Inverter and batteries

Guys, I've said it before here, you have 2 considerations when it comes to batteries:
1 - Rands per cycle.
2 - What you can afford now, especially first time setups.

You can setup a system with LA batteries now and migrate to Li in 3 or 4 years if you don't have the cash to buy Li now (remember you are buying all the other stuff).
 
If I want 20% left ( IE use 80% of the battery ) does the 2 become 1.4285?

As was discussed in load shedding whats needed to keep watching I do need to add in how to calculate the battery pack you need

volts x Ah / watts / dod x 0.85

Volts is the battery. i.e 12, 24 or 48v
Ah is the Ah of the string, i.e 102, 105, 2000
watts is your load, i.e. 1500w
DOD = depth of discharge. 2 = 50%, 3 = 33.3%, 5 = 20%
2 is to account for a maximum target of 50% depth of discharge of the batteries
0.85 is the efficiency of the inverter, i.e loss due to converting ac to dc to ac and maintaining float charge in batteries

So in my case:
I have 4 x 12V 102Ah batteries in series and parallel making a total of 24v and 204Ah.
My load is around 500w but can go up to 1500w

24 x 204 / 500 / 2 x 0.85 = 4h10m
24 x 204 / 750 / 2 x 0.85 = 2h45m
24 x 204 / 1000 / 2 x 0.85 = 2h05m
24 x 204 / 1500 / 2 x 0.85 = 1h25m

(edit: type the equation into a calculator as written from left to right. Although the order doesn't matter as long as you don't start with one of the dividers, in which case you need to start with 1/divider.....)

So if I am running at 750w I can easily get 2h30m, or one load shedding block. I have yet to break 500w (without the fridge) average during load shedding and this reflects in the battery level indicator after 2h15m load shedding only dropping 25%. I'm about to install a new plug point for the fridge that will be on the inverter side of the db board. This will add 300w when the fridge is actively working and an average of 100w at idle. I have made this decision to preserve the fridge motor rather than as a need to keep the fridge working during load shedding.
 
If I want 20% left ( IE use 80% of the battery ) does the 2 become 1.4285?

[24v x 204Ah / 0500w load] x 0.85 = 8.3232 decimal = 8h19m runtime, with no capacity/runtime left.
[24v x 204Ah / 1500w load] x 0.85 = 2.7744 decimal = 2h46m runtime, "

[24v x 204Ah / 0500w load] / 2 x 0.85 = 4.1616 decimal = 4h10m runtime, with 50% capacity/runtime left.
[24v x 204Ah / 1500w load] / 2 x 0.85 = 1.3872 decimal = 1h25m runtime, "

You want to only use 80% of the capacity?
[24v x 204Ah / 0500w load] / 1.4285 x 0.85 = 5.8265... decimal = 5h50m runtime,
[24v x 204Ah / 1500w load] / 1.4285 x 0.85 = 1.9421... decimal = 1h57m runtime,
Checking the maths, the above feels like 70% use 30% left.

80% would be 8.3232 / 100 * 80 = 6.65856 decimal + maths = It needs to be 1.25 rather

[24v x 204Ah / 0500w load] / 1.25 x 0.85 = 6.65856 decimal = 6h40m runtime,
[24v x 204Ah / 1500w load] / 1.25 x 0.85 = 2.21952 decimal = 2h13m runtime,

Yep, now it checks out.
 
[24v x 204Ah / 0500w load] x 0.85 = 8.3232 decimal = 8h19m runtime, with no capacity/runtime left.
[24v x 204Ah / 1500w load] x 0.85 = 2.7744 decimal = 2h46m runtime, "

[24v x 204Ah / 0500w load] / 2 x 0.85 = 4.1616 decimal = 4h10m runtime, with 50% capacity/runtime left.
[24v x 204Ah / 1500w load] / 2 x 0.85 = 1.3872 decimal = 1h25m runtime, "

You want to only use 80% of the capacity?
[24v x 204Ah / 0500w load] / 1.4285 x 0.85 = 5.8265... decimal = 5h50m runtime,
[24v x 204Ah / 1500w load] / 1.4285 x 0.85 = 1.9421... decimal = 1h57m runtime,
Checking the maths, the above feels like 70% use 30% left.

80% would be 8.3232 / 100 * 80 = 6.65856 decimal + maths = It needs to be 1.25 rather

[24v x 204Ah / 0500w load] / 1.25 x 0.85 = 6.65856 decimal = 6h40m runtime,
[24v x 204Ah / 1500w load] / 1.25 x 0.85 = 2.21952 decimal = 2h13m runtime,

Yep, now it checks out.



Love you sir.
 
Good deal on 7ah LiFePO4 batteries, R 489.00+ R70 shipping:
 
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