How many kilowatts to get an electric 747-8 airborne?

Question

Reading about the Airbus E-Fan and it's dual 30 kW thrust fans caused me to wonder: assuming the need for four electric fans, how many kilowatts would be required to power each fan with enough thrust to get a Boeing 747-8 airborne?

Answer 1

First, let's figure out how much power a 747 needs to takeoff:

Assume:

• Engine thrust = 284 kN
• Takeoff speed = 170 knots
• Takeoff power = 90% max power

Using $P=Fv$, converting the variables to SI units, we get $$Power=88,948,800 W$$

Or, in other words, around $90MW$.

_{EDIT: This calculation is only correct if the engine efficiency is 100%, because it is based on work done. For a realistic estimation, see Jan Hudec's answer. I'll leave the rest of this answer based on 90MW as I'm too lazy to update the numbers.}

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But just exactly how much is 90 megawatts?

An average laptop computer consumes 20W on daily use. You can power 4.44 million laptops with this power.

A high-end desktop consumes 300W under heavy load. You can power 300,000 desktop computers. If you stack 20 of them in a server tower and put 100 towers on each floor, you need a data center with 150 floors to fit those machines in.

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Let us take this a step further and evaluate the total energy needed for a flight. Assume all required energy comes from onboard energy storage (i.e. no solar panels / windmills). Let's also assume a full power climb to cruise altitude of 15 minutes, 50% power cruise for 4 hours, and a completely idle descent which does not consume any power at all.

Using $E=\int P dt$, we get the total energy needed for a flight is $$720,485,280,000 J$$

With that amount of energy, you can pull a 735 kiloton object up 100m. If all those weights are water, that's close to 300 Olympic swimming pools.

If you power the whole plane with batteries, you'd need 47 million AA batteries.

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Of course, if you manage to get 14 of these electrical 747s together, you can generate 1.21 gigawatts and get Back to the Future.

Answer 2

I can't quickly find the necessary numbers for 747-8, so I'm going to go with the numbers for Rolls-Royce RB.211-524G-T, one of options on 747-400, taken from here.

The engine can produce static thrust

$$ T_s = 58,000\ \mathrm{lb}_\mathrm{f} = 258\ \mathrm{kN} $$

using mass flow rate of

$$ \dot m = 1,604\ \mathrm{lb/s} = 727.5\ \mathrm{kg/s} $$

Now due to principle of action and reaction, the engine must apply the thrust to the air passing through it. We know that $ F = ma = m\frac{\Delta v}{t} $ and using $ m = \dot m t $ we get $ F = T_s = \dot m \Delta v = \dot m v_e $, where $\Delta v$ is change of velocity of the air inside the engine, which is equal to exhaust velocity $ v_e $, since we are starting with still air. We can solve for exhaust velocity:

$$ v_e = \frac{T_s}{\dot m} \approx 355\ \mathrm{m/s} $$

That's pretty fast. In fact, it is mildly supersonic, though in practice it is average of the just subsonic bypass flow and the faster hot core flow (in which the speed of sound is higher).

We also know that kinetic energy is $ E = \frac12 m v_e^2 $ and, deriving by time, power is $ P = \dot E = \frac12 \dot m v_e^2 $.

So we can substitute:

$$ P = \frac12 \dot m \frac{T_s^2}{\dot m^2} = \frac{T_s^2}{2\dot m} = 45.7\ \mathrm{MW} $$

That's one engine.

Total power is 183 MW.

For 747-400 with this particular type of engines. Different engine options will have slightly different powers, because they use somewhat different mass flow rates for the same thrust. And of course, 747-8 will have a bit more.

The above also used static case. At higher speeds, the pressure recovery allows turbine engines to produce even more power, but by the end of take-off run, the pressure recovery is not significant yet and at altitude, the lower overall pressure limits the output, so this does correspond to the maximum power the engine develops.

With some losses, we are looking at at least 200 MW power input. All the numbers in kevin's answer need to be multiplied by 2-and-a-bit.

Answer 3

Well, the GEnx-2B67 engines with which the 747-8 is fitted each produce a maximum of $ 284\ \mathrm{kN}$ thrust (according to its Wikipedia page). This translates to a total of $1136\ \mathrm{kN}$, since it has 4 engines.

Power, which has the units Watt, is the product of force (thrust) and velocity.

The 747 has a takeoff speed of about $290\ \mathrm{km/h}$, according to this page.

Assuming the 747 requires all its available thrust for takeoff, the power it requires can be found by multiplying its velocity by the force pushing it forward (in the case of an airplane, its thrust).

First convert these to SI units: $$1136\ kN = 1136000\ N$$ $$290\ km/h = 80.55\ m/s$$

We now multiply these to get the power:

$$P = 1136000 \cdot 80.55 = 91504800\ W = 91.5\ MW$$

So, for takeoff, the 747-8 will require approximately $91.5\ \mathrm{MW}$ of power.

This is a very crude estimate. Firstly, the maximum thrust listed on the engine's Wikipedia page is likely its maximum static thrust, when it is standing still. When it is moving forward at $290\ \mathrm{km/h}$ at takeoff, this thrust will be slightly less.

Also, if you want to provide this power with electric motors and fans, there are inefficiencies associated with both the electric motors and the fans, which will mean that the actual motors used will in total need to be able to produce quite a bit more than $91.5\ \mathrm{MW}$.

The weight of such motors will be monumental. And these motors will need to be fed by an energy source, batteries or hydrogen fuel cells come to mind, and the weight of these will be just as large.

Electric motors are quite efficient in converting energy, but the energy storage devices that will feed them have very low energy densities (read energy per weight).

Internal combustion engines are quite inefficient in converting energy, but the fossil fuels that they use have an extremely high energy density.

Answer 4

I know this isn't the exact answer to your question, but perhaps this will give you some perspective that other answers haven't.

The Airbust E-Fan electric motors produces 30 kilo-Watts of power. That is equivalent to about 40 horsepower. There are two of these, for a total of about 80 horsepower.

The first airplane I soloed in was a Cessna 150. It was fitted with a 200 cubic inch four cylinder internal combustion engine and could barely carry two people.

Additionally, the Wright Flyer was powered by a four cylinder internal combustion engine that produced 12 horsepower.

As others have shown, the total takeoff power of the 747 is about 90 Megawatts, which is about 120,000 Horsepower. In other words, you would need about 1,200 of the Cesnna 150 engines, or 3,000 of the Airbus E-Fan motors, or 10,000 of the Wright flyer engines to produce the same amount of power.

Answer 5

how many kilowatts would be required to power each fan

Apart from the calculation of the needed power, this is actually a very nice question to understand why all the fancy electrical-driven aircrafts that the bright future will bring in our skies have a lot of propellers on their wings, the so called "distributed propulsion".

As a first approximation, the power needed by 4 fans/propellers can be calculated, as usual, with the simple momentum theory:

$P=\sqrt{\frac{T^3}{2 \rho A}}$

Where $\rho$ is air density and $A$ disk area $\pi r²$. Each 747-8's GEnx-2B67 engine has a fan with some 2.8m diameter and delivers some 290kN of thrust $T$. Substituting these values in the previous equation gives a total needed power at sea level of:

$P=4\cdot \sqrt{\frac{(290'000)^3}{2 \cdot 1.125 \cdot \pi 1.4^2}}=168MW$

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An electrical locomotive needs only some 5MW so that's a lot... but it can be reduced: if we suppose that each propeller can be hung at the same height of the wing (instead of under the wing), then their radius can be doubled anyway respecting ground-clearance. But the total needed power becomes:

$P=4\cdot \sqrt{\frac{(290'000)^3}{2 \cdot 1.125 \cdot \pi 3^2}}=78.3MW$

Much better... but it can be further reduced: from root to tip, the 747's wing is some 30m long. That means that actually five of those propellers could be accommodated on each side of the wing, for a total of ten. Now each propeller must produce ⅒ of the total thrust i.e. $T=\frac{4 \cdot 290'000}{10} = 116'000N$, requiring this time a total power of:

$P=10\cdot \sqrt{\frac{(116'000)^3}{2 \cdot 1.125 \cdot \pi 3^2}}=50MW$

And what if we use contra-rotating propellers? Then we could pack twenty of them on the wing, each producing a thrust of $T=\frac{4 \cdot 290'000}{20} = 58'000N$ for a total needed power of:

$P=20\cdot \sqrt{\frac{(58'000)^3}{2 \cdot 1.125 \cdot \pi 3^2}}=35MW$

And if we duct each contra-rotating propeller? Then a 30% thrust increase can be expected, lowering the total power to:

$P=20\cdot \sqrt{\frac{(0.7 \cdot 58'000)^3}{2 \cdot 1.125 \cdot \pi 3^2}}=20.5MW$

Increasing the number of propellers, the power required can be reduced by a factor 8! Nice! And even nicer is that twenty of these electrical engines might be used, for a total engine weight of $20\cdot 100=2'000kg$ instead of $4 \cdot 6'150=24'600kg$!

So through the use of distributed propulsion, both the needed power and the engines weight can be roughly reduced by a factor 10.

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Obviously these are simple quick and dirty calculations which do not take into account any aerodynamic interference among rotors, any structural issue or any practicability of the concept. Anyway, considering that the energy density of lithium-ion batteries is 10 times lower than jet-fuel, the volume being the same and 40 times lower, the weight being the same, one sees that distributed propulsion is the only way to go towards aircraft electrification.

Answer 6

It's completely unfeasable. Without all the calculations (accurate as they may be) a very large Boeing (737 or 747) can't be make airborne with electricity/batteries alone. Maybe hydroelectricity could.

A Tesla mark 3 has about 6200 'laptop batteries'. They weigh about a 1000 kilo's. The electricity regenerated per battery is not that big: about a laptops requirement. The weight would be significant versus the weight of the plane, or the weight of the plane plus the weight of the kerosine.

For making it available for a very large jet (!) fueled engine, enormous techical achievements have to be made.

Can an elictrical powered engine be converted in a jet fueled one? Not with current technology, I think.

John

Answer 7

A single 777 engine develops 108 megawatts of power, with 2 engines, that is 216 megawatts...you are going to need more than 90 megawatts taking off in a 747 if you want to live.