# Why are Ion Thrusters so energy hungry?

Ion thrusters need a lot of energy. But I don't understand why. So, one thing I understand is that it takes a lot of energy to achieve a high exhaust velocity ( since kinetic energy is proportional to velocity squared) but we get little momentum. This is all great. But, the energy required is coming from the grids. The charge present on these grids doesn't change (due to ion optics), hence they can accelerate particles forever as long as the ionization chamber has sufficient pressure to let the ions enter the grid space. Now, it takes energy to ionize the gas. But when I calculated it, its not in kilowatts obviously.

The chambers' walls are positively charged. The walls act like an anode and take in the electrons from the plasma. So, the thing that I was wondering about was whether the constant charging of these walls could take up that much energy. I then tried to calculate the energy needed to do this through the knowledge of the metals work function and found out that it requires less than 100 watts of energy. I am pretty sure this is wrong, but I don't understand why ion thrusters are so energy hungry. I am talking about what we are using energy for, and how it translates into the ion beam energy. Like ionization, etc; not how to calculate the energy needed.

• Its not that clear whether your question is about losses or just adding up the basic beam power after all losses are taken off. One way of looking at this is Pe . eta = 0.5 .mdot Ve^2. Where Pe is the gross electrical power, and eta are the losses and the right hand side is the useful power in the beam (mdot is the mass flow rate, Ve is the exhaust velocity). The acceleration from the grids must provide the right hand side and a combination of inefficiencies results in eta. Perhaps the ionisation energy is a major component of eta. Could you clarify your question a little with this in mind? – Puffin Feb 3 '17 at 12:31
• If the charge on the grids was the only factor, ions would accelerate as they approached the grid, then decelerate an equal amount as they left it. It's that conservation of energy thing :-) – jamesqf Feb 3 '17 at 17:57
• Note to OP, try using *'s on each end of a line to be emphasized. Else you come across as yelling. – NZKshatriya Feb 3 '17 at 22:00
• the ions move fast, that takes a lot of energy: $E=0.5 mv^2$ but momentum is linear with velocity $p=mv$, – Jasen Feb 4 '17 at 10:49

The thing to notice about this type of drive is that the ions encounter the positively charged accelerating grid first. The ions that provide the thrust to the rocket are positively charged as well, so they will be repelled and contained within the plasma chamber. The ions' only means of reaching the interior of the accelerating grid is diffusion, meaning an ion must already have sufficient kinetic energy to approach and traverse the positive accelerating grid. If an ion does not have enough kinetic energy to get past the positive grid, it will be reflected back into the plasma chamber and not contribute to the acceleration of the rocket.

The ions in the plasma chamber behave like a gas in that they have a wide range of kinetic energies. Only the highest energy ions reach the accelerating field between the grids. There are two ways of increasing the rate of ions entering the accelerating field between the grids: increasing the temperature of the plasma and increasing the density. The first way increases the average kinetic energy of the ions, allowing more of them to get past the positive grid and reach the accelerating field. The second creates more ions, leading to more accelerated ions just as a matter of probability.

So, the power expenditure comes from keeping the plasma hot and dense enough to get enough ions into the accelerating field at a fast enough rate for the desired thrust. This would be true even in an ideal rocket with no losses due to heat loss to the walls of the chamber, charge loss on the grid due to ion/electron absorption, ion loss due to recombination, etc. If there were no power input, the plasma would cool off and thin out as it lost its highest energy ions to the exhaust (evaporative cooling). Eventually, the plasma chamber would have nothing but low energy ions contained by the walls and the positively charged grid, unable to escape.

• But in an ion thruster, how does the plasma get heated up? So, the energy is consumed to maintain the pressure. But how does it maintain the pressure? – Chandrahas Feb 4 '17 at 1:53
• @Chandrahas The plasma is heated by the impact from the electrons accelerated from the hollow cathode. The electrons impact the propellant atoms, ionizing them and heating them up. Pressure is not the quantity you want to think about for ion rockets. Plasma density (atoms/cm$^3$) and temperature are the relevant quantities. Density is maintained by injecting more propellant atoms into the chamber. Temperature is maintained by hitting these atoms with the ionizing electron beam. – Mark H Feb 4 '17 at 2:16
• Sorry, it's been a long time, but can you please do the calculation, that would help me very much. Thanks – Chandrahas Feb 13 '17 at 4:17
• @Chandrahas Which calculation? – Mark H Feb 13 '17 at 7:53
• The required kinetic energy of the particles to enter the acceleration grid space. (The ion temperature would be different from the electron temperature I think. How would we calculate this) – Chandrahas Feb 13 '17 at 9:02

A conventional thruster with two liquid propellants requires energy too. But it is chemical energy stored in the propellants. Ion thrusters use no chemical energy at all, all the energy of the ion beam is from the electrical energy used by the thruster. In fact, a conventional rocket engine with a lot more thrust than a ion thruster uses a lot more of chemical energy.

• Yes, spot on. I wrote an answer aimed at a different form of propulsion here space.stackexchange.com/questions/14201/… - I'm linking it just as it has nicely formatted equations. These basic equations apply equally to chemical and electrical power sources, the energy has to come from somewhere. – Puffin Feb 3 '17 at 12:38
• Yeah, the potential energy due to the ions and the grids just gets converted into converted into kinetic energy and the charge on the grids doesnt change. Hence, we do not have to 'supply' energy directly. Maybe indirectly... like the charging of walls so the ions could enter the grid space in the first place? So where are we physically supplying energy physically?? This is what I dont understand. I understand that the energy has to come from somewhere.( But does charging the walls take that much energy?) – Chandrahas Feb 3 '17 at 12:40
• Interesting, but I don't think this answers the question. – Hobbes Feb 3 '17 at 12:53
• @Chandrahas: The grids are not charged only, there is a current flowing. Without current,the ion flow could not be accelerated. The energy of the ion beam comes from the electrical energy used by the ion thruster. – Uwe Feb 3 '17 at 13:13
• @LocalFluff I believe pound for pound, it would be more efficient to just use chemical fuels rather than a fuel cell attached to an ion engine. The only way an ion engine makes sense is to use power obtained from some other source, like solar power or a nuclear reactor. – SGR Feb 3 '17 at 14:44

First, I recommend you having a look at the section 2.6 of the book "Fundamentals of electric propulsion (...)" from NASA researchers Dan Goebel and Ira Katz. It is free on the internet and they show a pretty nice energy balance for a conventional thruster. Here is the link for the pdf: Fundamentals of Electric Propulsion.

Second, the equation to calculate the overall power consumption would be:

$$P = \frac{g_0 T I_{sp}}{2\eta_T}$$

For example, for a thruster to produce 150 mN of thrust with a specific impulse of 2000 seconds, and a good efficiency of 0.8, the power consumption would be already 1.875 kW. This equation is demonstrated in the section 2.5 of the same book, and it is derived simply from the relation:

$$\eta_T = \frac{P_{jet}}{P_{in}}$$

The kinetic power of the beam, called the jet power, is defined as

$$P_{jet} = \frac{1}{2} \dot{m}_p v_e^2 = \frac{T^2}{2\dot{m}_p}$$

Thus, as discussed in the book, most of the power is simply used to the sheer acceleration of the beam. All the other power consumption (ionization, wall losses, collisions at the grids, etc) are accounted in the efficiency, and is minor in comparison to the electromagnetic acceleration of the beam.

• But my question is where are we physically supplying the energy to? To charge the walls? Ionize the gas?... – Chandrahas Feb 3 '17 at 16:58
• To accelerate the rocket. The energy is being used to create a field which accelerates the ions in one direction and the thruster in another. Every time an ion moves in that field, it takes energy out of the electric field. Most of it goes in an ion current between the positive and negative grids in the diagram. – pjc50 Feb 3 '17 at 17:14
• Yes, most part of the power is consumed by the acceleration itself. The rest of the consumption (the minor part) is due to ion losses, wall effects, cathode, etc. In the book there is a list of possible effects. Still, the most important power can be calculated from the idealized equation that I shown. – user17622 Feb 3 '17 at 17:56
• +1 for including math, can you describe the variables here also? Links tend to break over time, answers like this are much better if they can stand on their own as well, so any future readers can still learn from them. Just something like "...where $\eta_T$ is the..." – uhoh Feb 14 '17 at 4:34

There is a sense in which the wonderfully high ISP of ion thrusters is the cause of their energy hunger, independently of the detail of their operation.

It works like this: we need a certain amount of delta-v (that's what a thruster is for after all) or more formally a certain amount of impulse $J$, and we are going to get it by throwing mass overboard. $$J = \Delta m_\text{fuel} v_\text{exhaust} \;.$$ (Here we take $\Delta m_\text{fuel}$ to be small enough to neglect high order terms. This can be written in differential terms to formalize that, but it doesn't really change the argument.)

Now we compute the energy cost $W$ of this choice in term of how much mass we throw overboard (writing $K$ for kinetic energy of the exhaust relative the spacecraft): \begin{align*} W &= \Delta K \\ &= \frac{1}{2} (\Delta m_\text{fuel}) v_\text{exhaust}^2 \\ &= \frac{J^2}{2\,\Delta m_\text{fuel}} \;. \end{align*}

So, for a fixed impulse, using more mass of fuel requires less energy (neglecting the efficiency of the drive mechanism).

From this point of view, engines with very high exhaust velocities are worse rather than better!

But we use them anyway for several reasons. First, the tyranny of the rocket equation (with it's exponential dependence on mass ratio) overshadows other consideration, second launch from the surface requires high thrust at almost whatever cost and third the energy cost for chemical propellants is payed at manufacture rather than when rocket is operated and that means ground-side where energy is plentiful.

In terms of satellite and probe engines it is the mass-fraction issue that dominates.

• A description of how efficient the ion engine is would improve this answer. If this explains a large percentage of the energy use of the Ion Engine, this partly solves the OP's problem; if it only explains an extremely small percentage, this doesn't solve the OP's problem. – Yakk Feb 6 '17 at 20:06

The foundation of rocketry is Conservation of Momentum:

$p=mv$ (momentum = mass * velocity)

Since momentum is conserved, the change in momentum of the propellant is equal to the change in momentum of the rocket body.

Also: energy of an object in motion is defined by the Kinetic Energy equation:

$KE = {1\over2}mv^2$ (Kinetic Energy is equal to one half the mass times the velocity squared)

Note that the velocity term is linear in the first equation, and squared in the second.

If you, for example, doubled the velocity of your propellant and while you halved the mass, the change in momentum would remain the same, but the energy required to achieve that change would have doubled!

So Ion Engines appear "energy inefficient" because they are accelerating low mass propellant to very high velocities.

In some applications this is perfectly fine: it is expensive to put mass into orbit, so it can make sense to use a less "energy efficient" engine to save mass in propellant. Propellant speed also defines the upper limit in the rocketry equation, so higher end speeds are theoretically possible. In other cases, this characteristic is not desirable.

On top of what is the primary cause described by others: about all exhaust energy coming from the electric supply, as opposed to chemical energy of the propellant, in more advanced ion thrusters like VASIMR there's one more "sink" for all that electricity.

Your normal chemical engine is made from superior alloys resistant to high temperature and high pressure, with active cooling system to keep them from melting and being torn apart - and all that for propellant energy an order of magnitude lower than what ion engines handle.

The ionized propellant, accelerated to dozens of kilometers per second, would act as an extremely hot, extremely abrasive plasma that would make a short work of any nozzles, grids, generally all solid structural elements of the drive. Simpler drives deal with it simply - let it happen, shortening the drive's useful life and putting an upper limit on achievable delta-V because before you run out of propellant, the drive dies. The more advanced ones use the only "material" that can handle this kind of conditions, and is fully, immediately "self-healing": the magnetic field. In VASIMR, for example, at no point past the ionization system does the gas ever touch the structural elements; it's entirely trapped within a whole, rather complex "plumbing system" made entirely out of magnetic fields.

...and you need a lot of electricity to keep these electromagnets running. After all, they must be strong enough to stop and divert particles moving at several dozen km/s - completely independently from all the power-hungry systems that accelerate the particles to these speeds.

• Why cant we use electromagnets instead? The magnetic field is not time dependent right? – Chandrahas Feb 4 '17 at 3:33
• @Chandrahas: we are using electromagnets - did you mean permanent magnets maybe? I can't answer this one, plasma physics beyond my level of knowledge - likely same reason why we can't use them in a stellator though... – SF. Feb 4 '17 at 14:23
• Yeah. Sorry, I meant permanent – Chandrahas Feb 4 '17 at 14:50