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Old Jun 21st 2017, 07:27 AM   #1
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Question Convection in isolated box (HELP!!!)

Hi,

I do not have much knowledge about this stuff (just basic things), but I need to know something.

How do I calculate, how many cooling elements (bags of ice) do I need to maintain the temperature products (0C-4C) inside an isolated box for 24 hours? Depends on the cooling elements area.

Outside temperature: 25C
Cooling element size: L x H x W = 16 x 13 x 3 cm

Thank you.
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Old Jun 21st 2017, 07:35 AM   #2
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Isolated?

It depends on the insulating capability
of the box walls. This needs to be known, first.
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Old Jun 21st 2017, 07:39 AM   #3
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The thermal conductivity of the box wall is 0,038 W/mK
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Old Jun 21st 2017, 09:30 PM   #4
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so... thermal conductivity is 0.038 W/mK so if a cube 1meter square the heat conducted across would be 0.038W for 1* difference ... but your temp difference is 25* ...so 25x0.038 = 0.95 W for 1 meter cube ... but you don't have a meter cube ...

surface area of box.... you say the box is 13x16x3 .. this doesn't sound right !!!

We need the outside dimensions of the box and the thickness of the walls of the box.
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Old Jun 22nd 2017, 12:17 AM   #5
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It's the cooling elements size.

The box size is: 300 x 300 x 250 (outside)
After the insulation the box size (inside): 270 x 270 x 2000
L x W x H
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Old Jun 22nd 2017, 01:09 AM   #6
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Originally Posted by randomthings View Post
It's the cooling elements size.

The box size is: 300 x 300 x 250 (outside)
After the insulation the box size (inside): 270 x 270 x 2000
L x W x H
OK .... let's find the AREA through which the heat will be lost .

two faces are .3x.3 = 0.18 ...four faces are .3 xx.25= 0.3

total area which transmits heat is 0.48 sq meters

from post 2 we know heat loss through a one meter cube is 0.95W @ 25 deg difference .... but our surface area is less than 1 sq meter ... 0.48 , so 0.48x 0.95 = 0.456 W this is the rate of heat loss if the walls were 1 meter thick ... but they're only 15mm thick !

so the heat loss will be greater by 1000 divided by 15 = 66

so the rate of heat loss is 66x 0.45 = 28.5Watts

Latent heat of fusion of ice is 334 joules per gram, that means that one gram of ice can keep your box cold for 334 / 28.5 = 12 seconds

1 kg of ice will take about 3 hrs to melt , keeping the inside of the box at 0*c for 3Hrs

Check you have measured the wall thickness correctly , the figures you gave give the top and bottom as 25 thick ... if this is correct 1kg ice will last about 4 Hrs .. if you double the wall thickness you double the time the ice will last.

24 hrs will require 6 kg of ice (plastic water bottles will work OK ) ... rather a lot of ice perhaps think about thicker walls , or covering the bozx in a duvet or blankets.

Cooling elements ??? I guess you mean those blue plastic things containing jell , will not be the same as ice... I don't know if they will be better than ice or not??
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Last edited by oz93666; Jun 22nd 2017 at 01:29 AM.
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Old Jun 22nd 2017, 02:44 AM   #7
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Thanks. I get it now.

Can we calculate it depends on how big is the cooling element?
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Old Jun 22nd 2017, 03:58 AM   #8
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The size (dimensions) of the items releasing the cold does not really matter , it's the mass of frozen water which is important .

The insulating walls are not very thick , where you position this box will make a difference , if pushed tight into the corner of a room , then the walls of the room where they contact the box will help keep the cold in, covering the box with anything will help insulation.
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Old Jun 22nd 2017, 06:43 AM   #9
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Ok. Thanks for your help.
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Old Jun 22nd 2017, 07:11 AM   #10
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The bigger the surface area of the cooling element, isn't the faster the ice will melt?
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