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Old 12-21-2012, 07:18 PM   #21
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Originally Posted by bdjohns1 View Post
If you follow electrical code, thermal losses in wiring are negligible. Don't waste a brain cell.
Might want to measure before you make that assumption.

Loss due to resistive losses in power transmission:
I've measured the power (voltage and current) to my element. After line losses and control box losses, I get 5140 Watts out of my 5500W element.

Loss due to ambient:
Also, based on a bunch of empirical testing, I lose almost exactly 10% of the applied energy to air from my uninsulated boil pot (keggle - 60-degree day). So, I only get about 4600W of the power towards the boil. This 10% couple be recouped by insulating the keggle, which you can do easily. Not so easy with a flame though.


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Old 12-21-2012, 07:58 PM   #22
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Love that I have created some discussion around this. As i stated at outset...no doubt electric is cooler. Incidentally, a gallon of propane weighs 4.4# at 60* F. 1 gallon of Propane will produce 91k BTU's.

Passedpawn: like you said, how often (if using a burner) do we really have it going full bore? Rarely. Just got back from propane refill - 3.49/ gallon ended up with not quite a full fill. Will start off the brew day with 4.5 gallons and will measure along the way.



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Old 12-21-2012, 09:08 PM   #23
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Here's a good article explaining how to boil water.

http://www.treehugger.com/clean-technology/ask-pablo-electric-kettle-stove-or-microwave-oven.html

The electric kettle is by far the most efficient, but our argument is more about cost vs. efficiency. An argument can be made about on-site utilization of fossil fuels vs. burning at a power plant, then suffering further distribution losses before we plug in our pretty little tea pots.

Without natural gas on site, electric is the clear winner for me. However, it's going to take a heck of a lot of beer before I see a return on my investment.

That, my friends is what you call a win-win.

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Old 12-21-2012, 09:09 PM   #24
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The pennies difference on actual energy costs are of no real significance, the time and convenience factor are the savings. My current number for the cost of my time is $200 per hour when deciding on how to use it. If I have to spend a half hour getting a propane tank filled, I just spent $100 not to mention gasoline and whatever other errand my wife will sneak in on me :-). Having plenty of electric power to heat up my gear whenever I want in the comfort of my shop in a fraction of the time it takes on a burner in my driveway has been well worth the money I spent building it. I now have extra time to spend on other activities, it was a no brainer for me.

I can't believe people are trying to figure out how much heat they are loosing through their power cords. That's some heavy ****. lol

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Old 12-21-2012, 09:14 PM   #25
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If you're going to take a scientific and in-depth look at something, you may as well be scientific and in-depth the whole way. You can ignore parameters all you want, as long as you're doing it CONSCIOUSLY and for a reason. So yes, I maintain that 100% of the element's power doesn't go to actually heating the water (not considering losses from the kettle to the atmosphere) - some goes elsewhere. You can chose to ignore it in your calculations should you chose, but make the choice knowingly.

The other point of discussion - an X rated element and a Y rated burner do not actually produce X or Y for values. Measuring the amount of BTUs consumed by doing a fixed heating and then measuring the tank, and comparing that to the rated value of he burner, does not give you an accurate number. Again - you can chose to ignore this, but it's worth discussing. Same with the element, as [b]passedpawn[/b/] says - my 5500w element was likely not made in a high control, low tolerance laboratory... And my power supply isn't exactly 240.00v either.

Truly, the best way to do this would be to weigh a propane tank, complete a brew in whatever environment you normally brew in. Assuming it was an average day (IE, not snowing during a hurricane), you'd be able to weigh the consumed propane and know your cost. Do the same exact brew in whatever location you do your electric brewing in. Get one of those 'kill-a-watt' energy consumption meters and plug your panel into that. Measure the power consumption during your brew. Now you know exactly how much propane vs exactly how much electricity it takes to brew that one batch with your given equipment. Hypotheses are great, but there's no match for experimentally gathered empirical data.

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Old 12-21-2012, 09:16 PM   #26
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mateomtb - I'm guessing it took you at least 10 hours or more to research, acquire parts, construct, and test your electric setup, above what it would have taken to do a propane setup. At your rate of $200 / hour, that's $2,000 worth of money... or 20 some trips to get a propane tank filled. Don't forget to count it both ways...
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Old 12-21-2012, 10:32 PM   #27
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Get one of those 'kill-a-watt' energy consumption meters and plug your panel into that. Measure the power consumption during your brew. Now you know exactly how much propane vs exactly how much electricity it takes to brew that one batch with your given equipment. Hypotheses are great, but there's no match for experimentally gathered empirical data.
I've got a Kill-a-Watt (my keezer costs $1.22 / mo. to run!). It won't work for a 240V 30A appliance. I suppose you could juryrig it into one leg of the 240 though, but the current would burn it up pretty fast I'd guess.

It's not hard to figure out the cost of the electricity though. Once it gets to the element, ALL the energy goes into the water.
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Old 12-22-2012, 12:39 AM   #28
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wow, you guys pay a lot for electricity!!!

Last bill $0.074380/kwh

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Old 12-22-2012, 12:43 AM   #29
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wow, you guys pay a lot for electricity!!!

Last bill $0.074380/kwh
California, coming soon to a state near you....................
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Old 12-22-2012, 01:05 AM   #30
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it seems like the answer is to build a fume hood and exhaust system in my basement....so I can have a natural gas fired brew system???

The build would be cheaper....and the operating cost lower



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