Greetings homebrewers, Some of you who have requested this article are interested in immersion chillers. This article is about the other kind (what do you call them?) [Note: I now call them bath chillers] that has hot wort flowing inside the tube and a cool bath of water or ice water on the outside. As I state at the end of the article, a pretty serious caveat is that I have assumed that the outside water bath temperature stays constant. For the immersion chiller aficionados, I plan to work up a similar calculation in the future. I spent all day yesterday on this, though, when I should have been preparing for a conference presentation. Of course, this was more fun, given the beer tie-in. No promises on when I'll finish the immersion chiller analysis, but I'll let you know by a post to the HBD. This article is being sent out for review purposes: if you have any comments I would like to hear them (especially if you have found an error). Also, thanks to those of you who gave me tips about Zymurgy publication. Prost, Mike ------------------------------------------------------------------------ All rights reserved. No part of this publication may be reproduced or transmitted, in any form or by any means, electronic, mechanical, photocopying or otherwise, without the prior written permission of the author. Persons receiving this publication by electronic mail are authorized to make hardcopies and to store this publication, but only for their own personal use. The author makes no representations, express or implied, with respect to this publication or the information therein, including without limitations, any implied warranties of merchantability or fitness for a particular purpose, all of which are expressly disclaimed. ------------------------------------------------------------------------ Wort Chiller Calculations by Michael L. Hall Since my expertise is in the area of fluid flow/heat transfer, I thought that I would answer the inquiry about proper lengths for wort chiller coils. What follows is a simplistic analysis of the situation. (There is a nomenclature section at the end.) Given a tube with a constant wall temperature, the heat transfer to the wall per unit length is q' = pi D h (Tw - T) . The heat loss to the walls of the tube changes the temperature of the flowing wort by the following equation (neglecting many small effects): dT q' = d V Ax Cp -- . dz Putting these two equations together, integrating, and putting in the initial value yields -J z T(z) = Tw - (Tw - Ti) e , pi D h 4 h J = --------- = -------- . d V Ax Cp d V D Cp The heat transfer coefficient, h, is related to the Nusselt number (Nu) by Nu k h = ---- . D An empirical correlation for the Nusselt number of non-metallic fluids flowing in circular tubes is given by the Dittus-Boelter equation: .8 .4 Nu = 0.023 Re Pr . The velocity of the wort can be determined by equating the gravitational head to the kinetic energy and the frictional losses, 2 2 2 V L V V | L | g H = --- + f ----- = --- | 1 + f --- | . 2 2 D 2 | D | Note that H is the difference in height between the top of the wort that is being siphoned and the exit of the tube. Assuming laminar flow, the friction factor can be determined by 64 f = ---- . Re Combining equations yields 2 V 32 mu L V 0 = --- + --------- - g H , 2 d D^2 which has the solution -------------------- 32 mu L | |32 mu L|2 V = - ------- + | |-------| + 2 g H . d D^2 \| | d D^2 | To check and see if the assumption of laminar flow was correct, compute the Reynolds number. For Re < 2300, the flow is laminar, and for Re > 2300, the flow is turbulent (the 2300 break point is very rough). If the flow is turbulent, the friction factor can be determined by .2 f = 0.184 / Re . Combining equations yields 2 .2 V | mu L | g H = --- | 1 + 0.184 ------------ | . 2 | (d V D)^.2 D | The easiest way to solve this (short of a root-finding method) is to assume that the V^.2 part is slowly varying and use a guessed V for that part. Then, solving for the actual velocity gives ------------------------------------- | / | mu^.2 L | V = | 2 g H / | 1 + 0.184 ------------- | \| / | (d Vg D)^.2 D | As far as the thermophysical parameters go, we don't know them exactly, but we can guess: Water Wort Cp = specific heat 4180 4180 d = density 1000 1050 k = thermal conductivity 0.65 0.65 mu = viscosity 0.001 0.003 We now have all the information we need to proceed. The question that we can answer is this: "Given the siphoning height, the tube diameter, the length of the tube, the initial wort temperature, and the tube wall temperature (the temperature of the surrounding fluid), what is the final wort temperature?". I have programmed all of this up in an spreadsheet (available by email if requested). Here are some examples of the results. Assume that H = 0.5 m (19 in), Ti = 100 C (212 F), and Tw = 0 C (32 F). then, for different tube lengths and diameters, here are the exit temperatures: Length Exit Temperature Exit Temperature D = 0.0127 m (.5 in) D = 0.00635 m (.25 in) 3 m (9 ft) 54.6 C (130 F) 18.3 C (65 F) 4 m (13 ft) 43.5 C (110 F) 9.2 C (48 F) 5 m (16 ft) 34.2 C (94 F) 4.4 C (40 F) 6 m (19 ft) 26.7 C (80 F) 2.1 C (36 F) 7 m (23 ft) 20.7 C (69 F) 0.9 C (34 F) 8 m (26 ft) 15.9 C (61 F) 0.4 C (33 F) 9 m (30 ft) 12.1 C (54 F) 0.2 C (32 F) 10 m (33 ft) 9.2 C (49 F) 0.1 C (32 F) It would seem that a 1/4-inch tube is preferable, but the flow rate is *much* slower in the smaller tube (both the velocity and the cross- sectional area are smaller). In fact, for an 8-meter length of tube, the 1/4-inch tube takes 27.9 minutes to empty a 5 gallon pot, whereas the 1/2-inch tube takes only 4.4 minutes for the same job. But what if you don't use an ice bath, but only use your tap water, which is 50 F? Here are the results for that condition (ceteris paribus): Length Exit Temperature Exit Temperature D = 0.0127 m (.5 in) D = 0.00635 m (.25 in) 3 m (9 ft) 59.1 C (138 F) 26.5 C (80 F) 4 m (13 ft) 49.1 C (120 F) 18.2 C (65 F) 5 m (16 ft) 40.9 C (106 F) 14.0 C (57 F) 6 m (19 ft) 34.1 C (93 F) 11.9 C (53 F) 7 m (23 ft) 28.6 C (84 F) 10.9 C (52 F) 8 m (26 ft) 24.3 C (76 F) 10.4 C (51 F) 9 m (30 ft) 20.9 C (70 F) 10.2 C (50 F) 10 m (33 ft) 18.3 C (65 F) 10.1 C (50 F) Of course, there is one *big* caveat to all of this. I have assumed that the wall temperature (i.e. the temperature of the exterior water) stays constant, which may be far from true. If you have a big tank to put your tubing in, or you use a counter-current stream of water, then this may be a good assumption. Otherwise, if your water is not agitated well (causing an increase in temperature in the vicinity of the tube) or if you don't have an adequate heat sink (not enough water or ice), then your results will be worse than those given here. The results given here are the "best case" results. While doing this analysis, I noticed that the laminar flow exit temperatures were almost exactly equal to the turbulent flow exit temperatures, even though the velocities and heat transfer coefficients differed by a factor of 0.7 or so. After looking at the situation in more detail, I realized that the important thing in determining the exit temperatures was the h/V part of the exponent J. Since h is proportional to Nu, and Nu is proportional to Re^.8, and Re is proportional to V, most of the influence of the velocity cancels out, with only a V^{-.2} left. The bottom line is that if you are only interested in the exit temperature, using the laminar flow velocity (even if it is slightly wrong) won't change the answer much. That's all for the analysis part. Here is my technical *opinion*. If you are a homebrewer, brewing 5-gallon batches, and you are straining your wort through the hops before letting it siphon through the wort chiller (i.e. you have to wait a little while anyway), I would use one of these two options: a. 30 ft of 1/4-inch tubing with counter-flowing water (especially good in winter). Get as much water flowing through the other side as possible. b. 30 ft of 1/2-inch (or 3/8-inch) tubing in an ice bath. Try to stir or agitate the bath frequently. For good blocks of ice, I freeze old milk cartons full of water and tear off the cardboard when I need them. I don't think that 1/2-inch tubing with counter-flow water will give you enough cooling, and I think that 1/4-inch tubing in an ice bath is a little bit of overkill. The reason that I recommend 30 ft is that this gives you some extra cooling power when your ice bath has melted and the water is starting to warm up. These are just opinions; your mileage may vary. Dr. Michael L Hall hall@lanl.gov --------------------------------- Nomenclature: Ax = cross-sectional area of the tube (m^2) Cp = specific heat (J/kg/C) d = wort density (kg/m^3) D = diameter of the tube (m) f = friction factor (dimensionless) g = acceleration due to gravity (9.8 m/s^2) h = heat transfer coefficient (W/m^2/C) H = difference in liquid levels (m) k = thermal conductivity of the wort (W/m/C) L = length of the tube (m) mu = wort viscosity (N s/m^2) T = bulk wort temperature (C) T(z) = bulk wort temperature as a function of length (C) T(L) = exit temperature of wort (C) Ti = initial wort temperature (C) Tw = wall temperature (C) Pr = Prandtl number = (Cp mu)/k (dimensionless) Re = Reynolds number = (d V D)/mu (dimensionless) V = wort velocity (m/s) Vg = guessed wort velocity (m/s) q' = linear heat addition rate (W/m) z = length along the tube (m) Note: The vertical bars I have used in this ASCII version are solely grouping operators, and NOT absolute value operators.