# DIY battery heat generation

#### rexgoodwin

##### New member
Is there a suitable formula that I can use to estimate the heat dissipated from a DIY battery pack (48V, 300A using 30Q Samsung cells) when powering a 200A, 48V load? I want to run some simulation analysis based on the heat dissipation

#### floydR

##### Well-known member
300A or Ah?
Samsung might have a chart on individual 30Q cells heat of cell under various loads. Assuming there are enough in parallel I would think not much of a heat rise.
a 14s100p 30q battery would be a 300AH battery. A 200A load would be a .66C load. How long would you be pulling 200A out of the battery?
Later floyd

#### ajw22

##### Well-known member
Never really thought about it, but my guess would be the sum of:
1. NumberOfCells * IR(20mOhm?) * CurrentPerCell^2, so using floydR's values: 112 Watts of heat about 750Watts?
2. All wires and fuses that will heat up under high loads, using same formula ( Power = Resistance * Current^2 )

Only current matters, not voltage. Thus you'll want to maximize voltage and minimize current to reduce heat losses.

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#### rexgoodwin

##### New member
300A or Ah?
Samsung might have a chart on individual 30Q cells heat of cell under various loads. Assuming there are enough in parallel I would think not much of a heat rise.
a 14s100p 30q battery would be a 300AH battery. A 200A load would be a .66C load. How long would you be pulling 200A out of the battery?
Later floyd
300A current. The capacity is 45,000mAh. The config has 15 in parallel. I would be pulling 200A for about 13 minutes or so before the cells reach their minimum voltage and the bms shuts it down

#### rexgoodwin

##### New member
Never really thought about it, but my guess would be the sum of:
1. NumberOfCells * IR(20mOhm?) * CurrentPerCell^2, so using floydR's values: 112 Watts of heat
2. All wires and fuses that will heat up under high loads, using same formula ( Power = Resistance * Current^2 )

Only current matters, not voltage. Thus you'll want to maximize voltage and minimize current to reduce heat losses.
Ah I see so using that formula for a 13s15p pack, I get the following:

(assuming the max discharge current is 20A; some sites say 20A but others say 15A, but this is all theoretical so let's assume 20A for now)

P = 195*0.02*20A^2 = 1560 watts of heat?

when you say current per cell, does this mean how much current each cell will discharge for the 200A load, so:

200A load / # of parallel cells = 200 / 15 = 13.33A, then

P = 195*0.02*13.33^2 = 693 watts of heat?

#### ajw22

##### Well-known member
Yes. As you can see, going from 20A -> 13.33A reduces heat output to less than 1/2!
Not entirely sure about the actual IR of the 30Q. Some sites state as low as 12mOhm, others state 40+mOhm.

• rexgoodwin

#### rexgoodwin

##### New member
Yes. As you can see, going from 20A -> 13.33A reduces heat output to less than 1/2!
Not entirely sure about the actual IR of the 30Q. Some sites state as low as 12mOhm, others state 40+mOhm.
Thank you! So for my analysis I will probably take the second formula since it factors in how much current each cell contributes to the load and run the CFD based on the 693 watts of heat which is probably more accurate than taking the max discharge current for a 30Q cell 