**AWE lesson 1 - How much power from AWE?**

Airborne Wind Energy sounds really cool and fascinating, but being cool is not enough to be worth of studying. I always say to my friends that I wouldn't have believed in AWE if I hadn't done the basic math playing with the Loyd's model (1). This model will give you a quick way to estimate the power output of an AWE power plant. I decided to post here a guide hoping to persuade the more skeptical readers, helping the AWE community to grow further.

\[P = \frac{1}{2}\rho V_w^3 \frac{4}{27} E^2 C_L A\]

**How a 2-ropes kite flies**

This guide will assume that you know how a 2-ropes kite works. Anyway, here is a small summary.

A 2-ropes kite has nothing to do with the famous 1-rope kite that kids fly. The 2-ropes kite can be flown in a custom path in the so called wind window (the quarter of sphere that you have in front of you if the wind comes from your shoulders). For example, before take-off, you lay the cables on the ground along the direction of the wind with the kite in a ready-to-fly position. Then, with the wind coming from your shoulders, you pull both the cables and the kite takes off. The kite starts flying towards its leading edge and starts moving towards the top of the wind window. During this path you can pull for example the left cable and the kite will start steering counterclockwise. If later you pull for example the right cable and release the left cable, then the kite will start steering clockwise. If you alternate left and right steering properly, you can make the kite follow an horizontal-figure-of-eight path (like the infinity symbol: $\infty$) inside the center of the wind window.

If what you just read doesn't really make sense to you, then I would personally recommend buying a 40 dollars trainer kite and learn to fly it. Getting a real feeling of how a kite works not only will give you (or your kids) a lot of fun, but it will also give you a hands-on first feeling about Airborne Wind Energy. In my experience I can tell you that the people who tried kitesurf are much more likely to understand the potential of AWE technology before doing the math.

**How a ground-gen Airborne Wind Energy generator works**

During the figure-of-eight path described above the kite flies fast and generates high lift force. You can exploit this force to generate power by reeling-out the cables, i.e. unwind the cables from the drums on which they are wound, and finally you can extract energy from the drums rotation by means of an alternator. When the cables are completely reeled-out you can fly the kite at the edge of the wind window, where it does not generate much lift, and reel-in the cables by spending a certain amount of energy that is smaller than the amount you extracted during reel-out.

**The Loyd's model**

The Loyd's model is useful to understand how much power can be extracted from the kite during the reel-out phase described above. It is a physical model for which the first experimental results are already available in literature (2). In this guide I will explain the original model from (1) with the modifications from (3) and (4).

Let's start with the kite as you see in Fig. 1. The kite is flying horizontally in the direction inside the screen. The kite is flying at zero azimuth and at a certain elevation angle $\theta$. This means that the cables form a vertical angle $\theta$ with the wind direction and do not form an horizontal angle with the wind direction. The absolute wind velocity is $V_w$.

Fig. 1 The kite is flying horizontally towards the screen. Picture from (5).

Under the hypotheses of 1) steady-state flight (i.e. the kite is assumed to be flying constantly at zero azimuth and constant $\theta$ with no inertia), 2) negligible gravity force, 3) high aerodynamic efficiency, let's now impose the mechanical equilibrium at the kite as shown in the balloon of Fig. 2.

Fig. 2 Mechanical equilibrium at the kite under steady state flight. Picture modified from (5).

With reference to Fig. 2, $V_k$ is the absolute aircraft speed, $V_a$ is the apparent wind speed, $V_w$ is the actual wind speed, $V_r$ is the velocity of the cable in the direction of its own axis (the reel-out velocity), $T_k$ is the tether traction force, $L$ is the aircraft lift force, $D_{eq}$ is the equivalent drag force (i.e the drag force of the aircraft plus the equivalent cable drag force acting on the aircraft), and $V_w^*$ is the wind speed felt by the aircraft defined as $V_w^* = V_w \cos\theta - V_r$.

Notice that the force equilibrium at the aircraft makes the velocity triangle and the force triangle similar, thus yielding $V_k = E_{eq}V_w^*$.

The equivalent aerodynamic efficiency takes account of the cables aerodynamic drag and can be derived by computing the energy dissipated by the distributed cable drag and reads as:

\[E_{eq} = \frac{C_L}{C_D + \frac{C_{\perp} n_c r_c d_c}{4A}} = \frac{L}{D_{eq}}\]

where $C_L$ ans $C_D$ are the lift and drag coefficients of the aircraft, $C_{\perp}$ is the drag coefficient of the cable with respect to a flow perpendicular to its axis, $n_c$ is the number of cables, $r_c$ is the length of each cable, $d_c$ is the cable diameter, $A$ is the area of the aircraft, the same area to which $C_L$ and $C_D$ are referred.

Assuming $V_a$ $\cong$ $V_k$ (valid for a wing with high aerodynamic efficiency) and imposing the equilibrium of the aircraft it is possible to calculate the traction force as

\[T_k = \frac{1}{2} \rho {V_w^*}^2 E_{eq}^2 C_L A\]

The power output can be then computed as $P_k = T_k V_r$ thus resulting in

\[P_k = \frac{1}{2} \rho (V_w \cos\theta - V_r)^2 V_r E_{eq}^2 C_L A\]

This expression can be optimized in order to choose the reel-out speed that maximizes the power output, by setting $ \text d P_k / \text d V_r = 0 $. This simple optimization leads to the optimal reel-out speed $V_{ro}=1/3~V_w \cos\theta$ and to the optimal power output $P$

\[P = \frac{1}{2} \rho (V_w \cos\theta)^3 \frac{4}{27} E_{eq}^2 C_L A\]

that is what we were looking for.

Let's now put some numbers in the expression of $P$. Assuming:

$\rho = 1.225 kg/m^3$,

$V_w = 12 m/s$,

$\theta = 45 deg$,

$E_{eq} = 10$,

$C_L = 0.65$,

for an area $A = 122 m^2$, the same wing area of an Airbus A320 (my favorite airplane), the expression of $P$ yields $440 kW$. A bigger wing area, such as $A = 845 m^2$, like in an Airbus A380, would give a power output $P = 3 MW$.

Notice that the force equilibrium at the aircraft makes the velocity triangle and the force triangle similar, thus yielding $V_k = E_{eq}V_w^*$.

The equivalent aerodynamic efficiency takes account of the cables aerodynamic drag and can be derived by computing the energy dissipated by the distributed cable drag and reads as:

\[E_{eq} = \frac{C_L}{C_D + \frac{C_{\perp} n_c r_c d_c}{4A}} = \frac{L}{D_{eq}}\]

where $C_L$ ans $C_D$ are the lift and drag coefficients of the aircraft, $C_{\perp}$ is the drag coefficient of the cable with respect to a flow perpendicular to its axis, $n_c$ is the number of cables, $r_c$ is the length of each cable, $d_c$ is the cable diameter, $A$ is the area of the aircraft, the same area to which $C_L$ and $C_D$ are referred.

Assuming $V_a$ $\cong$ $V_k$ (valid for a wing with high aerodynamic efficiency) and imposing the equilibrium of the aircraft it is possible to calculate the traction force as

\[T_k = \frac{1}{2} \rho {V_w^*}^2 E_{eq}^2 C_L A\]

The power output can be then computed as $P_k = T_k V_r$ thus resulting in

\[P_k = \frac{1}{2} \rho (V_w \cos\theta - V_r)^2 V_r E_{eq}^2 C_L A\]

This expression can be optimized in order to choose the reel-out speed that maximizes the power output, by setting $ \text d P_k / \text d V_r = 0 $. This simple optimization leads to the optimal reel-out speed $V_{ro}=1/3~V_w \cos\theta$ and to the optimal power output $P$

\[P = \frac{1}{2} \rho (V_w \cos\theta)^3 \frac{4}{27} E_{eq}^2 C_L A\]

that is what we were looking for.

**Quick example**

Let's now put some numbers in the expression of $P$. Assuming:

$\rho = 1.225 kg/m^3$,

$V_w = 12 m/s$,

$\theta = 45 deg$,

$E_{eq} = 10$,

$C_L = 0.65$,

for an area $A = 122 m^2$, the same wing area of an Airbus A320 (my favorite airplane), the expression of $P$ yields $440 kW$. A bigger wing area, such as $A = 845 m^2$, like in an Airbus A380, would give a power output $P = 3 MW$.

**References**

- M. L. Loyd, Crosswind kite power (for large-scale wind power production), Journal of Energy 4 (3) 106 - 111 (1980).
- Damon Vander Lind, "Analysis and Flight Test Validation of High Performance Airborne Wind Turbines", Airborne Wind Energy, U. Ahrens, M. Diehl, R. Schmehl Eds. (Springer Berlin), Chapter 28, 473 - 490, (2013).
- I. Argatov, P. Rautakorpi, R. Silvennoinen, Estimation of the mechanical energy output of the kite wind generator, Renewable Energy 34 (6), 1525 - 1532 (2009).
- B. Houska, M. Diehl, Optimal control of towing kites, Proceedings of the 45th IEEE conference on decision & control, San Diego, USA, 2693 - 2697 (2006).
- A. Cherubini, R. Vertechy, M. Fontana, “Modeling Offshore High Altitude Wind Energy Converters”, (in review) (2015)