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Is there any Calculator or Guided formula for computing how many 18650 cell for a certain Kw Battery pack?
Like a formula if I want 1kw Battery Pack, how many 18650 cell depending also in their capacity (mah) will I need. Close to reality computations.
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This is just a matter of simple maths, a calculator isn't needed. And not helpful because you can reach the desired goal using many different approaches.
First of all, you can't have a 1kW battery pack. You can have a 1kWh battery pack, this is an important difference. 1kW is a power of 1000 watts, 1kWh is a power of 1000 watts over an hour, which is the energy stored in the battery.
A 1kWh battery can have all kind of different physical sizes/shapes and electrical configurations, there is no standard or THE single solution for this.
If you have, for example, 2000mAh 18650s then each of those stores 7.4Wh of energy and you need 136 of them (1000/7.4 ~ 136) for a 1kWh battery. 136 in parallel will give you a 1kWh battery with a nominal voltage of 3.7V.
If you want higher voltage, and you probably will, you have to put them in series as well. 7s is a typical minimum for a LiIon battery. 136 cells can't be evenly distributed over 7 packs in series, you then need 140 cells for a 7s20p setup. That will give you a battery with a nominal voltage of 25.9V (7x3.7V) and a capacity of 40Ah (20x2000mAh) which results in 1036Wh (25.9Vx40Ah).
And this is just one example, there are plenty other possibilities. It would be very hard to put this into an automatic calculator which then provides meaningful results because almost all of this depends on variables the calculator doesn't know or that have to be put in in the first place. It is easier to calculate this yourself.
For example you then have to figure out what your load is going to be and if 20p is enough to handle the current and if the runtime is what you want.
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Like DarkRaven said, no need for a calculator. Just take whatever voltage you need, divide by 3.7 to get how many cells in series you need and then add those batteries in series to get the desired Ah rating.
I did a basic Excel worksheet so I could play around with some numbers, namely mAh rating but it's very simple math.
I use 'desired aH' and 'mAh per cell' to figure out how many cells I need.
If I want a 12v battery with 200Ah of capacity, I need 448 cells in a 4s112p setup based on what I put in for mAh per cell which is 1800mAh right now.
200aH / 1800mAh = 111.1 cell packs with 4 cells per pack. Rounded up to 112 cells, that 112 * 4 = 448.
I tweak the mAh portion to reflect the quality of cells I'm using as we'll all get cells at different mAh rating when dealing with used batteries.
If I drop the mAh rating from 1800 to 1500, I now need 536 cells in a 4p134p configuration for the same total Ah rating of 200Ah.
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To the FAQ with that one Mr Raven!
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(11022017, 12:03 PM)DarkRaven Wrote: First of all, you can't have a 1kW battery pack. You can have a 1kWh battery pack, this is an important difference. 1kW is a power of 1000 watts, 1kWh is a power of 1000 watts over an hour, which is the energy stored in the battery.
But what if he wants to run power tools? How much power can his battery deliver before the voltage drop causes LVD on his inverter?
When I read the first part of OP that was my immediate thought, and I've done that math. Granted, the second part brings up capacity, but the intentfor asking the question wasn't clear to me at all.
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the Op asked for a calculator to find number of cells needed for certain kwh
desired kwh/system Voltage= ah needed ,
ah x 1000/desired cell mah= (T)otal cells needed per Parallel pack (S)eries number of packs in the battery
TxS= cells needed
later floyd
jek10 and Eddy like this post
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(11032017, 05:20 AM)floydR Wrote: the Op asked for a calculator to find number of cells needed for certain kwh
desired kwh/system Voltage= ah needed ,
ah x 1000/desired cell mah= (T)otal cells needed per Parallel pack (S)eries number of packs in the battery
TxS= cells needed
later floyd
Not exactly. OP's question is ambiguous. He might have been asking about power, or he might have been asking about capacity. if he was asking about capacity, that's too simple for anyone who understands how to multiply and divide.
Power is slightly more involved. If one needs 1kW of power from a 24V battery, it involves the acceptable voltage drop, the average cell internal resistance, resistance across intercell connections, fuses, etc., and the number of cells in parallel. This is a question that I asked myself. How much power can I get from an 8s32p LFP battery bank with 5Ah LFP cells at C/2? I measured the internal resistance of a cell. Direct measurement of voltage drop at two different currents gave me one answer. My PL8 gave me a different answer, because it uses a different method and also measures It during charging instead of discharging. I did the math.
There is at least one calculator online that estimates Crate from capacity and internal resistance or internal resistance from Crate. A calculator for estimating the number of cells necessary to deliver a given power might be handy, and that was how I read OP's question .
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You all have great information that I learn today. I'm sorry my questions was so noob, I'm just new in this stuff and I have Idea but I don't know where to start.
My idea here is to create a modular battery pack/power wall connected to solar panels that can power small devices like lights in house and usb devices maybe 250 or 500 watt per module coz here in my country we don't consumed much power per day. If I want more capacity I will just connect another module.
I don't know how to arrange those 18650 Battery if I want that capacity and what voltage configuration is ideal and efficient for home, i know there's also an inverter needed, charge controller and BMS. But I need to focus first on Batteries.
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11032017, 07:46 AM
(This post was last modified: 11032017, 08:38 AM by skyfridge.)
(11032017, 06:26 AM)jek10 Wrote: You all have great information that I learn today. I'm sorry my questions was so noob, I'm just new in this stuff and I have Idea but I don't know where to start.
My idea here is to create a modular battery pack/power wall connected to solar panels that can power small devices like lights in house and usb devices maybe 250 or 500 watt per module coz here in my country we don't consumed much power per day. If I want more capacity I will just connect another module.
I don't know how to arrange those 18650 Battery if I want that capacity and what voltage configuration is ideal and efficient for home, i know there's also an inverter needed, charge controller and BMS. But I need to focus first on Batteries.
500 watts? Holy cow! I rarely need 500 watts of power!
Let's use 500W at 12V with Lithium Iron Phosphate (LFP or LiFePO4 for short) cells.
NOTE: Your constants and variables probably will be different, unless you're using the same cells that I'm using.
LFP cells have a nominal voltage of 3.2 Volts Per Cell (VPC), so for a 12V system you need 4 cells in series for 12.8V. For LFP, you should go no lower than 3VPC, or you go into the steep part of the discharge curve, where falling over the cliff can damage your cells. That means that you have a 0.8V drop before you're in danger when your SOC is at 3.2VPC. If you want 500W of power, Ohm's Law gives 500/12.8=39.1 amps, so that's how much current you need. Now if you have 5Ah cells and the recommended Crate is C/5, you allow yourself 1 amp per cell, so the simple answer is that you need 40 cells in parallel, giving you 4s40p.
However, you aren't limited to C/5, and if you want a modular system you might not want to design it for C/5. You don't use a lot of power, so you're probably not using 500W all day. And actually your LFP cells can deliver 1C without a problem, as long as it's not a full time job. If 1C is the maximum power that you'll need once in a while, now you only need 40/5=8 cells in parallel (4s8p), a 12V 40Ah battery.
But we have to consider your LVD(low voltage disconnect) of 12V, which is dependent on the complex resistance in your battery. If you want to allow yourself a 0.8V drop at 40A, you can only have 20mΩ of total resistance in your battery. Let's see if 8p works, and for this you need to understand resistance in parallel circuits. Rtotal=sR/p, where R is the resistance in each branch, assuming they're all identical. Rearranging the equation tells you, p = sR/Rtotal. Now, the manufacturer claims that each cell has an internal resistance of about 12mΩ, and a Powerlab hobby charger comes up with something close, so let's go with that. Also, you'll have contact resistance and hopefully fuses, so let's bump that up to 40mΩ per branch. p=4 x 0.04 / 0.020=8 cells in parallel, for a rough estimate, which agrees with the simple calculation. So far so good. However, the manufacturer and your Powerlab might not agree with reality. This is only an estimate. You really need to take some measurements and to experiment. It's not an easy question to answer.
Do you understand my explanation?
By the way, that was a quick and dirty calculation, and it only looks at Power requirements. Usually when designing a battery bank you start with daily energy usage reuirements. The example of a 12V 40Ah battery only gives you about 500Wh, not accounting for the 80% DoD rule of thumb, which gives you 400Wh. For 1kWh, you'll double it to 4s16p, but that only gives you 1 day of autonomy if your charge controller breaks. The rule of thumb is 3 days of autonomy, for various reasons, so you'd triple that, giving you 4s48p. That assumes that you really need 1kwh daily AND that you're using 5Ah LFP cells which you're probably not. BUT that doesn't include charging efficiency, inverter efficiency, etc., etc. There are many variables to consider in designing a system.
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(11032017, 07:46 AM)skyfridge Wrote: (11032017, 06:26 AM)jek10 Wrote: You all have great information that I learn today. I'm sorry my questions was so noob, I'm just new in this stuff and I have Idea but I don't know where to start.
My idea here is to create a modular battery pack/power wall connected to solar panels that can power small devices like lights in house and usb devices maybe 250 or 500 watt per module coz here in my country we don't consumed much power per day. If I want more capacity I will just connect another module.
I don't know how to arrange those 18650 Battery if I want that capacity and what voltage configuration is ideal and efficient for home, i know there's also an inverter needed, charge controller and BMS. But I need to focus first on Batteries.
500 watts? Holy cow! I rarely need 500 watts of power!
Let's use 500W at 12V with Lithium Iron Phosphate (LFP or LiFePO4 for short) cells.
NOTE: Your constants and variables probably will be different, unless you're using the same cells that I'm using.
LFP cells have a nominal voltage of 3.2 Volts Per Cell (VPC), so for a 12V system you need 4 cells in series for 12.8V. For LFP, you should go no lower than 3VPC, or you go into the steep part of the discharge curve, where falling over the cliff can damage your cells. That means that you have a 0.8V drop before you're in danger when your SOC is at 3.2VPC. If you want 500W of power, Ohm's Law gives 500/12.8=39.1 amps, so that's how much current you need. Now if you have 5Ah cells and the recommended Crate is C/5, you allow yourself 1 amp per cell, so the simple answer is that you need 40 cells in parallel, giving you 4s40p.
However, you aren't limited to C/5, and if you want a modular system you might not want to design it for C/5. You don't use a lot of power, so you're probably not using 500W all day. And actually your LFP cells can deliver 1C without a problem, as long as it's not a full time job. If 1C is the maximum power that you'll need once in a while, now you only need 40/5=8 cells in parallel (4s8p), a 12V 40Ah battery.
But we have to consider your LVD(low voltage disconnect) of 12V, which is dependent on the complex resistance in your battery. If you want to allow yourself a 0.8V drop at 40A, you can only have 20mΩ of total resistance in your battery. Let's see if 8p works, and for this you need to understand resistance in parallel circuits. Rtotal=sR/p, where R is the resistance in each branch, assuming they're all identical. Rearranging the equation tells you, p = sR/Rtotal. Now, the manufacturer claims that each cell has an internal resistance of about 12mΩ, and a Powerlab hobby charger comes up with something close, so let's go with that. Also, you'll have contact resistance and hopefully fuses, so let's bump that up to 40mΩ per branch. p=4 x 0.04 / 0.020=8 cells in parallel, for a rough estimate, which agrees with the simple calculation. So far so good. However, the manufacturer and your Powerlab might not agree with reality. This is only an estimate. You really need to take some measurements and to experiment. It's not an easy question to answer.
Do you understand my explanation?
By the way, that was a quick and dirty calculation, and it only looks at Power requirements. Usually when designing a battery bank you start with daily energy usage reuirements. The example of a 12V 40Ah battery only gives you about 500Wh, not accounting for the 80% DoD rule of thumb, which gives you 400Wh. For 1kWh, you'll double it to 4s16p, but that only gives you 1 day of autonomy if your charge controller breaks. The rule of thumb is 3 days of autonomy, for various reasons, so you'd triple that, giving you 4s48p. That assumes that you really need 1kwh daily AND that you're using 5Ah LFP cells which you're probably not. BUT that doesn't include charging efficiency, inverter efficiency, etc., etc. There are many variables to consider in designing a system.
I know it's lame 500w Battery pack. I will just introduce the technology and prove that it will work and not expensive. They see Solar panel system here in the Philippines is a luxury item(they even use lead acid battery which is we all now is not efficient, etc).
I will try to understand first what you are trying to explain to me before I ask again to you.
I'm just more on Computer programming and not familiar with Electrical even though I'm using Arduino and raspberry pi, I will use my knowledge to create software or app for the system.
Thank you all for replying to my questions.
