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First solar setup, last step before putting the pieces together: wire and fuse sizing! Need help.

tripcode

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Hi, like the title suggests, I have finally received all my components and am in the process of putting the pieces together. This got me thinking a bit more about the wire size that came with my inverter, and raised some general questions about the inverter as such. I hope somebody can help me clarify those grey areas in my brain.

Here is my setup:

I have some questions reg. the inverter and the wiring/fuses in general in my setup:

Inverter
I have an inverter that is rated at 1000/2000W ..i.e. it should be able to handle a surge of 2000W, that is 2000W/12 = 166.xx A? I highly doubt that, what about the inverter efficiency that keeps popping up when I read about inverters online. Where does the 80% come from I havent seen anything in my manual? This is my inverter btw: http://www.ebay.com.au/itm/231544077381, turns out in the product description page it says 90% max efficiency, if I do the maths with that, I end up with a max current of 150A (2000*.9/12).

The follow up question to this would be, how do you suggest to size the cable/fuse for this, according to the maths with an optimal surge power, I should get 166 A, thus the wire should be at least 1/0 AWG if not 2/0 ? And the fuse around 175A or even 200A? Isnt this overkill??? Of course, one can say, just size to your needs.. well.. what if I decide to treat myself with a blender or sorts knowing that my inverter could handle it.. and then realize, whoops... I only sized my cables to my 'needs'. So my goal here is really to size to the maximum power consumption that is basically possible and safe. Which with this inverter certainly wont go over 2000W.

Starter/Solar Batt / DC Charger:
Following up on sizing question, is the 80A between my DC DC charger and the starter battery not overkill as well, as my DC DC charger can draw a maximum of 50 A from a single source or 25 A each if running on alternator and PV. Thus a fuse of, lets say 60A with a cable that can handle 70A should be fine, i.e. 6awg? Or simply put, can it not be the same as I have between the DC DC and solar batteries. 4-6AWG wires and a fuse of 70 A?

Solar Panels / Current Rating:
As I am running two 250W rated at 12V nominal, 18.6 Vmp. I should get a current of maximum of 26.88ish amps from the PV source. Therefore I decided to put in a 30A fuse. But something - maybe im just overthinking - got me thinking is, can I be certain that the DC DC charger wont try to draw, lets say 500W/12v = 41A. I am pretty sure I made a mistake in my thinking, and I am kind of realizing my mistake, just as I am writing this.. i.e. I need a 30A fuse PER solar panel, and not for the whole bunch where the branch connectors are. Is this correct? So 2x30A, or 1x 30A per panel. That makes more sense.

To summarize, this is what I am planning on doing, can somebody comment on its 'correctness' and point out the flaws?

solar panels in parallel (2x250W)-> 1x30A fuse per panel, 10AWG wire from branch connector to -> DC DC charger (max current shouldnt exceed 41A)
starter batt -> 4/6 awg with a 70A fuse -> DC DC charger (max current shouldnt exceed 50A)
DC DC charger -> 4/6 awg with a 70A fuse -> solar batteries 2x 105Ah in parallel (max current shouldnt exceed 50A, because the DCDC charger wont handle more)
solar batteries -> 1 awg, with a fuse of 175A (distance from batt to inverter is less than 1m) -> inverter (max current shouldnt exceed 150A as, because 2000w * 0.85 / 12 = 141ish A)

If you have read this, thank you, regardless if you actually try to answer my questions. Of course I would highly appreciate a bit of clarification as this is my first setup and I am planning to sleep in the same place as this setup will run (campervan) therefore, trying to avoid setting anything on fire :).

EDIT:
I amended my setup sketch a bit and here is the new version, please let me know if you also approve this:
 
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"max efficiency 90%"
That's not efficiency at full load. There should be a graph showing a curve. Probably 90% efficient at 10% to 20% load, may drop to 80% efficient at full load. That is fine if full load is only for a couple seconds or a minute.

So long as fuses aren't fast-blow they will handle a brief surge during start-up of loads.
Good fuses are expensive, so I have oversized fuses at the battery to deal with dead shorts, and a circuit breaker built into the inverter. If your only protection is the fuse, then it does need to be sized both for the wire and whatever the inverter says is max fuse size.

1 AWG, is not kosher for a 175A fuse if it is the only protection, but 1/0 AWG should do:


It will protect the wire in case of a short, so long as the fuse has "interrupting capability" higher than short-circuit current of the battery bank.

But you assumed 2000W. You can fuse for 1000W which is the steady state. Divide, don't multiply, by 85% efficiency. Assume low battery voltage.

1000W / 0.85 / 10.5V = 112A

Now 1 AWG 90 degree wire is OK, some headroom to fuse it without having the fuse blow.

Unbundled wire so it cools better, "free air". If you used 105 degree wire even better.


Fuses ought to be 25% above continuous load (and no greater than wire ampacity other than rounding up to next standard size.)
112% x 1.25 = 140, so standard 150A fuse should be good.
 
"max efficiency 90%"
That's not efficiency at full load. There should be a graph showing a curve. Probably 90% efficient at 10% to 20% load, may drop to 80% efficient at full load. That is fine if full load is only for a couple seconds or a minute.

So long as fuses aren't fast-blow they will handle a brief surge during start-up of loads.
Good fuses are expensive, so I have oversized fuses at the battery to deal with dead shorts, and a circuit breaker built into the inverter. If your only protection is the fuse, then it does need to be sized both for the wire and whatever the inverter says is max fuse size.

1 AWG, is not kosher for a 175A fuse if it is the only protection, but 1/0 AWG should do:


It will protect the wire in case of a short, so long as the fuse has "interrupting capability" higher than short-circuit current of the battery bank.

But you assumed 2000W. You can fuse for 1000W which is the steady state. Divide, don't multiply, by 85% efficiency. Assume low battery voltage.

1000W / 0.85 / 10.5V = 112A

Now 1 AWG 90 degree wire is OK, some headroom to fuse it without having the fuse blow.

Unbundled wire so it cools better, "free air". If you used 105 degree wire even better.


Fuses ought to be 25% above continuous load (and no greater than wire ampacity other than rounding up to next standard size.)
112% x 1.25 = 140, so standard 150A fuse should be good.

Thanks a lot for the response, a lot is clearer to me now! I would like to follow up on some of your responses though... first being the inverter with the advertised surge power of 2000W

Q1; I am trying to understand the idea of dividing the continuous power rating of my inverter, by the efficiency figure. Whats the idea behind this formula and we assume low batt voltage because this will provide us with a 'high amp scenario, correct' ?

Here is my take at demystifying this formula: The inverter efficiency tells us basically, how much power can I put out, for the power I draw from the batt banks. If we look at an example: my computer operates with a charger rated at 165W, therefore I would divide this by its 'average efficiency' of 0.85 (again ..this seems to be a 'guess' number.. I dont see where this is specified as I dont see a efficiency/power output curve for this inverter in the manual). But anyway, the formula would look like this: 165W/0.85 = 194W ..that means, my batteries need to provide 194W of power, rated at 12V.. correct? This would leave us with a current of maximum 16.2A? Now the formula remains the same, regardless of how many appliances are running from the inverter, the bottom line is.. the efficiency tells us, how 'costly' the conversion from DC to AC is. Am I interpreting this right?

On the other hand, the maximum output power would show us, given the required power consumption, how much power the inverter can output given the efficiency levels. This should result in a efficiency/power output curve.

Q2; Furthermore, why are we not using the max output (2000W) as a baseline to calculate the max current in this circuit?

What would occur if I had a device drawing 2000W from my inverter, basically its a no-go? With a 90% max efficiency, I should provide my inverter with 2,222.22W, with a 12V battery bank it would look like this: 2,222.22W / 12V = 185A. As I have only a 2x105Ah battery bank (50% max discharge). I would / should be able to provide this power surge for roughly 0.56h (given the batts are both fully charged and for the sake of simplicity lets assume, my batts can manage this discharge rate, although they probably cant.. ), is this correct?

Q2.1; How long can inverters operate on their max power output, what does the 'surge' condition mean exactly? It sounds to me like they can only operate for a few seconds on this scenario?

Also, I have received an answer regarding the wire size they provided in the box, from the shop that sold me the inverter, here is their answer:

Hi
Our cable is 5AWG. We suggest you use our standard cable, otherwise the electrical appliance may not start or the inverter may be damaged. Our inverter has enough fuses, no additional fuses are needed.
The inverter may require 3-4 times the wattage of the appliance to start, so peak power needs to be sufficient. For example, your coffee machine is 500w, peak power may reach 4*500=2000w. Therefore, pure sine 1000w/2000w is suitable for you.


Regards

This is pretty confusing to me, according to their message the efficiency is really bad? They say it needs 4-5 times the power, how should this be interpreted? And also, their suggestion is to use 5AWG (the provided cable) when our maths clearly points towards a 1/0 AWG? But looking at Wills video wiring up a 400w system with a 1500/3000w inverter, he is also using 4awg wires for the inverter/battery connection on a 170A CB:
. As I have a smaller inverter 1000/2000w makes me question... what is correct now, I go for a 1/0 AWG wire or will de 5AWG provided with the inverter 'do its job'?

?????
 
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The surge power is output the inverter can make for a couple seconds. A light bulb is about 1/10th the resistance, 10x the current draw, when filament is cold. Coffee pot not as big a multiple because the element doesn't glow white hot. Motors often draw 5x to 10x. So this doesn't refer to poor efficiency of the inverter.

Not using the 2000W surge to size wires or fuses because it'll only be for a second. The inverter won't deliver more than 1000W long enough to warm up a wire, fuse, or breaker. If asked to it will hopefully shut off. One guy linked a video of cheap 1000W inverters running a water pump. Most wouldn't, some burned up, some worked.

Here's the data sheet for my inverter:


About double power for 30 minutes. About triple peak current for 0.060 seconds.

Manual, page 231 has efficiency curve:


Yes, divide output power by efficiency (at that power) to get input power. Divide by low battery voltage to get the higher current at that voltage.

5 AWG? Between 4 AWG and 6 AWG on the chart for "in free air" that would be about 120A, just about 1 kW with our 85% efficiency and 10.5V figures.
For three conductors in a cable or conduit (the way we usually do house wiring) it would only carry 85A. So 5 AWG could work if short and individual wires able to cool themselves in the air. Thicker is better. I use 2 AWG in conduit for my 100A or 125A circuits.

So they say they have sufficient fuses. That may be, but it depends on how big a battery bank you have and if the wire could short to ground, like inside a vehicle or your house. With a cheap fuse blows with a massive battery (or the grid), the fuse just keeps on conducting (plasma). Solution is a fuse rated to interrupt 4000A or 20,000A, whatever the source can generate. Some just bolt onto the battery terminal, to be cheaper and simpler.


Looking at your sketch, I see you have 2x250W PV and 2x 12V, 105Ah batteries. About 20A per battery charge, 0.2C
I've been having a discussion with another member who thinks anything over 0.1C will wreck the battery. My vendor's documents recommend at least 0.2C or a different charge profile to ensure full charge. Review your battery charging instructions, make sure you follow them.
 
The surge power is output the inverter can make for a couple seconds. A light bulb is about 1/10th the resistance, 10x the current draw, when filament is cold. Coffee pot not as big a multiple because the element doesn't glow white hot. Motors often draw 5x to 10x. So this doesn't refer to poor efficiency of the inverter.

Not using the 2000W surge to size wires or fuses because it'll only be for a second. The inverter won't deliver more than 1000W long enough to warm up a wire, fuse, or breaker. If asked to it will hopefully shut off. One guy linked a video of cheap 1000W inverters running a water pump. Most wouldn't, some burned up, some worked.

Here's the data sheet for my inverter:


About double power for 30 minutes. About triple peak current for 0.060 seconds.

Manual, page 231 has efficiency curve:


Yes, divide output power by efficiency (at that power) to get input power. Divide by low battery voltage to get the higher current at that voltage.

5 AWG? Between 4 AWG and 6 AWG on the chart for "in free air" that would be about 120A, just about 1 kW with our 85% efficiency and 10.5V figures.
For three conductors in a cable or conduit (the way we usually do house wiring) it would only carry 85A. So 5 AWG could work if short and individual wires able to cool themselves in the air. Thicker is better. I use 2 AWG in conduit for my 100A or 125A circuits.

So they say they have sufficient fuses. That may be, but it depends on how big a battery bank you have and if the wire could short to ground, like inside a vehicle or your house. With a cheap fuse blows with a massive battery (or the grid), the fuse just keeps on conducting (plasma). Solution is a fuse rated to interrupt 4000A or 20,000A, whatever the source can generate. Some just bolt onto the battery terminal, to be cheaper and simpler.


Looking at your sketch, I see you have 2x250W PV and 2x 12V, 105Ah batteries. About 20A per battery charge, 0.2C
I've been having a discussion with another member who thinks anything over 0.1C will wreck the battery. My vendor's documents recommend at least 0.2C or a different charge profile to ensure full charge. Review your battery charging instructions, make sure you follow them.

Hey, thanks again for your answer, much appreciated.. you really help me understand the details much better. I think I finally understand the big picture of inverters. And also cable / fuse sizing seems to make more sense now. I will dig a bit more into the charge/discharge rate for my batts.

Speaking of batts, one question that I had already clarified but would like your personal opinion on as well is: I have a battery that is about 4 months old and has barely been used, also the alternator keeps it charged whenever I drive. The second battery was purchased recently and is brand new. Do you see any issues hooking them up in parallel? I spoke to a couple of people and they all said it ll be fine.

I have decided to use our formula and go for 1/0 awg wires for anything inverter related. I will use 4 awg for anything batt / charge controller related. I am currently working on putting it all together and have created a board with the respective outline for all the individual pieces. Would appreciate if you can have a look and let me know if it looks alright to you. I have based it on wills blueprint for the RV/alternator setup. Its pretty much a copy. Except that I have slightly more PV power. See here:
Based on: https://www.mobile-solarpower.com/simplified-400-watt-fewer-wires-and-alternator-charging.html
 
Yes, paralleling to similar age batteries should be OK. Series would expose differences more than parallel. (in that case aging the new one with same charge while driving conditions might help.)

4 AWG for 70 A charge path should be more than enough. I use 6 AWG in house wiring for 70A.

I think your sketch shows + wire taken from one battery and - from the other (best way) rather than connecting charger/inverter to a single battery and then wires to the second one (not great, imbalanced)

You have to figure out how AC should be hooked up safely. Does the inverter connect neutral to chassis? Is that even allowed by the design of the inverter? Will you have access to AC while outside? Need GFCI? Does it need to be a both-poles GFCI because neutral is not handled like in house wiring?

You should determine what charging current is appropriate for those batteries.
I like to have multiple strings of panels (two separate panels in your case) oriented toward morning and afternoon sun. That reduces peak current and spreads it over the day, less cycling of the battery. What you can do depends on whether that charge controller needs two panels in series, or if one is sufficient voltage even on a hot day.
 
Yes, paralleling to similar age batteries should be OK. Series would expose differences more than parallel. (in that case aging the new one with same charge while driving conditions might help.)

4 AWG for 70 A charge path should be more than enough. I use 6 AWG in house wiring for 70A.

I think your sketch shows + wire taken from one battery and - from the other (best way) rather than connecting charger/inverter to a single battery and then wires to the second one (not great, imbalanced)

You have to figure out how AC should be hooked up safely. Does the inverter connect neutral to chassis? Is that even allowed by the design of the inverter? Will you have access to AC while outside? Need GFCI? Does it need to be a both-poles GFCI because neutral is not handled like in house wiring?

You should determine what charging current is appropriate for those batteries.
I like to have multiple strings of panels (two separate panels in your case) oriented toward morning and afternoon sun. That reduces peak current and spreads it over the day, less cycling of the battery. What you can do depends on whether that charge controller needs two panels in series, or if one is sufficient voltage even on a hot day.

Glad to hear paralleling with similiary aged batteries is ok, also according to you :). I just put the second battery where it belongs, and have prepared the board with the sketch today so tomorrow I will look into wiring. I am really thinking about sizing my wires down by the next lower awg. I usually dont have more than 30-45cm to cover with the wires. Also, see the answer from the inverter shop:

Hi
5AWG already has 16.7 square meters. Our standard cable is 65cm, which meets the requirements. Of course, you can use a thicker thread according to your own understanding. If you want to change to a cable with a length of 2 meters, your cable diameter is more appropriate. Each time the wire length doubles, the wire diameter doubles.

The efficiency decreases by about 1% with the increase of temperature. At room temperature, the efficiency is 85 to 90 percent.

How many watts the inverter can run depends on the electrical appliances you use. Different appliances and different brands also have different starting wattages. Usually 3 to 4 times the wattage is the peak power of more electrical appliances. The continuous power of your inverter is 1000w. If your appliance only needs 1-2 times the starting wattage, then your appliance’s wattage can be 1000w. But we don’t recommend using full power, the inverter is easily damage.

Regards

I have a much better understanding of how my inverter works now, basically I can hook up anything to it as long as I dont go over 1000w and if I have appliances that exceed the 1kW threshold, than it can only be for a few seconds (starting up a coffee machine for example), but at all times under 2kW. Correct?

Regarding your raised points, see my answers below:
>> Does the inverter connect neutral to chassis? Is that even allowed by the design of the inverter?
That is a good question and one I had asked myself today, I will write the shop and see what their answer will be.

>> Will you have access to AC while outside?
As in, for conviencence purposes? I was planning to use a power board that I will have plugged in to the inverter at all times, that is easily accessible from the back and middle section of my van. However, only turning the inverter on when I actually need it.

>> Need GFCI? Does it need to be a both-poles GFCI because neutral is not handled like in house wiring?
No idea! Should I research this, what would be a good approach to find this out?

Regarding my batteries, here is a picture I took today:
There is no special mention of a charging/discharging rate. So I would assume its the usual for lead acid batteries? I found this on batteryuniversity.com:

The sealed lead-acid battery is rated at a 5-hour (0.2) and 20-hour (0.05C) discharge. Longer discharge times produce higher capacity readings because of lower losses. The lead-acid performs well on high load currents.

Basically I can safely withdraw 105*0.2A, that is 21A? Using two batteries, that would be 2*21A (because in parallel), thus my DCDC charger sizing of 50A seems to be well suited for this job (assuming discharge and charge rate, are both rated at 0.2). Or do I miss something?

As for the inverter, and the appliances, I should probably avoid hooking up super heavy loads to not exceed 1000W / 42 = 23.8V. Otherwise I would potentially harm the longevity of the batteries. I suppose its ok to use, for instance, a coffee machine once in a while that runs on 500W but surges to 2kW at the startup. Although I am not planning for it, but it could be done.
 
The answers from the shop are a bit confused, possibly due to their English or their math.
Wire "16.7 square meters" should have been 16.7 square millimeters.


Two main limits on wire gauge (beside fitting in the screw terminals):
Wire has resistance, so there is voltage drop V=IR, dissipating watts per meter. It heats up with current flow. That happens very slowly due to thermal mass, so it won't overheat at 10x rated current for perhaps a minute, or 100x for perhaps a second. At rated current (from Ampacity chart for free-air or bundle/conduit), it can reach 90 degrees C - almost enough to boil water - and reach steady-state as the heat conducts through insulation in maybe 30 minutes.
1) Ampacity charts show current to not cause overheating from steady current. Allowed temperature depends on type of insulation, and wet or dry as some insulation would be affected by water.
2) Voltage drop, which would leave less voltage for the inverter (making it draw more current.)

Light bulbs dim at lower voltage, which you'll observe in the house when a refrigerator kicks on. The light bulb draws less current at lower voltage.
Motors draw more current for the same load at lower voltage, and take longer to start. They can pull down the supply as low as 50V and burn themselves up because they never start.
Inverters will draw more current at low voltage trying to deliver the same power, voltage, and current to output as normal. Some commit Hari-Kari.
Can't find the link to video now, but one sat there spitting out smoke. One ran the pump but made a pop/flash and kept on running (probably won't start the pump a second time.)

"Each time the wire length doubles, the wire diameter doubles" - bad math. Each time wire length doubles, wire cross sectional area would have to double to keep resistance the same, so diameter increases by square root of two. Four times the length, use twice the diameter, four times the area.

Tables give resistance of wire. Consider round trip, out to the inverter (or appliance) and back. Keep voltage drop < 5%. This is where peak starting current should be used, not operating current.

So size battery cables and fuses the larger of both ampacity rating of wire and continuous current, and the larger of that and whatever limits IR drop to 5% at peak current.

" If your appliance only needs 1-2 times the starting wattage, then your appliance’s wattage can be 1000w. But we don’t recommend using full power, the inverter is easily damage. "
It may be this is one of those inverters that will deliver the power you ask, if it is the last thing it ever does. Better if it just said, "No! I wont!"

"(assuming discharge and charge rate, are both rated at 0.2). Or do I miss something?"
What you are missing is that batteries can deliver high discharge currents, but charging current has a smaller limit. Wet cells (which you don't have) want a minimum current to stir the electrolyte. Sealed batteries (which you do have) are more subject to losing their electrolyte due to outgassing, and to puffing up from overheating. People generally recommend 0.1C, although some sealed batteries are different.

Your charge controller doesn't know what your inverter is doing. If the charge controller delivers 0.2C and the inverter draws 0.1C, battery would be OK with the remaining 0.1C. But not if all 0.2C went into the battery.

My system presently has PV withcharging capability around 0.5C to 0.75C, but the inverter decides how much current goes to the battery, which I have set to 0.2C (due to vendor's recommendations.) The battery inverter tells the charge controllers when it wants more power to deliver AC, and when it wants less. My system is meant to keep the batteries bumping along 100% charge, with PV delivering whatever my air conditioner and other loads want. Then at night I run small loads off the battery.

Can't find that particular battery.
Label says "Initial current 31.5A Max" (meaning brief peak charging with a voltage)
"Battery may gas if excessively charged" Producing an explosive atmosphere, and after a few times the battery is dead.
Try to keep charging current down to 10A per battery, 20A for the pair.

Can't make out the model number of your Renology charger.
Use a temperature sensor on the battery. Program for 20A or lower charging current, if possible. Review voltage settings and try to use 14.4V Absorption phase, 13.5 to 13.8V float.
If the two PV panels are in parallel, orient one toward 10:00 AM sun an one toward 4:00 PM (give or take, about 90 degree angle between them so peak current is 0.7 as much as both same orientation, and power is spread out over the day.)

Your charger can deliver up to 50A, but PV is 2x 250W (STC) but when hot will deliver 80% to 90% of that, 200 to 225W. Charger loses a few percent efficiency. Around 17A each into 12V, 14A into 14V (after going through an MPPT charger). So two in parallel would deliver 28A to 34A, but at 90 degrees to each other 20A to 24A.
The charger could handle 1.5x to 2x as many panels if the battery back was large enough for that.
 
I just texted a bunch and the board seems to have struggles with my text, it wont let me post it, so I cant reply before I have figured out what the actual issue is !
 
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