Water mineral additions, losses to mash, and evaporation in the kettle. I.e., does water lost to the mash result in a proportionate loss of ions?

Question

Rephrasing my question:

I see three separate ways of calculating the final ppm of various minerals in the end product.

Say I begin with 9 gallons of DH20. After the mash, I lost 1.5 gallons and end up with 7.5 gallons in the boil kettle. After boiling, I lose 1.5 gallons to evaporation and end up with 6 gallons of wort.

Let's pretend I'm aiming for 200ppm of Sulphate. CaSO4 gives 61.5ppm of Ca and 147.4ppm of SO4 for one gram in one water. Therefore, 1.36g/gallon of CaSO4 will result in ~200ppm of SO4 and ~84ppm Ca (200/147.4ppm = ~1.36g CaSO4).

Do I multiply 1.36g by 9 gallons of water - my total starting water? This assumes both that I lost 1.36g*1.5gal to the mash and also 1.36g*1.5gal to evaporation. (Obviously salts don't boil off so this is incorrect)

Do I multiple 1.36g by 7.5 gallons of water - my starting water less that which I lost to the mash? This assumes that by losing 1.5 gallons of water to the mash, I am also losing 1.36g*1.5gal, leaving me with 1.36g*6gal CaSO4/gal.

Do I multiply 1.36g by 6 gallons of water - my starting water less that which I lost to the mash less that which I lost to evaporation? This assumes that the 1.5gal I lost to the mash did not also lose the 1.36g*1.5gal CaSO4, leaving me with 1.36g*6gal CaSO4/gal.

I'm reading Brewing Better Beer and Gordon Strong states "Homebrewers should be aware that you don't have to put all the salts in the mash, and that not all salts in the mash carry over to the kettle." He doesn't however make explicit that which I am asking -- namely if I lose X gallons in the mash do I precisely lose the ratio of salt to X gallons as well.

Answer 1

You are targeting 200 ppm for your final 6 gallons of wort, but you need that same amount of mineral present in 7.5 gallons of wort, pre-boil. Not the same concentration (200 ppm), mind you, but the same absolute amount, since it will concentrate as evaporation occurs.

To calculate ppm in 7.5 gal:

200 ppm x 6 gallons / 7.5 gallons = 160 ppm

To get 160ppm (with 1 gram CaSO4 per gallon giving 147.4 ppm):

160 ppm / 147.4 = 1.085 grams per gallon

For the total addition:

Here's where you account for your predicted 1.5 gallons left behind in the mash. Because the pre-mash water and the post-mash wort will both be 160 ppm (so will the water left behind) you have to have your full pre-boil volume + loss equal 160 ppm, hence:

(7.5 gallons + 1.5 gallons) x 1.085 grams/gallon = ~9.77 grams CaSO4

EDIT:

In regards to the last point in your post:

"if I lose X gallons in the mash do I precisely lose the ratio of salt to X gallons as well"

Sulfates and chlorides remain soluble during mashing so you should see a proportionate loss (~16.7% water retention would mean ~16.7% retention of Cl or SO₄ in the spent grain, i.e. the ppm won't change). Calcium, however, reacts with phosphates released by the malt and forms an insoluble precipitate which will be left behind in the mash or precipitate later. So you would hypothetically see a higher loss for Ca than 16.7% in your finished beer.

EDIT 2:

To elucidate the point above a bit more, here's the reaction that occurs between CaSO₄ and malt-derived phosphates that produces insoluble calcium phosphate:

3 CaSO₄ + 4 K₂HPO₄ --> {Ca₃(PO₄)₂}* + 2 KH₂PO₄ + 3 K₂SO₄

{...}* = insoluble/precipitate.

_{This formula is from The Practical Brewer}

I believe the reaction is basically the same with chlorides. You can see that calcium phosphate is the only insoluble substance produced, so you should only see a higher ratio of calcium being lost to the mash, and it will depend very much on your water and malt composition.

Answer 2

AIUI, none of the minerals will boil off, so they will be concentrated by the boil, and you should factor that into your computations.

I don't believe you need to take the mash-captured volume into account in any particular way. If you're treating the mash liquor, then you want that liquor to have a particular profile uniformly. I don't believe that mash-captured liquor has any more or less of the ions, compared to the mash runnings.

Answer 3

Your equation should only be adjusted for water loss due to boiling: for the above target use 1.088g per gallon assuming 7.5g of wort will boil down to 6g.

Losses in the mash are simply due to water sticking to the grains, the is no appreciable 'sticking' of salts to the grain, other than surface tension of water.

Furthermore even if your sparge water didn't have any salts, you could still recover the salts from the mash because the unidirectional flow through the grain bed will carry the salt down much faster than it can diffuse (using you don't do something silly like stirring the mash while lautering).

Answer 4

Most people put salts in pre-mash, according to the quantity of the final volume. So if you were brewing a 6 gallon batch, you'd calculate your salts based on 6 gallons, even though pre-mash you may have 9 or 10 gallons, and pre-boil you may have 7.5 gallons.

Answer 5

This is an interesting question, however I suspect there is no good answer because of the number of variables involved you would have to factor in reactions, precipitation, flocculation as well as additions of new ions from the grains and hops. Its also quite difficult to determine the exact concentration of various ions in your water supply. So why not determine the ion concentration after the boil (even though it will most likely change again after fermentation) if you are not determining it yourself then I would suggest you stick with your approximate formula.
(7.5/6)*1.36 = 1.7g/gallon of CaSO4