I left the Physics department for Literature 20 years ago, confident that the mysteries of thermodynamics would never torment me again. Alas, now I must dip my big toe in it again. I am constructing a 70 litre test model of a full mini-brewery for assembly in central Russia. To this end I have been pestering engineers, beer in hand.
Basically, some of the SMM converts to DMS in the boil and is carried away. Part of *what is left* is converted to DMS in solution, about 25% of which is then broken down in fermentation provided no DMS producing bugs are present.. So the final Sulphur flavour is determined by:
Using George's examples I have created two tables, the first to calculate the concentration of SMM/litre Wort, and the second to calculate the % of this left as DMS in solution *after* fermentation.
MICRO GRAMS of SMM/LITRE WORT BEFORE BOIL * Grain Bill/Vol* * Micro Grams SMM/Gram of Malt * Grams/Litre Lbs/Gal 3 4 5 6 7 8 9 ----------------------------------------------------------- 150 1.125 450 600 750 900 1050 1200 1350 200 1.5 600 800 1000 1200 1400 1600 1800 300 2.25 900 1200 1500 1800 2100 2400 2700 400 3 1200 1600 2000 2400 2800 3200 3600 500 3.75 1500 2000 2500 3000 3500 4000 4500This gives the initial concentration of SMM/litre in your mash.
PERCENTAGE OF INITIAL SMM LEFT AS DMS AFTER FERMENTATION *Boil Time* *Chilling Time* (100 C to 20 C) 5 10 15 20 30 60 Minutes ----------------------------------------------------------- 120 .5% .9% 1.3% 1.7% 2.4% 4.2% 105 .6% 1.2% 1.7% 2.2% 3.1% 5.5% 90 .8% 1.5% 2.2% 2.8% 4.1% 7.1% 75 1% 1.9% 2.8% 3.7% 5.3% 9.2%Example: We have a 6 micro gram/gram SMM content malt, and a grain bill of 2.25 lbs/gal. We decide to boil for 90 minutes and our Supercyberchiller does the trick in 10 minutes.
DMS = 1800 x 1.5% = 27 micrograms/litre
(DMS is below the flavour threshold of 30,
in lower secondary range).
If we cool for 15 minutes:
DMS= 1800 x 2.2% =40 micrograms/litre
(A slight malt/sulphur tone-quite agreeable).
Now we try some high SMM malt at 8 /gram and a high SG brew using 3lb/gal, a
short boil of 75 minutes and a bathtub chiller which takes 30 minutes.
DMS= 3200 x 5.3%=170 micrograms/litre
(This would clear your sinuses!!)
I am not confident of these figures, could they be checked by Fix or someone more familiar with the labyrinths of chemistry? I will repost new tables if corrected.
The kettle in a snow drift (my Russian grandmother-in-laws method) is a non flow system and fraught with temperature stratification problems and the transmission coefficient of water. (stir vigorously and it has the same properties as crossflow.)
The first trap I fell into was confusing "coeficients of heat transfer" with "coeficients of heat transmission". Heat travels through things and most high school kids could find out that Cu is 7 times better at this than steel. But the heat *transmission* properties of water are lousy and in a laminar flow situation a boundary layer forms on the wall of you tube, becoming an insulator. As well, the interface between liquid/metal and then metal/liquid has it's own resistance called the coeficient of heat *transfer*. The actual bit of metal (if its under 3mm) contributes much less than 10% to the "overall coeficient of heat transfer".
The rate of heat transfer depends on a few vital factors:
Without trying to understand Prandtl and Renyolds numbers, turbulent flow is achieved in a 1/2 " tube at .11 m/sec, in a 3/8" tube at .14 m/sec, and in a 1/4 " tube at .22 m/sec, which quite simply means in each case:
*put 5 gallons through your (1/2":3/8":1/4") exchanger in under (22:31:46) minutes and you have turbulent flow*, (Charlie's Law)
The overall coeficient of transfer jumps from about 1.25 KW/m2 K to about 12.5 KW/m2 K ! Fast flow systems rule, OK?
Example:
Now 5 gallons of wort at 100 C cooled to 20 C needs to loose about 6300 KJ of
energy,
If the temp difference (dT1) at exit of coolant and entrance of wort (same end)
is 30 C (run the hose fast) and at wort exit (dT2) = 5 C the the log mean temp
difference O = (dT1-dT2)/(loge (dT1/dT2)) = 13.8 C
Now Q=U A O (Q=heat transfer rate, U= coeficient, A= area, O=
log mean temp diff)
If we want to cool in 10 minutes we need a rate of 10.5 KW which translates theoretically into about 8 feet of 3/8 tube in a hose. Extra area in heat exchanges can't hurt and gives more control, so buy 12 feet. But tap water is seldom 5 C below the desired final temperature, especially for lager brewers. A crossflow "finishing" exchanger of 4 feet of tube in a ice bucket will pull it down a further 8 > 12 C without need to change the ice coolant.
The "tap water in a hose around a tube of wort" and "tap water in a coil in stirred wort" are thermodynamically efficient, but suffer from the limitation of the tap temperature as some thermal pressure is needed to drive the heat exchange.
The "coil in an ice bucket/bathtub" systems on their own need coolant changes and constant stirring.
The combination counterflow, crossflow exchanger avoids most of this except the stirring, the tap water receives most of the energy, preventing the rapid heating of the ice bucket.
To control you cooling for DMS results, the siphon height or outlet constriction will adjust your flow rate of wort and thus your cooling time. The tap water flow rate will adjust your final temperature to the desired figure. Record your results and taste the difference.
Pubs force tennis-size foam balls through their beer lines under pressure.
I'll work out how to send my drawings on this Internet thingame.