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Ask the Experts: Mash Efficiency

Homebrew expert Brad Smith, author of the Beersmith homebrewing software and the voice behind the Beersmith podcast, tackles a question about mash and brewhouse efficiency.

Brad Smith Dec 14, 2017 - 6 min read

Ask the Experts: Mash Efficiency Primary Image

What’s the difference between mash and brewhouse efficiency in all-grain brewing, and how can I improve my efficiency?

I’ll start with an explanation of mash efficiency. First, you need to understand that each individual grain you use has its own “yield.” The yield is a percentage that represents the percent of the weight of the grain that can be converted into sugar under ideal laboratory conditions. An average pale malt might have a yield of about 80 percent, meaning that 80 percent of the sugars could be converted and extracted in an “ideal” laboratory setting.

A real-world brewing system outside of a laboratory does not achieve this perfect number. In fact, a typical brewing system is only going to get perhaps 80–90 percent of the potential sugars extracted during the mash, meaning that 80–90 percent of the yield will be achieved in the runnings coming directly from your mash tun. The portion of the “ideal” number your system extracts during the mash phase is called the mash efficiency.

Of course, the mash itself is only one step in the brewing process. After mashing, we may add top-up water, boil, chill, and transfer the wort, all of which results in some further losses due to trub loss, chilling, and transfer losses into the fermentor. These additional losses result in a lower original gravity of the wort into the fermentor than we achieved coming out of the mash tun.

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