# Was DAWN's power-limited propulsion really limited by power used to accelerate the ions?

The discussion here uses the DAWN mission as a reference for the argument that the power used to accelerate the ions is what's important.

I'm curious how many watts were actually used to accelerate DAWN's ion thrust (e.g. $$P = IV$$), and how it compares to how many watts were actually used to sustain the plasma and unrelated to accelerating the ions.

My hunch is that the power for acceleration was actually trivially small compared to the power used to ionize and sustain the plasma, but I'd be happy to shown otherwise quantitatively.

## 1 Answer

According to wikipedia each Nustar engine on Dawn used 2100W, and achieved 92mN of thrust with an exhaust velocity around 30 km/s. So from the thrust and the exhaust velocity we can compute the Xenon mass flow as $$0.092/30000 = 3\times 10^{-6} kg/s$$.

Now each kilogram of Xenon has a kinetic energy of $$1/2 \times 30000^2 = 4.5\times 10^8 J/kg$$ so the power converted to KE of the ions is simply the product of these: 1350 W.

A little more power than that will go into accelerating ions in total, as some of them hit the grids, but roughly 2/3 of the power input ends up in the KE of the exhaust.

• This is way more efficient than I'd expected, you have indeed shown that my hunch was way wrong quantitatively! Your back-of-the-envelope calculation agrees very nicely with the total efficiency values given in Table 3 of Performance of the NSTAR ion propulsion system on the Deep Space One mission which max out around 0.63 well done! – uhoh Apr 13 '20 at 23:22
• so (though slightly out of context here) energy-limited missions (e.g. solar-electric) would benefit from propellants of heavy elements, while propellant-limited missions would benefit from light elements and their higher Isp. – uhoh Apr 13 '20 at 23:26
• @uhoh. Yes. This doesn’t really relate to in propulsion. To optimise total energy use over a mission you want to leave your spent propellant at rest in the original rest frame of the rocket. (Ignoring among other things orbital mechanics) so you want an Exhaust velocity equal to your rocket velocity at each point in the mission. – Steve Linton Apr 14 '20 at 6:53
• Well that's an interesting perspective, is that true when the object uses a significant fraction of its mass as reaction mass? Either way it sounds like something that could be demonstrated mathematically. I'll go off and think about it... – uhoh Apr 14 '20 at 6:59
• it seem's I've returned from that just now, thanks! – uhoh Sep 18 '20 at 2:46