Stainless steel cooling container, need some help with the maths

AI Thread Summary
The discussion focuses on designing a stainless steel cooling container to cool 250ml of boiling water to 20°C. The user initially struggles with calculations for the passive cooling system, estimating an unrealistic cooling time of 0.059 seconds, which is later corrected to approximately 5.868 seconds after adjusting surface area calculations. Suggestions include ensuring consistent unit usage and considering the heat transfer coefficient for improved accuracy. The user concludes that with proper ventilation and forced convection, cooling times could be reduced to around 12 minutes and 3 minutes, respectively. Accurate calculations and design adjustments are essential for effective heat transfer in the cooling system.
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Hi guys, I'm designing a cooling device as part of a project and need a little help with the maths.

OK, so the scenario is this, take boiling water, put it in a stainless steel container with Thermal Conductivity of 8.7 Btu/h ft °F and dimensions of 15.811cm x 15.811cm x 1 cm to store 250ml of water with a wall depth of 4.7625mm.


What I'm struggling to find out right now is how long it will take to cool the water to 20c.



I'm looking to cool the water as quick as possible so I'm looking at a passive system as described above and then to implement a forced system and add fins to both sides to improve heat transfer rate. I've got to cool the water to room temperature in around 7 minutes and the maths is killing me even to figure out the passive system so i guess with a heat sink type design and implement a forced air flow it's going to get even worse but I'm hoping you guys are in the mood to be helpful if you can.
 
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Thanks I've tried this equation but end up with a figure that seems too large.

heres my workings

energy difference of 250ml of water between 20 and 100 degrees= 83600 J

8.7 Btu/h ft. °F - watt/meter/K [W/(m·c)]= 15.47

area of container = 563.219cm^2

depth 4.7625mm

A = 5.63219m^2

k = 15.47 w/(m*c)

dT = 80

s = 0.0047625m

Q=1424590.431 watts?

So that means that I'm transfering 1424kj/per second? which means that the water will cool in 0.059 seconds?
 
Well, first off, check your surface area calculation. You are finding the area of the material contacting the water. If it's a box, you have 4 sides plus a top and bottom. Keep units consistent.
 
ahhhhhh, that's quite a big typo... thanks so now I'm down to .0563244m^2 I'm getting 14245.904 watts which makes takes me up to 5.868 seconds. still seems pretty low.


The surface area i found by taking the two larger sides 250cm^2 each and then the other 4 sides 15.811cm^2. 563cm^2.
 
Plus, as I said, that's just a starting point. Here's another couple:

Heat Transfer Look at the insulated box example.
HTC You'll need to figure out the Heat Transfer Coefficient, too.
 
It should also be noted that just filling a box with water isn't exactly the best way to transfer heat. Typically you want the fluid to be flowing over the contact surface for good heat transfer.
 
And here's another tip: 1 m^2 = 10000 cm^2, not 100 cm^2.
 
Thanks guys so I've got down that if i use it in a well ventilated container it will take around 12 minutes and if i use a fan to help with forced convection it will take around 3, that's using fouriers law and a heat transfer coefficient of 21 and 100 for the air respectively and a heat transfer co-efficient of 0 on the water side. sound about right?
 
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