Electrical Busbar help

b0re

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Aug 31, 2018
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Hello,

I have created a 7s pack 120 cells per pack. The max the pack will ever seewill be 3500w thats140amps if I divide this by 7 this gives me 20amp per pack.

Question - Each pack has 5 3mm copper rods going to a sold copper 3mm plate, the plate isabout an inch wide. Does the total amps/current run through the 3mmplate only?

Thanks
 

ajw22

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Nov 16, 2018
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b0re said:
I have created a 7s pack battery 120 cells per pack. The max the pack battery will ever seewill be 3500w thats140amps [...]

The math is correct up to this point, but the assumption not quite. The highest Amps will flow when the cells are near empty (3.0V), so you should expect as much as3500W / (7 * 3.0V) = 167 Amps to flow.


b0re said:
[...] if I divide this by 7 this gives me 20amp per pack.

Nope, 167 Amps has to flow through all 7 packs.



b0re said:
Question - Each pack has 5 3mm copper rods going to a sold copper 3mm plate, the plate isabout an inch wide. Does the total amps/current run through the 3mmplate only?

Assuming the 18650 cells are connected to the 5 rods (Pictures would help!),
the 5 pieces of 3mm copper rods share the 167 Amps more or less equally, so each one carries 33.4 Amps.
What the plate carries depends on where exactly the rods are connected, and where on the plateyou measure the current. It's basically like 5 small rivers flowing into a large river. Sections that are "down stream" will be carrying the full 167 Amps, butparts more "up stream"will be carrying less, depending on how many "small rivers" have merged into the main river.

In any case, the plate with3mm x 25mm (=75mm^2) cross section is plenty for carrying 167 Amps, according to this AWG table:
https://www.powerstream.com/Wire_Size.htm

You can also calculate the expected energy loss ( =heat generation) under max load by putting in the values in this calculator:
https://photovoltaic-software.com/solar-tools/dc-ac-drop-voltage-calculator

The "3mm copper rods" on the other hand might be a little tight...is that3mm diameter or 3mm^2 cross section area? Quite a difference there.
 

b0re

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Aug 31, 2018
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I made a picture forgive me for my paint skills.

If 33.4amps runs down the in rods under full loadI think I ill need something thicker for heat and efficacy reasons on all 7 packs



image_bnxdic.jpg
 

ajw22

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If the rods have 3mm diameter (ie. 7mm^2 cross section), then there should be very little heating.
If the rods have a cross section of 3mm^2, then it's pretty close to the "Maximum amps for chassis wiring" (AWG table). So still safe, but expect noticeable heating and energy loss when used over longer periods.
 

daromer

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That pack is to small to handle that max load comfortble IF the cells are 2nd hand... Only IF they are new cells i would do that.

IF 2nd hand you should not max out more than the tested current or else you Will have issues over time.
Example: IF testing per cell is 500mA then your max on the battery is 60a.
60a at 3v per cell is 14*3*60= 2500watt on the ac Side with lets say20% losses you have max 1800watt. So your inverter should not be specced to more than max that!
So in your case i highly recommend to double Up the cell count if its 2nd hand cells tested at0.5. IF you tested Them at 1a you have a tas more to play with :)
 

b0re

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The max is 3500w it will rarely see this. 99% of the time is will be only pulling 600w to 1000w.

Thanks for your feedback I think I will make the bars a little thicker just because I can at this stage.
 

daromer

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You always design for the max energy that could be transferred!

Its important that the fuses/breakers take the first hit. Secondly i would say that the inverter should be limiting and lastly the cables and then batteries. This will ensure a good setup in the long run.

Where is that max 3500W? is it calculated on the battery side or where? Many people tend to forget the nominal on the inverter is far from max at battery side.

Yes get it beefier now directly its better. Voltage drops is not dangerous as such but its important to know where you have them and how much energy is dissipated into heat :)
 

b0re

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The induction cooktop can pull 3200w AC, the chances of setting both hobs to 100% are very unlikely but I feed I should plan for this in case other items are running as well.
 

daromer

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You must plan for it. Ie max what the inverter is rated for. Thats the key. To mnay plan for what they think they might use and thats wrong :)
 

OffGridInTheCity

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Agree with @Daromer - planning for max inverter (power) draw is key.

My goal was a'whole house' setup - e.g. heat-pump (4500w),cooktop (4500w), dryer (4500w), kitchen circuits (can spike to 4000w), plus all the other smaller 'stuff' around the house.

At first I tried to think in terms of... "if I just remember to only run 1 thing at a time..." then I won't need to worry about 17,500w listed above. I can make sure to avoid the cooking while running the dryer kind of thing. BAD IDEA. Invariably as things work over time you just stop thinking in micro terms and forget and turn everything on at once!! So I now have 24,000w of inverter and the battery bank to back it up andall is smooth.

Conclusion - inyour planning, I suggest that you face up front the max load you plan to run and build the system to accommodate it OR reduce expectations and make sure the wiring is consistent with that:) Remember you can do things in steps... maybe this year its only 8000w max and then next year you expand to 16,000w etc - which can lead you to buy equipment that can be put in tandem to expand and leave room for more batteries etc.
 

b0re

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Aug 31, 2018
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This is what I have gone for the bars over the top is rated at 45amps per line. I have not cleaned up the solders or shrink-wrapped it yet. The out put cable is 400ams just because thats what i had.


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