On Feb 26, 9:24=A0am, CharlesRCap...@[EMAIL PROTECTED]
wrote:
> In regards to your suggestion to place the thermoelectrics on the
> radiator directly:
> That's an interesting idea about mounting the thermoelectrics on the
> radiator directly. I might use that if the operating temperature
> wouldn't be too high to melt them or worse, reduce their efficiency.
Note that your radiator's operating temperature must be lower
than the temperature of the reactor (and much lower in order to
get any sort of efficiency). You're getting bogged down in the
details of specific heat engines and not looking at the big
thermodynamic picture.
> What I am thinking is a radiator with a gold (or some unobtanium
> alloy) reflector set up to direct the radiation in a small cone. The
> actual cone itself does not really matter because I'm assuming that
> all of the energy is hitting the reflector and not radiating directly
> to space. So the narrowness of the cone is only important for
> determining the size of the reflector which I have not gotten to
> calculating yet.
Generally, the reflector is really big. For example,
let's suppose the initial radiator is a 10m diameter circle,
and the desired cone is 2 degrees wide. This means
the distance from the radiator to the mirror needs to be
about 280m. The mirror's radius of curvature to focus at
280m is 560m, resulting in an overall diameter of around
1000m. For this example, the overall diameter needs
to be two orders of magnitude bigger than the initial
radiator (four orders of magnitude greater area).
> The reflector will be actively cooled at to about
> 133K (see below for details) and the coolant pumped to the interior of
> the ship for compression back to operating temperatures. (On the way
> it passes through some thermoelectric that radiate directly to space.)
> The hull at 50K with an emissivity of around 0.99 is roughly 0.35w per
> square meter radiated. The reflector at 133K with 0.02 emissivity is
> roughly the same.
The amount of power is not a good measure of how
easy it is to detect. For temperatures like these,
significantly greater than the 3K background, the
detectability will be roughly proportional to the
number of photons emitted.
> So if we actively cool the reflector to 133K and
> pump the coolant through the electrothermal devices radiating directly
> to the 3K background (does not need to be on the back side of the
> reflector since we would need to actively cool it anyway and can pump
> the helium somewhere else) then we might not get 60% efficiency but
> it's still something... 130K differential is what kind of efficiency,
> 50% maybe? I don't know the math, I've been making it up for now.
At this point, efficiency isn't the problem. Power level
is the problem. You don't just get a 3K heat sink for
free. You need a really really really big radiator to dump
heat into the 3K background, and this radiator will be
radiating away at some temperature above 3K. Unless
you choose a temperature significantly higher than 3K,
the radiator will be HUGE. In any case, it's this radiator
which will be the most visible thing on your ship.
I get the feeling that your essential design approach is
flawed in some sort of fundamental way. You seem to
be looking at sources of heat as potential places to
get energy from, but the real problem that you need to
be tackling is radiating away waste heat. You can't
just turn waste heat into energy, unless you use an
even bigger more visible lower temperature radiator
as a heat sink.
The directional radiator in your design seems to just
be a complex unneeded and unwanted distraction.
Isaac Kuo
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