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I am trying to figure out if Xenon or Krypton produces more thrust for a given weight flow rate of fuel when the engine operates at the same discharge voltages. The information I find is mixed or I am not interpreting it right.

Some concrete info: Krypton has lower atomic mass and higher ionization potential/energy.

Here they indicate that the lower atomic mass could potentially produce a 25% increase in specific impulse due to the increased propellant exit velocity of lighter ions.

There are 2 stackexchange articles that are related to this: here and here.

They indicate that the higher velocity of Krypton (at a given acceleration potential) means higher Isp, but lower overall thrust and/or energy efficiency. (I guess because you need more power to ionize Krypton).

Does this then mean that while Krypton needs more power to operate, it achieves higher specific impulse because the electrons will have higher velocity when ionized? But then why is the overall thrust for Krypton lower?

Wouldn't the above mean that for the same flow rate of fuel, at the same discharge voltage, Krypton produces more thrust?

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    $\begingroup$ I recall reading (cannot find the source now, so won't answer) that Krypton overall is less efficient, but much cheaper and more available. $\endgroup$
    – geoffc
    Dec 21, 2022 at 14:32
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    $\begingroup$ @geoffc "Less efficient"? In which sense? Energy-wise, Mass-wise, Cost-wise, Lifetime-wise? $\endgroup$
    – asdfex
    Dec 21, 2022 at 15:34
  • $\begingroup$ Wouldn't the increased exhaust velocity be due to the higher charge-to-mass ratio of krypton ions as opposed to xenon (because of the lower proportion of neutrons in lighter elements), rather than directly due to krypton's lower mass? $\endgroup$
    – Vikki
    Dec 22, 2022 at 21:21
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    $\begingroup$ @Vikki That would be the case if ion thrusters used fully ionized ions (bare nuclei) - but they use single ionization only (1+ ions) $\endgroup$
    – asdfex
    Dec 23, 2022 at 9:43

2 Answers 2

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All your analysis is fully correct. At the same voltage and mass flow rate, Krypton produces more thrust. But you're missing one very important point: None of the existing applications is limited by flow-rate or voltage.

The limiting factor is always the power available for propulsion. And, as power scales with the exhaust speed squared, it needs to be higher for Krypton than for Xenon to get the same thrust. Or vice versa, for a given amount of power, Krypton produces less thrust.

On top of that, there is the additional factor of the higher ionization energy which needs more power - but these ~2eV are only a minor factor compared to the ~2keV kinetic energy per ion.

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    $\begingroup$ Thank you that is very helpful. Could you clarify what you mean by "compared to the 2keV E/ion? Do you mean the additional power needed to ionize is not comparable to the overall extra power you need to reach the same thrust? I got a bit confused. What do you need the extra power for if not to ionize? $\endgroup$ Dec 22, 2022 at 10:33
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    $\begingroup$ The extra power goes into kinetic energy of the exhaust. Lighter particles means higher kinetic energy to get a fixed amount of thrust. $\endgroup$
    – asdfex
    Dec 22, 2022 at 10:54
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    $\begingroup$ Note that despite xenon's 12 eV ionization energy, about 200 eV are needed to ionize one Xe+ and maintain the plasma. The extra 2 eV ionization energy of Kr could go much further than what you'd think. $\endgroup$ Dec 22, 2022 at 23:36
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    $\begingroup$ @Playstation_waifu: Thrust scales with exhaust momentum (m v), but power scales with exhaust KE (m v^2 / 2) beyond the ionization energy / power. $\endgroup$ Dec 23, 2022 at 0:13
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    $\begingroup$ @asdfex I got it from this book: descanso.jpl.nasa.gov/SciTechBook/series1/… I'm realizing I gave you the figure of 200 eV of the top of my head but that was from ion thruster plasma generators. I'm not sure right now whether this is also true for Hall thrusters but I believe it is similar. $\endgroup$ Dec 23, 2022 at 18:21
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You'll see everything clear if you use the expression for the total efficiency of the thruster, which is defined as the ratio between the propulsive power and the input power $P$:

$$ \eta = \frac{\frac{1}{2}\dot m c^2}{P} = \frac{\frac{1}{2}T I_{sp}g_0}{P} $$

Note that $T=\dot m c$, and $c = I_{sp} g_0$, and $c$ is the mean exhaust velocity of the particles. From conservation of energy, particles of mass $m$ and charge $q$ will be accelerated to a velocity $c$ (which we take roughly as the mean exhaust velocity) when falling through a voltage difference of $V$.

$$ \frac{1}{2} m c^2 = q V $$

Krypton and xenon have roughly the same charge in plasma thrusters (a bit more than the elemental charge) but the mass of xenon is much higher. Thus, for the same accelerating voltage $V$, the exhaust velocity and Isp is higher for krypton.

Efficiencies for krypton are lower slightly for the reason you mentioned that the first ionization potential is higher. But take a constant value of 0.6. For the same voltage, krypton will have a higher Isp than xenon, so the thrust $T$ of a krypton thruster must be lower if we use the same power for both thrusters.

I'm not sure what you were meant about the electron energy influencing the Isp but it doesn't influence in this.

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    $\begingroup$ Isn't this argument too complicated? You could just write P~.5mc²=.5 T Isp g0. But for a full proof you also need to argue why Isp is higher for Krypton. It might also make it clearer to specify that 'c' is not the speed of light, but the exhaust velocity, more commonly name v_e $\endgroup$
    – asdfex
    Dec 22, 2022 at 9:42
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    $\begingroup$ You're right, I didn't have to mention the efficiency, but I added it so it was clear where the expression of the power came from. I've added the other things you mentioned. $\endgroup$ Dec 22, 2022 at 15:17
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    $\begingroup$ I get the first part, but when you explain why the efficiency for Krypton is lower I don't get it. T = m_dot c, c is higher for Krypton, so why is thrust lower? It must have to do with the rate of mass flow then. Is the mass flow for Krypton lower considering the same V ? $\endgroup$ Dec 23, 2022 at 12:45
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    $\begingroup$ @Playstation_waifu Yes, you just don't have enough power to accelerate the same amount of mass. $\endgroup$
    – asdfex
    Dec 23, 2022 at 18:34

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