Name: Are Electric Planes Possible?
Uploaded: 2021-06-05T17:57:10.000Z
Duration: 10 min 24 s
Description: Thousands of YouTube videos with English-Chinese subtitles! Now you can learn to understand native speakers, expand your vocabulary, and improve your pronunciation...

Basic passive voice

This episode of Real Engineering is brought to you by Brilliant, a problem solving website

Over the past decade we have seen multiple industries looking to transition to renewable

fuel sources, and while we have been making huge strides in the production of renewable

energy, the technology required to allow every industry to use it has not kept pace. In theory

we could replace every coal burning power plant in the world in the morning, and manage

just fine, IF we had a reasonable way of storing that energy cost effectively and efficiently.

This energy storage dilemma is slowing our adoption of renewable energy, and one of the

industries this is most apparent is the aviation and aerospace industry. Elon Musk is running

around pushing electric vehicles and solar powered homes, yet every time a Falcon 9 launches

it burns 147 tonnes of fossil fuel. Boeing and Airbus are in a constant battle to create

the most fuel efficient plane, allowing their customers to save on ever increasing fuel

costs and increase their bottom line, yet they are still using kerosene, when energy

So what gives? Why isn't every industry on earth clawing at the prospect of transitioning

to renewable fuels? The aviation industry has one massive hurdle to overcome before

it can successful adopt renewable energy. The energy density of our storage methods.

Energy density is a measure of the energy we can harness from 1 kilogram of an energy

source. For kerosene, the fuel jet airliners use, that's about 43 MJ/kg. Currently even

our best lithium ion batteries come in around 1 MJ/kg. Battery energy is over 40 times heavier

So why is this such a huge problem. A plane flies when lift equals the weight of the plane,

so when we increase the weight, we have to increase the lift, which requires more power.

Needing more power means we need more batteries, which increases the weight again. So are caught

We could end the video there, but going by the demographic breakdown of this channel,

we can go a little deeper. To really understand why this is such a difficult problem, let's

do some back of the envelope calculations to convert two planes, the Airbus a320 and

a small personal aircraft like a Cessna, to battery power. Ultimately, we want to know

the power requirements of flight and how it will draw on the energy supply of the battery.

Animation 5 The work-energy theorem tells us that Work

= F × ∆x, where delta x is the distance over which a force acts. Power is work per

unit time, so P equals work divided time. (Work/∆t). Inserting our equation for work

and we get an equation for power that equals Force multiplied by distance divided by time,

otherwise known as velocity. Here delta v is the velocity of whatever is getting worked

on, in this case it's the air. When a plane is flying at a constant height, we know that

the the force of lift and the force of gravity are balanced. That means the upward force

of lift (Flift) has to be equal in magnitude to the downward pull of gravity, which equals

the mass of the plane multiplied by gravity So, the power required for lift equals the

mass of the plane multiplied by gravity and delta V.

The question is, what is delta v? It's the downward velocity of the air that the plane

pushes downward. So let's call it ∆vz. To find its value, we have to think about

The lift an airplane provides is equal to the rate it delivers downward momentum to

the air it displaces.This means that the force of gravity must be equal in magnitude to the

downward velocity of the deflected air, times the rate at which air gets deflected:

The mass of air that the plane affects is simply the volume of the cylinder that it

sweeps out per unit time, times the density of air. If we call the relevant cross sectional

area, Asweep, then the volume it sweeps out per unit time is A sweep times the velocity

of the plane. Therefore the mass flow rate equals the density of air times the cross-sectional

Now the only outstanding quantity that we don't know is the area of air affected by

the plane, Asweep. This isn't the cross sectional area of the plane, it's the area

of influence the plane has on the surrounding air. This changes with the relative velocity

of the plane and the air around it, but at cruising speed, the plane dissipates vortices

that have roughly the radius of the length of the plane's wings. Approximating this

circle as a square because we don't have enough ridiculous assumptions in this calculation,

the relevant area becomes L squared at cruising speed. Putting it all together, we have the

force lift needs to provide with this equation. This equation is simply telling us the plane

is sweeping out a tube of air and shifting it downwards, and that downward acceleration

of air is equal to the downward pull of gravity on the plane. So the plane avoids falling

by constantly paying the tax of streaming momentum downward via the air.

Rearranging this equation, we can now solve for ∆vz in terms of quantities we can easily

measure. And plugging this into our power equation, the power needed for lift is given

With this equation at hand, we can start noticing what variables really impact the energy requirements

of the plane. Notice that as the plane flies faster the power drawn by the engine actually

gets smaller, but this equation neglects to consider drag. It just so happens, that the

total power needed to fly is minimized when the force of lift and the force of drag become

equal, so we simply to to double our power requirements to get our total power requirement

Now we are getting a real picture of why increasing the mass of a plane is such a huge issue.

The mass component of this equation is not only squared, but also doubled. Doubling the

mass will increase our power requirements 8 fold.

With this knowledge in hand, let's start calculating the real world consequences of

converting an Airbus a32 To start, we can take the battery weight to be the usual mass

fraction that's devoted to fuel, about 20% of the planes mass for both. We also need

to take into account the fact that at the cruising altitude, the atmosphere is much

thinner than at ground level. For the Cessna, the density falls by factor of 2, and for

the Airbus, a factor of 3. Let's be generous, and take the specific

power of leading edge Lithium-ion systems, at about 0.340 kilowatts per kilogram kW/kg.

To meet the power demand, the Airbus and would need 31 tonnes of batteries:

10 500 kW / 0.340 kW/kg ≈ 31 000 kg (10)

while the Cessna would need just 100 kilograms:

35 kW / 0.340 kW/kg ≈ 100 kg (11) For the Cessna, this compares very favorably

with the typical weight of fuel it would carry otherwise, and it isn't terrible for the

Airbus, but this is just the power the plane needs at any one moment in time. What we are

really interested in is the weight of batteries we would need to match the typical range of

For the Airbus that' s a 7 hr flight from JFK to LHR and for a Cessna, that might be

a four hour flight from New York to South Carolina. The energy capacity required for

a trip is given this equation, multiplying the power required for flight by the duration

Again if we use leading edge figures for Lithium ion battery capacity, we can store about 278

For the Cessna, the equivalent battery weight is around 500 kg or just less than two thirds

the weight of the plane without fuel. For the A320, the required battery weight is around

260,000 250 000 kilograms or about 4 times the weight of the empty airplane! Compared

to the typical 20% that's allocated to fuel, this is devastating.

Now that we have a base figure for how heavy the batteries are going to be, we can re-calculate