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I am trying to predict the litres/hour vaporised during the boil, so I can adjust the amount of sparging water accordingly. Is there an easy way of doing this a priori by utilising other known parameters, like for example the surface area of the kettle, or must I resort to post-hoc empirical methods?

Edit: I realise from comments that the common way to do this is establishing a percentage of expected fluid loss for your individual set of equipment, but i feel that this is a somewhat unsystematic approach to it all. If there were some known constants roughly representing water at a boil at an altitude of say 0-100 metres, with a relative humidity of 30-40 %, that would be a step in the right direction (after all, homebrewers doesn't need very accurate estimates).

I have looked at some equations online (like https://physics.stackexchange.com/q/91996), but haven't managed to figure them out properly (I am not a physicist).

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    I'm too lazy to anything other than this: (1) measure volume before the boil (2) measure volume after the boil (3) subtract. I think any "a priori" calculation would have to include the vigor of your boil, which would be hard to quantify. – Jeff Roe May 15 '16 at 15:30
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I've not seen a calculator that's reliable. Could be calculated but too many variables: surface area, altitude, actual wort temperature etc. It's much easier to just do a boil test to establish your brewhouse boil off rate.

For example my brewhouse boil off rate with a keggle is 1 gallon an hour at modest rolling boil. So for 90 min boil, 12 gallon batch I start with 13.5 gallons.

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What you are actually looking for here is not the amount of evaporation you will be working with it is the amount of vaporization. This is a technical but important distinction the overwhelming amount of water loss is from the boil not from evaporation. Now this can be calculated but unfortunately I won’t be able to give you a magic bullet for this there are simply to many variables such as ambient temp the thermal conductivity of your pot how windy it is that day and many more besides but this equation will get you most of the way there.

((BTUTEF)/38.5)18=ml lost

The Three variables in that equation are BTU which is the BTUs produces by your burner in an hour, EF the efficiency of transfer of those BTUs into your pot and T the time in hours of the boil.

Now how to determine your efficiency. I have made a formula that should let us do that with one empirical measurement. Now you will want to do this in conditions similar to your planed boil day or ideally you could do it during brew day and then add additional water to get you to your pre boil volume.

(BTU*T)/(((100-((ST/1.8)-32))*mL*4.2)/1055)/100=EF

Where BTU is the BTU per hour output of your burner ST is the starting temp in F of your water mL is the volume of the water you are heating in milliliters and T is time in hours that it takes for your water to reach a boil.

This should be done while stirring if not constantly then often and the final time should be counted when your water just starts to boil.

As an example let us say I bring 5 gallons of water to a boil over this burner rated 210000 BTU/h.

http://www.homebrewsupply.com/bayou-classic-banjo-cooker.html?gclid=Cj0KEQjwvOC5BRCb_8yNmZ_ls9IBEiQACTz8vqK1aBUVKy4a-gsW36KYFN-Zruopcm3O0JzHqAtKo3EaAgel8P8HAQ

Let us say that this water just finished mashing and is at 160 F and that it took one hour to reach a boil.

So our efficiency calculation would look like this.

(210000*1)/(((100-((160/1.8)-32))*18927*4.2)/1055)/100 = EF =.646

So we would put that EF value in the first calculation.

Now we plan on doing a 90-minute boil so we set T equal to 1.5.

((210000*1.5*.646)/38.5)18=95138mL=~25gal

So in this example we would need to add 25 gallons of liquid to the pot bringing total volume at boil time to 30 gal. Suffice it to say my example got a bit away from me.

Now all of this math assumes you are using a constant heat source in this instance we would have that burner running at full heat the entire time if you reduced the heat at all you would need to adjust the calculations accordingly by reducing the BTU value in the last equation.

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    Interesting, but to be fair the BTU ratings on those burners are generally all over the place. Burners clearly made from the same casting from different vendors will have listed wildly different BTU #s. And there is no standard for determining the BTU rating in that part of the market. There is no standardization of the regulator on the propane tanks usually either. That alone can wildly effect the BTU output. Tough to use equations when your inputs are all over the place. – brewchez May 16 '16 at 10:32
  • @brewchez indeed, it's much easier to just test your burner / kettle combo and establish its actual BTU rather than figuring its "efficiency". It takes 1 BTU to raise 1lb of water 1°F. Even still it's just a novelty to have this data and doesn't replace a boil test on a brewsystem – Evil Zymurgist May 16 '16 at 12:11
  • Yea this method is impractical.l as all hell but it should work. Also the efficiency that I am calculating is exactly that I'm just using c instead of f. – Gremwatch May 16 '16 at 12:33
  • Also I'm expressing it as a fraction of total theoretical btu output. – Gremwatch May 16 '16 at 12:34
  • @Gremwatch I built a high efficiency hlt, calculated it was 118% efficiency btu of propanes potential, boiled 20gal from 55 to 205 with 1lb of propane. I decided I needed a better scale lol, haven't restested since. – Evil Zymurgist May 16 '16 at 12:49
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Despite the fancy scientific answers, the correct answer is you brew a few times and measure how much you start and end with.

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  • I'm sure observation works, but wouldn't it be great to be able to predict this before the fact? – DJHellduck May 27 '16 at 11:55
  • And it would also be great if I was Superman! – Denny Conn Jun 12 '16 at 9:38
  • I agree, that would be awesome. I still want to get this to work, though ;) – DJHellduck Jun 12 '16 at 13:55

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