Re: Obliterating the Rocket Equation with a Torusail

> If we're talking about a future where fusion power
> is used, I'd think the best way to get large amounts
> of 3He would be to manufacture it. I'm not sure what
> the most economical way would be. One possibility
> is to simply use the "waste" products of lithium
> deuteride fusion. About half the lithium gets converted
> to tritium which instantly fuses with deuterium (producing
> energy). The other half turns into helium-3, which
> mostly ends up as "waste".
>
> Lithium deuteride fusion power could actually be
> done today, with underground cave detonations
> powering something similar to geothermal.
> Political opposition to this form of nuclear
> power in today's world would be...hmm...significant.

Most of the proposed energy comes from He3 and 3H
reactions with D. So in effect these are manufactured
on-board from deuterium. Considered this way,
as a fuel deuterium has the highest energy per unit mass
if all the reactions are limited
to non-atomic-unit-conversions, and ignoring anti-matter.
That is neutrons and protons are not converted.
Also neutrons are absorbed so deuterium is recovered.

> Photovoltaics are subject to the same laws of
> thermodynamics as everything else. The
> theoretical thermodynamic limit to photovoltaic
> conversion is the same as a traditional heat engine.
> However, other forms of heat engine plausibly
> have better performance.

When a photon causes an electron to jump an energy gap,
in general there will be an energy surplus,
and a much longer wavelength photon will be emitted.
Eventually after successive transitions the remaining
photons will be of such long wavelength and
low "voltage" that they will thermalize into the
convertor, creating waste heat.

With a cyclic heat engine, a fluid is compressed,
heated by the heat source, decompressed into kinetic
energy, and cooled in a radiator. If the fluid is compressed
and heated too much, it will ionize and emit bremsstruhlung
radiation, losing energy. At one petawatt, delT = 5000 degrees K,
Cp = 1.4e4 J/Kg/degK for H2, get a mass flow rate of 14000
tonnes/second.
The maths doesnt work out at all. Given the limiting factor
is bremsstruhlung, make a virtue out of necessity and take
the light-bulb option, which uses this radiation as the power
transfer mechanism.

> Among all gases, hydrogen gas is the WORST
> insulator. At a given temperature, hydrogen
> molecules move faster than any other molecules.
> Of course, gases in general transfer heat mainly
> through convection. In order to turn a gas into
> an "insulator", you need to constrain it with some
> solid stuff to prevent convection.

The whole thing is spinning and the colder gas
is on the bottom. At the ends of the
spheres, about the axis, post more fluid
inlets to protect this surface more than
round the "equator". Convection requires heating
from "below". Neglecting fluid turbulence and
mixing, there won't be convection.

> I'm not sure why you think some things are not subject
> to thermodynamic efficiency limits. Certainly, the nuclear
> light-bulb concept is a heat engine. What you're describing
> is obviously a heat engine.

I think i've explained exactly what I mean above, without
getting into a semantic argument. I find an efficient
photo-voltaic convertor a lot more plausible than a
fluid cycle engine. I strongly considered this option.
When I calculated bremsstruhlung losses, it just blew
it away.

> Peltier's have low efficiency, and this isn't going to
> change, because of thermodynamics. There's just
> no free lunch in pumping heat in the opposite direction
> it wants to go.

I'm only proposing using such units to cool things, not
to gain efficiency. It's ok to move heat around, even
if it costs some energy. I find the photovoltaic
units of the nature I am suggesting more
implausible at 3000 degrees K then at 300 degrees K.

> For a journey this long, magbrake is easily enough
> to provide practically all of the braking. Roughly,
> a magbrake reduces ship's speed by one order
> of magnitude per decade. For this long journey,
> you could plausibly use the magbrake for three
> decades, bringing a cruise velocity of .45c down
> to .00045c.


I assume the interstellar medium to be largely un-ionised,
so a magbrake wouldn't work. But a laser beam might ionize it.
Assume 1 proton per centimeter-cubed, =1000000 per meter**3
Payload is 85000 tonnes.
Assume constant velocity of 0.167C
Assume time of 30 years
The amount of interstellar medium per square meter
of spaceship "frontage" is
density x velocity x time
= 75.3 milligrams per square meter.
To equal the payload, would need an area of 378 kilometers diameter.
It just seems improbable.

I have reconsidered the sound situation. Thanks for that
objection, as it means I have developed the concept
somewhat.

If a pellet is exploded in cold gas, then a shock
wave would certainly result. So the trick would be
to heat up the gas in the middle so it is such low
density that the penetration of the reaction products
will be over a significant distance. 10 meters would
be ideal. Considering the speed of sound, it would
be possible to achieve such a rapid rate that small
wavelength high frequency shocks would not result,
but it would seem like constant fusion over time
scales where the speed of sound is significant.

The overall rate of explosions should be such that the
radiating zone doesnt lose more than 10 percent of
its energy between explosions, so I get about 100000
explosions per second.
Each explosion has the energy of several tonnes of
TNT, but because it has such radiative penetration, it will
not cause shock, although the heating and cooling
of the radiating layer between explosions will cause ultrasound
of amplitude at source of pressure amplitude
of 10 percent of the pressure. This
will not impact efficiency, though it will cause
other problems.

On to the bombtrack. I get equations such as
Power = Mass * acceleration * velocity
distance = (velocity**2)/(2*acceleration)

Accelerating out of the solar system at 1 g,
to reach 0.17C, fuel has to be lined up for
a distance of 20 times the orbit of Pluto.
Nearing the end of the run, the power transfer
is 0.51 gigawatt per kg.
Given the 85000 tonne payload, this would
be 40 petawatts, which is 40 times the power
of my engine. Multiply the power and
payload by whatever factor you
conceed is necessary to slow it down.

Be away until the new year. Happy holiday.

.

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