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Load Shedding Impulse Buy - Massive Win so far


Natal_Nic

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Good Day,

Went out and picked up a Axpert KS 2KVA inverter, x2 105ah lead acid deep cycle batteries, summoned my electrician mate and voila. Load shedding a thing of the past.

The unit is not connected to any PV panels.

During load shedding the inverter is operating between 50 - 75% 

After the 2 hours of load shedding is up and mains kick in the inverter is now charging the batteries. 

Some questions:

1) Every now and then upon charging and after I hear 2 very quick beeps. As well as a single beep. These are few and far between. What are these beeps about?

2) Do I need to adjust any settings to optimise battery performance and longevity or will the factory ones suffice for this super simple setup?

 

Thanks in advance.

NN

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9 hours ago, Natal_Nic said:

1) Every now and then upon charging and after I hear 2 very quick beeps. As well as a single beep. These are few and far between. What are these beeps about?

You need to check the warning indicator on the LC Display.

9 hours ago, Natal_Nic said:

2) Do I need to adjust any settings to optimise battery performance and longevity or will the factory ones suffice for this super simple setup?

I would set the low DC cutoff voltage (setting 29) as high as possible, if you want to have your batteries last. If you prefer to have the lights on as long as possible, leave it as is (and replace the battery modules as needed).

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Wow thanks for letting me know gents. I had no idea. This is why this forum rocks

Please can you assist further. 

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Question 1

If I want to keep within a max DoD at 30% to save batteries. Should I only be utilising 378w's every hour 

24*105ah = 2520w (This is the total watts available from batteries?)

2520*30% = 756w 

756w / 2 hours of load shedding = 378watts I can use per hour?

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Question 2

How do I add more batteries to the 24v system to get more AH or Watts out of it.  Parallel or series? or is this not possible?

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Question 3

If I had a 5kva inverter and had say for example TEN 105ah batteries would there still only be 5040watts available? Is there a link for further understanding on connecting batteries up?

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Question 4

During last nights load shedding I went through the menu on the inverter and discovered that load in watts was around 350w. Is this measured per hour so its an average of what the load will draw within that hour?

 

Thanks once again.

 

NN

 

 

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Q1: Correct. (Ignoring losses, but it should be fine, you might be at 33% DoD if the inverter is 90% efficient)

Q2: The best would be to add 2 more batteries of exactly the same type. Add them in the same arrangement as the other 2. Perhaps simplest is to put each of them in parallel with each of the other 2.

Q3. I am not sure I understand this question. 5kva inverters are usually 48V, so the number of 12V batteries should be multiples of 4. The Ah rating of the batteries relate to the Wh available. Wattage is more-or-less independent of that.

Q4: I believe that would be the instantaneous Watt reading. If it was that for 1 hour, that would mean you used 350Wh, since it was for 2 hours, that would equate to 750Wh. But that is assuming your consumption was constant during that time, which is unlikely.

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27 minutes ago, P1000 said:

Q1: Correct. (Ignoring losses, but it should be fine, you might be at 33% DoD if the inverter is 90% efficient)

Q2: The best would be to add 2 more batteries of exactly the same type. Add them in the same arrangement as the other 2. Perhaps simplest is to put each of them in parallel with each of the other 2.

Q3. I am not sure I understand this question. 5kva inverters are usually 48V, so the number of 12V batteries should be multiples of 4. The Ah rating of the batteries relate to the Wh available. Wattage is more-or-less independent of that.

Q4: I believe that would be the instantaneous Watt reading. If it was that for 1 hour, that would mean you used 350Wh, since it was for 2 hours, that would equate to 750Wh. But that is assuming your consumption was constant during that time, which is unlikely.

Q1 - Noted thanks

 

Q2 - So leave 2 in series and connect the other 2 up in parallel?

 

Q3 - Trying to wrap my head around on how is it possible to have a back up system with a huge amount of AH. EG: say for a 48v system you have to have x4  12v batteries. Assume each of these batteries is 100ah. If you connect these in series as you previously explained earlier you still only have 100ah. What happens to all the other AH's if that make sense? How does one get a 48v system or 24v system with 400ah or 200ah respectively? Does this only happen when connected in parallel? I assumed if youve got x2 105ah batteries you will have 210ah to work with. Hope this makes sense. haha

 

Q4 - Noted thank you

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15 hours ago, Natal_Nic said:

@P1000 is it not 24*210ah = 5040wh?

It's either 12V * 210Ah or 24V * 105Ah, depending on whether the batteries are connected in series or parallel.
Regardless though of how they are connected, they still contain the same Wh, ie. 2520.

 

44 minutes ago, Natal_Nic said:

Question 2
How do I add more batteries to the 24v system to get more AH or Watts out of it.  Parallel or series? or is this not possible?

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Question 3
If I had a 5kva inverter and had say for example TEN 105ah batteries would there still only be 5040watts available? Is there a link for further understanding on connecting batteries up?

Question 2.
You would add another string of 2 batteries that are identical to what you have now. The new batteries would be connected in parallel to your current string (which are connected to each other in series btw).

Question 3.
With a 5kva inverter, you would be able to supply 5kW of power to devices, ie. 2x 2.5kW kettles, or 5x 1000W microwave ovens.
Your batteries determine how long you can power the load for. Having 8x 105Ah batteries would give you a total capacity of 210Ah @ 48V (2 parallel strings of 4 batteries (connected in series)) which is 10080Wh total. You shouldn't really be drawing more than 50% of that if you want the batteries to last, so 5040Wh. With 5040Wh, you can power 5kW of devices (remember the 2x kettles, or 5x microwaves) for 1 hour, or you could power 1x microwave for 5 hours, or you could power something using 350W for nearly 14.5 hours.

Not answering the other questions as they have been answered already.

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Q2 Old battery 1 in parallel with new battery 1, old battery 2 parallel with new battery 2. This should be the simplest.

 

Q2+Q3:

2 Batteries in series = voltage * 2

2 Batteries in parallel = Ah * 2

Wh = voltage*Ah <- if you double the amount of batteries, you can only double Wh, which means you can only double one of the variables on the other side.

That way the physics lawyers won't sue you.

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8 minutes ago, Natal_Nic said:

Q3 - Trying to wrap my head around on how is it possible to have a back up system with a huge amount of AH. EG: say for a 48v system you have to have x4  12v batteries. Assume each of these batteries is 100ah. If you connect these in series as you previously explained earlier you still only have 100ah. What happens to all the other AH's if that make sense? How does one get a 48v system or 24v system with 400ah or 200ah respectively? Does this only happen when connected in parallel? I assumed if youve got x2 105ah batteries you will have 210ah to work with. Hope this makes sense. haha

When connecting batteries in series, you add up the voltages, so 4x 12V 105Ah batteries gives you a 48V 105Ah string.
When connecting batteries (or strings) in parallel, the voltage remains the same, but the Ah increases., so 2x 48V 105Ah gives you 48V 210Ah. Not recommended to go with more than 2 strings in parallel.

To get the really big Ah capacity battery banks, the easiest way is big Ah batteries with low voltages, so a single string of 24x 2V 500Ah batteries = 48V 500Ah in a single string.

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1 hour ago, Natal_Nic said:

Question 1

If I want to keep within a max DoD at 30% to save batteries. Should I only be utilising 378w's every hour 

24*105ah = 2520w (This is the total watts available from batteries?)

2520*30% = 756w 

756w / 2 hours of load shedding = 378watts I can use per hour?

Your questions have been well answered. But I thought I'd correct the units for just this first question, hopefully to make it easier to think about these things.

Quote

If I want to keep within a max DoD at 30% to save batteries. Should I only be utilising 378w's every hour 

Your load  should be an average of 378 W (378 watts). Hours have nothing to do with this part.

Quote

24*105ah = 2520w (This is the total watts available from batteries?)

No, it's 24 * 105 Ah = 2520 Wh (This is the total energy available from battery)

Quote

2520*30% = 756w 

2520 Wh * 30% = 756 Wh 

Quote

756w / 2 hours of load shedding = 378watts I can use per hour?

756 Wh / 2 hours of load shedding = 378 W (watts, average) you can use for however long the battery lasts. Notice how the hours above and below the dividing line cancel out: Wh / h = W.

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