Re: The Dual-stage 4-grid Ion Thruster

Rémy MERCIER <Rmy.MERCIER.1zt49n@xxxxxxxxxxxxxxx> wrote:

> meiza Wrote:

>> Rémy MERCIER Rmy.MERCIER.1zqsxn@xxxxxxxxxxxxxxx wrote:
>> with isp = 19200s
>>
>> Huh, quite a great isp, but where do you need such a thing?
>...
> To go far and fast (Mars) we need a great tug: 70% mass = solair cells
> and structure, propellant=20% and cargo=10%

I actually did some calculations.
Let's assume we go to Mars and get back and can do that with 20 km/s of
delta v. If you assume a certain solar cell mass efficiency k(I assumed
100 W/kg), and required initial acceleration a_i, you can solve the mass
of the solar cells and the mass of propellant analytically as a function
of exhaust velocity. The rest is then left for the non-propulsion-related
mass which is proportional to payload. I've marked this as m_u. I made
some pictures:

http://users.tkk.fi/~tmaja/ion_optimality/20km.png

With very low acceleration, 36 days for a kilometre a second (going
10 km/s would take thus a whole year, not smart on a mars trip)
demonstrated by the black curve, the optimal exhaust speed is about
10^5 m/s, or in other words, specific impulse is around 10000 s.

Useful mass is a whopping 65% of the whole mass!
Fuel mass fraction would be exp(2*10^(4-5))=exp(0.2)= 1.22 ->
18 percent of total mass would be propellant.
Solar cell mass percentage would be 100% - 18% - 65% = 17%.

With higher accelerations, the isp optimizes to a lower value, as can be
seen from the red and blue curve. So it's not useful to increase solar
cell mass to 70% and put a high isp engine, at least not in this delta v
and with these assumptions.

Isp=2500 is about optimal for a craft which is a quicker accelerater,
only 40 days to 10 km/s. Useful mass would be about ten percent of the
craft.

The next picture shows values for a lesser, 12 km/s deltav.
http://users.tkk.fi/~tmaja/ion_optimality/12km.png
Isp:s optimize even lower.
And so on.

So I actually then maximized useful mass as a function of exhaust
velocity and deltav. I solved for the zero point of the derivative and
used some positivity constraints. Exhaust velocity is hard to get out of
this thing though, so I used a shortcut, using a variable
s = v_ex/deltav.
The optimality points lie on this curve:

s^2 * e^s = 2k*deltav / a_i


Resulting picture is here:
http://users.tkk.fi/~tmaja/ion_optimality/max_r_u.png
Here we can see that for a deltav of 10 km/s (10^4 m/s), an optimal
exhaust velocity would be about 1.3 - 2.5 times the deltav, depending
on required initial acceleration. (higher acceleration, lower isp).
That would mean isp of 1300-2500.

The part where the curves go below 1 is when any isp isn't sufficient to
give that deltav with that acceleration and power source density so the
payload goes negative.


So the original question goes: where is the isp 20000 s or exhaust
velocity of 2*10^5 m/s useful? The answer is here:
http://users.tkk.fi/~tmaja/ion_optimality/100km.png

Only in very low acceleration very high speed probes you get
least total mass with high isp:
Useful mass: 30%
Fuel mass: 40%
Solar cells / nuclear reactor 100W/kg mass: 30%
Final speed: 100 km/s
Isp: 20000 s
Initial acceleration: 1 km/s / 36d
Approximate time to full speed: 10 years.

Is ESA planning something like this? I didn't know at least.
Oort cloud explorer? Or have they invented better solar cells?
The power density assumptions are the least sound here, and
I'm happy to add to my knowledge.

All the Matlab source code is in the directory if you want to
look. I might have done some mistakes.
.

Relevant Pages

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