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-   -   Microwave Electric Backup Heat Source? (https://ecorenovator.org/forum/showthread.php?t=1905)

pinhead 11-18-11 07:33 AM

Microwave Electric Backup Heat Source?
 
I often see electric resistance heaters as a backup. What about a microwave system?

Patrick 11-18-11 09:26 AM

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.

pinhead 11-18-11 10:32 AM

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.

Daox 11-18-11 10:41 AM

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.

pinhead 11-18-11 11:59 AM

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)

Ryland 11-18-11 12:05 PM

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.

pinhead 11-18-11 12:23 PM

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

pinhead 11-18-11 12:35 PM

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.

AC_Hacker 11-18-11 03:05 PM

Quote:

Originally Posted by pinhead (Post 17704)
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.

You will absolutely not get a COP of 1 or higher with a microwave oven.

Re-check your assumptions & calcs.

-AC_Hacker

pinhead 11-21-11 07:37 AM

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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