Over the past decade, we have seen multiple industries looking to transition to renewable fuel sources, and while we have seen making huge strides in the production of renewable energy, the technology required to grant every industry to use it has not kept balance. 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 that are most apparent in the aviation and aerospace industry.
Elon Musk is running around pushing electric cars and solar-powered home development. Every time of launching of Falcon 9, burns 147 tonnes of fossil fuel which top Boeing and Airbus are in the constant battle to create the most fuel-efficient plane allowing customers to save on every increasing fuel cost and increase the bottom line yet take are still using kerosene energy from the grid is cheaper. So what gives?
Why isn't the Aviation industry transitioning to renewable fuels?
The aviation industry has one massive hindrance to cross before it can successfully adopt renewable energy. The energy density of storage methods. Energy density is a measure of the energy which we can harness from 1 kg of an energy source. For kerosene, the fuel Jet Airlines use that's about 43 megajoules per kg. Currently, Even our best Lithium-ion batteries come in around 1 megajoule per kg. Battery energy is over 40 times heavier than Jet fuel.
So why is this such a 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.
To understand why this is such a difficult problem let's do some back-of-the-envelope calculations to convert, the Airbus a320 and a small personal aircraft like Cessna, to battery power. Ultimately we want to know the power requirement of flight and how it will draw on the battery's energy supply. The work-energy theorem tells us that work equals force into delta x (W=F*∆x) where ∆ X is the distance over which a force acts. Power is work for unit time so P equals work divided by time(P=W/t).
Inserting our equation for work we get an equation for power that equals Force multiplied by distances divided by time (P=F*∆x/t) otherwise known as velocity(P=F*∆v). Where ∆v is the velocity of whatever it is getting worked on.
In this case, it's the air. When a plane is flying at a constant height we know that the force of lift and the force of gravity are balanced. That means the upward pressure of lift 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 requirement for lift equals the mass of the plane multiplied by the gravity and the ∆v.
So the question arises what is ∆v?
It's the falling 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 mechanism of the lift. The lift of an airplane provides 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 time the rate at which air is deflected; the mass of air that the plane of effects is simply the volume of the cylinder that it swept out per unit time, times the density of air.
If we call the suitable cross-sectional area Asweep, then the volume it sweeps out per unit of time is the sweep time of the plane's velocity. Therefore, the mass flow rate is equal to the density of air times the cross-sectional area times the velocity of the plane. 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.
Approximately this circle square because we don't have enough ridiculous assumptions in the calculation the relevant area becomes L2 at cruising speed. putting it all together we have the force lift needed to provide this equation. this equation is simply telling us the plane is sweating out a tube of air and shifting it down and the downward acceleration of air is equal to the downward pull of gravity on the plane. So the plane awards falling constantly paying the types of streaming Momentum downward via the air system. Rearranging the equation we can now solve for ∆vz in terms of quantities we can easily measure. And plugin this into a low power equation the power needed for lift is given by this equation
With the equation in hand, we can start noticing what variables really impact the energy requirements of the plane.
Imagine 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 need to double our power requirements to get out total power requirement at cruising speed. Now we are getting a real picture of why increasing the mass of a plane is such an issue. The mass component of this equation is not only squared but also double. doubling the mass will increase our power requirement 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 a usual mass fraction that is devoted to fuel about 20% of the Planes masses for both. We also need to take into account the fact that at the flying altitude, the atmosphere is much thinner than at ground level. For Cessna, the density falls by a factor of 2, and for Airbus a factor of 3. Let's be generous and take the specific power off leading-edge lithium-ion systems at about 0.340 kilowatts per kg.
To meet the power demand Airbus would need 34 tons of batteries (10500kw/0.340 = 31000kg). While the Cessna would need just 100 kg (35kw/0.340lw/kg =100kg). For the Cessna, this compares very favorably with the typical weight of field it would carry otherwise and it isn't terrible for the Airbus but this is just the power the plane needs at any interval of time.
We are really interested in the weight of batteries that we would need to match the typical range of these planes. For Airbus, that's a 7-hour flight from JFK to LHR, and for Cessna that might be a 4-hour flight from New York to South Carolina. the energy capacity required for a trip is given by this equation by multiplying the power required for the flight by the duration of the flight.
Again if we use leading-edge lithium-ion battery capacity we can store about 278 watt-hours per kg. For the Cessna, the equivalent battery weight is around 500 kg or just less than two birds the weight of the plane without fuel. For the a320 the required battery weight is around 260000 250000 kg or about four times the weight of the empty airplane. compared to the typical 20% that are located to fuel this is Devasting.
Now that we have a base figure for half having the batteries are going to be we can recalculate the actual range taking the added weight of the batteries into account let's assume at the very least we are not going to accept the reduction in flight speed or increases in Total energy used per flight.
How much is the range diminished for flights of similar speed and Total energy? As expected this downgrades Cessna’s flight time from 4 hr to about 2 hr. Not nominal but cleavable? A two-seater Cessna usually holds about 150 kg of fuel and another 100 kg for passengers and luggage.
It is easy to imagine endowing the Cessna with the required battery capacity through a combination of lowering the carrying capacity lowering speed increasing the wingspan with lighter parts and a more efficient electric engine. in fact, this is exactly what we are seeing with small electric aircraft coming to market in the past few years like the Alpha Electro. However, the downgrade is marked for the a320 taking us from 7 hours down to just 20 minutes less than 120th of the way across the Atlantic. If we plot the flight duration as a function of our battery mass for both planes we can see that the Cessna is already sitting around the optimum and could increase a battery capacity and improve the flight range.
It's a different story for their Airbus where we overshot our optimum battery capacity significantly. Reducing our battery weight to 60 tonnes will increase flight duration by about 15 minutes. We could last a little bit longer before crashing into the ocean assuming we could find a place to fit those 60 tons of batteries in the first place.
But we have been seeing great strides with short-range small aircraft coming to market and if we fly very slowly with lower drag wings we can even build a solar-powered drone that never has to land. We won't be seeing Airlines using electric engines anytime soon unless we can find a more energy-dense medium for storing that energy.
