mliben
Joined 6 years ago
Joined 6 years ago
In the aircraft world, Magnix has been doing some really awesome work. Their motors + inverters are about 4 kW/kg continuous. Our multiple areas of technology development should put us >12 kW/kg, which would be a 3x improvement. We are excited to build and iterate on a rapid pace to get there as soon as possible (hardware is, of course, hard :)
1) Considering megawatt-class machines are necessary for many future applications, the mass of the motor+inverter+gearbox (especially using best current technology) definitely adds up.
2) With a very distributed propulsion system, motors that end up near the wing tips have a big moment arm compared to the ones typically tucked under the wing root
D = A*v^2 + B/v^2 (D is total drag, first term is parasitic drag, second term is lift-induced drag)
dD/dv = 0 where v = (B/A)^(1/4)
Plug in v = (B/A)^(1/4)
D = sqrt(AB) + sqrt(AB), aka dD/dv = 0 exactly when parasitic drag is equal to the lift-induced drag
Just using the drag polar approach and neglecting second-order effects (assume negligible dependence on Re, sufficiently subsonic so negligible impact of M, and linear lift coefficient region aka no stall), we get the following (I'm skipping a lot of intermediary steps):
Cd = Cd0 + k*Cl^2 -> Cd0 is the parasitic drag coefficient -> k is the lift-induced drag coefficient -> Cd is the overall drag coefficient
Range is maximized when Cd0 = k*Cl^2 (parasitic drag = lift-induced drag) -> Cl is a function of speed: since the required lift is constant, more speed = less Cl required = less lift-induced drag -> maximum L/D is achieved at this range-optimal speed
This speed can be calculated exactly from the total weight (W), air density (rho), lifting area (S), and drag coefficients:
range-optimal speed = sqrt((2*W/(rho*S))*sqrt(k/Cd0))
As long as you always operate at this range-optimal speed (aka speed for maximum L/D) which is a function of air density (and therefore altitude), the equation for range reduces significantly:
R = endurance*velocity, where endurance = battery energy / drag power, and we know the equation for drag power...
Simplifies to:
R = E*eta/(2*sqrt(Cd0*k)*W) -> R is range -> E is battery energy -> eta is total system efficiency
Dimensionally, this equation is of course the same as yours, with an energy being divided by a force to get a distance. The key point I am trying to make is that if you just look at that equation with no context, speed and air density are not present anywhere. But what is hidden in the assumptions is that you are assuming that you are operating at the maximum L/D speed given the air density at any particular altitude. Going back to my other comment, range at the range-optimal speed does not depend on air density or velocity directly, but lower air density at higher altitudes will result in a higher range-optimal speed, and hence less travel time for a given range.
If you go through the analysis, the air density drops out of the range equation if you assume are operating at the range-optimal speed (which is higher at lower air densities).
The characteristic current of the machine is roughly the same as the maximum inverter current, so at maximum power the power factor is between 0.7-0.8. This also means the flux weakening capability (CPSR) of the machine is very good, although that isn't particularly useful in these propeller-load applications. In many applications the machine would spend most of the time at ~0.6pu speed and ~0.5pu torque (cruise), where the power factor is > 0.9.
I can't discuss the specific rotor design, but I will say it is a very high flux machine.
The inverter and motor are on opposite sides of the coolant, and so heat from each flows into the coolant. The mechanism by which the motor would heat the power electronics is via increasing the coolant temperature, which to some extent is mitigated by maintaining an appropriate flow rate given the loading conditions.
I can't go into much detail, but we are working on addressing this in a couple different ways in our insulation system design.
The biggest difference is the total mass specific power (including housing, bearings, etc) usually gets worse at much lower powers (1s-10s kW), because these components become a more significant fraction of the total mass.
The 12 kW/kg number is continuous output power / total system mass (active + inactive, including motor, inverter, gearbox, housing, bearings, etc). If you isolate just the motor to compare, it is much higher than 12 :)
We do have plans to develop a ~100 kW (maybe a bit smaller) unit in the future, but when is TBD.