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Resistojets heat a compressed gas or vaporize a liquid to expand it through a nozzle for thrust. They improve a bit on the performance of cold gas thrusters. This experimental water resistojet managed an Isp of 140 s.

I'm thinking about what sort of simple rocket engines might be feasible in the initial stages of a permanent lunar base, made on-site mostly from local materials. Oxygen is the only volatile that can be extracted in volume anywhere. Obviously the example shows water is a fuel option, but the paper cited above answers most of my questions regarding that, and also I wonder if oxygen might have advantages.

What materials that might be available early on for such complex fabrication is unclear, however I'm going to suggest that something can be come up with, that can handle storing liquid oxygen and gaseous elemental oxygen at the nozzle up to, say, 500 °C. Let's say iron combined with metal oxides or basalt. The resistance heater and other complex bits come from Earth. The LOX flows to the resistance heater and then out the nozzle.

Could such a system work? What sort of Isp and thrust might it have?


For what its worth, the idea is that these engines only need to circularize an orbit around the Moon, and do other orbital maneuvers or a burn to enter orbit around EML1. Launch is by a sling launcher using basalt fiber cable, that imparts orbital or escape speed. So low Isp isn't a problem in itself.

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This response addresses just the last part of your question "what sort of $I_\text{sp}$ and thrust might it have", and does so in the sense of an ideal resistojet. I hope this isn't too basic an approach but its worth stepping through it to develop a more intuitive view of the very high level physics. There are similar equations in the link you gave though, given your question, I'm guessing that the meaning is lost in lower level details. If not, it will be useful to someone out there!:

  1. The power in a rocket exhaust, the "beam power", is the rate of flow of kinetic energy, i.e.

    $$\frac{1}{2} \cdot \dot m \cdot v_e^2$$

    where $\dot m$ is the mass flow rate and $v_e$ the exhaust velocity. Also $v_e = I_{\text{sp}} \cdot g_0$. I wrote it in that form to start with, as kinetic energy is a relatively familiar concept.

  2. More usefully for your question: $\text{Thrust} = \dot m \cdot v_e$ (Eq 7 in your reference) and so:

$$\text{Beam power} = \frac{1}{2} \cdot \dot m \cdot v_e^2 = \frac{1}{2} \cdot \text{Thrust} \cdot v_e$$

  1. A pure resistojet, i.e. without any other energy source such as from chemical reactions, means that we can directly relate the electrical power in the resistance heater to the beam power thus (Eq. 1 in your reference):

    $$P_e \cdot \eta = \frac{1}{2} \cdot \dot m \cdot v_e^2$$

where $P_e$ is the power dissipated in the heater and $\eta$ is some efficiency factor allowing that some power will be lost, radiating to space and conducting back into the structure, e.g. $50 \text{%}$ (guessed). In the case of $\require{mhchem}\ce{LOX}$ storage, this will also need to allow for the $\ce{LOX-O2}$ phase change, but don't worry about that for now.

  1. Putting this all together we can choose our values of Thrust and $I_\text{sp}$ for any non-reacting propellant, and then work out the corresponding $dot m$ and $P_e$. This is the basic concept, it is meant to be as simple as it sounds.

  2. OK, OK, it's bound to be more complicated, but you have to start somewhere. This will equip you to begin to very roughly understand someone else's thruster. Check out your reference though also try this, as it just gives the headlines without getting lost in $C^*$ (characteristic exhaust velocity), chamber pressure, etc. It gives $\text{Thrust} = 0.045 \text{ N}$, $I_\text{sp} = 152 \text{ s}$, which I calculate gives a beam power of $33 \text{ W}$ for $100 \text{ W}$ electrical, i.e. $\eta = 33 \text{%}$. In this particular case, they are paying over the odds, because they are supplying heat for the water-steam phase change, too. You can check the published mass flow rate of $0.03 \text{ g/s}$ and it neatly comes back to about $45 \text{ mN}$, too.

  3. A problem: You can't arbitrarily increase the electrical power without finding that you have losses from the gas becoming so energetic that the power goes into exciting rotational and vibrational modes of the gas molecules, rather than keeping them zinging around (translational modes). When this begins to happen for $\ce{O2}$ as a propellant, is something you'd have to look into, but I'd guess not until you were up beyond an $I_\text{sp}$ of $500 \text{ seconds}$.

  4. Back to the idea of $\ce{LOX -> O2}$ as a propellant, there are a couple of other concerns:

    • $\ce{O2}$ specifically might be quite a hard choice for materials compatibility as it's very reactive, more so the hotter it is. As a first pass at temperature, assume the energy per molecule is $\frac{3}{2} k \cdot T$ where $k$ is Boltzmann's constant, $\ 1.38 \cdot 10^-23$.
    • As mentioned earlier, by storing the propellant as $\ce{LOX}$, you are directly creating more of a challenge as the electrical power input has to boil the $\ce{LOX}$ (a phase change) and then raise its temperature. This is another step you will need to do sums on. This area, and particularly the logistics of $\ce{LOX}$ production and storage aren't my territory at all, so I'll leave it there.
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  • $\begingroup$ @TildalWave Thank you for the formatting improvements. I'd like to know how to do this myself. You seem to have used two approaches. For the method with the codes like $$\text{ is there a reference somewhere? I've been looking on Help but I can't see a starting place. There is also some pasted in pre-formatted text (MathJax or MathML, I don't know its all new to me), again, is this a feature of the site or something I need to add to my browser? $\endgroup$ – Puffin Feb 25 '16 at 13:03
  • $\begingroup$ I had thought that the expansion ratio of oxygen would come into this (1:860). True, i didn't go properly through the reference - it's one of the things i don't feel i have time for right now and i sought a shortcut - so thanks for going through it so that is easier later on. But i'd also wondered if the materials issue washes out the idea from the get go too. My hope was maybe metal oxides are non-reactive enough to make that manageable. $\endgroup$ – kim holder Feb 25 '16 at 14:45
  • $\begingroup$ Volume sounds like a good point, thinking on my feet I suppose it will be related to the ease of heating the fluid before its out of the nozzle and off. On the material front its perhaps of interesting that there have been oxygen rich launch vehicle engines for a while. Leads me to suspect that small scale will be ok and larger will get expensive at some point but not obviously insoluble. It sounds very interesting though I really should leave this for someone else better appointed to take this from here. $\endgroup$ – Puffin Feb 25 '16 at 15:36
  • $\begingroup$ Dumping LOX into a vacuum vaporizes it immediately. Dumping it into a vacuum and then heating it sounds problematic though. $\endgroup$ – Joshua Aug 22 '17 at 16:09

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