Question about Thermophotovoltaics (thermal heat engine)

In summary, Thermophotovoltaics (TPV) is a technology that converts thermal energy into electricity using photovoltaic cells. It is a type of thermal heat engine that utilizes high-temperature heat sources, such as combustion or solar radiation, to generate electricity. TPV has the potential to be a clean and efficient energy source, but it is still in the early stages of development and faces challenges such as low efficiency and high costs. However, researchers continue to work on improving TPV technology and it holds promise for the future of renewable energy.
  • #1
Devin-M
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https://news.mit.edu/2022/thermal-heat-engine-0413

This team that developed this device claims it is capable of generating electrical current from a 1900C heat source via the photovoltaic effect.

I was wondering if the generated electricity was put through an initially 1901C electrical conductor, would it be possible for the I^2R=W copper losses in the conductor to result in a net heat transfer from a 1900C reservoir (the heat source) to a hotter 1901C reservoir (the electrical conductor)? Assume the conductor is well insulated thermally so convection and conduction are minimal and the current through the conductor is sufficient to increase its temperature via the I^2R=W copper losses.
 
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  • #2
Interesting idea, this energy storage/extraction device. I feel like there would be practical problems both in generating and maintaining the temperature. Also, 40% extraction efficiency isn't great for a battery. If that's what they want to use it for they're comparing it to the wrong thing(a steam engine).

But your question doesn't seem to be about this device?
Devin-M said:
I was wondering if the generated electricity was put through an initially 1901C electrical conductor, would it be possible for the I^2R=W copper losses in the conductor to result in a net heat transfer from a 1900C reservoir (the heat source) to a hotter 1901C reservoir (the electrical conductor)? Assume the conductor is well insulated thermally so convection and conduction are minimal and the current through the conductor is sufficient to increase its temperature via the I^2R=W copper losses.
If the wire is being heated by electricity that's electrical energy, not heat (thermal energy).

You're just describing an incandescent light bulb as far as I can tell. What do you want to do with this long, skinny, insulated light bulb?
 
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  • #3
A light bulb is a good description of the circuit.

According to this page, a standard light bulb filament reaches temperatures well above 2000C:

https://www.e-education.psu.edu/egee102/node/2035

So if we have a 1900C heat source, and a photovoltaic device is producing electrical current from that, and the current heats the light bulb’s filament above 2000C, is that even possible or would it violate the 2nd Law of Thermodynamics? The bulb’s filament appears to be spontaneously achieving higher temperatures than the 1900 C heat source in the scenario. Wouldn’t it mean the net heat flow is flowing from a cold reservoir (the 1900C reservoir) into a hot reservoir (the over 2000C light bulb filament)?
 
  • #4
The efficiency of a thermovoltaic cell (TPV) cell is given by ##\eta_{TPV}=P_{out}/(P_{out}+Q_c)=P_{out}/(P_{inc}-P_{ref})##.

##P_{out}## is the electric power generated by the TPV cell. The total heat absorbed and generated in the cell is denoted by ##Q_c##, which is made up of the heat generated by parasitic absorption in the semiconductor or metal reflector, thermalization losses due to excess incident photon energy, Joule heating losses due to current flow and non-radiative recombination losses. ##P_{out}+Q_c## can also be expressed as ##P_{inc}−P_{ref}##, where ##P_{inc}## is the incident energy and ##P_{ref}## is the reflected energy (see https://www.nature.com/articles/s41586-022-04473-y#Sec3).

Mere temperature specifications are not of help to calculate the relevant quantities.
 
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  • #5
Devin-M said:
So if we have a 1900C heat source, and a photovoltaic device is producing electrical current from that, and the current heats the light bulb’s filament above 2000C, is that even possible or would it violate the 2nd Law of Thermodynamics? The bulb’s filament appears to be spontaneously achieving higher temperatures than the 1900 C heat source in the scenario. Wouldn’t it mean the net heat flow is flowing from a cold reservoir (the 1900C reservoir) into a hot reservoir (the over 2000C light bulb filament)?
Again, it's not heat transfer so no it wouldn't violate the second law. Nor does a light bulb.

However, it's worrisome that you are searching for second law violations and, presumably, first law violations.
 
  • #6
So if we have a 1900C gas reservoir, the photovoltaic device is generating some current from the radiation off the walls of that reservoir, the current heats a filament above 2000C, and the more than 2000C filament heats a small amount of gas to approximately 2000C… isn’t that working similarly in principle to a “Maxwell’s Demon?” In the end the hot reservoir gets hotter essentially by cooling the cold reservoir, correct?

https://en.m.wikipedia.org/wiki/Maxwell's_demon

7D599568-72E6-46D7-A146-6E6AF20C3991.png

Wiki: “Maxwell's demon is a thought experiment that would hypothetically violate the second law of thermodynamics.

“…the demon's actions cause one chamber to warm up and the other to cool down.
 
  • #7
"...for every process that accumulates thermal fluctuations, there is a counter process in which further fluctuations undo the accumulation. The fluctuations that gave us hope that a demon could be built also dash those hopes."

John D. Norton in "All Shook Up: Fluctuations, Maxwell’s Demon and the Thermodynamics of Computation"
(Entropy 2013, 15, 4432-4483)
 
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  • #8
Devin-M said:
So if we have a 1900C gas reservoir, the photovoltaic device is generating some current from the radiation off the walls of that reservoir, the current heats a filament above 2000C, and the more than 2000C filament heats a small amount of gas to approximately 2000C…
Where did this "small amount of gas" come from? You didnt mention it before. You can't violate the Laws of Thermodynamics by being vague. They know what you're up to even if you don't say it. Tell us exactly what your system/setup is and what you think it is doing. A diagram would help.

It also strikes me that you don't understand the conditions under which PV works.
isn’t that working similarly in principle to a “Maxwell’s Demon?” In the end the hot reservoir gets hotter essentially by cooling the cold reservoir, correct?

https://en.m.wikipedia.org/wiki/Maxwell's_demon

View attachment 300509
Wiki: “Maxwell's demon is a thought experiment that would hypothetically violate the second law of thermodynamics.

“…the demon's actions cause one chamber to warm up and the other to cool down.
I see no relationship whatsoever to Maxwell's Demon. Please post a diagram that shows exactly what your system/setup is, not just a diagram of something else that isn't it but may or may not be similar.

I'm near certain I know what you want to do here, and it doesn't work/would be a violation, but I'm not guessing. Out with it!
 
  • #9
Suppose there are two well thermally insulated chambers of gas, both 1900C. The photovoltaic device is pointed at or within one chamber to absorb infrared radiation and the heating filament is inside the other, connected by wires to the photovoltaic device in the other chamber. When some energy is lost by radiation from the enthalpy of one chamber into the photovoltaic device, the electricity heats the 2nd chamber via the filament above the initial temperature of both chambers.
 
  • #10
Devin-M said:
Suppose there are two well thermally insulated chambers of gas, both 1900C. The photovoltaic device is pointed at or within one chamber to absorb infrared radiation and the heating filament is inside the other, connected by wires to the photovoltaic device in the other chamber. When some energy is lost by radiation from the enthalpy of one chamber into the photovoltaic device, the electricity heats the 2nd chamber via the filament above the initial temperature of both chambers.
No heat transfer between chambers = no violation.

What is making you think there could be one? You don't think the heat added to the last reservoir is equal to what is removed from the first, do you?

As someone else pointed out and deleted, much of the world's electricity is generated with steam engines at a few hundred C. We really can't fathom why you think there might be a violation.
 
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  • #11
Am I mistaken that there would appear to be no issue then with doing useful work with no temperature difference (ie utilizing a single temperature reservoir…)? Suppose the photovoltaic device is in a cavity surrounded on all sides by the 1900C reservoir… seems like it’s essentially a heat engine generating useful electrical power without a cold reservoir or a temperature difference unlike other heat engines like a stirling engine attached to a generator… a stirling engine can’t do any work from a 1900C reservoir without a separate cold reservoir… I thought that was a 2nd Law Violation.

Also this other photovoltaic sensor (linked below) is sensitive to 3.5 micron photons which are emitted as black body radiation by room temperature objects (even in the dark with no light source) while the sensor itself is also at room temperature… so with a large enough room temperature reservoir (like the underground of my back yard) and a large enough sensor, I could conceivably generate enough useful power to boil some water or charge my phone, day and night, rain or shine, from a photovoltaic panel simply buried underground (the underground in this case acting as a room temperature heat reservoir):

https://www.researchgate.net/publication/343856156_Higher_Operating_Temperature_IR_Detectors_of_the_MOCVD_Grown_HgCdTe_Heterostructures
 
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  • #12
russ_watters said:
As someone else pointed out and deleted, much of the world's electricity is generated with steam engines at a few hundred C. We really can't fathom why you think there might be a violation.
Sorry @russ_watters that was me. Everything got very screwy...I don't know wtf happened. Anyhow you have faithfully repesented my point. What's the big deal here.
 
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  • #13
Suppose I had a very well thermally insulated 1900C chamber of gas. In the center of this chamber is a walled cavity with a photovoltaic device inside absorbing infrared radiation off the walls of the cavity with an incandescent light bulb (also within the central cavity) as an electrical load. The light bulb runs off the electricity from the from the photovoltaic device. But since the visible and infrared emissions from the light bulb ultimately get converted back to heat that gets reabsorbed into the surrounding reservoir, the question becomes how long could that light bulb run for assuming the losses through the outer insulation of the 1900C heat reservoir are negligible?
 
  • #14
Devin-M said:
Am I mistaken that there would appear to be no issue then with doing useful work with no temperature difference (ie utilizing a single temperature reservoir…)? Suppose the photovoltaic device is in a cavity surrounded on all sides by the 1900C reservoir… seems like it’s essentially a heat engine generating useful electrical power without a cold reservoir or a temperature difference unlike other heat engines like a stirling engine attached to a generator… a stirling engine can’t do any work from a 1900C reservoir without a separate cold reservoir… I thought that was a 2nd Law Violation.
That would be a violation, yes. The PV cell is at least implied to be sitting in a cold reservoir in your prior scenario - as it must be. Obviously, radiative heat transfer requires a temperature difference.
Also this other photovoltaic sensor (linked below) is sensitive to 3.5 micron photons which are emitted as black body radiation by room temperature objects (even in the dark with no light source) while the sensor itself is also at room temperature… so with a large enough room temperature reservoir (like the underground of my back yard) and a large enough sensor, I could conceivably generate enough useful power to boil some water or charge my phone, day and night, rain or shine, from a photovoltaic panel simply buried underground (the underground in this case acting as a room temperature heat reservoir):

https://www.researchgate.net/publication/343856156_Higher_Operating_Temperature_IR_Detectors_of_the_MOCVD_Grown_HgCdTe_Heterostructures
I think you linked that before and we did not agree that you understood its function properly. Let's not rehash that please, especially since it is very different from what we were talking about. Jumping around between different scenarios without fully dealing with and understanding any is a sure-fire way to just confuse yourself without learning anything
 
  • #15
russ_watters said:
The PV cell is at least implied to be sitting in a cold reservoir in your prior scenario - as it must be. Obviously, radiative heat transfer requires a temperature difference.
I wasn’t assuming the PV device was necessarily operating from a cold reservoir. The reason for this is the other PV sensor I linked to (the one sensitive to 3.5 micron photons) is able to generate useful electricity from 3.5 micron photons emitted by 300k objects while the sensor is also 300k.
 
  • #16
Devin-M said:
I wasn’t assuming the PV device was necessarily operating from a cold reservoir.
You should have said so when describing the scenario. I feel like you described a multi-part scenario full of mundane processes for no reason while omitting the description of the one process you actually wanted to talk about. What are we doing here? Why even bring the rest of it in?

Please state clearly and explicitly the purpose of this thread. I don't want to play games.
 
  • #17
I appreciate the input and feel my original questions have been sufficiently answered. My only remaining side-question that was risen in my mind is do you dispute the findings in the paper that the other PV panel that is sensitive to 3.5 micron photons can absorb them and produce DC current while the sensor is 300k? Objects which are 300k emit some 3.5 micron black body radiation in the dark.
 
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  • #18
Devin-M said:
I appreciate the input and feel my original questions have been sufficiently answered. My only remaining side-question that was risen in my mind is do you dispute the findings in the paper that the other PV panel that is sensitive to 3.5 micron photons can absorb them and produce DC current while the sensor is 300k? Objects which are 300k emit some 3.5 micron black body radiation in the dark.
We've discussed this before. We don't dispute the paper we dispute your understanding of it. And I'm not interested in re-hashing that here or worse trying to apply that wrong understanding to a different device. The 2nd law of thermodynamics is what it is and you can't violate it by soft-pedaling the violation past us. You are reminded of our rules against personal theories and perpetual motion machines. Thread closed.
 
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1. What is thermophotovoltaics?

Thermophotovoltaics (TPV) is a technology that converts thermal radiation (heat) into electricity using photovoltaic cells. This process involves using a thermal emitter to generate heat and a photovoltaic cell to convert the heat into electricity.

2. How does a thermophotovoltaic system work?

A thermophotovoltaic system consists of three main components: a thermal emitter, a photovoltaic cell, and a heat engine. The thermal emitter heats up and emits thermal radiation, which is then absorbed by the photovoltaic cell. The photovoltaic cell converts the thermal radiation into electricity, which is then used to power the heat engine. The heat engine produces more thermal energy, which is then used to heat up the thermal emitter, creating a continuous cycle of energy conversion.

3. What are the advantages of thermophotovoltaics?

Thermophotovoltaics has several advantages over other renewable energy technologies. It can operate at high temperatures, making it more efficient than traditional photovoltaic cells. It also has a higher power density, meaning it can produce more electricity in a smaller space. Additionally, TPV systems can be used in a variety of applications, including power generation, waste heat recovery, and solar energy conversion.

4. What are the limitations of thermophotovoltaics?

One of the main limitations of thermophotovoltaics is its low conversion efficiency. Currently, the efficiency of TPV systems is around 10-20%, which is lower than other renewable energy technologies such as solar panels. Additionally, the materials used in TPV systems are expensive and not widely available, making it more costly to produce compared to other energy sources.

5. How is thermophotovoltaics being used in real-world applications?

Thermophotovoltaics is still a relatively new technology and is currently being used in research and development projects. However, there are some real-world applications of TPV, such as powering remote sensors and small electronic devices. Researchers are also exploring the potential use of TPV in power generation and waste heat recovery in industrial processes. As the technology continues to advance, we may see more widespread use of thermophotovoltaics in the future.

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