Chill

I have never liked the process of chilling the wort at the end of the boil. You have to do it, of course, but it comes after creating the recipe, cracking the grains,  mashing in and boiling, which are all much more fun and rewarding aspects of the brewing process in my opinion.  In addition one usually stands in front of a wide open boil kettle just waiting for airborne contaminants to land in your carefully prepared wort. It is a situation that is not conducive to the “relax, don’t worry” mantra of the home brewing community.

Even more disturbing, chilling the wort involves taking beautiful, pure, clean drinking water and using it once and often just sending it down the drain.  Obviously you can recycle.  In the past I have recaptured the hot water from the chiller (I use an immersion chiller) for use in the clean up process, but there always seems to be much more water than is really required to do the job.  In the summer you can capture the water and use it for watering the garden, but trust me, during January in Pittsburgh the need in that regard is pretty low.

I have been on a trek to decrease water waste in my brewing while not compromising the quality of the suds. I have tried a few ideas, and I thought I would share what I’ve found. But since I am a geek first, let’s talk some numbers in order to get  an idea of the magnitude of the problem.

Heat can be transferred in a number of ways.  Conduction might be the most easily recognized form.  Put your hand on a hot stove?  Heat conducted from the stove to your hand activates the nerves that causes your hand to recoil; the heat conducted into your hand might cause redness or even blisters.  Heat transfer by radiation is also a very common experience.  Stand outside on a sunny day and you will feel the warmth of solar radiation warming your body.  In fact you might begin to sweat, and then heat will be transferred from your body by evaporative cooling.

In chilling wort, the principle method of heat transfer is conduction.  Heat energy in the wort is transferred to the copper and then to the water flowing through it.  The water exits the chiller carrying heat away.  The rate of heat transfer depends on the temperature difference between the wort and the water and that is why the wort temperature drops rapidly in the first few minutes of chilling and then slows noticeably, especially as the wort temperature approaches the temperature of the water leaving the tap.

Water (and wort is mostly water) can hold a lot of heat energy.  The ability of a substance to store heat is called heat capacity, and it is determined by measuring how much energy must be put into a material in order to raise its temperature by a set amount (usually one  Kelvin, which is essentially one degree Celsius). Just for reference, compare water and steel.  It takes about ten times more energy to raise the temperature of water, versus steel, by one Kelvin (4.2 Joules versus 0.4 Joules for steel).   If you want to compare various materials here is a link to a table of various materials and their heat capacities.

But this still doesn’t get to the question of how much energy really needs to be removed from the wort in order to reach pitching temperature.  Suppose you have one kilogram of water at boiling temperature (100° C) and you want to cool it down to 20° C  (that’s 68° F for you non metric types, about typical ale pitching temperature).  If your tap water was at 10 °C, how much would you need in order to cool your wort?
One can get a good approximation by approaching this as a mixing problem.  If you mix two liquids together, the final temperature is given by the following equation:

Final Temp =  Tf = (T1 x M1 + T2 x M2) / (M1 + M2)

Where T refers to the temperature of the liquids and M to the mass.  In reality this equation is a little more complicated if you are using materials of different heat capacities, but since we are using water in both cases we can use this simplified version.

What we really want to determine here is M2.  By using a little algebra we can rearrange the equation to solve for M2:

M2 = M1 x (T1 – T) / (Tf – T2) = 1 x (100-20) / (20 – 10) = 80 / 10 = 8 !!

So in order to cool down one kilogram of wort you need about 8 kilograms of cold water.  Applying that factor of eight we can estimate it requires about 40 gallons of water to cool down 5 gallons of wort.  That seems like a lot, and I know that some heat is lost from the wort due to evaporation and radiation, but still that is a significant amount of water.

Can we use less?  Certainly, and I aim to tell you how I am doing it.

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