How do you create a brewing recipe?
As home brewers the more we brew the more experience we gain and the more our confidence in our own ability to brew great beer grows. We perhaps start off brewing using malt extract, we then progress to full grain mashing and ultimately onto developing our own recipes. For me the most satisfying and interesting part about home brewing is developing my own recipes rather than using those that are published in most home brewing books. Although these published recipes are perfectly acceptable developing your own brewing recipes really gives you the freedom to innovate and express yourself as a brewer.
The question is how do you create a brewing recipe? How do you move from using someone else’s recipe to developing your own recipes for your own unique beer. Hopefully in this series of articles we hope to show you how easy it is to create a recipe and give you the confidence to try it for yourself.
Let’s get started and show you how to create a brewing recipe.
First off we will start looking at the malt side of the recipe and how we calculate the quantity of malt we need to use to meet a target original gravity (OG).
For me the first part of generating a recipe is to consider the type of beer I want to brew, which indicates things such as target OG, bitterness and colour. To calculate the quantity of malt required we need to decide on the final alcohol we want to achieve, which provides us with an initial target OG which we want to achieve in the fermenter.
For example if I want to produce a 4% abv beer, as a rough rule of thumb, I am looking to target an OG of 1.040. If it is a 5% abv beer I am looking to achieve a target OG of 1.050.
The calculations that we employ will therefore all target the initial OG of the beer that we want to brew. Let’s use a worked example for a 5% abv beer.
Having decided on an OG, in this case 1.050, how do we work out how much malt to use? The answer to this question is actually easier than you think and involves something often referred to as the LDK value.
What is the LDK value of your malt?
If you take a look at your malt analysis certificate you will find the analytical parameter IOB extract. Usually this is measured on a coarse or 0.7 mm grind and can be reported on a dry basis or an as is basis. If you take a look at the units that the value is reported in you will notice that it will say something like L deg/kg which stands for Litre Degrees per Kilogram of malt and hopefully you can see why some people refer to this as the LDK value of malt. The figure gives us an idea of the extract that is possible per kilogram of malt used. But how do we use this figure?
The LDK value is actually a very useful figure and provides us with a simple way of calculating the amount of malt required to produce a certain volume of wort at a specific OG.
Going back to our worked example where I want to produce a 5% abv beer. In this example I am targeting an OG of 1.050 and let’s say I want to produce 25 litres of wort. To keep it simple I only want to use a single malt type, let’s say ale malt, for my recipe. To summarise:
Beer Type – Ale 5% abv
Target OG – 1.050
Target volume – 25 litres
Malt type used – 100% Ale Malt
The calculation that we use is:
(((OG – 1) x 1000) x Volume)/ LDK
This might look a bit daunting but it is probably a lot easier if we break this calculation down into bite size chunks.
The first thing you need to do is check your malt analysis certificate and find the IOB coarse extract or LDK value. Generally you should be looking to use the as is value for the coarse extract as this takes into account the moisture content of the malt. It is generally around the 300 litre degree/kilogram or 300 l°/kg.
For this example we will use an IOB coarse extract or LDK as is of 302 l°/kg. To work out the malt required the calculation is as follows:
Malt required = (((Original Gravity – 1) x 1000) x Volume wort required)/ Coarse extract as is
Therefore for our recipe for 25 litres of wort with a gravity of 1.050 the calculation is:
Malt required = (((1.050 – 1) x 1000) x 25)/ 302
Malt required = ((0.050 x 1000) x 25)/302
Malt required = (50 x 25)/ 302
Malt required = 4.14 kg
This calculation assumes that we are 100% efficient when we are mashing but this is very rarely the case. Many commercial brewers would aim for at least a 90% brewhouse efficiency which means for every kg of malt that they use they are able to solubilise 90% of all the extractable material in the malt. They compensate for this inefficiency by adding additional malt to achieve the desired wort volume and gravity. If we work on a brewhouse efficiency of 90% for our home brewing this would mean multiplying the kg of malt figure by 100/90 or 1.11. Therefore the kg of malt required is:
Malt required = weight of malt x brewhouse efficiency
Malt required = weight of malt x (100/ brewhouse efficiency)
Malt required = weight of malt x (100/90)
Malt required = 4.14 x (100/90)
Malt required = 4.6 kg
Using this simple method we now know that the amount of malt required to produce 25 litres of a simple ale with a starting gravity of 1.050 is 5.175 kg. You can make a more complicated grist by introducing coloured malts and other grains such as wheat and maize as long as you know the coarse extract or LDK value for each of these grain types.
We hope that you enjoyed this article. If you would like to understand the basics of creating a brewing recipe then check out our new book, details below.
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