Part Cooling
Part Cooling
(OP)
I am working on a project were I have to get rid of 56000 BTUs in one hour. This is a process on a conveyor. There are 800# of parts passing through in one hour. I'm not sure what temperature drop the BTUs were figured out at, but I have to drop the temperature using air knives. Does anyone know how many CFM I would need to blow on these parts at probablly 80-100 PSI, and for how long?
Thanks,
Simon
Thanks,
Simon





RE: Part Cooling
RE: Part Cooling
RE: Part Cooling
Thomas J. Walz
Carbide Processors, Inc.
www.carbideprocessors.com
RE: Part Cooling
TTFN
FAQ731-376: Eng-Tips.com Forum Policies
RE: Part Cooling
RE: Part Cooling
BTU/hr = 1.1 X CFM x (temp rise of the air)
However, this is useless for you, as it assumes that convection from the springs to the air is the only heat transfer mode. Also, you need to know the convection coefficient to figure out the rate of heat out of the springs and into the air.
RE: Part Cooling
MintJulep - would the "temp rise of the air" be the temp reduction i am trying to acheive? 552 F - 105 F = 447 F
RE: Part Cooling
RE: Part Cooling
RE: Part Cooling
It is the temperature rise of the air passing over the parts as a result of heat exchange between the parts and the air.
RE: Part Cooling
If you need a real answer, find someone who does lots of heat transfer.
RE: Part Cooling
I was going to refer you an earlier thread that seemed very similar ... and then noticed you were the originator of that one too. Is this the same problem?
thread391-178013: Part Drying
Patricia Lougheed
Please see FAQ731-376: Eng-Tips.com Forum Policies for tips on how to make the best use of the Eng-Tips Forums.
RE: Part Cooling
RE: Part Cooling
Standard heat transfer:
htc*area*(Tstuff-Tambient)
Anyways, it looks like you'd need some VERY COLD air and VERY LONG exposure times to cool these things off.
Your best bet may be to run them through cold water
TTFN
FAQ731-376: Eng-Tips.com Forum Policies
RE: Part Cooling
I was previously trying to see if we could run the parts thru a tank of water and cool the fluid during the process. I was told that you need one gallon of fluid per pound of material going thru the system in one hour. So, since we have 800 lbs/hr in this process we need to have a tank that can hold around 800 gallons of fluid. The tank would be too big.
RE: Part Cooling
RE: Part Cooling
RE: Part Cooling
However, the big problems are the boiling that would lose water and the heat that needs to be removed from the outlet water. So, somewhere in the plant, you'd still need to get something that can dissipate >16 kW.
The advantage of a secondary heat exchange is that you can amplify the area to be cooled, i.e., like through an car radiator, which has way more area than your parts.
TTFN
FAQ731-376: Eng-Tips.com Forum Policies
RE: Part Cooling
I think this might be one of those, it worked before so it'll work again, things for the gallons to pounds per hour relationship.
RE: Part Cooling
You've got a unit problem in your earlier post.
Heat capacity in C and delta-T in F.
C/F = 1.8 ~=2
RE: Part Cooling
Or are they invisible conversions?
RE: Part Cooling
1. Figure out how much heat you really need to remove from the parts. IRStuff's calculation earlier has the basic method, lack of unit conversions notwithstanding.
1b. Figure out how much heat you will have to remove from the conveyor, unless you have that special air that can selectively remove heat only from the product.
2. Figure out how quickly heat can transfer from the parts to the air. This will be a function of velocity. Renyolds, Nessault, Prandlt and probably a few other numbers are involved. Easier to do some tests I would think.
3. Figure out what temperature air you have available, and what temperature air you want to exhaust.
4. There you go.
Please post your completed calculation here. We're all curious.
RE: Part Cooling
Lets me blithely mix units ad nauseum.
TTFN
FAQ731-376: Eng-Tips.com Forum Policies
RE: Part Cooling
TTFN
FAQ731-376: Eng-Tips.com Forum Policies