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Edit: I rewrote my post. To downvoters, can you comment on your reasons? As a kind reminder:

Downvote questions that don't show any research effort or don't contain enough information to be clear and answerable. These questions may also need to be closed.

The Seebeck effect is used to power space probes that travel too far for solar panels to be sufficient. In that scenario, the heat source is provided by radioactive decay, which lasts for decades. More info on Radioisotope thermoelectric generator and the example of Voyager 2.

The voyager probes had 3 RTG - 99.6kg total without the plutonium - for a rated power of 470W. Keep in mind that this device can have a structural value and therefore an inferior net weight. Also, those are values from the 1970s.

The Seebeck effect is the first option that came into my mind to produce electricity from the waste heat, of turbofans or other areas with unwanted heat (pumps, radar?). The point is to recover wasted energy (other ideas welcome).

Electricity consumption figures of the airplane systems (link provided by Bianfable) are currently around 90kW for essential load and utility systems (basis for comparison, keep in mind these values can be reduced).

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    $\begingroup$ Related: How would exhaust heat recovery into the combustion chamber affect the efficiency of a jet engine? $\endgroup$
    – Bianfable
    Commented Mar 3, 2021 at 11:22
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    $\begingroup$ I thought aeroplanes didn't have separate onboard generators with fuel? I thought the engines were connected to generators? $\endgroup$ Commented Mar 3, 2021 at 11:44
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    $\begingroup$ Powering the electric needs of a jet aircraft with thermocouples? The supply & the demand are mismatched here by a few orders of magnitude, it seems. It takes some pretty serious amps to run the 2 electric hydraulic pumps in a 737, plus radar & various computers."Silly to have an on-board generator"? Ah, no. -1 $\endgroup$
    – Ralph J
    Commented Mar 3, 2021 at 17:45
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    $\begingroup$ @FlorentHenry This answer has a nice diagram illustrating required electrical power for a single aisle aircraft. It averages ~100kVA! $\endgroup$
    – Bianfable
    Commented Mar 4, 2021 at 14:21
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    $\begingroup$ Both electric hydraulic pumps run full time throughout the flight as long as any generator is operating. They do not, however, run in the (emergency) case of all generators failed - they'd drain the battery far too quickly. The power generated by a thermocouple is wildly out of proportion to the power used by the aircraft. Until the understanding of the math involved enters the picture, this is all in the realm of science-fantasy. $\endgroup$
    – Ralph J
    Commented Mar 4, 2021 at 14:38

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The amount of current flow produced by a seebeck effect device is tiny relative to their size. You would need a plane load of them just to power the lighting system in a passenger jet.

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  • $\begingroup$ Any figure of the lighting system power requirements? $\endgroup$ Commented Mar 4, 2021 at 13:17
  • $\begingroup$ an estimate would be around a thousand watts. $\endgroup$ Commented Mar 4, 2021 at 18:17
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You can't escape the tyranny of the law of conservation of energy and the second law of thermodynamics (with it's corollary limiting thermodynamic cycle efficiency).

The energy taken from cooling in a reciprocating engine is truly wasted, so the thermal gradient there can be further used without affecting efficiency of the engine itself. But turbines have very little external cooling; only a bit for the oil. Most of the heat is kept in the working fluid by cleverly directing the air around and into the combustors.

And if I understand the thermodynamic right, the heat in the exhaust gas is the part that must be ‘transferred to the low temperature reservoir’ for the cycle to work, and therefore tapping into it will reduce the efficiency of the engine.

Now WP says about the efficiency of thermoelectric generators:

Currently, ATEGs are about 5% efficient. However, advancements in thin-film and quantum well technologies could increase efficiency up to 15% in the future.

That's very, very poor compared to an alternator.

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  • $\begingroup$ They're of relative poor efficiency but they'd be running on waste heat anyway. $\endgroup$ Commented Mar 4, 2021 at 13:25
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    $\begingroup$ @FlorentHenry Most waste heat is contained in the exhaust flow and not really wasted. If you cool down that flow, you would extract most of the momentum it has gained in order to push the airplane forward. So, no, that's a reservoir you better leave untapped. $\endgroup$ Commented Mar 4, 2021 at 17:57
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The context is that commercial airplanes make up most of the traffic.

Let's take the A320 as an example of commercial airplane.

http://a320dp.com/A320_DP/electrical/sys-3.7.1.html

The main AC electrical power sources are:

  • Left and right engine integrated drive generators (IDGs) (90-KVA each)

KVA is kiloVolt-Ampere, roughly equivalent to KW.

So we need about 180KW. From you question:

The voyager probes had 3 RTG - 99.6kg total without the plutonium - for a rated power of 470W.

So $180000/470 \approx 383$, thus $383 \cdot 99.6 \approx 38145 kg$.

That's extra 38 TONS.
For an aircraft whose maximum payload is just shy of 20 tons.


Addendum: thanks to Brianfable for finding this other question with extra data on power generators.

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  • $\begingroup$ (I edited the question, figures are from the 70s). 180kW is for redundancy, 100kW would be a more accurate starting point. Supposing we can crank it up to 600W, that's 16.6 tons. Using 1/3 of it as structural elements, we're looking at 11.1 tons, from which we should remove generators and some of the fuel weight (I have no plausible value to offer here). And we've lost redundancy. But it changes many more things not taken into account here. Providing only part of the electricity with a smaller system could prove more efficient overall anyway. Q remains, how much can be practically recovered? $\endgroup$ Commented Mar 4, 2021 at 16:22
  • $\begingroup$ @FlorentHenry nothing. nothing is practically recoverable. seebeck effect requires too much weight per watt to be practical for aviation. solve that first and then the answer might change. $\endgroup$
    – Federico
    Commented Mar 4, 2021 at 16:45

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