Estimating Requirements

An article series on how to estimate the required power, energy storage, etc. to build an electric aircraft.

Energy Storage E-mail

In order to power the electric motor which will propel our aircraft, we need to store energy on board the aircraft for it to use. In traditional aircraft with internal combustion engines chemical energy is stored in the form of gasoline or jet fuel. We will have to store electrical energy, with the most obvious medium being batteries. There are a number of battery chemistries available today, as well as other energy storage schemes, but for the most part I will focus on Lithium Polymer (LiPo) batteries in this article.

The amount and type of batteries to be used varies with the goal of the project. For example, an aircraft designed for aerobatic displays would require a small amount of energy storage to conserve weight, but with a high power rating. An electric aircraft designed purely for endurance on the other hand would use fill all available weight capacity with high energy density batteries. A practical airplane would be somewhere between these two extremes.

We can use our trusty C172A from the previous article as an example. The airplane has a maximum gross weight of 2200 lbs (1000 kg), and an empty weight of about 1300 lbs (585 kg). By replacing the continental engine and all associated plumbing and controls with an electric motor, we can save about 200 lbs, bringing us down to 1100 lbs (495 kg) empty weight. While this is normally a four place aircraft, we will reduce it to a two place electric to make room for the batteries. If we want to have usable load of 500 lbs (225 kg), we are left with 600 lbs (270 kg) for batteries.

Lithium Polymer is not a single battery chemistry but rather a category of chemistries, therefore the cell chemistries and their respective energy densities vary from one manufacturer to the next. Energy density claims range from about 90 Wh/kg to 160 Wh/kg. As a practical example, I will use the Kokam 100 Ah (SLPB 80460330H) cells used by ProEV's Electric Imp race car. These cells weight 5.95 lbs (2.68kg) each and have a rated energy capacity of 370 Wh, giving us an energy density of 62 Wh/lb (138 Wh/kg). With our battery weight allowance we can install 100 of theses cells, which results in an installed capacity of 37 kWh nominally.

So how long can we fly with these batteries? The reader will recall that in our previous article we found that the minimum power required to stay aloft in our C172A was about 45 kW. One would expect then that we could fly for 37 kWh / 45 kW = 0.82 hours, or about 49 minutes. There are some other factors though which reduce our flight time. For one, discharging a LiPo battery all the way to 0% state of charge (SOC) all the time will degrade the battery rather quickly. It is usually recommended that an 80% depth of discharge be used, leaving 20% SOC in the battery unless it is absolutely required. This takes our available flight time down to 0.66 hours (39 min) due to only having about 30 kWh available.

This is all good if we are launched from an airship, but a normal airplane also has to take off and climb. We determined that to get a good climb rate we need about 108 kW. So lets say we do a full power takeoff and climb for about 3 minutes before settling into a low level cruise (this should get us a good 2000 ft above ground, which is adequate for a sight seeing flight). This will use about 108 kW * 0.05 h = 5.4 kWh of energy, leaving us with 24 kWh for cruise. This results in a realistic cruise time of 0.54 hours, or 32 minutes.

From this example, it can be seen that the C172A does not make a very good candidate for an electric conversion while leaving 500 lb useful load. More batteries could be added by reducing the useful load and making it a single person aircraft, but that would only extend the usable cruise time by another 20 minutes or so. It is much better to start off with a more efficient airframe which requires less power to cruise and climb.

Required Power E-mail

One of the first things that people think of when choosing an alternate powerplant is how much power will actually be needed. Power required is proportional to the speed of the aircraft multiplied by the thrust (force) required to overcome drag.

P = v * f(drag)

This formula by itself doesn't do us much good unless we know the drag produced by an aircraft in a given configuration, which is hard to measure directly without a wind tunnel. It can, however, be estimated from the lift to drag ratio (L/D) and the weight of the aircraft.

f(drag) = weight
(L / D)

It just so happens that the lift to drag ratio of an aircraft happens to be approximately equal its glide ratio when in a zero thrust configuration. A zero thrust configuration means engine off in a jet aircraft, propeller(s) feathered in an aircraft with variable pitch props, or throttle just slightly cracked with a fixed pitch propeller (you can get a close, albeit slightly pessimistic, estimate using closed throttle with a fixed pitch propeller). So the lift to drag ratio can be found by dividing your horizontal air speed by the rate of descent. Horizontal air speed will be close to true airspeed in a flat glide, but in a draggy aircraft with a steep descent, you will need to do some trigonometry to figure this out.

v(horizontal) = sqrt(v(slant)^2 - v(vertical)^2)

L/D = v(horizontal) / v(vertical)

Make sure to use appropriate units when using these formulas. When using SI units (speed in m/s, weight in newtons, power in watts) no conversions are necessary, so I recommend you use them.

Let me give you a quick example. I'm flying along in my trusty C172A at a gross weight of 2200 lbs, (or about 10 kN). I pull the power back to almost nil, and set up a glide at about 65 KTAS (33.44 m/s slant speed). This results in a vertical speed of about 700 ft/min (3.56 m/s). Using the Pythagorean theorem (formula above) we find the horizontal speed to be 33.25 m/s (close to the slant speed). This results in a glide ratio of 33.25/3.56 = 9.34, which is also our L/D at this speed. With this, we can calculate our drag as 10000/9.34 = 1071 N. This means that the power required to overcome drag at this speed is 1071 * 33.44 = 35.81 kW, or about 48 hp. This is the minimum power which must be produced by the propeller to maintain altitude at this airspeed and in these conditions. Propellers, however, are not totally efficient, especially at low airspeeds. At best, their efficiency is about 80%, so the minimum power produced by the engine would have to be at least 35.81/0.8 = 44.76 kW, or about 60 hp.

This is all good if all we want to do is just barely stay afloat at a low airspeed that's close to L/Dmax, but a practical airplane has to be able to climb and cruise as well. The power required to climb at a certain rate is a sum of two things: the power to overcome drag at the climb airspeed, and the power to change the potential energy of the plane due to gravity. The first part is the same as the calculation we did above, whereas the second part is simply the change in energy divided by time. Potential energy of an object due to gravity is the product of its mass, its height above ground (or some other reference), and its acceleration due to gravity (9.81m/s^2).

P(climb) = E(new) - E(old)
E = m * g * h

Since an airplane's mass stays relatively constant during a short climb (neglecting fuel burn) and since the acceleration due to gravity is constant over small changes in height, we can say the product of these terms is equal to the plane's weight and the change in energy is proportional only to change in height, so power is proportional to climb rate.

P(climb) = v(vertical) * w

To continue the above example, say we want to climb at the same 65 KTAS (to save us recalculating the drag power), and have a climb rate of 1000 ft/min (5.08m/s). Our weight is approximately 10 kN, so our power required for the vertical ascent is about 50.8 kW over the constant altitude power requirement. This, again, is the power that needs to be transfered from the propeller, so if we again assume an 80% efficient prop, the engine has to produce an additional 63.5 kW, or 85 hp. Add the drag power of 60hp, and we find that in order to climb at that rate and airspeed, the engine must be able to produce at least 145 hp.

It should be noted here that this calculation is in fact a bit optimistic. Most propellers only reach 80% efficiency when being operated at their most efficient airspeed and rpm which, for a most fixed pitch propellers, does not correspond to peak engine power and low airspeeds. A C172A does not climb at 1000 ft/min fully loaded. In fact, the only time you'd get that climb rate with that plane is lightly loaded on a cool day at low altitude. Another reason that the C172A does not see those climb numbers is that although the engine is rated at 145 hp peak, it does not produce that kind of power at altitude. A constant speed propeller is able to operate closer to its peak efficiency throughout the rpm/power range, and an electric motor (or turbonormalized engine) does not lose power with altitude, so the estimate is a little more accurate for that configuration.

We can repeat the calculations for cruise power requirements at various airspeeds in order to get an idea of how much continuous power output we need for a practical airplane. Although, in the above example, we could get away with a motor that supplies only 45 kW continuous power, we would need to cruise around at 65 KTAS all the time, which is close to this plane's L/Dmax airspeed. From a practical standpoint, it would be wise to choose a motor with a continuous power rating at least a third higher than that in order to account for varying atmospheric conditions and imprecise airspeed control. If you want to cruise faster (understandably so) you must find the power requirements at your desired airspeed and then leave some margin on top of that. Most electric motors have peak power ratings that are several times higher than their rating so it is not usually necessary to size the motor according to the climb power requirement, but you should pay attention to the length of time that the motor can maintain peak power and compare it to your climb requirement.