Part Cooling

[simoncad]

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

 

[MintJulep]

Not with the information provided.

 

[simoncad]

the parts need to drop from 552 F to 105 F. the parts are springs 1.5" OD, 1/4" wire diameter, 6-1/2" long, 10 spirals in the 6-1/2". each part weighs 11.3 oz.

 

[tomwalz]

I always go to "industrial Ventilation" first for something like this.

Thomas J. Walz
Carbide Processors, Inc.

http://www.carbideprocessors.com

 

[IRstuff]

OK, so figure out the amount of heat that needs to be removed from each spring and determine the air flow required to do that.

TTFN

FAQ731-376

 

[simoncad]

would you happen to know the equation for removing kw or btu determining on the air flow?

 

[MintJulep]

The relation between sensible BTU and airflow is:

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.

 

[simoncad]

i just need to approximate this and it would be ok to overshoot the actual CFM needed.

MintJulep - would the "temp rise of the air" be the temp reduction i am trying to acheive? 552 F - 105 F = 447 F

 

[MintJulep]

Are your parts made of air?

 

[simoncad]

no, i don't believe anyone makes springs completely out of air.

 

[MintJulep]

Then no, "temp rise of the air" is not the desired temperature reduction of the parts.

It is the temperature rise of the air passing over the parts as a result of heat exchange between the parts and the air.

 

[willard3]

This is a complicated heat transfer problem and not nearly so simple as you think. Any estimate you do may be off by orders of magnitude.

If you need a real answer, find someone who does lots of heat transfer.

 

[vpl]

simoncad

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

Patricia Lougheed

Please see FAQ731-376 for tips on how to make the best use of the Eng-Tips Forums.

 

[simoncad]

no this is not the same problem. it's pretty close in difficulty but this is for a different application.

 

[IRstuff]

BTW, your BTU numbers seem to be 2x higher than what I calculate for (800 pieces/hr)*(11.3oz)*(0.46J/gm*ºC)*(552F-105F). I get about 28000 BTU/hr, assuming stainless steel as the material.

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

 

[simoncad]

I believe the 56000 BTU was figured if we were removing the heat from water.

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.

 

[MintJulep]

Flow through a tank and the amount of fluid in a tank are not the same thing.

 

[simoncad]

thank you for pointing that out. I was told that this is a relationship you use to select the appropriate tank size for the cooling system.

 

[IRstuff]

Right, what you need is 800 gals/hr of flow, which is not insubstantial, but you could get by with 5 gallons, if it were able to recirculate 160 times/hr. Note that it looks like the flow is sized for about a 5ºC rise in water temperature, so if you have a more efficient cooling stack somewhere, you can get by with less water flow.

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

 

[simoncad]

that is the type of cooling system that the fluid would go through (car radiator type) if we cooled the parts with fluid. the application engineer from the company that would supply the cooling system told me that the tank would have to hold around 800 gallons to efficently cool the parts.

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.