# What is the theoretical maximum velocity of electric aircraft?

I'm wondering about theoretical maximum velocities of supersonic electric aircraft. I think the fastest demonstration of any type of aircraft has been close to mach 10. This was done using a rocket engine, so propulsion was not limited by air characteristics as much as an electric motor + compressor's propulsion would be (I think).

Under best conditions; Assuming an incredibly high power to weight ratio of motors, extremely aerodynamic design, strongest materials, etc.

Theoretically, does anyone know what the limiting factors in the velocity of an electric compressor propelled supersonic aircraft would be (barring energy use)? Does it depend on size, etc?

Could it reach mach 2+?

Thanks a lot in advance for any input.

• Just to clarify, you're limiting this specifically to electric motors driving a compressor and not some (for example) yet-to-be-invented scramjet-like electric engine, right? – reirab Oct 2 '16 at 9:38
• Well yeah, electric motors to compressor is what I mainly want to know about since it theoretically exists and probably has defined limitations. Though the latter would be increasing the domain from what current technology can do to what is possible given the laws of physics, regardless of current technology. That would be awesome to know about too. – BuggyLemon Oct 2 '16 at 9:56

Since you seem not to be concerned about the usability of this hypothetical electric aircraft, let's make some simplifications:

• The aerodynamics are so good that it does not create any drag. This is completely unrealistic but helps a lot to simplify things.
• Structure, controls and propulsion are so advanced that they leave one third of the take-off mass for batteries. Who needs payload anyway?
• Efficiency is 80%. Not even electric propulsion can be lossless.
• Energy storage is the best current technology has to offer. This would be about 1.8 MJ per kg of battery for non-rechargeable lithium cells.

To accelerate a battery of the mass m to a speed v needs an amount of energy $$E_\text{kin} = \frac{m}{2}\cdot v^2$$ Flying Mach 2 at sea level is impractical; it helps to climb to at least 6.000 m (20.000 ft) to do so. This adds potential energy: $$E_\text{pot} = m\cdot g\cdot h$$ Conveniently, the result is independent of mass, so the potential maximum speed is $$v_\text{max} = \sqrt{2\cdot\frac{1}{3}\cdot0.8\cdot1{,}800{,}000\ \mathrm{m^2\ s^{-2}} - g\cdot h}$$ If we solve this for 6000 m, the speed at the end of the acceleration is 949.3 m/s, which is already Mach 2.966. If you prefer to use rechargeable batteries (1 MJ/kg), the maximum speed drops to Mach 2.15.

If you start to add a reasonable figure for drag, you won't even make it to the speed of sound before the batteries are empty. Currently, propellers are the most sensible way to convert electric energy into thrust, and large, low-speed propellers are the most efficient type. If you think to drive a conventional compressor with an electric motor, its subsonic efficiency is already far worse than the assumed 80%. Even its mass is not yet competitive with conventional gas turbines.

• @ymb1 Thank you for telling me. No apologies needed. – Peter Kämpf Oct 4 '16 at 19:46
• The density of energy storage is also tempered with the rate at which it can be extracted from the material. Therefore the power available to the supply will have temporal capacity limits, however that of course doesn't prevent the possibility of entirely depleting the primary cells over many hours into the supply and then being consumed in seconds by the engines. – jCisco Oct 7 '16 at 22:49
• @jCisco: Absolutely right, I did not cover any of the constraints. A more complete answer should include this, but I felt that just looking at how much energy it needs to accelerate would show how far away supersonic electric aircraft are. – Peter Kämpf Oct 8 '16 at 9:16
• Rather than write a contrary minority answer, how does your answer extend to electrohydrodynamic thrust. (Ionic Wind). I don't have the link but Steven Barrett of MIT had a paper on this you might find very relevant. – jCisco Oct 19 '16 at 23:27
• @jCisco: Thanks for digging this up! – Peter Kämpf Oct 20 '16 at 23:20