Heat fin design - these numbers look wrong 1

[istocrabin]

I'm designing fins for a heat exchanger to transfer heat from an enclosure to environment using natural convection, and I want to ensure that the heat exchanger will transfer enough heat out of the enclosure. The heat exchanger has two sets of fins – one set pointing vertically down into the case and another pointing up vertically out of the case, making the wall horizontal. I’ve checked my formulas against several heat transfer textbooks, dimensions, and math for errors and don’t find any, yet I also don’t trust the numbers I’m getting.

Here are my specs for the bottom row of fins:
Temperature Ambient – 80 C
Temperature Wall – 70 C
Height – .03 m
Width – .059 m
Thickness – .0019 m
Number of fins - 38
Thermal Conductivity – 172 W/m-K
Heat Transfer Coefficient (estimate for air inside enclosure) – 2 W/m^2-K
Perimeter – .122 m
Cross Section Area - .00011 m

With m = (hP/kA)^.5, I calculated m = 3.549 m^-1 and m*L = 0.1065. If heat flux continues to increase as mL approaches 3, then my mL value is very small and should be increased. Yet if I use mL=1.46 as an optimized design value the resulting height would be 0.4 m, which is nearly the size of the enclosure! Intuitively this number seems wrong.

The strange numbers get stranger when calculating optimal fin spacing.

If
Dynamic Viscosity (of air) – 2.03e5 N*s/m^2
Gravitational constant – 9.81 m/s^2
Air density – 1.029 kg/m^3

and
(Optimum fin spacing/Length) Grashof No.×Prandtl No.=50, which simplifies to
Optimum Spacing=0.29 (L^0.25×Dyn. Visc.^0.5×Tair^0.25)/(g^0.25×density^0.5×temp. rise^0.25)

then the optimum fin spacing would be .7 mm, which sounds optimum for a brick, but not a finned array. So what am I doing wrong?

 

[ione]

Your cross section area Ac is equal to 0.0011 m^2 and not 0.00011 m^2

 

[ione]

Yup, your Ac is correct (my bad)

 

[ione]

Well, it seems your approach is correct.

The thermal conductivity you’ve reported is in line with that typical of aluminium alloys (i.e. alloy 6061).
To be picky the only thing I’d probably have changed is the heat transfer coefficient h.

IMHO your guesstimate for heat transfer coefficient is too pessimistic. Just considering a single fin and only natural convection heat transfer mechanism the heat transfer coefficient 5 W/(m^2 K) is a more appropriate value. Then m would be approx 5,68 m^(-1) and this would lead to a smaller fin height for optimal m*L = 1,46 (H = 0.26 mm against 0.4).

I have also checked optimal spacing Dopt using the following formula

Dopt = 2.71*H * Gr^(-0.25)

From Y. Jalura, Natural Convective Cooling of Electronic Equipment, Proceedings of NATO Conference on Cooling of Electronic System, Smirne-Turchia, June 1993

And I’ve got Dopt = 8.3 mm

 

[istocrabin]

Thank you for your help Ione. It gives me more confidence that in fact my methodology is correct. Based on the solutions you provided I have a few follow up questions.

1) What is a good resource for determining the heat transfer coefficient of air? So far I've only come across valid ranges but no definitive charts or formulas.

2) Even if the optimal fin height in this scenario is in fact near .26 m, this value is still much greater than what my current design is (currently .03 m), not to mention much more expensive for material and machining. Looking at the geometries of other natural convection heat sink I notice that none actually come close to that optimized height. Beyond ensuring that the heat flux of the sink adequately removes heat from a surface, how do others determine an appropriate length?

3) Going through your spacing formula, it appears that your beta coefficient for air is 1.24e-3. Is this correct? Looking at

http://www.engineeringtoolbox.com/air-properties-d_156.html,

I thought 2.83e-3 would be more appropriate, for an optimum spacing of 6.7 mm.

 

[ione]

1. The site at the link below offers both a free calculator for different heat transfer mechanisms and also a theoretical background.

http://www.thermal-wizard.com/tmwiz/default.htm

You might also want to consider

http://heatsinks.wordpress.com/2010...hermal-management-of-electronics-part-1-of-3/

http://heatsinks.wordpress.com/2010...hermal-management-of-electronics-part-2-of-3/

http://heatsinks.wordpress.com/2010...hermal-management-of-electronics-part-3-of-3/

2. Maybe this report can help you

http://www.hpl.hp.com/techreports/Compaq-DEC/WRL-86-4.pdf

And also this site has got good stuff

http://www.frigprim.com/online/natconv_heatsink.html

3. My mistake. Thermal expansion at Tfilm = 75 °C is 2.87*10^(-3) [1/K]. This would give Dopt = 3,94 mm (as Gr = 181950).

 

[IRstuff]

I have some difficulty understanding why your wall temperature is a given. Normally, that's supposedly fall out of the simultaneous solution of the heat flows from the interior and to the exterior.

And fin heights taller than maybe 1.5 inches would seem to be too tall for practicality. Perhaps, what that's saying is that your attempt at a purely passive solution is not practical.

Also, nowhere have I seen a value given for the power dissipation. This is what drives the temperatures, not the other way around.

You also seem to be only solving for one heat sink, which is not optimum, since that means that you have to force a temperature to a specific value to get a solution. Using a lumped parameter model, you should be able to solve both heat sinks together, which will allow you to just use the external ambient temperature.

For the interior heatsink, I'd caution against using the normal heat transfer coefficient, especially if the channel sides of the heat sink do not have sufficient access to open air. Convection is about air movement, and the ability of the air to reach the wall end of each channel is critical to achieving the stated heat transfer coefficients. With channels on the order of 1/4 inch, air access from the sides is mandatory. Only when the gap exceeds about 1/2 inch would there be free convection directly within the channels.

Given all the difficulties, you might want to look at either using a fan to crank up the heat transfer coefficient, or using a more direct method of getting the interior heat to the walls of the enclosure, such as using heatpipes or thermal straps.

TTFN

FAQ731-376

 

[istocrabin]

Thanks for sharing those links ione. They look like good reading. IRstuff, you raise some good questions.

I haven’t gotten solid numbers on power dissipation yet, so for now I’m guesstimating what kind of temperatures we might see in the upper third of the enclosure for the mean time. Once I get solid power dissipation numbers I can get a better idea of what wall temperatures to expect.

Concerning fin height, the optimal dimensions of individual fins are temperature and convection independent. Temperature and convection factor into spacing.

I do like the idea of doing a lumped parameter model to calculate total heat flux of top and bottom fins together. I’ll have to do some review on the topic but it is the approach I need to take.

Honestly speaking, I’m not sure exactly how the bottom interior fins are going to work because I haven’t worked on the numbers to predict circulation from natural convection in the enclosure. While there will probably be some convection, since the majority of heat will be trapped under the enclosure wall on top I imagine conduction will also be significant. I’m am curious if fins will help in a conduction scenario, either for taking heat out or pulling cold in (same thing).

Unfortunately the specs I’m working with require no moving parts in the enclosure and needs to survive 60C ambient temperatures with passive cooling. Fortunately I don’t anticipate power dissipation in the enclosure going beyond 10 W, so there should be a feasible solution. If not there are always heat pipes. It’s a good design challenge to say the least!

 

[ione]

If moving parts are not allowed as per specs requirements, you could evaluate the option of using a “chimney”. This tall configuration should boost a bit your convection coefficient and even if it asks for available room above the heat sink, it is not an expensive solution.

 

[IRstuff]

A reverse way to look that the problem, assuming a more conservative htc for the interior at 3W/m^2-K:

10W/htc = 3.33 m^2-K

So, the product of the effective convecting area and the delta T needs to be at least 3.3 m^2-K. If you design for a 10°C deltaT, then you need at least 0.33 m^2 of convection area, or roughly 510 sq. in. of heat sink area, which, divided by 38 fins, gives 13 sq in per fin. If they're 0.059-m wide, then the fins would need to be about 2.9 inches tall, assuming convection from both sides.

I can't tell from your previous postings how big the box itself is, but it sounds like it's nowhere near able to provide 0.33m^2 of surface area on the board end of the problem. That would potentially make the interior temperature air temperature substantially higher, which would potentially force the design deltaT to be a much smaller value across the two heat sinks, since your external ambient temperature is unchanged.

TTFN

FAQ731-376