In the last discussion on why ion thrusters are so energy hungry(Why are Ion Thrusters so energy hungry?), people said that the system is not only supposed to ionize the gas but also heat it to sufficient temperatures needed to overcome the potential and enter the acceleration space. This perfectly makes sense.. But then I was wondering, If the heat is getting converted into the kinetic energy (in a particular direction, not random), then the temperature of the plasma is supposed to drop as soon as it gets converted into beam current and its kinetic energy. Does this happen.. Does the temperature of the plasma drop after being accelerated?
2 Answers
To a degree. In a typical converging-diverging nozzle, pressure is the force that pushes against the nozzle. This pressure is the result of both the density and temperature of the gas. As the gas flows throughout the nozzle and kinetic energy is recovered, the temperature (and pressure) drop. There are also reactions in the gas which can affect the temperature.
The "nozzle" profile of an ion thruster, on the other hand, is quite small. There is very little time for the plasma to exchange heat with the ion thruster during its transit through the acceleration grid. The "optics" of the grid are actually designed to avoid having the ions interact with the grid wall. Moreover, the propulsive force is the Coulomb force, rather than pressure. We rely upon the ion having a charge and interacting with an electric field. There is no attempt to recover the thermal energy of the plasma as there is in a rocket nozzle.
It is likely that the plasma will cool in the plume as it interacts with other ions or matter, but as to the order of magnitude, it depends on the density of the plasma and the surrounding system. Low density of matter results in little heat transfer (as there is nothing to transfer it to).
There are examples of electric propulsion that do attempt to turn thermal energy into kinetic enery: electrothermal and microwave electrothermal thrusters. They use a traditional converging-diverging nozzle to expel a heated gas (either heated by a resistive metal or by a plasma). In the latter case, the gas will be partially ionized as it travels through the nozzle region and the thermal energy will be recovered, assuming the nozzle is well-designed.
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$\begingroup$ So, we need kinetic energy to push the plasma into the acceleration grids space but then if the temperature doesn't drop? where is the energy for the acceleration coming from? $\endgroup$ Feb 25, 2017 at 5:06
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$\begingroup$ I mean, if somehow we can extract energy from the temperature of the plasma, and also from the kinetic energy (beam's kinetic energy), and the only energy we are supplying for the gas to enter the acceleration space is thermal energy, Can't we produce more energy than that is given into the system? Doesn't this violate the conservation of Energy? Where am I going wrong? $\endgroup$ Feb 25, 2017 at 5:09
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$\begingroup$ No. The source of all the energy in the system is electrical energy supplied to the thruster. It is just transformed into thermal energy (thru joule heating), a charge for each ion produced (thru ionization), and an electric field (thru running a voltage bias across the grids). Total Power Input Into System = Ionization Energy + Thermal Energy + Beam Energy Total Power Output: Beam Energy We lose the ionization energy and thermal energy we put into the system. You could attempt to extract them, but you need a real nozzle. But then you have frictional losses. There is no win / win here. $\endgroup$ Feb 25, 2017 at 8:47
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$\begingroup$ Basically, I recommend you take a gander at Goebel & Katz's "Fundamentals of Electric Propulsion." It's a free JPL publication and probably the most thorough book on ion thrusters: descanso.jpl.nasa.gov/SciTechBook/series1/… $\endgroup$ Feb 25, 2017 at 8:51
This post is, strictly, a detailed comment, but one that requires more space than the comment box allows. Please be nice to me on this, its done in a positive spirit and it seems unavoidable as the subject line contains a key assumption that is probably clouding the issue. (EDIT - the question title has since been changed so this post is more of an orphan now). This same point came up in several answers to the first version of the question. Its not meant to be a complete answer, clearly the detailed questions on the plasma temperature are still of interest to the OP; I aim to be helpful rather than win up-votes.
I want to examine the phrase "so energy hungry". My point is that ion thrusters have losses, so do all other forms of propulsion. In the following example the losses are about 35%. That is to say that from the start of the manoeuvre to the end only 35% of the DC electrical energy does not make it into useful beam kinetic energy.
Ion thruster example
The XIPS thruster given here has the following envelope characteristics in low power mode:
Total Input Power (W) 2067
Thrust (mN) 79
Specific Impulse (s) 3400
Putting the latter two bits of data together with the beam power equation shown in this answer to a different question ( but the equations are valid for any type of thruster, chemical or electric ) gives a beam power of 1317 W. Thus the efficiency is 1317 W / 2067 W and losses in this case are 36%.
Clearly there is ~700 W spent on support processes such as plasma generation. It would indeed be interesting to identify these more clearly though the aim of this post is to put that 36% in context.
Chemical thrusters
Are chemical rocket motors substantially different? The context there is that the starting point is stored chemical energy and the losses are largely thermal losses trapped the exhaust and radiation from the nozzle. This link suggests that the ceiling is commonly 70%, i.e. >=30% losses, from a thermodynamics perspective.
Overall
Thus the two thruster types may not be so different overall. Clearly at a system rather than thruster level the trade-off has many factors such as the availability of the power in electrical form and often becomes a trade-off of energy limits vs power limits for chemical and electric respectively.