I didn’t see the option at Southwire for battery cable calculations.
Sol-Ark 15k will only invert 13k of DC PV to AC
- Thread starter Clint-L
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On the first screen of the app, you change it from AC to DC. Then on the calculate screen instead of 120 or 240 you select other and you can define your voltage e.g. 52v. Then define your cable length and voltage drop you would allow and it will calculate it.
I believe 4/0 pure copper cable is rated the same, no matter the type or number of strands. I suppose if there were any difference, that is why the app lets you select DC vs AC - possibly?
Right now you're only using one of the breakers so you're limited to 200A*80%=160A.
Not sure about voltage drop, but amps go up with battery cable.
I’ll check it out.
Well, 4/0 on your chart shows 440 amps. I’m good with the single cable from a wire perspective, I believe.
Prince Of Darkness
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Couple of comments;
Historically, 10 feet of cable between inverter DC input terminals and battery terminals has been considered the absolute maximum distance.
It may be a bit difficult to do with the existing 4/0 battery cable runs, to the Sol-Ark, but, TWISTING each of these two cable sets together, should help the dynamics (one existing pair of cables, and the second pair, to be added).
Have had some concerns about the Ripple voltage on the AC connections to the Grid, and at the battery terminals of the Sol-Ark. (SolrGuppy covered the voltage drop, but thinking about ripple V, too). An oscilloscope ccould reveal quite a lot, but would be very careful trying to use scope to measure 240 VAC ripple, as most probes are not rated for this ... A DMM set to AC could give some clue of ripple voltage, but most DMMs are S L O W.
Have not read the SA manual, but, to me, 35-ish kWh seems a bit small (even with Li batts, and their probable lower Z, compared with LA batts).
Had also wondered about thermal rise possibly contributing to some Throttling (spec seems to set 45 C, as the beginning of some throttling).
Just wondering, some of this might have been covered. FWIW, 'Luke'
Historically, 10 feet of cable between inverter DC input terminals and battery terminals has been considered the absolute maximum distance.
It may be a bit difficult to do with the existing 4/0 battery cable runs, to the Sol-Ark, but, TWISTING each of these two cable sets together, should help the dynamics (one existing pair of cables, and the second pair, to be added).
Have had some concerns about the Ripple voltage on the AC connections to the Grid, and at the battery terminals of the Sol-Ark. (SolrGuppy covered the voltage drop, but thinking about ripple V, too). An oscilloscope ccould reveal quite a lot, but would be very careful trying to use scope to measure 240 VAC ripple, as most probes are not rated for this ... A DMM set to AC could give some clue of ripple voltage, but most DMMs are S L O W.
Have not read the SA manual, but, to me, 35-ish kWh seems a bit small (even with Li batts, and their probable lower Z, compared with LA batts).
Had also wondered about thermal rise possibly contributing to some Throttling (spec seems to set 45 C, as the beginning of some throttling).
Just wondering, some of this might have been covered. FWIW, 'Luke'
That’s an interesting perspective I might try to dig in to more on the DC battery cables.
I would love to have more battery storage. I would only need the batteries if the grid was down, and there’s no natural gas supply to me, and I run out of propane. I have the batts I suppose for some TOU supplement from time to time and in the event things get really really bad (could happen ehhhh). ?
I might buy a lottery ticket next week. If I were to win, I’d for sure get more batts!!!! ???
Temperature rise is something we should observe. I agree with that 100%. I can say, that the capping out for me, is at exactly just over 13k and it happens immediately, event if I’ve had PV disconnected for a bit and the inverter temp is cool.
Today I did some testing and phase swapping on the CT’s just for giggles, but didn’t seem to make any difference. We only had a few minutes where the sun popped out today, so I didn’t see the clipping.
I currently have about 15ft of battery cables between my 1000A busbar and sol-ark. Not ideal but not too bad. I won't be able to reduce that until I move my installation to the basement in a year or two.
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Hedges
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"Ripple voltage on the AC connections to the Grid" - it is all ripple, at 60 Hz; what is the concern?
Battery, I've measured large rectified 60 Hz of a LF inverter. HF, likely much less ripple because only HF boost converter drawing from it. Doubt that varies it's draw to follow the AC, but it is possible. Would expect more like what I saw on PV input of a GT PV inverter, which was about 1% or so ripple, on the HV bus rather than on the battery.
Many scope probes are good for 300V, so 120Vrms 170Vpeak is OK. (Transients could be a problem if you get unlucky.)
Usually ground wire of scope probe goes to ground pin of power plug, so can't measure 240V with a single probe.
I have a high voltage differential probe so mostly use that.
At work we have a handheld R&S scope that happens to have isolated BNC driving differential inputs.
The ferrites as common-mode choke would avoid saturation with both positive and negative cables through them, as was mentioned.
Their purpose would be to prevent high frequency common mode of the switchers and microprocessor clocks from using battery cables as an antenna to transmit. Inverter should work fine with or without, but some radio communications might get interference.
Hedges
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SMA says 33' (page 45)
that's 48V, so 1/4 the current, 1/16th the loss of 12V for same wattage.
that's 48V, so 1/4 the current, 1/16th the loss of 12V for same wattage.
Interesting.
I was planning ~13’ of AWG 2/0 for battery cables.
so around 0.5 V drop at 100 A. But being that I may send up to 200A, 1.0 volt drop?
Hedges
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2/0 0.000256 ohms/m
4m one way? or round trip?
4 x 2 x 0.000256 = 0.00205 ohms for 8m of cable
200A x 0.00205 = 0.410V drop
200A x 0.410V = 82W dissipated in 8m of cable.
If I did the math right.
For 12V system, 200A is 2400W, and cable dissipates 3.4%
Hopefully lower current most of the time.
A reasonable drop.
How about surge? is 2400W surge, or is there a 5000W surge to start a motor?
In that case, 0.82V drop, probably still tolerable.
4m one way? or round trip?
4 x 2 x 0.000256 = 0.00205 ohms for 8m of cable
200A x 0.00205 = 0.410V drop
200A x 0.410V = 82W dissipated in 8m of cable.
If I did the math right.
For 12V system, 200A is 2400W, and cable dissipates 3.4%
Hopefully lower current most of the time.
A reasonable drop.
How about surge? is 2400W surge, or is there a 5000W surge to start a motor?
In that case, 0.82V drop, probably still tolerable.
I thought I was following what you're saying until the last 3 lines. If you don't mind using lay man's terms for dummies like me.
Are you saying it is a good thing to have this ferrite toroid as close to the other components of the inverter or not?
In this picture would be it be a good idea to the put the toroid around the positive and negative 4/0 cables on the inverter side (red Circle) or would that cause the overheating issue that you mentioned?
I’ll add, I’ve tested this output with no batteries connected. The solark does not require batteries being connected to operate in “on grid” mode and to sell back. Even without batteries, my 15k will only invert 13.2kw to AC.
Hedges
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Yes, good to have ferrite close to inverter (rather than close to battery), and put it around both positive and negative so it reduces radiated RF power.
Two separate issues:
1) If the inverter generates common-mode noise on the wires, that radiates. Assuming the ferrite is effective blocking the noise, the length of cables from inverter to ferrites is what acts as an antenna, so ferrite closer to inverter is more effective.
2) AC current in wire causes changing magnetic field, which the ferrite attempts to oppose. The ferrite is suppose to do this by storing and returning energy (an inductor) but also dissipates some power so could get hot with large AC current.
The ferrite is quite small compared to a power transformer, can store very little energy before it "saturates" (becomes as strong a permanent magnet as it is able to be.) A small current through it, and it can be magnetized first with one polarity then with the other; that works as an inductor. Once it is fully magnetized (all the atoms' spinning electrons are oriented one way) it ceases to absorb magnetic field from the wire, and current flows unimpeded. The current to saturate could be on the order of milliamps.
With 100A DC flowing through a cable, the ferrite would saturate and be useless. With positive and negative cables through it, The sum of +100A and -100A = 0A, no net current and the DC current (and any differential current) cancel and don't interact with the ferrite.
What the ferrite can do is block common mode current (only if differential mode is canceled by having both wires go through it.) If both battery terminals are swinging up and down together in voltage, and the battery is not (relative to earth), current will propagate as a wave. Just like in the antenna of a radio transmitter. Just like a digital signal in a wire, over a ground plane. Often, wires used to carry data are dimensioned to have 50 ohms characteristic impedance. The pair of battery cables dangling in the air (small diameter and far above the ground) could be more like 100 ohms, just a guess. Consider if the battery terminals have 1V RMS common mode on them. That is, negative is at zero volts and swings +/-1.4V, positive is at 48V +/-1.4V. 1Vrms/100 ohms = 10 mArms. That is what the ferrite sees, and its impedance reduces the voltage amplitude on the cable after it. Maybe zero +/-0.1V and 48V +/-0.1V, so less power is transmitted.
That was about preventing radiated power from an electric "E field" antenna.
3) Third issue, preventing radiated power from a magnetic "B field" antenna:
Current flowing in a loop is differential, and radiates power. That depends on how large the current is, and how much loop area. If you lay the positive and negative wires a foot apart, there is several square feet of area. The 100A DC current doesn't matter (makes a DC magnetic field) but the pulses of current drawn by switcher, to the extent not filtered by capacitors, are AC current in the wires. Choke can't help that. Inductors in the inverter design may help. Twisting the wires reduces loop area (and positive to the left, negative to the right for one twist cancels filed from negative to the left, positive to the right of next twist, at a distance.) Batteries themselves form a loop; if you were really trying to minimize radiated power, might arrange batteries so they also form a twisted loop.
For the most part, minimizing differential current is the job of inverter designer. Simply cable tying positive and negative together should be good, or twist if easy.
Different topic, those cable terminals are exposed and close. Make sure they can't come in contact. A piece of of insulation between them might be good.
Thanks a Lot! Just the clarify, The overheating issue mentioned above would be a problem if you have only one of the battery wires going through the ferrite, instead of both?
