing...
(One of only two engineers selected nationwide for the program.)
* 2019 State of Texas Energy Advisory Council Member
Professional Profile: https://www.linkedin.com/in/dan-lepinski-p-e-5339bb22/
"Scientists investigate that which already is; engineers create that which has never
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On 8/28/19 8:31 AM, Christopher Warfel wrote:
Hi Dan, Sorry for the confusion.
Maybe an example would help. Take a 2000 w dc multimode system that is
coupled with a 3600 w ac rated multimode inverter. What percentage of
the dc capacity is inverted into ac capacity. So, roughly 2000 x .9
(wiring and similar losses from array to battery) x .95 for an assumed
5% usage to charge/float the batteries x .98 for dc to ac inverter
efficiency makes me think that 2000 x .9 x .95 x .98 * (4.5 x 365)
yields = 2,752 kWh annually. The utility is assuming roughly 3600 x .98
x (4.5 x 365) {if they are assuming anything at all} = 5,368 kWh revenue
lost. The regulatory assumption overestimates the capacity factor by @49%.
I'll post to the WRENCHES too to help clarify hopefully. Chris
On 8/28/2019 9:10 AM, Dan Lepinski, P.E. wrote:
OFF-LINE REPLY
Hello Chris,
The wording of your question is confusing. You ask about the percentage of DC
capacity vs AC power output, then add batteries into the mix.
What system topology is the focus of your query?
I'll be glad to do what I can to assist. Just need clear direction on the
configuration.
Regards,
Dan
