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Grogster
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Posted: 02 February 2018 at 10:59am | IP Logged Quote Grogster

Now I'm confused.

So PP sheet would be a better idea then AL sheet - is that what you are saying?

Once heated through, the PP will hold the heat better then AL - I get that bit.
But plastic is a thermal insulator, generally speaking, so it would resist the transfer of heat from one side to the other, so would take much longer to heat up would it not? Am I making any sense?

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Warpspeed
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Posted: 02 February 2018 at 11:02am | IP Logged Quote Warpspeed

Red makes a very interesting point.

Some things like rock, stone, earth and concrete also have quite high heat storage but poor thermal conductivity. If you leave a brick in the sun it will stay hot for a very long time.

The problem is though, we want the heat to travel fairly quickly from the "thermal stored mass" to the air within the incubator if its going to be resistant to thermal shock, as when opening the door.

We need good thermal coupling into the air whenever there is a sudden drop in air temperature. Its not much good if the stored heat cannot escape fairly quickly as a backup heat source to the electric heater.

A fan would certainly help, especially if the PP had a very large surface area, as in the whole incubator housing being made from thick PP. But a small block of PP just by itself probably would not be up to the job.

I still keep coming back to the idea a big heavy chunk of aluminium with a lot of fins. If the heater fails, there is a good chance any fan will stop too...
So the reserve heat really needs to be passive.



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Grogster
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Posted: 02 February 2018 at 11:30am | IP Logged Quote Grogster

Why do you like the finned heatsink idea so much?
Would not the thick flat plate heat up to exactly the same temperature as a finned heatsink would.

The finned heatsink is designed to get rid of heat, so every time you open the box, the fins will actually help to cool down the situation I would have thought, more-so then a thick flat plate will.

I might be totally wrong in my interpretation of things though.

Having said that, I could not get a finned heatsink the size I wanted, so I am going for AL plate at the moment. I could always add fins later if you can convince me of a good enough reason to do it.

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Posted: 02 February 2018 at 12:17pm | IP Logged Quote Warpspeed

[quote]The finned heatsink is designed to get rid of heat,[/quote}
Yes exactly, that is exactly what we want !

As soon as the air starts to cool the finned heat sink really kicks in.

When the air is quickly back up to temperature the heat transfer from the fins drops to almost nothing, provided the enclosure is very well insulated.

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Grogster
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Posted: 02 February 2018 at 12:58pm | IP Logged Quote Grogster

OK, sounds good to me. I will design some fins I can screw or rivit to the plate.
I can get the engineering works to slice me up some 2mm or 3mm AL angle at the same time. Do you have any suggestions as to the height of the fins(30mm?), or is anything better then nothing? I do need to keep the overall height as short as possible so the thing is not a burden to carry cos of it's size.



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Posted: 02 February 2018 at 1:07pm | IP Logged Quote Warpspeed

I suppose whatever fits and looks about right will probaly do.

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redrok
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Posted: 02 February 2018 at 3:09pm | IP Logged Quote redrok

Hi Grogster;
I thought I should complete my study of the Polypropylene thermal mass calculations to include the conductivity comparisons with Aluminum.

materials
Conductive Heat Transfer
AL 205W/mK
PP 0.1W/mK to 0.22W/mK
Lets call it .16W/mK as an average
BTW, the "per m" part of the "units" may be a bit confusing. This is actually
W * Thick / Area / Delta T
W * m / m^2 / K ---> W / m / K ---> W/mK

200mm * 300mm = A of plate = 0.06m^2
12.0v * 3.2A = 39.4W At Maximum power

Thermal Conductivity Equation
(k / s ) A dT = q
(W/mK / 8mm) * Area * dT = W
W/mK / 8mm * Area * dT = W
Rearrange
W/mK * Area / 8mm * dT = W
Rearrange
W / ( W/mK / 8mm * Area ) = K
AL = 39.4W / ( 205W/mK / .008m * .08m^2) = .019K
PP = 39.4W / ( 0.16W/mK / .008m * .08m^2) = 24.6K

Since the plate will be the source of heat when reheating the case
the average resistance to heat flow escaping from one side is 1/2.
AL = 39.4W / ( 205W/mK / .008m * .08m^2) / 2 = .010K
PP = 39.4W / ( 0.16W/mK / .008m * .08m^2) / 2 = 12.3K

If we consider the plate is sourcing heat from both sides
the average thermal resistance to heat flow is 1/4.
AL = 39.4W / ( 205W/mK / .008m * .08m^2) / 4 = .005K
PP = 39.4W / ( 0.16W/mK / .008m * .08m^2) / 4 = 6.2K

This was calculated at the given max heater power.
I assume the running power will be much less so the actual
delta T would proportionally be lower.

BTW, another thermal mass material could be ceramic floor tiles.
Relative Specific Weight is around 2.8g/mm a little heavier than Aluminum.
Relative Thermal Mass would be greater than Aluminum.
Thermal Conductivity is in the neighborhood of 5W/mK Quite a bit better than PP.

Ok, I'll give up on this line of thought.
It was interesting to think "Way out of the box".
redrok

Edited by redrok on 03 February 2018 at 8:10am
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