Re: Obliterating the Rocket Equation with a Torusail


pgarrone@xxxxxxxxxxx wrote:
>>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".

>Most of the proposed energy comes from He3 and 3H
>reactions with D. So in effect these are manufactured
>on-board from deuterium.

I am addressing your objection to using D-He3 fuel
on the basis of lack of availability. Obviously,
you can manufacture He3 from Deuterium by your own
assumptions that D-D fusion is a viable power source.
An even cheaper way might be get the He3 "for free"
from the waste products of lithium deuteride fusion.

>Considered this way,
>as a fuel deuterium has the highest energy per unit mass

But not necessarily the highest energy available from
your available natural resources. If you've got
lithium around and you've got deuterium around, then
you can get more energy from lithium + deuterium than
you can get from deuterium alone.

Considering how much easier lithium deuteride fusion
is than either D-D or D-He3 fusion (we can do it today),
it's not crazy to suppose future power reactors which
use lithium deuteride fuel and produce He3 as waste.

So basically, your objection to the use of He3 fuel
based on natural scarcity is not at all a given.

>>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.

If you're clever about it, the longer wavelength
photons get radiated out into space directly. This
is possible with a solar panel because the Sun only
subtends a small fraction of the sky.

But in any case, you're limited to the laws of
thermodynamics. In order to get any net result
from radiating a photon, you have to be radiating
it in some direction where you are NOT also receiving
radiation of the same (or higher) temperature.

>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.

Where do you think that radiation goes? It doesn't
just magically disappear into nothingness, it's
turned into photons. Those photons can be reabsorbed
by the working fluid, if you want to set it up that way.

>>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.

You can't neglect fluid turbulence and mixing,
least of all with hydrogen gas. Convection is
typically the dominant mechanism of transfering
heat in a gas, even if the "hot" side is "above".

>>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.

You reached completely wrong conclusions based on
completely wrong assumptions.

>>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.

The cost in energy is roughly equal to the amount
of heat energy being moved--and the result is that
you STILL have to remove heat from whatever is on
the "hot" side of the peltier.

Peltiers are just a horrible way to cool things,
if you can use just about anything else instead.

>>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.

Doesn't matter. A magbrake creates its own
magnetosphere, which ionizes interstellar medium
with its bow shock.

>To equal the payload, would need an area of 378
>kilometers diameter. It just seems improbable.

A magbrake creates its own magnetosphere, which
the particle "wind" blows up to many times its
own diameter.

>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.

The great thing about bombtrack propulsion is
that it is self-cooling. You simply don't absorb
energy in the first place. The thrust is produced
by interaction between charged particles and a
FIXED magnetic field. To a rough approximation,
this is a conservative interaction so no energy
is absorbed by the starship.

Most of the waste energy of the bombs in the
form of neutral radiation is simply vented into
space, never even reaching the starship. It's
the tiny fraction which hits the starship which
it must contend with. But this could be many
orders of magnitude less than the amount of
power going into thrust, so it's a huge win.

The fact that most energy from a bomb is in the
bomb products cuts out one order of magnitude.
The distance between the detonation of the bomb
and the starship cuts out another order of
magnitude (as the starship goes faster, the
detonations need to be further forward from
the ship). The fine geometry of the starship
cuts out another three orders of magnitude.

So, the waste heat the starship must reject is
on the order of 1/100,000 as much as the amount
of power going into thrust. What's more, the
overwhelming majority of this energy is absorbed
by the sail lines themselves--already a nearly
ideal geometry for naturally radiating the
waste heat.

Amazing power levels are plausible when most of
your waste energy is simply vented directly
into space.

Isaac Kuo

.

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