**Introduction**

OK, we have had a lot of background posts but we are finally getting to the crux of the biscuit ( ‘ ) !

We are now going to calculate how figure out how much water and grain will be needed to execute a recipe, using a batch sparge approach. One thing to know about batch sparging is that the highest efficiency will be achieved if both the first and second runoffs are the same volume. A tip of the hat to a home brewer by the name of Ken Schwartz, who not only figured this out a long time ago, but wrote down the explanation. If you are interested in the math behind this assertion, please click here.

As a reminder, here are the “givens” that will be used in this example:

Target Original Gravity for Wort | 15° P / 1.061 |

Final Wort Volume | 20 liters |

Water Boil Off Rate | 4 liters per hour |

Boil Time | 1 hour |

Water Absorbed by Grain | 1.5 liters per kilogram |

Water to Grain Ratio for Mash | 3.0 liters per kilogram |

Expected FGDB Yield | 78.9% |

Target Wort Mass (pre-boil) | 25.22 kilogram |

Sugar in Wort | 3.18 kilogram |

Water Required (in wort, pre-boil) | 22.04 kilogram |

**The Algorithm**

Here is the steps that will be followed:

- Estimate grain weight
- Calculate water requirements
- Calculate how much sugar will be extracted in first runoff
- Calculate how much sugar will be extracted in second runoff
- Add the results of steps 3 and 4
- Does the total from step 5 equal the amount of sugar called for in the recipe?
- If the answer to 6 is no, then go back to step 1, otherwise you are finished!

**Water Requirements**

In my approach to this you need to start with a guess as to how much grain will be needed. Since we need 3.18 kg of sugar and since the grain is supposed to yield 78.9% of its weight as sugar let’s guess:

3.18 kg / 78.9% = 4.03 kg of grain

The recipe requires 22.04 kilograms of water in the boil kettle, so we will need that plus how much water is absorbed into the grain. The amount absorbed by the grain is estimated to be 4.03 (grain weight) times 1.5 (estimated waster absorbed by grain) or 6.04 kg.

Since we want to runoff approximately one half of the water in each of the two runoffs, we will need to initially add one half of the water required in the kettle before boil plus the water absorbed by the grain, or 22.04/2 + 6.04 = 17.06 liters. After we finish the first runoff we will add the second half of the water requirement (22.04/2) or 11.02 kg water to the lauter tun before conducting the second runoff.

**Note:** If you mash in with 17.06 kg of water you will have a water to grain ration of about 4.2, which is probably about 2 quarts per pound. This might be higher than you want, if so, just mash in at the ration you want and add additional water just prior to runoff.

**Sugar Calculations**

Assuming you actually convert 78.9% of the grain to sugar (which you probably won’t but we will discuss that in a moment) then before the first runoff there will be 3.18 kg of sugar dissolved in 17.06 liters of water. The gravity is then:

(100 x 3.18) / (3.18 + 17.06) = 15.7 °P

Once the first runoff is drained into the boil kettle there will be a wort with 11.02 kg of water and a gravity of 15.7 °P. The sugar mass is therefore

Ms (1st runoff) = -(ºP x Mw) / ( ºP – 100 ) = – (15.7 x 11.02) / (15.7 – 100) = 2.05 kg

The sugar remaining in the grain is now:

Ms (grain) = 3.18 kg – 2.05 kg = 1.13 kg

When the second addition of water is made to the lauter tun, there will again be 17.06 kg of water but only 1.13 kg of sugar. The new gravity is

(100 x 1.13) / (1.13 + 17.06) = 6.2 °P

When this is runoff, the mass of the sugar that will be collected is:

Ms (2nd runoff) = -( ºP x Mw ) / ( ºP – 100 ) = – (6.2 x 11.02) / (6.2 – 100) = 0.73 kg

So the total sugar collected is

Ms (total) = Ms (1st runoff) + Ms (2nd runoff) = 2.05 kg +0.73 kg = 2.78 kg

While this represents a brewhouse efficiency of 87% (2.78 kg actual sugar / 3.18 kg targeted sugar) we have not achieved our required 3.18 kg of sugar in the boil kettle. What to do?

**Recalculate**

We needed 3.18 kg. We got 2.78 kg. So why not increase the grain by the ration of 3.18/2.78 (or 114.4%) ?

114.4% x 4.03 kg = 4.61 kg of grain.

What changes? Well since we have more grain we will lose more water to the grain. The new water loss is 4.61 x 1.5 = 6.91 kg. So the first water addition is:

11.02 + 6.91 = 17.93 kg water.

The sugar that should be converted in the mash is now

78.9% x 4.61 = 3.64 kg

Initial gravity is

(100 x 3.64) / (3.64 + 17.93) = 16.9 °P

And

Ms (1st runoff) = -( ºP * Mw ) / ( ºP – 100 ) = – (16.9*11.02) / (16.9-100) = 2.24 kg

Ms (grain) = 3.64 kg – 2.24 kg = 1.40 kg

New Gravity = 1.40 / (1.40 + 17.93) = 7.2 °P

Ms (2nd runoff) = -( ºP x Mw ) / ( ºP – 100 ) = – (7.2 x 11.02) / (7.2 – 100) = 0.86 kg

Ms (total) = Ms (1st runoff) + Ms (2nd runoff) = 2.24 kg + 0.86 kg = 3.10 kg

Closer, but not quite there. What if we scale up by the factor of 3.18/3.10? That is 4.73 kg of grain. If you run through the calculations a third time, the sugar in the kettle will be 3.16 kg. One more iteration results in the the target of 3.18 kg in the boil kettle, starting with 4.76 kg of grain. Since 4.76 kg of grain can be expected to yield up to 3.76 kg of sugar, the brewhouse efficiency is 85%.

That is not really the whole story. At the beginning of this example a conversion efficiency equal to the FGDB yield of the grain, and that is unlikely. Around 4-5% of the grain weight is water, and achieving perfect conversion may not be practical. I assume a mash conversion efficiency of 90%, typically, which means that instead of 4.76 kg of grain I would need perhaps 5.29 kg. But even that represents a brewhouse efficiency of 76%.

**Summary**

Obviously an iterative calculation such as this are well suited to an Excel spreadsheet. I plan to create one and use it to illustrate factors which can impact efficiency.

Stay tuned!