Re: 2.2 KW sustained thermonuclear power?

In article <tY8Bl.285823$OT2.194588@xxxxxxxxxxxxx>,
ns_cjrs@xxxxxxxxxxxxxxxxxxxx says...
I now have dioreah. Have I got radiation sickness from my reactor?

I had a problem with my eyes from UK light but that has cleared up now.

I did the heat experiment again. At a distance of 220 mm from the reactor
the temperature registered was 214 C. According to stefan's law of radiation
that works in reverse the heat falling on it is given by 5.6E-8 (T1^4-T2^4)
= 5.6E-8 (5.62491E10-0.737E10)=5.6E-8 x 4.8879E10 = 2737 watt /m^2 over an
area of 0.0025^2 *3.14*2737 = 1.9625E-5 = 0.0537 watt. So over a sphere at
radius 0.22 meter the total area is 4/3 pi R^2 (I think) = 0.2 m^2 so the
total power is is 0.2/1.9625 E-5 * 0.0537 = 546 watt and this is the power
out so the excess power is 346 watt. Someone please correct me on the
surface area of a sphere.

The area of a sphere is 4*Pi*r^2.

But you're misinterpreting your ir thermometer's data; it gives the
temperature of the surface of whatever you're measuring (after taking
account the emissivity of the surface). There's no distance or area
calculations involved in it.

Then take that temperature and use the radiation law to get the heat
flux _of the apparatus_.

Then multiply your heat flux by the surface area _of the apparatus_ to
find your total heat output.


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