Extract to All Grain conversion help?

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michael_mus

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Hey HBT,

Recipe I have calls for 6lbs Munich LME. How do I convert that to grain?
 
The way to do this that seems most logical to me is to convert into gravity points and then convert the points back to grain compensating for efficiency.

6 lbs of LME = 210 gravity points
Munich malt is 34 ppg. At 75% efficiency that's 25.5 ppg
210 / 25.5 = 8.2 lbs

Or you could solve it algebraically and then plug and chug.

6 lbs * 35 = n * 75% * 34
n = (6*35)/(75%*34)
n=8.2

If you want to assume 75% efficiency then just multiply by 1.4.

EDIT:
That's half base malt and half Munich. See the data sheet here:
http://www.brewingwithbriess.com/Assets/PDFs/Briess_PISB_CBWMunichLME.pdf
 
Also a general rule of thumb that is often repeated here:
1 lb grain = 0.6 lb DME = .75 lb LME

So in this example you'd get 8 lb grain for 6 lb LME.
 
6 lbs of LME = 210 gravity points
Munich malt is 34 ppg. At 75% efficiency that's 25.5 ppg
210 / 25.5 = 8.2 lbs

Also a general rule of thumb that is often repeated here:
1 lb grain = 0.6 lb DME = .75 lb LME

So in this example you'd get 8 lb grain for 6 lb LME.

Brilliant. Thank you both. In case you are wondering, I'm AG engineering based off the partial-mash recipe for the Sunset Blvd Amber Ale in my sig. It calls for 6lbs of Munich and then about a pound of specialty grains.

I had seen the simple grain to DME/LME math before but what I've been reading regarding Munich LME made it sound like a mixture of at least two different malts. I searched around like crazy but couldn't make sense of it.

Exciting to hear that it's simply Munich grain.

Thanks!
 
Ahhh, I understand the question now. This link from Briess says their Munich LME is 50% Munich malt and 50% "base malt" (pale 2-row, I guess):
Munich LME

Thanks - I eventually found that too. Looking forward to putting all this extract talk behind me real soon.
 
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