Is using thermal energy an effective method to powering an implanted tracking device?

Using thermal energy for power generation is a thing, but you need a temperature difference to make it work. You implanted device is likely to be essentially uniform in temperature meaning that it won't be able to generate any power this way.

Even if you do have a minor thermal gradient you can't get much from it. The Carnot limit sets the maximum possible power to the thermal energy flowing across the gradient times an efficiency given by $$\eta = \frac{T_\text{high} - T_\text{low}}{T_\text{high}}$$ (with the temperature expressed in an absolute scale where zero is absolute zero). For the small thermal gradient available to an embedded device both the raw flux and the efficiency will be very small, so the available power is vanishing.

Short answer: nope.

If you want to diddle the biology in hopes of changing the answer you should note that you restrict the part of the anatomy where the thing will work to the part with a steep thermal gradient.

Much longer answer

Let's do an estimate of how much power would actually be available.

Assumptions:

• Working area of $1\,\mathrm{cm^2}$.

• Thickness $1\,\mathrm{mm}$. We might make the device thinner, but then it won't be able to use the full thermal difference available.

• We place the power-generation zone between the muscle and the top of the dermis/bottom of the epidermis (which gets us fairly consistent temperatures over a range of external conditions: don't want to be at the mercy of the weather for this thing to keep working ) And we'll pretend we get the full nominal core temperature to skin temperature difference: $34$"“$37^\circ\mathrm{C}$.

  This is rather more than a little optimistic (especially as thin as I have proposed to make the device), but let's go with it.

• Perfect Carnot efficiency. (Note that this is a really big ask: the current leading solid state technology for this get about 1/6 of the theoretical maximum.)

• Selecting a thermal conductivity is the hard part. A higher value results in more heat flow if the source region can be replenished and the waste heat rejected fast enough; otherwise the temperature difference drops as the hot end cools or the the cool end heats up (or both). So we use a value close to that of flesh: $k = 0.5 \,\mathrm{W/(m \cdot K)}$.

The available thermal power is then \begin{align*} P_{th} &= k A \frac{\Delta T}{\Delta x}\\ &= \left(0.5\,\mathrm{\frac{W}{m \cdot K}}\right) (10^{-4}\,\mathrm{m^2}) \left( \frac{3\,\mathrm{K}}{10^{-3}\,\mathrm{m}}\right)\\ &\approx 0.15\,\mathrm{W} = 150\,\mathrm{mW}\;. \end{align*} That's pretty low, but if we're assuming a lot of technological advance it might not be disastrous. After all a 3G phone can transmit with only about $750\,\mathrm{mW}$.

But now let's look at that Carnot efficiency (recalling that we have to use absolute units here): \begin{align} \eta &= \frac{(310\,\mathrm{K}) - (307\,\mathrm{K})}{310\,\mathrm{K}} \\ &= \frac{3}{310} \\ &\approx 0.01 \;. \end{align} So the actual power available in our highly optimistic model is around $1.5\,\mathrm{mW}$.

That's awfully low for something that is suppose to keep tabs on the wearer and we've made a lot of highly optimistic assumptions to get it that high. Image what happens if we let that temperature difference sag even a little bit. Or we have to make it thicker to access the full temperature difference. Or we can only get halfway from where we are now to the theoretical best efficiency.

posted over 6 years ago

dmckee‭

21 reputation 0 7 2 2