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I'm currently sourcing Li-Po batteries and need to find an average mAh per cm3 to implement it into my product. From what I read (wikipedia), the energy density of Li-Po ranges from 250 to 730 W.h/L

I also read that : $$Q(mAh) = E(Wh) x 1000 / V(v)$$

What I did (be careful, you might facepalm), in order to get a mAh/cm3 value for a 3,7v battery was :

N.B : I didn't really understand the "W.h/L" unit, I expected it to mean W.h-1.L-1

$$Q(mAh)/L = (E(Wh) x 270.27 ) / L$$

Which gave me, assuming the energy density was 750 W.h/L :

$$Q (mAh/L) = E (Wh/L) x 270.27 $$

$$Q (mAh/L) = 197297.1 $$

I then converted this to cm3, considering 1dm3 = 1L :

$$Q (mAh/cm3) = 197.297 mAh/cm3$$

The lower value gives me $67.56 mAh/cm3$

Is this calculation correct ? I tried to check this, by comparing the common sizes of Li-Po batteries on the market. I have results ranging from $50$ mAh/cm3 to $110$ mAh/cm3 for these, which looks normal.

If it correct, how come I didn't find any Li-Po battery, which goes over $110$ mAh/cm3 on the market ?

I guess this belongs to physics, as it deals with unities and energy. I din't practice maths and physics for a very long time now, so I might have just made something stupid :-)

Thank you very much for your help.

Stanislas.

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  • $\begingroup$ On the advanced formatting page you'll find some advice on using MathJax (a latex math-mode like formatting engine that is active on the site) to typeset readable math in posts. $\endgroup$ – dmckee Apr 9 '15 at 15:35
  • $\begingroup$ Thanks dmckee, never used it yet, I will reformat for readability. $\endgroup$ – Stanislasdrg Apr 10 '15 at 8:08
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I didn't really understand the "W.h/L" unit, I expected it to mean W.h-1.L-1

Think about the meaning of units. You are trying to find the volumetric energy density, that is you want $$ \frac{\text{energy contained}}{\text{volume occupied}} \,.$$

Now Watts ($\mathrm{W}$) are a measure of power which is $\text{energy}/\text{time}$, so $\mathrm{W \cdot hr}$ is a measure of energy. On the other hand $\mathrm{W/hr}$ has units of $\text{energy}/\text{time-squared}$.

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Your calculations are fine, and the fact that they come out close to what you find on the market (good for you for checking) supports that. The difference in charge/volume ratio may come from the fact that the quoted energy density is for just a cell, while the commercial batteries have cases, charging electronics, etc. increasing the volume and decreasing the charge/volume ratio.

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  • $\begingroup$ Thank you very much for your answer Ross. I picked accepted the other answer, as it will be more explanatory for others. Will upvote when possible. $\endgroup$ – Stanislasdrg Apr 10 '15 at 8:20

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