HERMS Coil

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I am currently building a HERMs system, using keggles, to make the jump to AG. I am in a very similar situation as most of us DIY'ing brewers are in, in that I often have to strike a balance between cost, time, and relative difficulty to get projects completed. Here's my problem; I have (or had) an old IC that I cut done for dip tubes and was hoping to leave what was left for my HERMs coil. There is approx. 15 feet of tubing that I can work with. This is appealing for two main reasons, 1) It will keep the lowest possible profile in the keggle in order to keep the water level down and 2) It's f-ing free and reduces my waste of old parts to about 5% of the old part. Reduce, re-use, recycle, you know.

I would like to maintain the low profile to keep water level down and increase the performance of the heat exchanger. Right now, it would take 6-7 gal. to submerge the coil and that is about already a little over my goal. I wanted to do it in 5 or less. My main question is, will I get enough heat transfer to to maintain mash temps and what formula can I use to determine this?

As long as you maintain turbulent flow inside and outside the coil, 15' is enough for it to reach equilibrium. BTW, that is with either 3/8" or 1/2" OD copper. Turbulent flow through the coil is easy (You do have a pump?). Turbulence on the outside of the coil requires stirring or recirculating the water so that it is in constant motion.

Hope this makes sense.

Yup, it makes sense. The turbulence in the HLT would keep stratification to an absolute minimum and keep the hottest possible temperature against the coil at all times.

I am using 0.5" refrigerant tubing, so 0.375" ID. and a pump. I am also using a false bottom and want to achieve the maximum flow rate allowed by the grain bed while still getting the heat transfer. How important is flow rate to this equation?


Thanks.

Ok, now that I know that you are using 1/2" OD tubing:

(Excerpts from an article at The Brewery.org)

" ~~ turbulent flow is achieved in a 1/2" OD tube at .14 m/sec which quite simply means:
*put 5 gallons through your 1/2" exchanger in under 31 minutes and you have turbulent flow*, (Charlie's Law)
The overall coefficient of transfer jumps from about 1.25 KW/m² K to about 12.5 KW/m² K ! Fast flow systems rule,~~~ "

I think you are really all set IMHO.