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About Battery Temperature

A project log for A 110 kWh Powerbucket

The lead-acid batteries of my off-grid solar system are dead. I will replace them by a 18650 batteries stack housed in a big wooden box.

Michel Kuenemann 02/18/2017 at 08:380 Comments

Several hackers asked me about temperature management of the powerbucket, so let me give some info about this aspect.

Currently, I have put one sensor in the middle of the string. This sensor is attached to a 35 mm² bus bar, tight to the stack. I have a very good confidence in the measurements it provides. Of course I assume that the individual temperature variations between the 1000 cells are very small and can be neglected. This assumption has to be discussed, of course.

Thanks to my Raspberry Pi based solar data base I have produced a chart showing the temperature of the stack since its start of service in novembre 2016.

This chart shows clearly that the battery temperature remains close to the average room temperature of my house, namely 21 °C (70 °F). One can say that these temperature conditions are very safe for the stack. One can also assume that none of the 1000 elements gets a lot warmer or colder than the measuring point.

Why are the temperature variations so small ?

Batteries heat up when they are crossed by a current, regardless if it is a charging or discharging current. This heat comes (mainly) from the power (Joule effect) dissipated in the internal resistance of the cell:

P = R * I²

Where

P = dissipated power (Watt).
R = Internal resistance of the battery element (Ohm)
I = current flowing through the element (Ampere)

Let us assume a current of 140 A flowing through the stack. This represents a power of nearly 8 kW. In my system, this can last only for a few minutes, due to inverter power limitation.

The internal resistance of a module, composed of 70 cells wired in parallel is about 1 milliOhm.
Power dissipated by one module = 0.001 * 140 * 140 = 19.6 Watt

Power dissipated by one 18650 cell = 19.6 / 70 = 0.28 Watt. This is the power rating of a small through hole resistor. Such a small power cannot heat up a 18650 element significantly.

Keep also in mind that if the current is of 70 A (half of 140), this power is divided by 4 !

This is why my Powerbucket needs no cooling system.

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