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Open circuit battery in a series string while midpoint connected to a battery in other string: how much current will flow?

I am trying to understand the path of the electons flow. Let's say we're charging with some current (200A) and that the battery Y tripped open:
View attachment 129125
So the electrons flow (follow the blue line) through the main pos conductor to the buss bar. From there the flow is divided equally to 100A between the two strings. Then the path is shown unti. the point where I am able to understand it, see two blue dots. Where does the flow goes from there?
 
Sorry not following?
I want to understand what is the current flow in such configuration when battery Y trips open:
1673474331734-png.129125

Where and how much current will flow when Y is open.
 
To learn. This config was adviced in a product manual. I was told that is bad. I want to know why it is bad. To identify other bad configurations if I see them in the future.
 
To learn. This config was adviced in a product manual. I was told that is bad. I want to know why it is bad. To identify other bad configurations if I see them in the future.
The configuration I showed you makes the batteries completely independent.
I value that.
 
This product line is the current state of the art for high current power distribution.
The overall design pattern is to connect up to 4 batteries to the lynx power-in.
A lynx distributor is attached to the power-in via their backplanes.
The loads and charge sources are attached to the distributor.
The lynx shunt can optionally be placed in between the power-in and the distributor.
If you wanted more batteries then add another power-in to the line side of the shunt.
 
This product line is the current state of the art for high current power distribution.
The overall design pattern is to connect up to 4 batteries to the lynx power-in.
A lynx distributor is attached to the power-in via their backplanes.
The loads and charge sources are attached to the distributor.
The lynx shunt can optionally be placed in between the power-in and the distributor.
If you wanted more batteries then add another power-in to the line side of the shunt.
Thanks for sharing. I saw it, but I don't think I need it. I have a relatively small battery bank.

View attachment 129149
If that's right, then it is clear where 200A is coming from and which battery takes that load. Although looking at this I am concluding that interconnects should be able to carry 100A and not 200.
@Ampster do you think that's right?
 
I know this was done for a short period of time with 12V VRLA batteries on large UPS systems, but it is a bad practice in many, many ways. IIRC it was done because early battery monitoring systems (like Alber) were really expensive and it was a belt-and-suspenders way protect against an open-circuit within a jar.

In that type of system you have a (relatively) high fault current and no inherent protection, so you need to make sure you are over-sizing the connectors so you don't have them melt. You also have reverse charge potential around the open circuit, which can eventually lead to other problems.

As to the oversizing logic, it is to ensure that you don't melt a wire with a single open circuit before you address the problem, so you size the paralleling conductor at (1+1/(n-1) )* OCPD Amps.
 
it was done because early battery monitoring systems (like Alber) were really expensive and it was a belt-and-suspenders way protect against an open-circuit within a jar.
the problem is that modern battery monitor systems (Victron) suggest doing it in their manuals.

I am trying to understand the mechanism of that fault. I've been told that if one battery trips open, than the other battery receives a full stack charge/discharge. I want to understand how that happens in terms of current flow.
The only way I can explain how that could happen is as following, the battery to the right takes 200A when battery Y trips open.
1673484283947-png.129149
 
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