Is there an ideal ratio of evaporation area to condenser surface area?

Been doing some thinking about ethanol recovery systems, falling films and others and I’m curious about something - more than a few fruitless google searches showed a lot of info but most of it seemed directed at the microdistillery markets. From what I can tell those guys aren’t really interested in stripping ethanol from whats already a reasonably pure (from a water:ethanol ratio standpoint) ethanol product as fast as humanly possible, they want some predictability and controllability so that they can separate their heads and hearts out from the body of their alcohol distillation. So I’m left wondering:

Is there an ideal ratio between how much surface area you have on the sum of your evaporation surfaces and how much surface area you have on the condenser?

I know that in most people’s setups the bottlenecks probably aren’t the ratio of evaporation SA: condensation SA, I’m guessing for most its the heating and chilling appliances they are using, but the question remains.

Thinking about limits no matter how cold you get it, you aren’t going to recover 100 gal/minute with a couple square inch condensing contact area, right?

Would this dynamic change any with different styles of evaporators? My gut feel says no, the gaseous ethanol amount is going to be what it is regardless of whether its a WFE, FFE or pot still producing the gaseous ethanol but I was curious if anyone with more experience can chip in

once you break the heat delivery bottleneck, it moves into your pipes and condenser. surface area for cold delivery becomes increasingly important. shoot for 1:1 or better.

Generally, chillers are less energy efficient than heaters, so maybe a larger condenser to heater surface area (think about how condensers seem over sized in large rotovaps). But really it’s about the rate of heat transfer across the surface you are working with. Stainless steel will always be faster at heat transfer than glass. So a stainless steel coiled condenser will be much more efficient for a similar sized glass condenser. But a shell-in-tube design for a condenser is going to be probably more efficient depending on size, and you can have one custom built for you if you give them the specs it needs to perform to, by a company that builds these things kinds of things.

In an ideal situation you will remove exactly as much heat (by condensing) as you added (by evaporation). Qc = Qe, right?

So let’s say you’ve already calculated one of these, now you know both. Assuming you know the temperatures of inlet heating medium and inlet cooling medium, and assuming the outlet temperature of both heating and cooling mediums are the boiling temperature of EtOH, we can calculate Log-Mean temperature for both the evaporator and condenser. DIFFERENT THAN LMT FOR HEAT EXCHANGERS. HEADS UP.

Anyways Qe = UeAeTLMe
Qc = UcAcTLMc

Qc = Qe (in a perfect situation, which is what we’re assuming)

We know TLMe and TLMc.

Ue and Uc are super, SUPER difficult to find…that’s something I would charge for if it were requested of me, but there are decent approximations online for evaporators/condensers. Google: overall Heat Transfer coefficient evaporator

Ac/Ae = (UeTLMe)/(UcTLMc)

There’s your ratio.

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