# Context

To understand what the current level of electric airplane propulsion is with respect to large passenger airliners I developed the following question:

# Question

How many battery swaps would be required to fly the same route as the first non-stop transatlantic flight with an electric equivalent of a passenger aircraft able to transport +-600 passengers at once with a cruise speed of Mach 0.85?

# Assumptions

To minimize the ambiguities regarding the question I made the following assumptions:

1. The first non-stop transatlantic flight took place from Lester's Field, near St. Johns, Newfoundland on June 14,1919, and landed June 15,1919, at Clifden in Ireland. The time for the crossing was sixteen hours, twenty-seven minutes.
2. It is assumed there is currently no battery technology that is able to perform the flight for 600 passengers without reloading during the flight, a mid-flight battery swap system is assumed.
3. The battery swaps are assumed to be possible at 34.000 feat altitude at mach 0.85.
4. The battery is assumed to be fully charged upon placement in the aircraft.
5. A suitable runway can be built/is available at/near the departure and arrival destinations.
6. Max thrusts required for take off and landing are ignored based on assistant launcher aircraft/rialguns. I.e. the aircraft is only considered in (optimal) cruise conditions.
7. The batteries can be located, in fully charged state on any position on the surface of the globe along the route.
8. Current (2020-06-30) battery technology is assumed. In particular the specs for the Pipistrel Velis Electro with 11.0 kWh, capable of delivering a constant 49 kW is assumed.
9. A lithium battery density of 500 W/kg and 900 W/l is assumed. (Optimistic assumption of portability of technology that is not been proven in aviation applications despite cyclic acceleration loading, vibrational loading, temperature loading).
10. The equivalent electric engines are assumed to be required to produce the same thrust in newtons as their combustion equivalent engines at 4x348=1392 kN.
11. Speed of sound at cruising altitude of 34.000 ft of 240 m/s is assumed.
12. A constant power usage of 1392.000x0.85x240=283968000W=283968 kW =~ 284 MW is assumed required to be continuously delivered by the electric jets.
13. The mass penalty of the battery swap system is assumed 5000 kg. (Random unfounded optimistic estimate)
14. The mass penalty is electric jet-engines is assumed to be twice the mass of the A380 Trent 970-84/970B-84 engine of 6246 kg is used, resulting in a mass penalty of 4x6246= 24.984 =~ 25000 kg.
15. An additional mass penalty of 10.000 kg is assumed due to requirement of aircraft reconfiguration requirements. E.g. concerning electric cabling, lack of aircraft design optimization experience with large replaceable batteries in stead of wing-integrated fuel tanks. (Random optimisitic estimate/guess)
16. The flight should be reproducible at least 95 % of the days of the past decade, to prevent estimating on unlikely favorable weather conditions.

# Notes

Different assumptions can be made, please use comments to argue why assumptions could/should be adjusted, and keep answers for explaining how one can come up with the estimated amount of required battery swaps.

• It's really not this complicated. You need about 50 because electric motor is about 2x more efficient (near 100% vs around 50%) than a turbine and battery is about 100x less energy dense (0.4MJ/kg vs 40MJ/kg) than jet fuel. – user3528438 Jun 29 at 23:30
• 500Wh/kg is kind of overly optimistic as your linked source it's typically 90Wh/kg. – user3528438 Jun 30 at 0:07
• @user3528438 thank you, I have updated the reference (to the reference used in the claim for the 500 W/kg in the article). Your approach provides a nice sanity check, which also appears to indicate I made either a computation mistake or too conservative assumptions. – a.t. Jun 30 at 0:21
• This question does not make any sense at all. If you want to recreate the first transAtlantic flight, you do not use a 600 passenger airliner travelling at Mach 0.85. (If indeed you could even get a 600 passenger electric plane to that speed.) You use something equivalent to the twin-engine biplane used by Alcock & Brown: tps://www.history.com/news/first-transatlantic-flight-nonstop-alcock-brown – jamesqf Jun 30 at 4:14
• With so many assumptions and handwaving I feel this question has no real value or broad appeal. – Sanchises Jun 30 at 6:43

# Assumptions

1. The total mass penalty is assumed of 5000+25000+10000=40.000 kg is assumed w.r.t. a conventional A380. This is assumed to go at the cost of fuel tank/battery costs.
2. The A380 has 253,983 kg fuel capacity, leaving 213.983 kg of battery capacity.
3. The distance of the journey is 3040 km.
4. The A380 cruises at 903 km/h.

# Computations

With a power density of 0.5 kWh/kg, a total capacity of: $$213983\cdot 0.5=106991.5 kWh=106.991MWh$$. Since the engines require a continuous flow of 284 mWh, the complete battery set would have to be swapped every $$\frac{284}{107}=2.65..$$ times per hour, which is approximately every 1356 seconds. Since the transatlantic flight covers 3040 km at a velocity of 903 km/h, the flight takes $$\frac{3040}{903}=3.37$$ hrs. Therefore, according to the set of assumptions, $$\frac{3.37\cdot 3600}{1356.22..}$$ = 8.94..=~9 battery swaps would be required, with a station approximately every 337 kilometers.

# Doubts

The assumptions made in the question appear to be highly inaccurate, hence a more detailed analysis of the assumptions, in partciular to the battery swap system mass and scalability of the battery, engine and thermal systems could perhaps illustrate the (lack of-) accuracy of the estimate.

• 213983⋅0.5=106991.5Wh=106.991kWh= 0.107mWh . the second term should be kWh, thrid MWh (M not m), forth GWh. so your answer is off by 1000 times. – user3528438 Jun 30 at 0:01
• I confirm the mistake in the math. – h22 Jun 30 at 5:51
• @user3528438 Thank you, I corrected the mistake by switching the units from Wh to kWh and propagating the impact of the correction through out the remainder of the estimate. – a.t. Jun 30 at 9:25
• Power is in MW (megawatt), not mWh (milliwatt hours). Also you completely neglect that MLW is much smaller than MTOW so unless you ditch the batteries before landing you're in trouble. – Sanchises Jun 30 at 9:40
• @Sanchises Thank you, I corrected the units for power in the question from mW to MW. Yes, those neglections are implicit to the assumptions stated in point 5 and 6, pertaining to assuming a suitable runway can be built, and that the aircraft is only considered in cruise condition. – a.t. Jun 30 at 10:34