Heat Transfer in a Compound Parabolic Dish

In summary, the conversation discusses modeling heat transfer by conduction from the aperture to the base of a compound parabolic dish (CPE). The participants discuss various equations and functions to use for the calculations, such as θ, a', a, L, T1, T2, y(Φ), z(Φ), A(x), Q'•dx/A(x), and Φ(z). The main goal is to determine if a polyethylene mesh can limit convection enough to create a passive refrigerator. Suggestions for other surface models, such as spherical, cosine, and Fourier transform approximations, are also mentioned. The conversation ends with confusion about a solution for a truncated cone and the potential complications of calculating the length of the
  • #1
skyfridge
9
0
I'm looking to model the heat transfer by conduction from the aperture to the base of a compound parabolic dish (CPE). Here's what I have so far, thanks to CFDFEAGURU's solution for a cone at https://www.physicsforums.com/threads/heat-transfer-by-conduction-in-a-truncated-cone.368972/

let θ = the acceptance angle;
let a' = the radius of the base;
let a = the radius of the aperture, where a=a'/sin(θ);
let L = the height of the CPE from base to aperture, where L=(a+a')/tan(θ);
let T1 be the temperature of a';
let T2 be the temperature of a;
where T2>T1;

y(Φ)=((2a'(1+sin(θ)sin(Φ-θ))/(1-cos(Φ))-a, gives us the radius at Φ;
z(Φ)=((2a'(1+sin(θ)cos(Φ-θ))/(1-cos(Φ)), gives us the height at Φ;

Solving z for Φ gives us,
Φ(z)=2cot-1((sec(θ)sqrt(zcos(θ)a(sin(θ)+1)+a(sin(θ)+1)2)-tan(θ)a(sin(θ)+1))/a(sin(θ)+1));

A(x)=π•(y(Φ(x)))2;

Q'•dx/A(x)=-k*dt;

Integrate the left side from 0 to L. Integrate the right side from T1 to T2.

The calculus gets very hairy at the function composition (y ° Φ)(x). It's even too dense for Wolfram Alpha, which times out, given such a monstrosity. It looks to me like a differential equation, but that looks almost as hairy. Is there a better integral technique to address this mess?
 
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  • #2
Yikes! Forget about z() and Φ()! I can just integrate over Φ! Duh!

Let me see how that looks.
 
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  • #3
So I have,

Q•δz(Φ)/δA(Φ)=-k•dt;
Where A(Φ)=π•y(Φ)2;

I couldn't find the exact character for the partial derivative, so I used δ.

Am I close?
 
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  • #4
I've read this thread and your previous one on the same subject several times over and I still can't actually see what there is to analyse in the set up described .

You just have a small shallow bowl with a wind blowing over the open top of the bowl . Even a moderate speed wind will just circulate out the entire contents of the bowl and replace with new air many times over in a short space of time .

Please try to describe more clearly what it is that you are trying to do .
 
  • #5
Nidum said:
You just have a small shallow bowl with a wind blowing over the open top of the bowl . Even a moderate speed wind will just circulate out the entire contents of the bowl and replace with new air many times over in a short space of time .

That was my assumption for the uncovered case. However, looking at cold air pooling (meteorology), a valley or basin can retain a cold air pool under certain conditions. Of course, we're talking a difference of scale, but I wondered what the the dynamics were, for the purpose of calculating a threshold (windspeed) at which convection would dominate in both the uncovered case and in one covered by a polyethylene mesh.

Please try to describe more clearly what it is that you are trying to do .

Gentle and Smith[1] described results of an experiment that compared the performance of radiative cooling having the radiator uncovered and having the radiator covered with polyethylene meshes of different colors and porosities (to minimize heat gain by convection). The radiator was in an insulated box, and the mesh-covered radiator outperformed the uncovered radiator. Having the radiator in an insulated box exposed the radiator to infrared radiation from the entire sky and other sources of radiative heat.

Smith[2] described limiting the radiative input from the sky and other objects by limiting the viewing angle with an aperture. Doing the math shows that limiting the viewing angle will substantially reduce the radiative input from the sky to the radiator. The compound parabolic shape can do that very nicely.

That leaves conduction and convection as other sources of heat input. My current estimate for conduction puts it in the energy budget, so it's not a great concern. I asked the question about calculating conduction in this thread to build a better model for conductive losses, which I previously estimated with the shape of a cylinder. Since the cold radiator will be lower than the aperture, free convection isn't an issue. That leaves forced convection from wind.

US Patent 4624113 describes a radiative cooling system with a compound parabolic aperture that can achieve up to 80°C below ambient, for making ice. To minimize convective losses, the radiator is isolated from the atmosphere with a solid IR-transparent cover. However, the cover introduces the problem of condensation, when the temperature of the cover falls below the dewpoint, which decreases radiative output. To overcome that obstacle, the invention in the patent thermally isolates the radiator and cover with a vacuum. However, to achieve that thermal isolation, the mean free path must be greater than the size of the container, which requires a very high vacuum, which is impractical, due to practal materials. Besides, I don't need that kind of ΔT!

So it's an engineering problem. If a polyethylene mesh limits convection enough, then we have a passive refrigerator. I don't have much knowledge about fluid dynamics, so I asked the question.

[1] Performance comparison of sky window spectral selective and high emittance radiant cooling systems under varying atmospheric conditions, Dr Angus Gentle, Prof Geoff Smith
[2] Amplified radiative cooling via optimized combinations of aperture geometry and spectral emittance profiles of surfaces and the atmosphere, G.B. Smith
 
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  • #6
One calculus complexity with a parabola is the length of the curve.
You might consider another surface model for the thermal analysis.
That could be a spherical surface, or it could be a rotation of a cosine function.
You might also consider the Fourier transform approximation of the parabola.
Hyperbolic surfaces may be more tractable than parabolic surfaces.
 
  • #7
Baluncore said:
One calculus complexity with a parabola is the length of the curve.
You might consider another surface model for the thermal analysis.
That could be a spherical surface, or it could be a rotation of a cosine function.
You might also consider the Fourier transform approximation of the parabola.
Hyperbolic surfaces may be more tractable than parabolic surfaces.

Thanks for the ideas. I did consider using a truncated cone for an estimate, but I was aiming to look at the compound parabolic shape for comparison, just to see how it looked.

How does the length of the curve complicate it? Calculating the radius at any axial length or angle parameter is straightforward, so it looks very similar to Matt's 1D solution for a truncated cone.

I seem to be confused about that solution for the truncated cone. For the CPE, I integrated z(n)/y(n) dn on the left side and came up with ~63000 on the left side and -253 on the right. Dividing gave me .004W, which doesn't look right. The estimate for a cylinder with a radius of the aperture came to just shy of 2W.

I'll look more closely at the solution for the truncated cone, to see if I correctly understand the calculus, by coming up with a matching answer in that example.

EDIT: Obviously, I'm not understanding the notation in the solution for the truncated cone, because my numbers aren't matching.

EDIT 2: I figured out the truncated cone, and now I might be able to solve the CPE.

EDIT 3: Re the strikethrough, yeah, I already ran into the hairy problem with the length. And that's why the math didn't work with by integrating over the angle (I think!)
 
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  • #8
Turn the system upsidedown. Use a wide-angle Compound Parabolic Concentrator to funnel the heat up a narrow conduit to a small narrow-angle Compound Parabolic Emitter, with a base made of polyethylene mesh, and then neither wind nor condensation is a problem, as long as you have a condensation trap. It is all about the optics. With a low-emittance reflector, the energy budget can absorb the losses. The cross section of the reflector would allow many square meters of surface area.

I was right. It was an engineering problem. Now, the answer is clear. Use optics to limit the losses, and you have an energy budget of about 100 watts with the radiator at -18°C, which is a mighty fine freezer temperature. That's how you create a passive refrigerator.
 
  • #9
(1)

For practical solutions this type of heat transfer problem is much easier to deal with using numerical methods . . There are standard procedures and there is extensive software available .

Indeed much of what you are trying to analyse overall would be easier to deal with numerically .

There are Finite Element systems which can deal with very complex scientific and engineering problems .

Finite difference and improvised numerical calculations may also be helpful for detail problems .

(2)

There may be merit in controlling the flow path of the wind over the cooling vessel .

(3)

There are many practical difficulties to consider . How to deal with rain , hail , snow , sandstorms , leaves and crisp packets for example ?
 
  • #10
Nidum said:
(1)

For practical solutions this type of heat transfer problem is much easier to deal with using numerical methods . . There are standard procedures and there is extensive software available .

Indeed much of what you are trying to analyse overall would be easier to deal with numerically .

...

(2)

There may be merit in controlling the flow path of the wind over the cooling vessel .

(3)

There are many practical difficulties to consider . How to deal with rain , hail , snow , sandstorms , leaves and crisp packets for example ?

Thanks.

(1)

A spreadsheet works for the numerical analysis. I've already modeled it numerically, using Stefan-Boltzmann, Berdahl-Martin and G.B. Smith. All that's left is modeling the insulation.

Whereas numerical analysis software would be fun, there can be a steep learning curve. Also, the only internet access here in the boonies is wireless mobile, so I don't have the bandwidth to download all of the OSS goodies, or I would set up a Linux system to do so.

(2)

I considered a windbreak. However, turbulence was a concern. Also, it would increase the footprint and heat input. Although the acceptance angle of the CPE would only allow radiation within the acceptance angle to the base, if the reflector has an emittance of .03 - and absorbitivity equals emittance - the reflector is only reflecting 97% of the heat radiated from above the height of the aperture and outside of the acceptance angle. Of course, it could be minimized by lowering the emittance of the windbreak, but that's a lot of surface area for a windbreak that is high enough to minimize turbulence enough to be effective, unless you increase the footprint even more with a sloped surface.

(3)

I've already considered precipitation. That's another benefit of using mesh for the base of the emitter. Instead of being a water collector, it's now a conduit, which can drain the water before it reaches the radiator.

Debris shouldn't be too much of a concern.

Edit: Also, the footprint of the CPE is small enough that volume of precipitation wouldn't be a concern, though I might have to clean out the snow once in a while.
 
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1. How does a Compound Parabolic Dish work?

A Compound Parabolic Dish (CPD) is a reflective surface that is shaped like a parabola in the vertical direction and like a hyperbola in the horizontal direction. This unique shape allows it to capture and concentrate sunlight onto a smaller focal point, where the heat is transferred to a fluid-filled receiver. The receiver then carries the heat away to be used for various applications.

2. What are the main applications of CPD?

CPD is primarily used for generating heat, typically for cooking, heating water, or producing steam for industrial processes. It can also be used for electricity generation by using the heat to power a turbine.

3. How efficient is the heat transfer in a CPD?

The efficiency of heat transfer in a CPD is dependent on several factors, including the quality of the reflective surface, the design of the receiver, and the intensity of sunlight. Generally, CPDs have an efficiency of 50-80%, which is much higher than traditional flat solar panels.

4. Can CPD be used in all locations?

CPD can be used in most locations, but its effectiveness may vary depending on the amount of sunlight and the angle of the sun. CPDs are most efficient in areas that receive high levels of direct sunlight and have a clear sky. They can also be adjusted to track the sun's movement for maximum efficiency.

5. What are the advantages of using CPD compared to other solar technologies?

CPD has several advantages over other solar technologies, including its high efficiency, compact size, and ability to generate heat for various applications. It is also less affected by cloudy weather and can be easily integrated into existing heating systems. Additionally, CPDs have a longer lifespan and require minimal maintenance.

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