Voltage drop very fast on load

[acdoctor]

I am assuming you are referring to battery amps. Not line voltage amps.
That would be about 50 amp hours. Should be at 75% SOC. Or around 49.5 volts.
What is the current specific gravity of cell or two from each battery??

 

[Montanat]

I am assuming you are referring to battery amps. Not line voltage amps.
That would be about 50 amp hours. Should be at 75% SOC. Or around 49.5 volts.
What is the current specific gravity of cell or two from each battery??

I'm referring to the inverter amps load on the screen.

I Consumed 14a for 2 hours and 24a for 45 min. Then inveter shut down coz I reached 48v (cutoff setup)

I Removed 10a (turned off the heater), inverter showed 14a load and lasted another 1 hour).

I also removed 9a (shutting down another heater) now the screen shows 6amps load and battery voltage is 48.8v.

So final results : it lasted 3 hours on 14a load + 45 min on 24amps load + I'm consuming 6amps now and batteries still have 48.8v (measuring voltage of batteries by voltmeter)

Gravity is 1.200 / 1.198 on all cells (voltage is 48.8v) but I have a cheap 2$ Chinese hydrometer I bought 2 days ago dunno if it's accurate or defected.
This gravity was the same when batteries were at 57.6v as well.

 

[acdoctor]

So that is :14 amps x 120 volts / 48 volts x 3 hours = 105 amp hours
Then 24 amps x 120 volts / 48 volts x .75 hours = 45 amps hours
Then 6 amps x 120 volts / 48 x ? Hours = ?
So 150 amp hours and going.
If those are battery amps:
14 x 3 = 42 amp hours
24 x .75 = 24 amp hours
6 x 1 = 6 amp hours
So that is 72 amp hours and going.
So we need to define which you are looking at.
And those are low calculations the 24 is probably 24 amp hours because of the Peukert affect.
Specific gravity at 1.20 is somewhere around 50 to 60 % SOC so it lines up with the voltage.

 

[Montanat]

So that is :14 amps x 120 volts / 48 volts x 3 hours = 105 amp hours
Then 24 amps x 120 volts / 48 volts x .75 hours = 45 amps hours
Then 6 amps x 120 volts / 48 x ? Hours = ?
So 150 amp hours and going.
If those are battery amps:
14 x 3 = 42 amp hours
24 x .75 = 18 amp hours
6 x ? = ??
So that is 60 amp hours and going.
So we need to define which you are looking at.
And those are low calculations the 24 is probably 24 amp hours because of the Peukert affect.
Specific gravity at 1.20 is somewhere around 50 to 60 % SOC so it lines up with the voltage.

Good morning. We have 220v electricity not 120v.

 

[acdoctor]

Oh that’s right, sorry it was late for me. How long did it go at the 6 amps?

 

[Montanat]

Oh that’s right, sorry it was late for me. How long did it go at the 6 amps?over

I replaced the inverter today. It's a Sumry 5.5k.
It has a maximum bulk 58.4v, maximum float 58.4v (previous one has 61v / 61v)

Setup I did = 58.4v bulk / 58.4v float / charge @30amps.

Charge started at 30 amps (previous one always starting at 27amps)
In 1 hour voltage reached 57.6v and charging amps was 25amps. And kept increasing till 58.1v while charging amps decreasing but not as fast as the previous one.

Now I have 58.1v / charging at 2amps since 3 hours and it didn't go any further.

Both fans didn't stop and still on, which is annoying (it's a defected thing as Chinese tech of "Sumry" told me).

I will keep it charging to see if it goes more than 58.1v, I still have 4 hours before power goes off in here.

Ps: the 6amps lasted 1 hour more last night but I went to sleep it was 5am.

 

[acdoctor]

Where did you end up when power went down?

 

[Montanat]

Where did you end up when power went down?

No improvement. Voltage stayed 58.1v and charging kept blinking at 2amps.
So it stayed around 7 hours at this point (58.1v). then I decreased the float to 54.4v and let it sit there.
All voltages at screen + pins + batteries were the same during those 7 hours.

Then electricity went off and I went on batteries. I tried 20amps it lasted 2 hours, cutoff was at 48v.

CS of the Indian battery "Japston Battery" sent me a special calculation for my batteries and how they should work (attached are the docs he sent).
And he said I should set the cutoff to 46v and not 48v so they will last longer . He said 48v is the SOC.

I will get back to the other inverter tomorrow this one sounds like a washing machine about to take off.. M

 

[acdoctor]

These are the same ones you presented before. Did you get anything similar on discharge? It should show something like a graph with C ratings.
C 10 is 180 amp hours. 180 / 10 = 18 amps discharge with a usable 90 amp hours to 48 volts.

You said you tried 20 amps. That should be at least 80 amp hours.
You said for 2 hours. That’s 40 amp hours.

 

[acdoctor]

Deleted battery specs

 

[Montanat]

It looks like these are your batteries.View attachment 89084View attachment 89085

No it's those ones (attached)
If I setup the cutoff to 50% dod (46v and not 48v like the tech guy said) I'll get 2 more hours or more at 20amps load. This will become 4 hours or more on 20amps load.
This will be close to normal duration of those batteries?

Again, sorry im not good at math at all.

And by the way we have cold weather now it's around 12 to 14 celcius at night. Last week was 8 celcius.

 

[acdoctor]

Where are the batteries? What is the battery temperature?

 

[Montanat]

Where are the batteries? What is the battery temperature?

Batteries are inside the house. Temp yesterday at night here was around 17 celcius indoor / 13C outdoor.
Now it's 21C outdoor, 25 indoor.

Tried with 2 medical thermometer now it doesn't read less than 35c I'll buy one to check the batteries

Just one question, in simplified English, how long should 20 amps load last till batteries reach 48v, and 46v.

 

[acdoctor]

Have you verified the load with a amp meter other than the one on the display?20 amp load on the batteries you have is a ( C 10 load ). Capacity of batteries 200 AH divided by 10 hours. the hours cancel and you are left with 20 amps. That is from 100% charge to 0% charge or from 51 volts measured after sitting for 4 hours To 10.5 volts.
But you have to pay a penalty for the rapid discharge faster than the C 20 rate of 10 amps. That penalty is a loss of 20 amp hours.
Answer: So the short answer is about 4.5 hours.
Now my experience is that you won’t get but 3.5 hours with new batteries.
Deep slow discharge batteries have plates in them. Car batteries have plates that look like a screen. The difference is surface area. Your batteries after being used for some months the plates will soften they call it. What is really happening is tiny pits, or erosion which causes more surface area. That allows more chemical reaction to happen at a faster rate than when new.

 

[Montanat]

Have you verified the load with a amp meter other than the one on the display?20 amp load on the batteries you have is a ( C 10 load ). Capacity of batteries 200 AH divided by 10 hours. the hours cancel and you are left with 20 amps. That is from 100% charge to 0% charge or from 51 volts measured after sitting for 4 hours To 10.5 volts.
But you have to pay a penalty for the rapid discharge faster than the C 20 rate of 10 amps. That penalty is a loss of 20 amp hours.
Answer: So the short answer is about 4.5 hours.
Now my experience is that you won’t get but 3.5 hours with new batteries.
Deep slow discharge batteries have plates in them. Car batteries have plates that look like a screen. The difference is surface area. Your batteries after being used for some months the plates will soften they call it. What is really happening is tiny pits, or erosion which causes more surface area. That allows more chemical reaction to happen at a faster rate than when new

I have a multimeter with a clamp meter. I measured the charging amps to match the screen while charging. The screen was zero, clamp meter showed 1.2 / 1.5 amps (clamp on + of the battery)

Where should I clamp it to measures the load? On the output cable of the inveter right?

You mentioned 3.5 hours on 20amps load, this is from 51v down to 10.v or down to 48v? (Sorry for asking just want to learn more)

Now I measured batteries with the multimeter pins (multimeter on A section) it gave me 39.2A. Each battery 9.8A. Saw this method of measuring amps on the net dunno if it's correct.

Clamp meter pic attached it's a 600v / 600A auto range.

 

[Bob B]

I have a multimeter with a clamp meter. I measured the charging amps to match the screen while charging. The screen was zero, clamp meter showed 1.2 / 1.5 amps (clamp on + of the battery)

Where should I clamp it to measures the load? On the + of the battery or the output cable of the inveter?

You mentioned 3.5 hours on 20amps load, this is from 51v down 10.v or down 48v?

Now I measured batteries with the multimeter pins (multimeter on A section) it gave me 39.2A. Each battery 9.8A. Saw this on the net dunno if it's correct.

Clamp meter pic attached it's a 600v / 600A auto range.

That is an AC clamp meter .... You will not be able to measure the current on the battery side with that meter.
You need on that will measure DC amps .... or use a shunt or meter than can be put in series with the current flow.

 

[acdoctor]

Yes 3.5 hours to about 48 volts. Maybe as low as the 46 volts with that C10 load.
I agree that meter only properly measures A/C amps. Diode and D/C volts is the only D/C functions.
You can check A/C output amps I would like to know that.?
If you had or can barrow a D/C clamp meter you would check main battery cables. So meters it is important which face of meter is towards battery, usually marked with + . I would like to know between each battery as well?

 

[Montanat]

Just did a test dunno if it's the right way.
Load on batteries was 11amps on screen (discharging) . I clamped the AC voltmeter on the output cable of the inverter it read 1.6amps .

Will buy a DC clamp one and check.

 

[acdoctor]

What was the voltage at the same connections?

 

[acdoctor]

You mean to say the same amperage screen you have referenced before that was 20 amps.This time was showing 11 amps?