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11-18-11, 07:33 AM | #1 |
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Microwave Electric Backup Heat Source?
I often see electric resistance heaters as a backup. What about a microwave system?
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11-18-11, 09:26 AM | #2 |
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Microwaves work by exciting the molecules within the food. To use it for heating, you will probably need some medium to absorb the radiation then transfer it to the air. If you have water coming in to your system, you could use a microwave to heat that. Magnetrons are about 64% efficient, but you could recoup some (all?) of the loss by mounting the magnetron in the air stream if it's a forced-air unit. IMO, it would probably be more trouble than it's worth.
Last edited by Patrick; 11-18-11 at 09:39 AM.. Reason: added info |
11-18-11, 10:32 AM | #3 |
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This number of 64%, do you know what resistance would be in comparison? I am thinking mostly of a toaster compared to a microwave. An amount of water: Toaster≈800W Microwave 1000W and I would think the microwave would be about 2 minutes to boil vs 10 minutes.
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11-18-11, 10:41 AM | #4 |
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Electric resistance heating is 100% efficient. In a toaster you are loosing tons of heat out the top is all, so not all of it goes to cooking.
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11-18-11, 12:05 PM | #5 |
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If you want to heat a room quickly, instead of heating water or something else first, then an electric infrared heater would be the way to go, or good old heat lamps because they both will warm the skin instead of warming the brain and other high water parts of the body.
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11-18-11, 11:59 AM | #6 |
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So I was thinking an electric kettle and a microwave would be a better comparison, but I do not use either often enough to guess at the time to boil similar amounts of water. I am off to test it now. (the boss took his kettle home, so I will not be testing this afterall)
Last edited by pinhead; 11-18-11 at 12:02 PM.. |
11-18-11, 12:23 PM | #7 |
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Okay, I will go searching for the SEER or COP calculations, but here is the data from a microwave.
1 Liter of water. 1.855Kw / 60minutes = 0.030Kwh Starting temp: 16.5°C T=0 33° T=1minute 42.6° T=2minutes 53.2° T=3minutes Last edited by pinhead; 11-18-11 at 12:24 PM.. Reason: volume of water |
11-18-11, 12:35 PM | #8 |
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So I am completely lost on the calculations, and at the whim of Google.
But I am getting about COP of 1.4 without the circulation pump required. |
11-18-11, 03:05 PM | #9 | |
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Quote:
Re-check your assumptions & calcs. -AC_Hacker
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11-21-11, 07:37 AM | #10 |
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So 1 liter is 1000 grams. The heat capacity of water is 4.175 J/gk.
Over the three minutes the rise was 36.7°C, 4.175j/gk x 36.7k x 1000g = 153222j 1855W= 6678000 j/hour and it was .05h -> 333900j Giving an efficiency of 45.8% Not too sure what math I was using friday. |
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