121–130 of 169 posts
Curious about whether you're looking at silicon carbide as a replacement for IGBTs for your inverters. Is the technology practical yet? (I'm a big fan of IGBTs but there's always room for improvement.)
Curious about whether you're looking at silicon carbide as a replacement for IGBTs for your inverters. Is the technology practical yet? (I'm a big fan of IGBTs but there's always room for improvement.)
Whoops. Just noticed you mentioned SiC on your website. Guess the answer is yes. Great minds.
How much noise will these make?
Earlier quoted context omitted.
A simple physics-based plane model (like the one we made to understand vehicle-level impact of our technology development) dictates that the range-optimal cruise speed is proportional to 1/sqrt(air density), so it makes sense that the blackbird was more efficient at high speed when at high altitudes (admittedly, this simple model is subsonic, and there are a lot of other factors for supersonic flight). Since having l…
That's really not too accurate. The most efficient aircraft are sailplanes (the high end ones usually have a motor, BTW), and they operate at lower altitudes typically. Lift-to-drag of 70 has been achieved. The SR-71's L/D is probably classified still, but probably around 7 or so. The issue is a certain aircraft has an optimum cruise altitude. If you try to fly fast at low altitude, it'll be horrendously inefficient.…
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.
Given that the batteries weigh a lot more than the motors, I would have thought that motor efficiency (which scales battery size) was much more important than motor weight. My back-of-the-envelope is: - Assuming 0.4 kWh/kg for batteries, and they have to run for 4 hours, then the total mass per kW is 10 kg (batteries) and 0.08 kg (motor). - A 1% increase in motor efficiency could eliminate 0.1 kg of batteries, which…
Keep in mind that an electric motor isn’t limited by the amount of oxygen in the air. As a result it can fly significantly higher where there is far less air resistance. Since air density is proportional to the square of the elevation this can lead to significant efficiency gains. Believe it or not, partly as a result of this, the SR-71 had it’s best mpg at peak speeds.
Earlier quoted context omitted.
Is the scale of the engine for marine applications very different to aviation? Ferries usually have huge hulking diesels so I'm curious what the equivalent electric powertrain is like. I'm also assuming that a scaled down version would be ideal for personal watercraft?
We are planning to add a MW-class machine to our portfolio in the next five years which could serve as a nice replacement for the dirty diesel engines on these larger ships. Looking at Taiga Motor's electric jet ski, 250kW would be a bit on the high-side. It would also be a very expensive jet ski :) https://taigamotors.ca/watercraft/
Earlier quoted context omitted.
I think most forms of assisted takeoff technology would reduce electrical power requirements for takeoff, but not put much of a dent in the climb power (unless you can catapult/ JATO all the way up to cruising altitude, which would be challenging). Since climb power is still significantly higher than cruise power, and is effectively thermal steady state for these components (10-20min), it would still drive the propul…
You still need enough reserve power at the end of a flight to do a rejected landing/go around, or you won't get certified. If you are relying on some kind of catapult, rail gun, or rocket assist to take off, not sure how that happens.
Earlier quoted context omitted.
A simpler alternative could be to just have an electrified runway. The plane draws power from power rails embedded in a runway, or something like that. So, it doesn't switch to batteries until it's in the air. You could even have a long cable that hangs behind the plane and keeps an electrical connection until you're a few hundred feet up. (I'm picturing it connected to something like a slot-car that travels in an el…
As a rule, whenever one feels tempted to say "just do ", it's time to wait and think. Because, if it's "just" about doing something, why isn't it being done already? In this case: let's say it's feasible to retrofit runways to use this system (it probably isn't) and look at a few issues. For instance: "the cable releases from the plane". No system is fail safe. What happens if the cable does NOT release from the plan…
Given that the batteries weigh a lot more than the motors, I would have thought that motor efficiency (which scales battery size) was much more important than motor weight. My back-of-the-envelope is: - Assuming 0.4 kWh/kg for batteries, and they have to run for 4 hours, then the total mass per kW is 10 kg (batteries) and 0.08 kg (motor). - A 1% increase in motor efficiency could eliminate 0.1 kg of batteries, which…
Wouldn't it be much simpler to state that a 1% increase in motor efficiency could eliminate 1% of battery weight? (trying to get the theory clear)
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Obs: This is only approx. valid if efficiency is already high. If efficiency was very low, e.g. 2%, then 1% more (going to 3%) would enable eliminating 1/2 - 1/3 = 1/6 = 16.7% of the batteries.
An equation to describe this situation, assuming constant energy need, is Eb = Em / n, where Eb is energy provided by batteries, Em the work of the motor, and n efficiency.
Also, the energy need should indeed decrease with decreasing battery weight, amplifying this effect even more, but at high efficiency the correction isn't too large. Equations omitted because there are too many assumptions (acceptable battery mass fractions, energy usage vs weight, ...).
(A starting model would be: Maircraft = Mbatteries + Mconst; Mb = aEb; Em ~ Ma^p ; Em = ( k(aEb+Mc) ) ^ 1/p; Is p~=1?; Eb = kMc/(n-a*k); )
So in principle an 1% increase in motor efficiency gives even more than 1% of less battery weight!
A complication however is that batteries have power constraints as well as energy constraints (how power constrained . If the peak power only has to be sustained over a very small period, this would allow complementing energy-dense sources (batteries) with power-dense sources (capacitors). However, some power-dense sources do not last long enough to cover the peak-power intervals, so they would not fit.
If the following diagram is to be trusted:
https://commons.wikimedia.org/wiki/File:Power_vs_energy_dens...
Then for my guess of 5 minute take-off constant peak power time lithium-ion still has the greatest power density, which means other sources should not be combined.
You can use variations in chemistry among Li-ion cells to achieve this tradeoff, but those limitations provide a slight negative correction (greater efficiency giving less mass gain).
Those effects would need to be combined.
Anyway, there is a lot of interesting performance and Operations Research (Linear programming) optimization here.
