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I'm designing a simple copper coil wort chiller. Digging formulas I'm having hard time to figure some things out, perhaps some simple guides may help me to go the right direction:

  • LMTD need inlet and outlet temperatures differentials. On Time = 0 this can be for the inlet the initial refrigerant temperature and wort initial temperature. But what about then $T$ = 5 mins? And Outlet temperature? How do I even calculate it on $T = 0$? I bet this is related to the refrigerant flow inside the tubing, but can't find any paper to help me out.

  • Wort Reynolds number. How the flow turbulence helps to improve the $U$-Number? Is there a way to calculate it (related to RPMs?) or I just use empirical numbers from tables?

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LMTD has nothing to do with time. If you need time, you're doing it wrong. An implicit assumption is that the temperatures of everywhere in the heat exchanger have equalized to steady state. If the system experienced some change, and if it's sufficiently at steady-state after the change, then use the new set of inlet and outlet (4) values. If you don't have enough information to determine these, then you have a larger problem. – Alan Rominger Feb 20 '13 at 19:55
I'm inexperienced with heat exchanging systems. But the wort mass does not change over time, it's constant so the temperature will go down over time and that's why inlet and outlet temps will also change with time. Are you saying that using LMTD in these cases is wrong? – Gabriel A. Zorrilla Feb 20 '13 at 21:28
Okay, that can make sense if your time dependence primarily comes from the decreasing wort temperature. Unless the time the refrigerant resides in the coil extremely long it's unlikely you'll be bothered by the steady state approximation for a DIY. The LMTD will relate $\dot{Q}$ (the heat transfer rate), the flow rate ($\dot{m}$), and the bounding temperatures, which you'll likely reduce to 3, given mixing in the wort. In order to apply the LMTD, you'll still be missing the heat transfer coefficient, and LMTD assumes Newton's law. That's slightly incompatible with turbulence... (cont.) – Alan Rominger Feb 20 '13 at 22:29
(continuing) but since the flow rate is constant and dictated by your refrigeration equipment pumps, you'll likely ignore that. In your question, however, you're asking about the wort Reynolds number and that's a confused question. Isn't the Wort the large volume of beer-like stuff? You don't know the flow rate? It's mostly natural circulation, which is doubtfully helpful. Maybe other people in the business do have tables for this sort of thing, but you would need so much more information to find the Wort Reynolds number on your own, so I don't really know what your starting point is. – Alan Rominger Feb 20 '13 at 22:34
My Reynolds question is because if the wort gets affected by turbulence of course will be better heat exchange between the water in the coil and the wort being cooled down, i just asked if can be calculated to bring some math to cuantify how much better it is. I'm using a water-copper heat transfer coefficient BTW. – Gabriel A. Zorrilla Feb 20 '13 at 22:56

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