Battery help

[rasen58]

1. A discharged lithium-ion battery is constructed using a thin slice of graphite as anode, a piece of

porous polypropylene sheet as separator, and a thin piece of lithium cobalt oxide as cathode. The

dimensions of the anode, separator, and cathode are 1 cm x 2 cm x 50 microns, 1 cm x 2 cm x 25

microns, and 1 cm x 2 cm x 50 microns, respectively. When charging the battery, lithium ions

deintercalate from the cathode and intercalate into the anode. Once the battery is fully charged, the

anode has expanded by 10% in volume, and the cathode has expanded by 20% in volume. Assuming the

overall volume of the battery is held constant, calculate the percentage in volume change for the

separator. Since the separator is porous, what does this mean to the pores and the ability for lithium

ions to pass through these pores?

So the anode is 100 x 10^-10 and separator is 50 x 10^-10 and cathode is 100 x 10^-10

Total Volume = 250 x 10^-10

Anode becomes 110, cathode becomes 120, so separator has to decrease to 20, so it decreases by 60%

I'm just not sure about the question about what happens to the pores and the lithium ions. Do the pores become smaller and the ions cannot pass through anymore?

But that wouldn't make sense because they need to pass through to discharge.

2. Due to space constraints, one of the battery packs is placed next to the internal combustion engine

and separated by 2 cm of foam insulation. During maximum power, the outside surface temperature of

the engine can reach 220°C. If the highest temperature that the battery can experience is 120°C before

going unstable, are the current placements of the battery pack and the engine safe? The heat transfer

rate through the insulation is 720 W/m², and the thermal conductivity of the insulation is 0.12 W/m °C

I think the answer is no because 220 is a lot more than 120, but how do I use the heat transfer rate and thermal conductivity to help me do the problem?

Edited March 1, 2014 by rasen58

[Opcom]

OK if the thermal conductivity of the foam is 0.12W/m, and you have 0.02m of foam insulation,

then the heat transfer through the foam would be 0.12W/m / 0.02m, right?

That is 6W going through the foam's thickness.

I agree with 720W/m^2.

The battery is so small, 0.02m * 250 x 10 ^-10, as you said.

So the heat to the battery is 6W - which is a lot to put nto a small volume like that. I would agree that the temperature rise would be excessive and damage the battery, but I don't see enough info for a complete solution. They didn't say how the battery was to dispose of excess heat. it could only do so through the wires and through the foam insulation, but the overwhelming amount of heat energy is flowing the other directioin, into the battery.

 

I agree with no.

 

Sorry it is soo late at night, or, 3AM. , my reasoning may be a little wweird at this hour.