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- Thread starter thorongil
- Start date

bump... wow... 45 views and nothin' huh...

Does anyone know how to calculate this?

Does anyone know how to calculate this?

"You can calculate this yourself using ProMash.

1. Go into ProMash and open the Strike Temp calculator.

2. For Mash Tun Thermal Mass, Enter 0.

3. For Total Grain enter 0.00001 pounds. (It won't let you go any smaller - entering 0 will give you a division error.) This should make your water:grain ratio 120000:1.

4. For Total Water, enter 16.00 quarts.

5. For Desired Strike Temp, enter 170. (You can use any temperature but this will more accurately reflect actual mash temperatures.)

6. For Grain Temp, enter the current room temperature.

7. Heat 16 quarts (4 gallons) of water to 170 degrees and pour it quickly into the mash tun. Close the lid and seal it up, and let it sit for about 5 minutes.

8. Measure the temperature of the water in the mash tun.

9. Enter the temperature of the water in the mash tun in the "Desired Strike Temp" field in ProMash.

10. Now look at the "Initial Strike Water Temp Should Be" window. You will note that, with Thermal Mass set to 0, this number will be the same as the "Desired Strike Temp" field.

11. Now, in small (say 0.1 or smaller) increments, increase the Thermal Mass field until the "Initial Strike Water" field reads 170 (or whatever temperature you started with, if you didn't use 170 for some reason.).

12. Ta-da! Whatever the Thermal Mass field says is your mash tun's thermal mass!

Weight: 9 lbs

Specific heat: 0.300

I think that the thing to remember is that this needs to be dialed into your rig and process. Its not just enough to take somebodies word for it, you should follow a consistent routine and take your own measurements.

The units on thermal mass are energy/degree so in this case it would be 3.55 Btu/degree F

The heat lost by the water :

Qwloss=41.7lbswater*1 Btu/lbm/degree F*(164-157)degreeF=292 Btu

The heat gained by the mash tun is equal to the heat lost by the water (assuming no heat loss to the air) :

Qmlt=[m*Cp]mlt*(temp change of the mlt)=292 Btu

where [m*Cp]mlt is the thermal mass of the mash tun, which represents the mass of the mash tun [m] involved multiplied by the heat capacity of the mash tun involved [Cp]. Involved implies that only a fraction of the total mass of the mash tun actually absorbs heat in the short term.

To get 3.55 your mlt was assumed to be at 75F

[m*CP]mlt=292 Btu/(157-75)=3.55 Btu/degree F

I find this number helpful in predicting my sparge water temperature to hit a 168F batch sparge temp in the mash tun.

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