I was gonna use 1.5kg of LME and then make another 1.5kg with some MO grain. I've only got a 5L pot but I could get another to make 10L total.
It sounds stupid but to get another 1.5kg would I just add 1.5kg of MO to the pot or do I need more?
Topic partial-mash homebrew
Category Mac
I am not sure I understand your question, so I am assuming you mean to ask, "How much Maris Otter do I need mash in order to get the extract equivalent of 1.5 kg of liquid malt extract?" I also assume you know that Maris Otter is a grain that must be mashed.
LME has an extract potential of 292 gravity points per kg per liter (PKL), meaning that one kg of LME dissolved in 1 liter of water will give you a wort with a specific gravity of 1.292. (Incidentally, DME has extract potential of 351 PKG).
Maris Otter has a theoretical extract potential of 317 PKL, but if you assume a mash efficiency of 65%, that is 317 * 0.65 = 206 PKL.
Thus, to replace/equal one kg of LME with Maris Otter, you would need 292/206 kg of Maris Otter, or 1.42 kg of Maris Otter.
Your answer: to replace/equal 1.5 kg of LME, you would need 1.42 * 1.5 kg, or 2.13 kg of Maris Otter.
65% efficiency is a good guess if you don't know your efficiency, but if you have a different, known efficiency, you could repeat the math above with your actual efficiency estimate.
I'm not sure you're asking the question in the right way, e.g. what your intended sort is and what OG you're targeting, your batch size etc.
For what it worth, accordingly to a Beersmith calculator, 1.5 kg LME should give you ~0.024 gravity points for 20 liters. To get the equal amount from grain, you need 2.5 kg of grain with efficiency 65%, give or take. A 10 liters pot would be handy.
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