Noob Question: Big grain bills & mash volumes

I am brewing Belgian quads, some of which required multiple step infusions during the mash. Most of the mashing descriptions I have read recommend a 1.5 - 2 quart per pound of grain for the mash. If I follow that formula for the 10 gallon batch I want (using almost 29 pounds of grain), I end up with almost 15 gallons of water.....

I need some guidance on what volume of water to start with & add for the following mash schedule:

Step infusion 1 144F 20 minutes
Step infusion 2 154F 20 minutes
Step infusion 3 162F 20 minutes
Mash-out 174F 15 minutes

Any help greatly appreciated, thanks.

 

What type of equipment are you using? Are you able to heat the mash tun or are you adding hot water to raise the temp? There are calculators out there that will show you how much water at x temp is needed to raise the temp of x amount of water.

 

I am using a 120 quart cooler as my mash tun. I'm aware that I can use readily available calculators to figure out much water to add @ what temperature.

I'm still not sure how much water I should start with, given my grain bill is so big.

 

Are you batch sparging, or fly sparging? If you batch sparge, you can drop the mash out step.

 

Batch. How much water do I start with, given such a big grain bill? 60 quarts (15 gallons!) seems way too much.

 

You'd normally use about 1 to 1.5 quarts per pound of grain. There are benefits to thinner mashes, but you can start a step-mash at 1 quart per pound.
You'd then need to add around 1/3 quarts per pound to raise each of the steps you listed, so you could still comfortably stay under 2 quarter per pound without a mashout step.

 

Thanks. Given that I am brewing a 10 gallon batch, then, this is what I calculate:

1) 29 lbs grain X 1 quart = 29 quarts. = 7.25 gallons
2) 29 lbs grain X 0.33 quarts = 9.57 quarts = 2.4 gallons
2) 2.4 gallons X 2 infusion steps = 4.8 gallons
3) 7.25 + 4.8 = 12 gallons

Does that sound about right? I'm assuming I'll loose a gallon during the boil, for a final of 11 gallons.

 

You're forgetting grain absorption, sparge additions, and kettle losses (trub).
Grain absorption will be around 1/2 quart per pound or a bit over. Thats around 4 gallons for 29lbs of grain. Kettle loss (trub) could be anywhere from 0 (if you tip everything in) to a gallon if you leave behind lots of hot and cold break.
If we say 4 gallons of grain absorption + a gallon of trub, you now only have 7 gallons into the kettle. That needs to be made up with 5 gallons of sparge water. That's not much of a sparge, and will likely see a lower than normal efficiency, which means more grain is needed, which further lowers efficiency! Without knowing much about your process, a ballpark figure for a beer of this size would be 60% efficiency. This is why big beers are difficult to brew, especially with infusion steps. You can improve things by having a two hour boil - an extra gallon of boiloff means an extra gallon for sparging, which improves efficiency. You could also choose to make up some of the gravity using Extract. Another option is to mash in a bag, so that you can give it a good squeeze to get more wort out (less grain absorption). Personally, I prefer Belgian Quads with a healthy dose of Dark Syrup, which reduces the grain bill and makes the beer drier and easier to drink. Most big Belgian beers use quite a lot of sugar of some sort to dry them out.

 

Are you quite sure about needing multiple step infusions? Several people have done experimentation on this and found that they really couldn't tell the difference between the beer with all the infusions and one with a single mash infusion. Unless you really like the process, try one batch with a single infusion. Most recipe instructions are simply a record of what someone else did, not a complete requirement.

 

Thanks for the clear descriptions & advice. My process uses a 120 quart square cooler. Based on comments here, I'm thinking of these adjustments. Assume 12 (yep, I was off by a gallon above) gallons of boiled/finished wort desired:

1) single mash.
2) recipe additions: 1 lb DME, 0.75 D-90 syrup
2) 10 gallon initial mash = 6.5 gallons wort
3) 10 gallon sparge = 6.5 gallons wort
4) 90 minute boil 13 gallons for roughly 12 gallons.

Reasonable?

 

Too complicated for me. I use 1.3 quarts per pound of grain for the mash. I then vorlauf and drain the tun. I measure the volume of wort collected and I know my boil off rate. I subtract the wort collected from what I need for preboil then I know almost exactly what I need to sparge with. For an example, for a 5 gallon batch boiled for an hour. I need 7.25 gallons preboil. If I collect 2.25 from the mash, I know I need 5 gallons from the sparge.

 

Is there something special about this recipe that requires a 90 minute boil? I've been doing only 30 minutes on my beers. I probably should add more hops for the slight loss in bittering but the beer tastes good to me.

When you know how much sparge water you need, divide it into 2 approximately equal sparges to get the highest efficiency. Cool water for sparging works about the same as hot so you can save the trouble of heating it. The hot, wet mash warms the water a lot so it doesn't even take much longer to get the wort to a boil.

 

This is my process exactly.

 

This is actually what I do but I wanted to keep my answer simpler. Just measure the wort collected, whether in one sparge or 2 step sparge and collect the amount you need for preboil. This requires that you know your boil off rate.