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 Home Brew Forums > Cost of Energy Experiment
12-21-2012, 06:21 PM   #11
wilserbrewer
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TOTAL COST OF ENERGY TO BREW A 5 GALLON BIAB BATCH:
kWh = \$1.37
LP = \$0.74
NG = \$0.19

Wow, 19 cents to brew a batch w/ NG, how much \$ was spent heating up the neighborhood while brewing the 19 cent batch. Excellent work, but your example assumes 100% of the combustion going into the beer. Happy accounting and brewing!

12-21-2012, 06:22 PM   #12
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Quote:
 Originally Posted by BadNewsBrewery As one of those engineers, I love the math. It was brought up, but the efficiency is important here. Near 100% of the energy from the electric element goes straight to the water. A burner pumps a lot of heat into the atmosphere before even heating the water.
BTW - been following your build very closely - very much like the design!
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12-21-2012, 06:30 PM   #13
passedpawn
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Just an FYI. This is a copy/paste from a post I made a couple of years ago here.

Some useful info:

• Burners are actually rated by "BTU/hr". "BTU" doesn't technically make sense since the longer you run the burner, the more BTUs expended.
• 1 pound of propane contains 21,591 BTU of energy.
• 1 gallon of beer weighs 8.34 pounds.
• (Pounds of beer ) * (Temp change in F) = number of BTUs.

To determine the BTU ( per hour ) of your burner:
1. weigh your tank (W1)
2. Run your burner full tilt for exactly 1 hour. Doesn't matter what you are using it for. Maybe heat up some water or something.
3. Weigh the tank again. (W2)

Burner Output (BTU/hr): (W1 - W2) * 21,591

To determine how many BTUs went into the beer
1. Determine weight of beer (see interesting info above) (W1)
2. Record the starting temp of your water. (T1)
3. Bring your water up to 200 (roughly) degrees. Best not to boil.
4. Record the final temp of water. (T2) Make sure to stir briefly before measuring to ensure homogenous mix.

Utilized BTUs: W1 * (T2 - T1)

Heating Efficiency
This is the efficiency of your burner / boilpot combination.

Efficiency = Utilized BTUs / Burner Output * 100%

Note: If using Natural Gas, replace the 21,591 with 20,161.

It's possible your burner is 56000 BTU. Flame burners throw most of their energy into the air and not into the pot.

Electric, BTW, is 100% effiicient: it all goes in the pot.
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12-21-2012, 06:31 PM   #14
kosmokramer
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And you also must figure intangibles.

Coolness factor of a blinged out electric system >= 5\$ per brew session credit
The look on peoples faces as they see said system = 2\$ per brew session credit
Sitting in the ac when its 105 outside or in the heat when its dumping rain = priceless

Im in socal as well and ran my first brew day last weekend.... it is seriously nice to brew in the kitchen with a laptop

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12-21-2012, 06:43 PM   #15
Effingbeer
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If time is money, it takes about 25 minutes for me to heat up 15g of strike water from 50-170 in my electric HLT. I guess that goes with the efficiency of heat transfer. No way I could do that with propane.

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12-21-2012, 06:46 PM   #16
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Electric may be 100% efficient on heating, though I'd argue some energy goes to heating the element itself, the cord, etc.. But I digress - even if we say it's 100% heat transfer to the water, you still must account for losses to the atmosphere via the kettle walls and lid.

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12-21-2012, 06:52 PM   #17
jeffmeh
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What about the cost of gasoline and the share of automobile maintenance attributable to trips to refill your propane tank?

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12-21-2012, 07:00 PM   #18
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I figured the cost awhile ago and found that if I brew after 10 at night I can use the stove for the strike water and sparge water. I use propane to boil but at worst it is a 90 minute boil and I get over 10 brews easy per bottle doing it that way. Any way it is dang cheap doing it after 10

It would be cheaper to brew it all on the stove but the boil I get with 6 gallons is rather wimpy to say the least. That and while I love the smell of the boil my wife just does not appreciate so much.

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12-21-2012, 07:02 PM   #19
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Quote:
 Originally Posted by BadNewsBrewery Electric may be 100% efficient on heating, though I'd argue some energy goes to heating the element itself, the cord, etc.. But I digress - even if we say it's 100% heat transfer to the water, you still must account for losses to the atmosphere via the kettle walls and lid.
Yes, but you are going to get the same losses for any heating method so it's kind of irrelevant if all you are concerned about is relative cost.
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12-21-2012, 07:08 PM   #20
bdjohns1
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Quote:
 Originally Posted by BadNewsBrewery Electric may be 100% efficient on heating, though I'd argue some energy goes to heating the element itself, the cord, etc.. But I digress - even if we say it's 100% heat transfer to the water, you still must account for losses to the atmosphere via the kettle walls and lid.
If you follow electrical code, thermal losses in wiring are negligible. Don't waste a brain cell.

All other things being equal, the thermal loss out of the kettle should be the same for identical kettles under identical ambient conditions. If I were to brew outside with NG on a day where the temperature and humidity were identical to my basement electric brewery (and no wind) all of those loss terms are identical and can be ignored.

Thermal mass of the element is similarly not significant when you have a ~1 lb element with 80+ lbs of water (1% error).
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