What energy density is required for the batteries in order to make an all-electric analogue of the Cessna 150 or similar plane?

Question

I think I have some general understanding that at the present moment even with the best possible batteries, a 2 seater training aircraft (like the C-150) would be very impractical. I'm not even talking about electric analogs of the Piper Seneca, let alone airliners. At the same time, compared to piston engines, electric engines are less heavy for their output and far more efficient (I think around 90+ %). Pipistrel claims that Siemens has a 85kW engine which is only 14 kg. I also like the idea that electric engines offer us some flexibility in the energy distribution.

What sort of progress do we need in terms of energy density of current batteries, to make a 2-4 seater aircraft real and equivalent in terms of range and performance to its piston analogues? Also, will it be cheaper to charge as opposed to the cost of liquid fuel? (Assuming the electric engine is similar to a piston one in terms of power.)

Answer 1

Short Answer

Batteries would need to be somewhere around 16.7MJ/kg to give the same range and performance as liquid fuels, this is about 18.5 times the capacity of the best lithium-ion batteries. Price-wise it will cost about 30-35% to charge your airplane as opposed to fill it with liquid fuels at today's prices.

Long Answer

This is a good question which is hard to give an exact answer to, so this will be more of a Fermi Approximation. To answer this you need to look at the energy contained in the fuel and the efficiency of the engine used.

Looking at the energy of the fuel I will use Specific Energy, which is the energy stored in a material per unit of mass. Specific Energy is related to Energy Density, which is the amount of energy contained per unit of volume. Often the terms are interchanged.

The specific energy of avgas and jet fuel is about 43 MJ/kg. The best lithium-ion batteries top out at about 0.9 MJ/kg (the batteries in a Tesla are about 0.7 MJ/kg), so they have a fraction of the storage of liquid fuels. The best battery technology in theoretical development (Lithium-air) has a theoretical maximum of 41 MJ/kg, more realistically they'll get 1/4 to 1/3 of that from the technology, which is still vast.

Internal combustion engines are about 35% efficient, the other 65% is wasted whereas electric engines are much closer to 90% efficient or more.

There are other factors to consider:

• Weight: fuel tanks, piping and pumps take weight, and electric motors are much lighter than internal combustion engines. The batteries would need a storage structure but an electric system would probably be lighter overall
• Battery inconsistencies: avgas supplies a consistent amount of energy in every single drop whereas batteries will sag as they discharge (this is regulated to give consistent power to the motor, however at some point levels will drop below what is regulatable, so not all the power in the batteries will be usable). Also, over time they lose capacity and become less efficient. Both these factors mean that you'd want to build in extra battery capacity to compensate

So I'm going to assume these two will compensate for each other, the weight savings from going electric will be offset by the need for extra capacity to ensure consistency. Assuming that all other factors are equal (propeller efficiency, etc) we can extrapolate the actual efficiency of the systems to get an approximate figure of an internal combustion on avgas: 35% of 43 MJ/kg = 15 MJ/kg of actual benefit. We can use that figure to determine what specific energy we would need from batteries to get the same amount by dividing by electric motor efficiency: 15 MJ/kg / 0.9 we get 16.7 MJ/kg.

So batteries would need to store 16.7 MJ/kg to give us the same energy as liquid fuels, but how does it compare to existing battery tech? Right now commercially available battery technology is about 0.9 MJ/kg, so it would need 18.56 times more storage capacity (16.7 / 0.9 = 18.56) to supply the same amount of energy.

As for energy costs this is going to vary a lot depending on fuel and electricity prices over time and place; we can use some of the same figures above to work the numbers. I'll assume an airplane that holds 40 US Gallons (150 liters) as the math is easy and it's about the capacity of a Cessna 172. I'm going to do separate calculations for the US and UK to see how they compare:

• In the US: Avgas is about \$5.20 per US gallon at the moment, that's \$208 to fill up your airplane of 40 gals. Avgas has about 34.2 MJ/liter of energy, multiply that by 150 liters for 5130 MJ of energy. Internal combustion engines are 35% efficient and electric engines 90%, so we can calculate that an electric airplane would need 5130 * 0.35 / 0.9 = 1995 MJ of electrical power to fill up. 1995 MJ is about 554 kWh, at somewhere around \$0.135 per kWh it would cost you \$75 to charge an airplane with the same amount of energy.
• In the UK: Avgas in the UK is about \$2.23 per liter (\$8.47 per US Gallon (ouch!)), so it would cost \$334.50 to fill up an airplane. 554 kWh of electricity costs about \$0.17 per kWh, so it would cost \$94 to charge the airplane

EDIT: I did consider the efficiency gain from weight loss as fuel is burned, i.e. a variable mass system, I left it out of the answer as it isn't significant compared to the other factors in what is already an approximation. It has to do with the fuel fraction, which is the percentage of the aircraft weight that is fuel, which on a light piston single is comparatively low. A Cessna 172 carries about 40 gallons of fuel, 38 of which is usable, weighing about 228 pounds as opposed to a typical take-off weight of 2200-2300 lbs. In other words its fuel fraction is about 10%, even on a long range flight using every bit of fuel you will only lose 10% of your weight, and I would approximate that you'd get around 5% benefit from it. This was not enough of a factor compared to the other considerations, and it would probably be offset by regenerative descents where the electric motor actually generates power from the spinning prop.

For airplanes where a more significant portion of the weight is fuel, i.e. a higher fuel fraction, efficiency gains from variable mass is a much more significant factor, for example an A380's fuel fraction is 44%.

Answer 2

An all-electric craft similar to what you described has actually been created (IEEE Spectrum did a nice article about it). The batteries are described as 260 Watt-hours per kilogram with a power plant output ratio of over 5 kW/kg. The two-seat craft is designed for training flights, and the fuel for each hour of flight time costs less than one-eighth the cost of a conventional-fueled aircraft. The aircraft's official specs indicate that the 2-seater version has about 3 hours of flight time (4 hours for the 4-seater). It uses the same sort of "supercharging" outlets that electric cars use, but there's no specific numbers regarding recharge time.

The article discusses the technical development of the craft and many of the scaling issues that you mentioned; some were able to be engineered around, but some (including battery storage density) remain a problem for the foreseeable future. To summarize, electric drive is currently competitive only in slower aircraft. Drag increases with the square of speed, more drag means more batteries needed to supply the power, and that means more weight. Battery energy density ultimately limits the size and speed of an all-electric plane, but current technology appears to be plenty sufficient to produce a practical aircraft.

Answer 3

You're missing the point. Electric vehicles don't even try to match energy density of petroleum.

This isn't a matter of yank out a Lycoming and drop in a VFD and induction motor and battery bank. Even electric cars take a blank-slate approach. They don't simply mimic the ratio of powertrain mass to rest-of-vehicle mass. They design a new vehicle that is workable. You bet you'd do the same thing on an airplane.

Keep in mind an electric motor is much smaller and lighter than an aircraft engine, and power can be distributed around the aircraft, e.g. a multi-motor electric plane is perfectly reasonable and almost inevitable. Whereas a multi-engine gas plane is a wildly different creature with very different certification.

That means your props are in better places, making more of their swept area, instead of stuck in front of bulky engines or otherwise wasting energy trying to bend air around a fuselage. They could blow the wing, or be centerline thrust using 2 motors per contra-rotating prop. You want more rudder authority at low speed? Blow the rudder. You can put 'em anywhere.

Also a factor is that aircraft have nice large flat surfaces that lend themselves to solar panels. This adds mass but also adds range in day-flight, which then raises the question whether this is a day-only or night aircraft. All that calculus has to go into vehicle design.

Another part of the calculus is lightweight materials like composites. It doesn't make sense to use Dreamliner or F-22 construction methods on a cheap basic GA aircraft, but when it is critical to range/performance, you revisit that. And it could become affordable in mass production.

You bet energy density helps but it may be possible to build a usable aircraft with existing tech. You just don't know until you iterate on vehicle design and see where it goes. That's not cheap.

Answer 4

Here's a rule of thumb: you can assume that the range of a practical electric aircraft, in nautical miles, is approximately equal to the energy density of its batteries, in Wh/kg. Today, that number is about 250, tops.

That rule of thumb assumes that the cruise L/D is 20:1. If your design gets 10:1, halve the range.

Is 20:1 realistic? Well, a Cirrus SR22, a modern all-composite airplane, gets about 17 at a best L/D around 90 kt. So, 20:1 is ambitious, but realistic.

If your idea of "practical" is a 160 kt cruise speed, you'll need an airframe with an L/D of 20:1 at 160 kt, that also has a big enough wing to slow down to 60 kt as required by Part 23. That's hard. Or, you can get 10:1 at 160 kt, meet Part 23 requirements, but halve the range.

If your idea of "practical" is a range of 600 NM, you'll need batteries with 600 Wh/kg. They don't exist.

If 90 kt cruise for 250 NM is your idea of "practical", the technology is good enough today. And, 120 kt cruise for 250 NM may be feasible with clever airframe design.

Let's turn to the system engineering behind this answer.

Energy required = Force x Distance = Drag x Range = [Weight / (L/D)] x Range = Energy stored in the batteries

$E_{req}= F \cdot x = D \cdot R = \frac {W\cdot D}{L}\cdot R = E_{bat}$

With:

• $E_{req}$ = energy required
• $F$ = force
• $x$ = displacement
• $D$ = aerodynamic drag
• $R$ = range
• $W$ = weight
• $L$ = lift
• $E_{bat}$ = energy from the battery

So,

$R \approx \frac{ E_{bat}}{W}\cdot \frac{L}{D}$

Weight = Payload + Electric power system weight + structural weight

For a practical aircraft, the structural weight is approximately half of the total weight, maybe a little less. Let's call it 0.5 if we include the weight of the electric motor, which will scale with the aircraft weight.

So, if the structure including the motor is half the total weight, we have

$W \approx 2 (W_{payload} + W_{bat})$

Let's define $k$ as the fraction of the lifted weight (i.e., Payload + Battery) that is battery.

So, $k = \frac{W_{bat}}{W_{payload}+W_{bat}}$, and therefore $W_{payload} + W_{bat} = \frac{W_{bat}}{k}$.

So, $W \approx \frac{2 \cdot W_{bat}} { k}$

Then,

$R \approx \frac{E_{bat}}{ W_{bat} }\cdot\frac{k}{2} \cdot \frac{L}{D} $

This needs one adjustment: the energy available from the battery in practice is not $W_{bat}$, but rather $U \cdot W_{bat}$, where $U$ has a value of about 75%. This is because if you fully charge and discharge the battery on each cycle, using the full amount of $W_{bat}$, the battery will not last for many cycles.

So, we adjust to show

$R \approx \frac{E_{bat}}{ W_{bat} }\cdot\frac{k}{2} \cdot U \cdot \frac{L}{D} $

Now, that's all in SI units, where Distance is in meters, energy is in joules, and weight is in Newtons (not kg!). Let's do a unit conversion:

$R = 1852 \cdot R_{NM}$

$E = 3600 \cdot E_{Wh}$

$W_{bat} = 9.8 \cdot M_{bat, kg}$

So,

$1852 \cdot R_{NM} \approx \frac{3600 \cdot E_{Wh}}{ 9.8 \cdot M_{bat, kg} }\cdot\frac{k}{2} \cdot U \cdot \frac{L}{D} $

and thus

$R_{NM} \approx \ 0.0743 \cdot \frac{E_{Wh}}{M_{bat, kg} }\cdot\ k \cdot \frac{L}{D} $

or, if we assume $\frac{L}{D} \approx 20$

then

$R_{NM} \approx \ 1.48 \cdot\ k \cdot \frac{E_{Wh}}{M_{bat, kg} }$

The maximum possible range is if $k = 1$, i.e., there's no payload, and the aircraft carries nothing but battery.

But, for a more practical design, if we set $k = \frac{1}{1.48} = 0.67$, i.e., the battery weighs twice times as much as the payload (think of it as 200 kg of battery, or 440 lb of battery, per person carried), then

$R_{NM} \approx \frac{E_{Wh}}{M_{bat, kg} }$

Which is the rule of thumb: range in nautical miles equals energy density in Wh/kg.

More precisely,

$R_{NM} \approx \frac{E_{Wh}}{M_{bat, kg}} \cdot \frac{\frac{L}{D}}{20}$

You could add more range by having a bigger battery fraction k, but going from battery weight of 2 x payload to 4 x payload only adds 20% to range - not very exciting.

Note that the basic rule of thumb assumes quite a high $\frac{L}{D}$ ratio of 20:1 in cruise. Note also that it says nothing about speed or altitude flown: ultimately, all that matters, for range, is cruising $\frac{L}{D}$ and battery energy density.

Answer 5

It all depends upon what range or performance profiles you would like out of the airplane. Electric aircraft - or at least prototype electric aircraft - which have similar performance in terms of speed, useful load, etc to petroleum powered analogs. It’s just that the energy densities of the batteries do not allow for a useful endurance. Current designs like Pipistrel’s Alpha Electro have about a 1 hour endurance plus a day VFR reserve of 30 minutes of power in an economy cruise. When compared to an analogous Rotax powered LSA with over 6 hours of endurance plus reserves, you quickly see how limited this is.

It would be better to ask as to what energy density would be needed to match the performance and endurance of existing petrol powered light aircraft while matching their useful load. As mentioned above a density of approx 15 MJ/kg would allow for this. This will require a considerable leap in electrochemistry technology to store and deliver that kind of energy reliably and safely.

Answer 6

This plane is comparable to a C150, altho less baggage area if you wanted to take a short trip. We upgraded from a C150 to a C177B when baggage area (and a child going on 2 years old) became a driving factor. Had even purchased some cable and turnbuckles to try and find a way to secure a car seat (which ended up in the basement, never used, as the C177 was purchased pretty quickly (lucky timing on the GA marketplace)).

https://www.pipistrel-usa.com/alpha-electro/

I don't see if they talk about battery energy density tho, might have to download one of the information booklets to find that.

High capacity, easy to replace dry-type 12V battery

aircraft fitted with 60 kW electric motor.

nominal battery capacity 21 kWh

engine 50 + kW @ 2100-2400 rpm

standard endurance, traffic patterns 60 min + reserve

standard range at cruise 80 kts 70 NM (130 km)

Standard Battery System

Maximum voltage 399 V

Minimum voltage 297 V

Recommended voltage range for storage 345 V - 365 V

How long does it typically take to charge the batteries with the different chargers? (20%-95% range)

6hrs with 3kW charger, 1h 40' with 10 kW charger, 1h 5' with 14 kW, 45 minutes with the 20 kW charger

How heavy are the batteries and can I swap them over myself? Each battery pack is 53 kg. Yes, you can remove the pack with no extra help

What kind of batteries are installed? Lithium ion. Cells are manufacturer by Samsung. Battery box design and assembly is Pipistrel, Battery Management System (BMS) is also designed and manufactured by Pipistrel

Enough info there to do the math?