Conveyed cosmetic cooling calculation... how many BTUs?

[natem84]

I am trying to calculate the required W/BTUs of cooling needed to cool a cosmetic product down from 140 F to room temperature at 72 F in a 30 ft tunnel.

The product is filled and then transported on a conveyor system. These products are lined one-right-after-the-other which means in a 30 ft. tunnel there are approximately 100-120 products at any given moment. The product weight is 15 grams and the SG of the product is ~ 2000 J/kg C. Also the conveyor belt speed is at 16 ft/min which means the product has ~ 1 min 53 sec in the cooling tunnel.

I performed some initial calculations and found the value to be just under a half-ton of cooling. I don't know if I adequately accounted for the transient flow of the products on the conveyor.

We are hoping to be able to tap into existing AC units in our facility in order to cool this tunnel. It all depends on our available cooling capacity if we have that option. Wost case scenario we'll have to buy a small residential AC unit to do the job.

Thanks in advance!

 

[IRstuff]

I get a total heat removal of about 4000 BTU/hr, which is about 1/3 ton. However, that assumes perfect heat transfer.

However, looking at the pure convective cooling problem:

(2000J/kg-K)(15gm)(70F)/(6in^2)(10W/m^2-K)(70F) = 775 s

That means that you'll need either gale force winds, or way cooler air temperature , or both to meet the time constraint. It may be possible to add cooling fins to increase the convective area, but that depends on your product and its container.

Running the containers through chilled water might speed things up.

TTFN

FAQ731-376

 

[natem84]

IRstuff:

Do you know what the air temperature of a standard AC Unit usually is? I have the delta T as 70 degrees because we want the product at that temperature. That does not reflect the actual temperature of the air.

 

[IRstuff]

Typically, A/C outputs are around 55F, I think.

TTFN

FAQ731-376