Heat Transfer Between Heated Upper Plate and Cooled Lower Plate

[skygerator]

The question relates specifically to US Patent 4624113, Passive-Solar Directional-Radiative Cooling System, University of Chicago representing Argonne National Laboratory on behalf of assignee, USA, DOE, 1986, in a 3D compound parabolic reflector assembly with a 3D compound parabolic housing, a circular plate for the radiator, an HDPE cover and a vacuum in the cavity between the radiator and cover.

The patent claims that the invention can achieve 40-80°C below ambient at the radiator without forming condensation on the cover. It does not quantify the vacuum, but let's assume that it can't be high vacuum, due to practical construction materials, as described by the patent. The patent also does not describe the model for determining this performance potential. The patent claims that dry air will not achieve the goal of preventing condensation.

The upper HDPE plate is heated by the ambient heat of the environment. This seems to be addressed in Eq. 47 at Convection from a Rectangular Plate. Eq. 47 deals with a heated circular downward facing plate. Given that the cover is well below the lip of the compound parabolic aperture, for the purpose of this question let's assume that the radiator and cover are the same size.

The lower plate is cooled by net radiative transfer with the sky through the compound parabolic aperture. That calculus is easy with GB Smith-2009, Berdahl-Martin, etc.

For estimating non-radiative losses, Smith is using air or dry air in the cavity and also looking at insulative losses of the compound parabolic housing but makes no mention of using a vacuum instead of air. Of course, Smith was looking at solving a different problem than US4624113.

I've looked at the answers to similar questions here, but the model eludes me.

The question: How does one model the heat transfer between the radiator and the cover with a reduced pressure in the cavity, to determine the necessary vacuum to achieve the aforementioned goal of preventing the cover from reaching the external dew point for a set of ambient conditions?

 

[racookpe1978]

Seems like they could assume anything they like, and would not need to (nor want to!) state any specific value.

Nor actually, would they need to declare any real economic value: For example, assume they needed 10^-6 vacuum (a meaningless number, right?) but needed a 10,000,000.00 air pump and a 100,000,000.00 vacuum chamber to actually get to that value.

They wouldn't care. They claim "Simulated in a space environment." And then run the model program to show a radiation heat loss to prevent somebody else from using that kind of radiator in space. And, more important, to get the "claim" on their government bonus check of having another patent application. For themselves, their boss, and their agency.

 

[skygerator]

racookpe1978, you beautifully cynical bastard,

We're not talking Stanley Meyer running a VW on water. Does the patent office give them a pass for being UofC and Argonne and the USA DOE? Why would the DOE want to prevent anything of this nature?

I noticed something about a computer model, but I missed the space environment. What koolaid are you drinking that you get condensation in space, with or without dry gas in the cavity?

Also, you need a thin enough HDPE cover before your assembly losses prevent the radiator from reaching anywhere near 40°C, much less 80°C, below ambient. Wouldn't high vacuum rupture saran wrap or painter's plastic, inside or outside the cover?

The math in Smith works out and closely agrees with Berdahl, Martin, etc, when you apply the inverse concentration ratio of the compound parabolic assembly to the input to Berdahl-Martin. If you don't want to do the calculus in Smith, Berdahl Martin is a simple equation.

Why are you pulling my leg?

If the air above the cover is absolutely still, the top of the cover will be much cooler than ambient. Otherwise, there will be forced convection above the cover.

The cover will have a certain thermal conductivity, moving heat across the very thin barrier.

The cavity is pretty much still, though it's not absolute, so you're dealing with mostly natural convection.

The thermal conductivity of air changes very slowly until you have high vacuum, so that thermal conductivity should be negligible.

That should leave us with convective coefficient as the defining factor.

Unless the radiator has an active heat source, such as waste heat from some device, the temperature should keep dropping, until the net radiated output reaches a stagnation temperature, at which the output equals the losses in the compound parabolic assembly (and insulative losses at the source).

You have heat coming downward through the cover. You have a positive net radiated output at the radiator. How are all of these factors tied together into one model? The answer eludes me.

 

[skygerator]

racookpe1978,

I apologize if my response above sounded somewhat snarly. I was trying to be humorous.

Since the patent makes it clear that the invention is for terrestrial environments, I concluded that you hadn't read the patent, nor were you trying to be helpful.

I have done the math with numerous models, but I'm stuck at the aforementioned model. I'd like to verify the patent with a sound model, before I build a prototype. The implications are huge, and i'd like to know if the only reason that the invention was buried in the patent office is because the device is susceptible to bird droppings and roof rats.

 

[chicopee]

I would look at the saturation pressure of water vapor at the temperature that the HDPE cover would incur and maintain the vacuum at or below the saturation pressure within that space.

 

[skygerator]

Thank you, chicopee. I'm confused. That would address the volume between the radiator and the HDPE cover. However, that would be covered by replacing the volume in the cavity with dry gas, which is supposedly not effective at preventing condensation externally (atop the cover), when the cover temperature equalizes with the radiator, via heat transfer from the cover to the radiator through the gas in the cavity.

In order for the claim in the patent to be valid, the reduced pressure must affect the heat transfer between the cover and the radiator. I'm guessing this involves natural convective heat transfer from the downward facing horizontal surface (47 from the link above), I'm just not grasping how to apply it, given the definitions of the constants (considering the closed volume, where the calculation of one of those constants involves the temperature difference at a far distance, whatever that means) and given that the radiator will be radiating heat simultaneously.

I'll have to dig further into Rayleigh and Prandtl to see if I'm on the right track, unless someone can point me in a different direction. However, combining the natural convective heat transfer of the cover with the net radiative output of the radiator is the real head scratcher for me.

 

[racookpe1978]

Thank you for the courtesy of your reply. 8<)

With no sketch or diagram attached to the original question, no - I did not look up the patent application.

My reference to "in space"? Radiation cooling is overwhelmed at the "usual" surface temperatures and around almost anything between two surfaces with an air flow between them. Unless the two (or more) surfaces are completely isolated, convection and conduction dominate until temp's get higher. (T^4th factor of course. Then at those higher temperatures, radiation is very, very much more important.) Thus, although it takes "furnace like", or "open flame" temperatures to make radiation the dominant energy transfer - until you hit space (vacuum) conditions: Then radiation provides the ultimate heat sink, and is the original heat source for earth and closer orbits.

A patent application needs to only record a workable solution - not an economic or profitable solution - for a process. So, the environment of a patent's process (a cooling method) may work under unusual conditions: since your question reverses the usual convection-assisted heat flow (hot air rises!) and involves deposits and evaporation of a liquid/vapor, I was assuming some either vacuum chamber where the vapor is being extracted (to improve the vacuum or to allow collection of a valued vapor from the gasses), or a space-based cooling system where the vapor collection is unimportant.

 

[skygerator]

racooke1978,

Maybe I'm talking apples, and you're talking oranges. I thought that I described the device adequately for the purposes of the heat transfer question.

I came up with the device before I found the patent. I'm just trying to verify the model.

The patent defines a device for making ice in a warm climate, using radiative cooling, through a compound parabolic aperture, with the radiator at the base of the 3D compound parabolic shape (CPS), separated from ambient by an IR-transparent cover, which is below the lip of the compound parabolic shape.

Example (earthly) scenario:
CPS viewing angle: 30°
Ambient temperature: 20°C
Clouds: none
Dewpoint: 10°C
Effective sky temperature: -79.63C
Radiator emissivity: .989
Radiator temperature: -18°C
Radiator output (Stefan-Boltzmann): 238.9W m^-2
Input from sky (Smith, adjusted with Berdahl): 79W
Net: 159.8W
Losses: 52.8W
Net minus losses:107W
Stagnation temperature of radiator: -60°C

It looks like a passive refrigerator to me.

 

[IRstuff]

"The patent claims that dry air will not achieve the goal of preventing condensation."

Air presents more issues than that. The thermal conductivity and convection of air would prevent the radiator from achieving its desired temperature. The vacuum is needed to ensure that what little thermal convection there is doesn't swamp out the radiative cooling.

However, the model will still be non-trivial, since the patent doesn't really address the radiative heating from the cover, and the environment. The necessary optical coatings needed to accomplish the desired results would appear to be daunting.

There's an old mathcad worksheet that's still floating around the internet that describes the process of radiative cooling. Attached is a PDF if that file

http://files.engineering.com/getfil...c2-8dff-92224e2df0ae&file=ClearNightFrost.pdf

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[skygerator]

Yikes! I had been playing with the model with the same spreadsheet and lowered the non-radiative losses and dropped the cabinet and thermosyphon losses. At -18°C radiator temperature, net radiated output minus losses would be approximately 50W for a 1m^2 radiator.

It still looks like passive refrigeration. Now to fix the bird nests and roof rats.

 

[skygerator]

Thanks, IRStuff,

I was aware of the issue that you presented. Smith addresses radiative losses from the cover, but Smith was only looking at 10°C below ambient. Of course, as Smith admitted, non-radiative losses would be a large portion of total aperture assembly losses, and Smith suggested improvements to insulation of the assembly housing.

The issue is precisely why I seek a model for the cavity at reduced pressure. Until I see the math and do some subsequent experimentation I am not convinced that it must be high vacuum or that it requires exotic materials.

Do I calculate the conductive heat transfer, add the convective heat transfer for a downward facing plate per Eq. 47 (linked above) and then subtract net radiated output (Smith + Berdahl)?

 

[skygerator]

Okay, looking at it again, I misunderstood the answer to a question that I saw somewhere here regarding heat transfer in a container at reduced pressure. That answer led me to believe that convective coefficient would be the dominant factor.

Looking at Nusselt, Rayleigh and Prandtl again, it seems that due to the direction of heat flow the heat transfer will be dominated by conduction and radiation.

I was unable to find the emissivity for natural HDPE, but I found a .1 for polyethylene (density not given). Black H(?)DPE is somewhere above .9. I'm not sure how I'll deal with the reflector, but the inverse concentration ratio feels right.

For the conduction component, it appears that I need to do some calculus. That solution should look something like heat transfer by conduction in a truncated cone, adapted for the dimensions of the compound paraboloid. I'll do the calculus for this subproblem and then move on to other parts.

Thank you for your assistance and patience thus far.

Are there any other suggestions?

 

[chicopee]

No, you would not have to replace with dry gas, just drop the pressure of the air within the cavity below that saturation pressure of the water vapor and the saturation pressure is based on the temperature that you would have on the surface of the HDPE cover which I would think is the coldest part of the reflector.

 

[IRstuff]

"For the conduction component, it appears that I need to do some calculus"

A picture of what you are trying to solve might help. Conduction should only be an issue relative to how the cold plate is physically supported in our gravity field. Generally, this can be accomplished using things line G10 (glass-epoxy legs), which work quite well down to liquid helium temperatures. The gas environment inside is purely a convective case, assuming that the spacings are sufficiently large.

I can give some general numbers; for a conventional vacuum Dewar, split-Stirling, cryocooled imaging device, the thermal load is almost entirely radiative, and for something that's about 3/4" across with a 3/8" coldfinger, the radiative heating load is on the order of 1/4 W. With appropriate coatings like GOLD, and multilayer insulation, that sort of performance can be achieved, even for liquid He Dewars that are quite large. although that performance does require He coolant. However, this is not cheap by any stretch of the imagination.

HDPE isn't necessarily an ideal substrate for high-performance dielectric optical bandpass coatings, so trying to do such a system in real life would be exceedingly challenging. Note that the 4-K cold sky is only possible above 100-km altitudes. The atmosphere radiates as a function its temperature, so in the daytime, 200 K or warmer sky temperature is more likely.

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[skygerator]

Chicokpee,

I am still perplexed by your reasoning. What does the saturation pressure of water inside the volume have to do with condensation outside the volume, if dry gas *without water vapor* is ineffective at preventing condensation.

Actually, the radiator should be the coldest part of the assembly. The cover is in contact with the ambient temperature above the cover.

The radiator has a positive net radiative output, so it's temperature falls. Heat flows from the volume to the colder radiator, which cools the volume, while the radiator continues to radiate heat. The entire volume cools. The cover starts to lose more heat to the cold volume to the radiator to radiation than it gains from the ambient temperature. If it is stagnant enough above the cover the boundary layer above the cover starts to cool. If the boundary layer cools below the dewpoint then water vapor condenses atop the cover, which blocks radiation from the radiator, which ceases the desired cooling.

The object of having the volume at a reduced pressure must alter conduction, convection or radiation between the radiator and cover if it is to prevent condensation atop the cover. Conduction of air doesn't fall much until pressure approaches high vacuum. Unless the object is to pull high vacuum from the volume, look at the other two modes. Radiation is a function of temperature and surface area. Does pressure inside the volume affect either? Look at our other option. Does pressure affect convection? Okay, then let's look there. Is convective coefficient of the fluid in the volume enough to keep the cover at the external ambient temperature at a pressure above the point at which conductivity changes enough to become the dominant factor? That's what I'd like to determine.

Could it be the density of the volume that affects heat transfer between the cover and the volume? Less density means less mass, which means less heat capacity. Temperature drops faster per watt of heat radiated by the radiator. How does that affect heat transfer from the cover to the volume? It would require less heat from the cover to raise the temperature of the volume.

How does that balance with the temperature of the radiator? Doesn't thermal conductivity vary more with temperature than with pressure? And in a fixed volume doesn't pressure drop with temperature? So the volume becomes an even better insulator between the radiator and cover. Is it enough to keep the radiator above the dewpoint of the environment that is external to the volume and radiator?

 

[skygerator]

IRStuff,

200K is a dandy temperature, only 6.5K above the effective sky temperature in my example. I'll radiate at 253K and need a coat to stay warm.

Who said anything about cryogenic temperature? Who said anything about 4K?

You're not playing fair.

 

[IRstuff]

You don't say anything about how much heat load you're trying to remove, so anything is fair game. Most HDPEs are not transparent in the bands with the most energy, which means they absorb. Additionally, most HDPEs are unlikely to be structurally strong enough to be a vacuum wall.

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[btrueblood]

Here is the patent, in pdf form, downloaded from uspto.gov (see link inserted below).

I'm not sure where the confusion is coming from. The space within the radiator cavity (labeled "26" in the front page figure) is presumably the one kept at a near vacuum, to reduce heat transfer within that cavity and preclude the formation of condensation that might affect the optical surfaces. The cover, labeled "28" in the figure, is not well described by the patent. Presumably, if there is vacuum present within the cavity, heat transfer to the inside of the IR-transparent material is dominated by conduction within the cavity, and convection from the outside air (28). Keeping that first term lower than the second will keep the outer surface of the cover above the local dewpoint, and keep the outside surface of the cover from forming condensation.

They aren't very specific about how exactly to do that, but that wasn't important enough for the patent office, apparently.

You might search public archives in the DOE website, to see if you can find reports authored by the patentees, which may give more details on if, and how, they made any working models. The fact that you can't find one of these devices for sale may give you another big clue.

 

[skygerator]

IRstuff,

I did say something. The example for radiative output above gave 50W/m^2 net radiated output, at -18°C (38°C below ambient), using Smith's model for radiative cooling through an aperture of 30° and non-radiative losses of 1.5W m^-2 °C (loss estimate per Smith).

You're stuck on the details. I used HDPE as an example because published IR spectroscopy shows it to be better than other *practical* materials. The question isn't about the suitability of HDPE for a vacuum wall. The question is about modeling the heat transfer in an enclosed volume at a reduced pressure between a cooled lower horizontal circular plate (radiator) and a heated upper horizontal circular plate, to determine the pressure reduction that would be required to prevent condensation on the cover, external to the enclosed volume.

It doesn't matter that the volume is a cylinder, a cone or a compound parabolodal section. Any one of the three would give a close enough approximation.

I gave too many details for the stated question. Now, everyone only wants to debate the details, going off on off-topic tangents about cryogenic chambers, etc., rather than answering the question.

 

[skygerator]

Thank you, btrueblood! The required vacuum might be too high for practical materials, but you might not find the devices for sale for other reasons. Maybe there isn't a market due to maintenance reasons, susceptibility to bird droppings, roof rats, UV degradation of practical cover materials, etc. That's what were're trying to answer, but the model is incomplete.