## Other peoples estimates of wind farm power per unit area

In the government's study [www.world-nuclear.org/policy/DTI-PIU.pdf] the UK onshore wind resource is estimated using an assumed wind farm power per unit area of at most 9 W/m2 (capacity, not average production). If the capacity factor is 33% then the average power production would be 3 W/m2.

The London Array is an offshore wind farm planned for the outer Thames Estuary. With its 1GW capacity, it is expected to become the world's largest offshore wind farm. The completed wind farm will consist of 271 wind turbines in 245 km2 [6o86ec] and will deliver an average power of 3100 GWh per year (350 MW). (Cost £1.5 bn.) That's a power per unit area of 350MW/245km2 = 1.4 W/m2. This is lower than other offshore farms because, I guess, the site includes a big channel (Knock Deep) that's too deep (about 20 m) for economical planting of turbines.

I'm more worried about what these plans [for the proposed London Array wind farm] will do to this landscape and our way of life than I ever was about a Nazi invasion on the beach.

Bill Boggia of Graveney, where the undersea cables of the wind farm will come ashore.

Queries

What about micro-generation? If you plop one of those mini-turbines on your roof, what energy can you expect it to deliver?

Assuming a windspeed of 6 m/s, which, as I said before, is above the average for most parts of Britain; and assuming a diameter of 1 m, the power delivered would be 50 W. That's 1.3 kWh per day - not very much. And in reality, in a typical urban location in England, a microturbine delivers just 0.2 kWh per day - see p66.

Perhaps the worst windmills in the world are a set in Tsukuba City, Japan, which actually consume more power than they generate. Their installers were so embarrassed by the stationary turbines that they imported power to make them spin so that they looked like they were working! [6bkvbn]

### Notes and further reading page no.

264 The maximum fraction of the incoming energy that can be extracted by a disc-like windmill... There is a nice explanation of this on the Danish Wind Industry Association's website. [yekdaa].

267 Usually, wind turbines are designed to start running at wind speeds around 3 to 5m/. [ymfbsn].

- a typical load factor for a good site is 30%. In 2005, the average load factor of all major UK wind farms was 28% [ypvbvd]. The load factor varied during the year, with a low of 17% in June and July. The load factor for the best region in the country - Caithness, Orkney and the Shetlands - was 33%. The load factors of the two offshore wind farms operating in 2005 were 36% for North Hoyle (off North Wales) and 29% for Scroby Sands (off Great Yarmouth). Average load factors in 2006 for ten regions were: Cornwall 25%; Mid-Wales 27%; Cambridgeshire and Norfolk 25%; Cumbria 25%; Durham 16%; Southern Scotland 28%; Orkney and Shetlands 35%; Northeast Scotland 26%; Northern Ireland 31%; offshore 29%. [wbd8o]

Watson et al. (2002) say a minimum annual mean wind speed of 7.0 m/s is currently thought to be necessary for commercial viability of wind power. About 33% of UK land area has such speeds.

Figure B.9. An Ampair "600 W" micro-turbine. The average power generated by this micro-turbine in Leamington Spa is 0.037 kWh per day (1.5 W).

Figure B.10. A 5.5-m diameter Iskra 5 kW turbine [www .iskrawind. com] having its annual check-up. This turbine, located in Hertfordshire (not the windiest of locations in Britain), mounted at a height of 12 m, has an average output of 11 kWh per day. A wind farm of machines with this performance, one per 30 m x 30 m square, would have a power per unit area of 0.5 W/m2.

### C Planes II

What we need to do is to look at how you make air travel more energy efficient, how you develop the new fuels that will allow us to burn less energy and emit less.

Tony Blair

Hoping for the best is not a policy, it is a delusion.

Emily Armistead, Greenpeace

What are the fundamental limits of travel by flying? Does the physics of flight require an unavoidable use of a certain amount of energy, per ton, per kilometre flown? What's the maximum distance a 300-ton Boeing 747 can fly? What about a 1-kg bar-tailed godwit or a 100-gram Arctic tern?

Just as Chapter 3, in which we estimated consumption by cars, was followed by Chapter A, offering a model of where the energy goes in cars, this chapter fills out Chapter 5, discussing where the energy goes in planes. The only physics required is Newton's laws of motion, which I'll describe when they're needed.

This discussion will allow us to answer questions such as "would air travel consume much less energy if we travelled in slower propellor-driven planes?" There's a lot of equations ahead: I hope you enjoy them!

### How to fly

Planes (and birds) move through air, so, just like cars and trains, they experience a drag force, and much of the energy guzzled by a plane goes into pushing the plane along against this force. Additionally, unlike cars and trains, planes have to expend energy in order to stay up.

Planes stay up by throwing air down. When the plane pushes down on air, the air pushes up on the plane (because Newton's third law tells it to). As long as this upward push, which is called lift, is big enough to balance the downward weight of the plane, the plane avoids plummeting downwards.

When the plane throws air down, it gives that air kinetic energy. So creating lift requires energy. The total power required by the plane is the sum of the power required to create lift and the power required to overcome ordinary drag. (The power required to create lift is usually called "induced drag," by the way. But I'll call it the lift power, Plift.)

The two equations we'll need, in order to work out a theory of flight, are Newton's second law:

force = rate of change of momentum, (C.1)

Before

After

Before

After

Figure C.2. A plane encounters a stationary tube of air. Once the plane has passed by, the air has been thrown downwards by the plane. The force exerted by the plane on the air to accelerate it downwards is equal and opposite to the upwards force exerted on the plane by the air.

Cartoon

### A little closer to reality

Figure C.3. Our cartoon assumes that the plane leaves a sausage of air moving down in its wake. A realistic picture involves a more complex swirling flow. For the real thing, see figure C.4.

Cartoon

A little closer to reality and Newton's third law, which I just mentioned:

force exerted on A by B = — force exerted on B by A. (C.2)

If you don't like equations, I can tell you the punchline now: we're going to find that the power required to create lift turns out to be equal to the power required to overcome drag. So the requirement to "stay up" doubles the power required.

Let's make a cartoon of the lift force on a plane moving at speed v. In a time t the plane moves a distance vt and leaves behind it a sausage of downward-moving air (figure C.2). We'll call the cross-sectional area of this sausage As. This sausage's diameter is roughly equal to the wingspan w of the plane. (Within this large sausage is a smaller sausage of swirling turbulent air with cross-sectional area similar to the frontal area of the plane's body.) Actually, the details of the air flow are much more interesting than this sausage picture: each wing tip leaves behind it a vortex, with the air between the wingtips moving down fast, and the air beyond (outside) the wingtips moving up (figures C.3 & C.4). This upward-moving air is exploited by birds flying in formation: just behind the tip of a bird's wing is a sweet little updraft. Anyway, let's get back to our sausage.

The sausage's mass is msausage = density x volume = pvtAs.

Let's say the whole sausage is moving down with speed u, and figure out what u needs to be in order for the plane to experience a lift force equal to

its weight mg. The downward momentum of the sausage created in time t is mass x velocity = msausageM = pvtAsu. (C.4)

And by Newton's laws this must equal the momentum delivered by the plane's weight in time t, namely, mgt. (C.5)

Rearranging this equation, pvtAs u = mgt, (C.6) we can solve for the required downward sausage speed, u- mg

pvAs

Interesting! The sausage speed is inversely related to the plane's speed v. A slow-moving plane has to throw down air harder than a fast-moving plane, because it encounters less air per unit time. That's why landing planes, travelling slowly, have to extend their flaps: so as to create a larger and steeper wing that deflects air more.

What's the energetic cost of pushing the sausage down at the required speed u? The power required is kinetic energy of sausage

time

2 pvAs

The total power required to keep the plane going is the sum of the drag power and the lift power:

where Ap is the frontal area of the plane and Cd is its drag coefficient (as in Chapter A).

The fuel-efficiency of the plane, expressed as the energy per distance travelled, would be energy distance

if the plane turned its fuel's power into drag power and lift power perfectly efficiently. (Incidentally, another name for "energy per distance travelled" is "force," and we can recognize the two terms above as the drag force jC^pApV2 and the lift-related force The sum is the force, or

"thrust," that specifies exactly how hard the engines have to push.)

Real jet engines have an efficiency of about e = 1/3, so the energy-per-distance of a plane travelling at speed v is energy 1 distance e

This energy-per-distance is fairly complicated; but it simplifies greatly if we assume that the plane is designed to fly at the speed that minimizes the energy-per-distance. The energy-per-distance, you see, has got a sweetspot as a function of v (figure C.5). The sum of the two quantities \c^pApP2

and j is smallest when the two quantities are equal. This phenomenon is delightfully common in physics and engineering: two things that don't obviously have to be equal are actually equal, or equal within a factor of 2.

So, this equality principle tells us that the optimum speed for the plane is such that

Pvopt pv2 As'

This defines the optimum speed if our cartoon of flight is accurate; the cartoon breaks down if the engine efficiency e depends significantly on speed, or if the speed of the plane exceeds the speed of sound (330 m/s); above the speed of sound, we would need a different model of drag and lift.

Let's check our model by seeing what it predicts is the optimum speed for a 747 and for an albatross. We must take care to use the correct air-density: if we want to estimate the optimum cruising speed for a 747 at 30 000 feet, we must remember that air density drops with increasing altitude z as exp(- mgz/kT), where m is the mass of nitrogen or oxygen molecules, and kT is the thermal energy (Boltzmann's constant times absolute temperature). The density is about 3 times smaller at that altitude.

The predicted optimal speeds (table C.6) are more accurate than we have a right to expect! The 747's optimal speed is predicted to be 540 mph, and the albatross's, 32 mph - both very close to the true cruising speeds of the two birds (560 mph and 30-55 mph respectively).

Let's explore a few more predictions of our cartoon. We can check whether the force (C.13) is compatible with the known thrust of the 747. Remembering that at the optimal speed, the two forces are equal, we just thrust (kN) 200

50 0

I optimal speed

100 150 200 250 300 350 400 speed (m/s)

I optimal speed

100 150 200 250 300 350 400 speed (m/s)

Figure C.5. The force required to keep a plane moving, as a function of its speed v, is the sum of an ordinary drag force \cdpApi>2 - which increases with speed - and the lift-related force (also known as the induced drag) \ ^r^— which decreases with speed. There is an ideal speed, voptimal, at which the force required is minimized. The force is an energy per distance, so minimizing the force also minimizes the fuel per distance. To optimize the fuel efficiency, fly at voptimal. This graph shows our cartoon's estimate of the thrust required, in kilonewtons, for a Boeing 747 of mass 3191, wingspan 64.4 m, drag coefficient 0.03, and frontal area 180 m2, travelling in air of density p = density at a height of 10 km), as a function of its speed v in m/s. Our model has an optimal speed voptimal = 220 m/s (540 mph). For a cartoon based on sausages, this is a good match to real life!

Bird |
747 |
Albatross | |

Designer |
Boeing |
natural selection | |

Mass (fully-laden) |
m |
363000 kg |
8kg |

Wingspan |
w |
64.4 m |
3.3 m |

Area* |
Ap |
180 m2 |
0.09 m2 |

Density |
P |
0.4 kg/m3 |
1.2 kg/m3 |

Drag coefficient |
Cd |
0.03 |
0.1 |

Optimum speed |
vopt |
220 m/s |
14m/s |

= 540 mph |
= 32 mph |

Table C.6. Estimating the optimal speeds for a jumbo jet and an albatross.

* Frontal area estimated for 747 by taking cabin width (6.1 m) times estimated height of body (10 m) and adding double to allow for the frontal area of engines, wings, and tail; for albatross, frontal area of 1 square foot estimated from a photograph.

need to pick one of them and double it:

force energy distance Cd P Ap ^pt cdpAp

## Renewable Energy Eco Friendly

Renewable energy is energy that is generated from sunlight, rain, tides, geothermal heat and wind. These sources are naturally and constantly replenished, which is why they are deemed as renewable.

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