So what? The battery in my Model 3 has the equivalent of 2 gallons of gas by that metric. And yet it goes 325 miles on a charge. My previous car would have only gone 60 miles on that. And it’s not nearly as range-optimized as it could be, if the price were right–it could be doubled, trivially, with only small changes and an increase in price.
I’ll see if I can whip up some numbers later to see how far off things are. However, I can start with this one:
A good lithium ion battery has a specific energy of around 250 Wh/kg, or 900 kJ/kg. It takes 9.8*10000 = 98 kJ to lift a kg to 10 km.
If the plane is 1/3 battery, then there’s a factor of 3 remaining. Planes are generally pretty efficient at producing lift, because they push a tremendous amount of air downward at a relatively low velocity. I’m not sure how it works out in practice, but if they’re 66% efficient, we’ve used half the capacity to get to 10 km.
That doesn’t sound half-bad as a ballpark. And the early planes on regional hops aren’t going to fly to 10 km anyway.
I think we all hope to ride an electric Beaver one day.
That’s pretty neat, at any rate. I expect there are lots of these little regional airlines like Harbour Air that serve only very short flights (30 min in their case).
I’m hopeful that the Beaver is the first electric in the air, for with floats and skis De Havilland’s Beavers and Otters opened up much of Canada. Bridging from one epoch to the next would be pretty nifty.
Yes, I probably have an Iron battery. I have to take it out and store it in the winter so it doesn’t freeze. It may be more temperamental than other Lithium batteries but there is no way to deal with a failure at altitude. A cascade failure would impossible to control. The article I linked to talked about mitigating the fire risks associated with lithium-ion batteries.
joema mentioned the ATR 72 as a good starting point, so I’ll use that as a baseline.
Although I can’t find a drag coefficient for the ATR, efficient aircraft can be in the ballpark of Cd=0.022. This Cd uses wing area as the reference area and includes both parasitic and induced drag (i.e., lift drag). The ATR has a wing area of 61 m^2.
The ATR has a cruise speed of 510 km/h, or 140 m/s, but I’ll reduce this to 125 m/s for efficiency reasons. We’ll also assume it flies at 5 km, which has an air density of 0.53 atm.
So, the drag force comes to:
F = 0.5pu^2CdA = 0.51.225 kg/m^30.53*(125 m/s)^20.02261 m^2 = 6807 N. At 125 m/s, that’s 851 kW.
Propellers are efficient; around 90% is achievable, but we’ll go with 85% as more typical. That puts us at 1001 kW.
If we want to sustain that for an hour (traveling 450 km), we need 1000 kWh. For the ATR at max takeoff weight, we need an extra 23000 kg9.81 m/s^25000 m / 3600000 kWh/J = 313 kWh for the potential energy in getting to altitude.
So we need 1313 kWh of cells, and with current cell tech (250 Wh/kg) that means 5252 kg of them. The dry weight of the ATR is 13,300 kg, with the max takeoff weight 23,000 kg. So there’s enough capacity. But it’s better than that, because if you’re doing a clean-sheet design then you will use the cells as part of the structure. Also, the electric motors should weigh less than the jet turbines, and you can get rid of a fair amount of other stuff.
Exactly how this balances out in the end depends on the design, but it’s not totally crazy to think that the net weight increase might only be, say, 3000 kg instead of 5300. Especially since there would be a strong motivation to reduce the weight in other ways, like using composite materials. Whatever the case, the cells have not completely killed the payload.
Note also that efficient commercial craft have a glide ratio of around 20. So, once at altitude, you get another ~100 km of range due to trading the 5 km altitude for horizontal distance.
Of course I’ve left out a bunch of stuff, such as safety factors in the charge level–this is just napkin math after all. But I made conservative assumptions in some areas and unconservative ones in other ones. Some of those will balance out.
I also didn’t go for somewhat more extreme approaches, like immediately flying to an extremely high altitude (say, 30 km) and then gliding for long distances. There’s a huge design space to explore here that’s completely new due to the differing characteristics of electric power. Some of these are bound to be improvements.
One thing I’ve always wondered is the effect on aircraft performance of hauling all the dead battery weight around. One big advantage, I would think, of jet fuel is that as you burn it off, the aircraft is getting lighter. As you get lighter, you can manage the power to generally give you greater range, a higher altitude, or more speed. With batteries, once they run out of energy, they are just extra weight. (There may be a more elegant way to describe this, but essentially, if I have a fuel tank with 1000lbs of Jet-A and burn off half of the fuel, I am now only hauling 500lbs of fuel and getting lighter by the minute. With 1000lbs of batteries and just half a charge left, the batteries still weigh 1000lbs and will weigh the same amount even at zero charge.)
Plus, you can’t lighten the aircraft by defueling a battery. So if you need to remove weight for performance (maybe to get in and out of a little bit shorter runway) it would have to be passengers or freight, which no airline is going to appreciate. That leads to not being able to dump fuel and the landing gear structure has to be even stronger because it always has to be able to land at max takeoff weight. (The Embraer ERJ140, a typical 44 seat jetliner, has a max takeoff weight of about 46000lbs, but a landing weight of only 41000lbs. So the gear would have to be stronger to handle the excess weight.)
Am I overstating that, or would batteries actually have to be some significant percentage more efficient at storing power then hydrocarbon fuels just to haul the extra weight around?
It’s a fair point. However, for the short ranges viable in the near term, the fuel weight wasn’t a significant factor anyway. The fuel burnoff matters a lot more after thousands of miles as compared to several hundred.
This is neither here nor there, but the Electron rocket by Rocket Lab uses electric turbopumps. As it depletes its batteries, it jettisons them so that they don’t have to be carried any longer than necessary (on a more conventional rocket, the pumps would be powered by rocket propellant, and thus be burning through the mass as it flies). Of course, the whole rocket is expendable, so there’s no real cost to doing this. Maybe we can have the batteries fly back to an airport on a little glider :).
Googling suggests that modern jet engines can achieve specific power around 5 to 6 hp per pound - about what state-of-the-art brushless electric motors can manage. But the jet engine is complete, whereas you must add a propeller (or ducted fan) to the electric motor.
Yes - same with all aircraft. You consume extra energy to climb, and “get it back” when you descend in a glide.
Note that one of the the main experiments on the X-57 is to greatly reduce the wing area and thus reduce drag. (The other one is to put the propellor at the wing tip to eliminate vortex drag.) We don’t know what the resultant Cd will be; that’s why they’re doing the experiment. But I think it would be safe to use a somewhat lower Cd in your calcs.
All this strikes me as trying way too hard to make electric aircraft work. You start with maximum theoretical efficiencies, then admit that you ‘left out a bunch of stuff’.
No aircraft comes anywhere close to maximum theoretical efficiency. There is drag everywhere. Wing roots, pitot tubes, engine pods, boundary layer tripping from antennas and other junk, all sorts of things contribute to the real-world drag of an airplane.
The climbing-then-gliding trick doesn’t work. Climbing not only requires higher power, but it creates a lot of induced drag. You never get that energy back on the glide down.
A major difference between airplanes and cars is that when cars are cruising at speed the engine is loafing. The only energy a car uses up at cruising speed is rolling resistance, driveline friction, and air resistance. That’s why electric cars can have pretty decent range. But an airplane at cruise is at 65%-75% of maximum power. Drag goes up with the square of velocity, so the faster airplane has to overcome more drag both from pushing the air out of the way, and drag created from lift.
You are generally handwaving away a lot of stuff to show that an ideal airplane would work. But we’ve never come close to an ideal airplane - not even sailplanes manage theoretically maximum values of efficiency. Not even close.
A much better, more reasonable way to analyze this would be to look at the most efficient commercial transport planes we have, figure out how much efficiency could be added with electric power, then work the numbers from there.
Take the ATR-72 you mentioned. You started out with the actual numbers for performance, but ignored them in favor of trying to recalculate the drag of the plane. Why bother with all that? Those numbers are baked into this published specification: At cruise, it burns 149 kg of fuel per 100km. Using that number, we can easily calculate the equivalent weight of batteries.
The energy density of Li-ion is about 1/50 that of jet fuel (well, 1/60 for Tesla batteries, but let’s do better). So if you could convert an ATR-72 to electric power, all else being equal you would need to haul 7450 kilos of batteries for each 100 km of cruise. Climb and descent equal roughly an hour of cruise, so to climb, fly 100km then descend to an airport would require 14,900 kg of batteries. Since 100km is far too short for even a regional airliner, let’s use a more reasonable range of 500km for an electric plane, and to be generous we’ll only require a VFR reserve of 45 minutes. For that, I get just over 50,000 kilos of batteries. That’s about twice the max takeoff weight of a fully loaded ATR-72, or about six times the useful load of an empty ATR-72.
Your job is to explain how you get from there to something usable. Just how will the switch to electric be SO efficient that you can reduce the above numbers to something reasonable? I could believe that electric motors will be smaller, the nacelles might be smaller, etc. But these don’t make up anywhere near the bulk of drag on an airplane. Much of the drag is an intrinsic function of frontal cross-section (form drag), skin area (parasitic drag), weight and wing length (induced drag). None of these change when you move to electric. Weight probably gets worse, meaning higher induced drag for the same performance.
You are also assuming that electric motors are superior in an airplane situation. But one of the characteristics that makes electric motors good for cars is that they make maximum torque at zero RPM, which really matters in cars. But airplanes are the opposite - you want maximum torque at higher RPM, and you don’t care how much there is when the props are at 0 rpm…
Overall engine weight including starters, oil and fuel systems, fuel lines, etc might be lower with a move to electric, but that’s going to be offset by the fact that you have to fly at maximum weight the whole trip, which will increase induced drag.
But hey, maybe you can point to some areas of inefficiency due to internal combustion engines that are so great that a shift to electric can overcome the problem of energy density in the batteries. But I really doubt it.
If you have a better example of a highly efficient airplane, we can start with that. But using numbers such as the minimum energy required to lift a weight a certain height is a total non-starter. No airplane comes close to that kind of theoretical efficiency. And as the Beech Starship showed, even highly documented theoretical gains in the design failed to materialize when real-world constraints were applied as they built the things.
Wikipedia has a decent list. It suggests that the Rolls-Royce T406 achieves 10.4 kW/kg, while the Emrax 268 brushless motor gets 10.05 kW/kg. In both cases that excludes the fan/prop.
So, pretty comparable I guess. The motor should be denser, which might make it more efficient on average–less blocking of the airstream. And motors are more granular, so you can choose to have a bunch of small ones or a few big ones, depending on what works best. Turbines are so complicated that you really don’t want to have more than two of them.
There’s been some research into superconducting motors. That could put motors well past where turbines are now.
Wrong, you are again comparing thermal energy of fuel to electrical energy in batteries.
Just like with cars, an electric car uses much less electrical energy for a given distance than a fuel car uses thermal energy.
Jet fuel contains 12kWh thermal energy per kg. Using numbers equivalent to yours, an ATR-72 consumes 750kg/hr of fuel cruising at 500km/h. That’s 9000kWh/h (9MW) of thermal energy.
But the PW127 only has SFC of 300g/kWh. It only makes 2400kWh of mechanical shaft energy from that fuel. (1200kWh/h or 1,2MW per engine, seems about right)
An electric motor is much more efficient, needing only 2700kWh of electrical energy to make that same mechanical energy.
The mechanical energy is what propels the aircraft.
At 4kg/kWh that’s 10,000 kg of batteries per five hundred kilometers or “only” 2000 kg per 100km cruise.
Still a lot of course. But not nearly as bad as your calculation.
Of course you do. That’s why it’s called gliding: traveling without power input. The altitude doesn’t come for free, which is why I included it in the calculation.
No, that’s the same dumb way of looking at problems that got us shitty electric cars before Tesla came around. My grandmother’s EV Focus only goes 70 miles on a charge because some schmuck figured the only empty place to put batteries was in the trunk and customers wouldn’t buy it if they filled it more than halfway.
Current aircraft have spent decades being optimized around the constraints of hydrocarbon power, and there’s no reason to think they’d be optimal for electric power either. In fact there’s lots of reasons to think they’d be distinctly suboptimal.
I don’t care what the ATR-72 burns per hour. I’m sketching out an electric plane, not a converted hydrocarbon plane. I used the ATR-72 only as a baseline because it meets the basic constraints for the kind of aircraft that will be interesting initially, and I don’t want to recalculate things like the required wing area. In practice, it will be possible to do better than that.
Again, all of this is ass-backwards. I can’t easily characterize all the ways in which some particular heat engine inefficiently converts chemical energy to thrust. EVs bypass huge swaths of that. The equations are very simple. My Tesla gets very close to the theoretical numbers just from looking at aero drag + rolling resistance. Airplanes are simpler yet, since there’s just the aero term.
You can design an electric motor to have whatever characteristics you want. The torque at zero RPM is handy in a car because you can get rid of the gearbox; that’s not useful on a plane, as you note. But you can still design them to be most efficient at a particular power level.
I *used *numbers from a real world craft. The Cd=0.022 is from a Learjet 24. A Boeing 787 is 0.024, so not much higher. This includes parasitic and induced drag in level flight.
It should be close. Airplanes achieve efficient lift for the same reasons why propellers achieve 85+% efficiency. It’s because they work by accelerating large volumes of air by modest velocities. In fact it’s better for wings than propellers because propellers have to accelerate air already at high speed to a higher speed, whereas wings redirect air with a zero vertical velocity component. The resulting air has very little kinetic energy in it.
It’s counterintuitive, because even in level flight the plane has to support its entire weight, and one gee of acceleration applied horizontally is enormous. But at typical cruise speeds, induced drag is less than parasitic drag–a lot less, at high enough speeds.
I’m certainly willing to be corrected here, but not by citing the inefficiency of craft already known to be inefficient. Unfortunately, I don’t have an aeronautics text handy which might give more insight into the problem.
Engine weight is not a particularly big part of overall airplane performance in large jets. A typical number for a regional jet is that about 10% of the weight of the empty plane is in the engines. So if you managed to come up with an amazing electric motor with twice the power per pound of a standard engine, you’d save about 5% of the weight of the empty plane. In the meantime, you’re adding huge weight in batteries - weight that has to be carried all the way to landing.
And in reality, we are 't likely to see much more than small incremental improvements in efficiency. Nothing like 2:1.
According to this PDF from ATR about optimizing fuel consumption, with some reasonable tricks you can optimize the block fuel for a 300NM (500km) flight to around 800kg.
That would be about 3000kg of batteries according to my earlier estimation.
The ATR-72 has a fuel capacity of 5000kg.
So replacing fuel one-for-one (by weight) with batteries, you can make an ATR that does 500km flights on batteries and just barely penalizes the useful payload (less than 1000kg out of 7500kg
as a rough guess).
There are a great many 500km ATR flights in my country.
That would be pretty badass for all kinds of reasons. You could accelerate the craft far past the usual point of takeoff. Not only would you save the energy of the takeoff and further acceleration, but you would avoid a very inefficient flight regime: induced drag (drag from lift) is dominant at low speeds, so if you can skip right to high speed you can avoid all that.
Best of all, you could avoid a a huge class of engine out accidents. The plane could have a high enough velocity to glide back to the airport without any additional power.
Specialized gliders do this already, with a winch on the ground. And of course there are carrier catapults. Just gotta scale that up a bit.
