Re: D-3He Fusion fuel tanks for space travel

Hello Rick!

Your question contains a number of interesting parts, so let's break these
down in turn. I'm sorry in advance for the long answer!

D-3He are good fuels for a futuristic fusion reactor, as they produce
little in the way of neutrons following fusion, which are difficult to
control and dangerous to humans. That said, the reaction would be harder to
initiate compared to "easier" fusion reactions.

This fusion reaction produces ten million times the energy output of the
equivalent mass, for example, of burning coal. An energy-dense fuel such as
D-3He therefore makes good sense as a power supply for a spaceship. To put
some numbers to it, 1kg of fuel would produce approximately 350 terajoules
(about 10 million kWh) of energy. A terajoule is a million million joules.

However, this energy needs to be used to generate thrust, in order to
change the momentum (speed) of the ship, to enable it to go anywhere.
Current (purely theoretical) designs suggest that the plasma by-products of
the fusion reaction could be used to heat up and expel a reaction mass at
high velocities in order to achieve this thrust.

One commonly available reaction mass would be to use hydrogen, stored
cryogenically onboard the spaceship. Now, all rockets have a listed
"Specific Impulse" for their output, and a google around for hydrogen and
fusion rockets suggests a probably realistic Isp for hydrogen of about
10000, compared to about 400 for the space shuttles' liquid-fuelled
chemical rockets. (This number is the amount of thrust in kilogrammes that
one kilogramme of reaction mass can produce. Clearly, the higher the number
the better.)

You state that the spaceship should mass as much as an ocean liner, so
let's say it masses about 75000 tonnes.

Now for some rocket science! There's a famous equation called the Rocket
Equation. Here it is:

V = g*Isp*ln(start mass/final mass)

"V" is the change in velocity you can achieve. "g" is the acceleration due
to gravity ("g * Isp" is actually the exhaust velocity of our hydrogen).
"ln" is the natural log: your calculator probably has a button marked with
this. The start mass is the fuelled mass of the ship, and the final mass is
the mass once some amount of reaction mass has been ejected.

The higher V is, the more places we can go. For example, to go from low
Earth orbit to Jupiter orbit takes about 14.4 km/s. To get back home again
will take another 14.4 km/s. (Google is good to find a list of these "delta
v"s for various locations within the solar system.) So let's use 30 km/s as
a baseline for our rocket before it needs to be refuelled (actually
re-reaction-massed, as the mass of fuel for the energy source of the rocket
is really quite low). Putting these results into our equation, where X is
the final mass of the ship:

30000 = 10*10000*ln(75000/X)

Rearranging this equation (this might seem hard maths, but it really isn't)...

X = 75000 / 2.7183^(30000/(10*10000))

X = 55561 tonnes.

i.e. a reaction mass of 75000-55561=19439 tonnes is required. While this
might seem like a lot, it is actually quite low! A chemical rocket, such as
one using shuttle-type engines with an Isp of 400, would mean that X would
be 41.5 tonnes...that is, the final payload, once the fuel had been burned,
would be only half the mass of the space shuttle, and that more than 99.9%
of the chemically-powered ship would have to be fuel and oxidiser. This is
one reason why missions to the outer planets don't come back, have to use
gravity assists, and take their time.

So how much fusion fuel would our fusion rocket require?

We've seen that the exhaust velocity of the hydrogen is 100000
metres/second, and that nearly 20000 tonnes of hydrogen is required. We can
calculate the kinetic energy of this amount of reaction mass by the
following equation:

E = 0.5 *m*v^2

E = 0.5*20000000*100000*100000

E = 100000 terajoules = 286 kg of D-3He fuel.

(In reality more fuel than this would be required, as the fusion process
and energy spent on accelerating the hydrogen wouldn't be 100% efficient,
but you can see that the resulting fuel cost is likely to be just a tonne
or so.)

Finally, you stated that a year would be acceptable as a travel time. Now,
the above equations don't care whether the reaction mass is shot out of the
ship in a short duration, or over a long period of time - the end velocity
purely depends on the mass of the reaction mass and the Isp of the rocket.

However, I've suggested a destination (Jupiter) and listed the required
delta-v's to get there and back again. These velocities are the most
efficient ones to get to Jupiter. For much of the trip, the ship would be
coasting along what is known as a Hohmann transfer orbit. A mission to
Jupiter following one of these paths would take 2.7 years, with a similar
time to return. To get there quicker, as you desire, would require less
efficient (higher speed) trips, which in turn would require more reaction mass.

How much more? Let's assume you want to get to Jupiter (5.2 AU away) in
four months. In this scenario, we need to accelerate up to a distance of
2.6AU in two months, and then turn around and decelerate the final leg of
the journey over a similar timespan. We will be going at our fastest speed
at the turn-around point. This speed is given by:

V=2*s/t

Where s is the distance and t the time. Converting to SI (metric) units,

V=2*2.6*150000000000/(2*30*86400)

V=150000 m/s (150km/s!)

But we also need to slow down the same amount, and do the same for the
return journey after a four-month stay at Jupiter. Here the total delta v
would be 4*150 = 600km/s, compared to our 30km/s for the most efficient
trip. Recalling the rocket equation earlier, the required reaction mass for
this high-speed mission would be:

74814 tonnes, leaving only 186 tonnes for the fusion rocket and payload. 

This is a big reason for taking your time!

I hope the above has given you some ideas.

Regards,

Andy Goddard