design of cooling tunnel

[M.Iskander]

thread391-46312
How we can design a cooling tunnel for
;1000 ib. aluminum part 6' Wide) X 10' Length.
Conveyor speed 12 FPM
Length of the tunnel 52 FT.
to cool the part from 400F - 70F.
Please participate as much as you can.
Thanks

 

[EdStainless]

It sounds like you already designed it.
Calculate how much heat you have to remove, allow for at least double that.
How thick is the part? will it possibly cool in 4.5 min?

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P.E. Metallurgy, Plymouth Tube

 

[M.Iskander]

The thickness is 3/16 ".
The tunnel itself is built but the cooling ventilation system are not selected or installed.
As an P.e. if you have any software to be used in calculation the cooling system c/w exhaust and or supply fan and air distribution system. Or empirical formulas , equations , or another tools will be highly appreciated.
Regards
Mamdouh Iskander, P.Eng

 

[EdStainless]

Pull your old heat transfer text off of the shelf and read up on forced convection heat transfer.
At some point this will involve the Heisler charts, the thermal conductivity of the metal could be a limiting factor.
You will need to know how much heat you are removing, how fast, and so on.
You will have to size the blowers to give you enough velocity and the coolers to extract the heat.


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P.E. Metallurgy, Plymouth Tube

 

[BronYrAur]

I'm just throwing some VERY QUICK numbers out here. I did not spend much time on this, but I am just trying to put a magnitude on it.

Looks like you want this part to travel through the conveyor in roughly 4 minutes.

1,000 Lb of Al
Specific Heat of Al is 0.215 BTU/Lb/F
Temp change from 400F to 70F

So in 4 minutes time,70,950 BTUs need to be removed. If this is a continuous process, that's 1,064,250 BTUH.

If you had 10,000 CFM moving through the tunnel, the temp rise of the air removing all the heat would be approx 100 deg F. Now, here is where the real challenge of heat transfer cones into play. How do you get the Aluminum to release all of that heat? Geometry is important for air flow patterns, and there are a lot of other heat transfer coefficients that come into play. What temperature air will you be introducing? Are you just ventilating, or will you actually be introducing cooling?

Lots of unknowns.

 

[M.Iskander]

[pre][/pre]

 

[IRstuff]

The online calculators say that it can't really be done. You'd need category 5 hurricane winds to generate enough forced convection to even come close. At 167 mph, I'm only getting 54% of the required convection to meet your timeline.

http://www.thermal-wizard.com/tmwiz/convect/forced/fp-isot/fp-isot.htm

Another approach is to mist the aluminum with cold water, or immerse it altogether, but you'd need to determine whether warping is an issue.

TTFN (ta ta for now)
I can do absolutely anything. I'm an expert!

https://www.youtube.com/watch?v=BKorP55Aqvg

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[EdStainless]

Thanks IR, I was looking at Heisler charts and realized that they didn't go to high enough air flows to cool the center of this part in 4 min.
Using atomizer nozzles (not mist or water droplets) you can accelerate cooling and minimize the distortion risk. But you need very clean water for these to work constantly.
Just using air, my guess is that you need high velocity (but not 100mph), cooled air (at least 50F), and about 10 min.
Usually this is done in stages, first just gas and since the part is hot the gas just need to be cooler (200F would work), then you go into a cooled gas section, and then a mist or spray system (the material is cold enough that it is strong enough to not warp, and the water needs to be ~20F below desired end point temp).

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P.E. Metallurgy, Plymouth Tube

 

[jari001]

Here's another page that has a calculator to replace Heisler charts for second point verification.

http://faculty.virginia.edu/ribando/modules/OneDTransient/index.htm