The Power of the WindThe power in the wind is not a mystical thing, it is simply the product of a mass of air running headlong into a stationary object (like a propeller). This is "kinetic energy" -- energy of motion.
Our intuition should tell us several things.
A faster wind has more power than a slower one, and still air has none. The faster the wind the better, right up to the point that something breaks and falls off the tower!
The wind that does not hit our propeller does not give us any power, so the bigger our propeller the better.
You ever notice how the wind in the tops of the trees seems stronger than the wind at your feet? This is not imagined. The wind near the ground is slowed by friction, so mounting our propeller on a tower far above the ground will put us in the real wind, undetered by friction with the ground.
And there are some minor and less obvious effects. The air at higher altitude has a lower pressure and thereby a lower density. A gallon of air in Denver simply weighs less, has fewer atoms, than a gallon of air in Panama. Likewise, air expands with temperature, so summer air is less dense than winter air -- except in Los Angeles, where the temperature is the same year 'round ;)
So, without recourse to any math at all, we know that the most productive wind plant will be high in the air, have a large propeller, and be located in a generally cool and very windy place at sea level. Maine for sure works! For the rest of the world we'll need math.
The kinetic energy (E) of a mass (m) moving at velocity (v) is:
E = (1/2) * m * v2
The mass of air we will get energy from is only that mass of the wind that actually hits our propeller, so we must have the density of air multiplied by the volume of air that hits the propeller. Given the area (A) of the propeller and the density of air (d) and a unit of time (t) the total flow-rate of mass through the propeller is:
m = d * A * v * t
Substituting this computation of mass (m) into the first equation and simplifying we obtain:
E = (1/2) * d * A * v3 * t
which is the total energy delivered in time t. Since power (P) is the rate of the delivery of energy, we have:
P = (1/2) * d * A * v3
Pause a moment and compare this with our intuition. Sure enough a bigger propeller, more area (A), creates more power. We see the density (d), which varies with temperature and altitude, changes our power output. And look at that velocity term! We knew that a faster wind should provide more power, but oh my, a doubling of wind speed produces eight times more power. It may not have been clear from intuition, but it is from the math, that wind speed above all else is the most important factor.
The equation for power (P) is the power of the wind, not the power our generator can produce from it -- the difference being the less than 100% conversion efficiency of our wind machine. Propellers are not perfect, there is gearing that is not perfect, and the electric alternator itself is imperfect. If we lump all of these together and call it 30% we will be pretty close to reality.
It's time we worked an example, but we first should change our equation a bit to use different units. Skipping the tedious algebra, we obtain:
P = .005 * e * A * v3
where P is still in watts, we have introduced an efficiency factor of e, A is square feet and v is miles per hour. [With apologies to our international browsers. You might prefer the original equation which is in "mks" units, so v is m/s and A is m^2. Remember to multiply by e.]
Assuming we use a propeller, then the area (A) can be rewritten in terms of the propeller diameter (D) in feet to yield:
P = .004 * e * D2 * v3
So, now for that example. Say we have a 10 foot diameter propeller in a 20 MPH wind and our conversion efficiency is 30%. What is our power output?
P = .004 * .3 * 102 * 203 = .004 * .3 * 100 * 8000 = 960 watts
What is our power output at 10 MPH?
P = .004 * .3 * 102 * 103 = 120 watts
Here's an interesting observation. A one hour storm that brings with it a 40 MPH wind provides more power than 2.5 days of 10 MPH winds. Wind speed is critical (we knew that) so let's talk about the advantage of a tall tower.
Height above ground does increase wind speed, but how much? The simple answer is there is no way to tell exactly, but that isn't a very useful answer, so let's see if we can tell approximately. Even there, not much can said unless we assume flat open ground with no "obstructions." What's an obstruction? Just about everything bigger than a cow.
How will obstructions effect the wind? You'll need a wind tunnel and an aeronautical engineer to even make a guess.
So, realizing that the question has no answer, here is the answer (huh?).
Wind speed is related to height above ground in by this rule of thumb:
(v / v0) = ( H / H0 )N
where v is the wind speed at height H, v0 is the wind speed at reference height H0 and N is a proportionality factor that we will set to 0.2 for our purposes here (wind speeds from 5 to 35 MPH).
Since power varies as the cube of the wind speed, we can write:
(P / P0) = ( H / H0)3N
to give us the ratio of power produced between a new height (H) and a reference height (H0). Using our prior example, let's tabulate the effect a tower should have. Remember we showed 960 watts of output in a wind speed of 20 MPH. Since we probably came up with the 20 MPH number by what it felt like on our own face, let us use a reference height (H0) of 5 feet. Here's the table.
Tower Wind Power Total Height Speed Factor Power 5 20 1 960 10 23 1.5 1455 20 26 2.3 2206 40 30 3.5 3343 50 32 4.0 3822 75 34 5.0 4875 100 36 6.0 5793 200 42 9.2 8780
In a nutshell, a 75 foot tower is as good as 5 wind generators. Well, sort of. You can't use the same generator. A 1000 watt generator will not produce 5 KW even if we have that much wind. To make use of all that wind we get by being 75 feet in the air, we must use a 5 KW generator.
And can we build and maintain a 75, 100, or 200 foot tower? Can we get FAA clearance? (just kidding)
There will be a trade-off, generally based on money, between tower height and propeller diameter. Tower height has the mathematical advantage. For our example case with a 10 foot diameter propeller, we get four times the power if we go to a 20 foot propeller. But we can do the same thing with a 50 foot tower without a larger propeller. Which is the better choice? Probably the tower. A 50 foot tower isn't any longer than a small manufactured home. Nearly any serious amateur radio operator has a tower that size.
But wait, what if we do both? What about a 20 foot propeller and a 50 foot tower? That's great! But just remember that a wind generator produces zero power after if falls apart. As propellers get larger and towers get taller, they become more expensive and much more exposed to damage from high winds.
How much more? The force exerted by wind we typically call "drag" and is proportional to the square of wind speed. So as we double the wind speed we get four times the drag (force) and eight times the power output. Not a bad tradeoff as long as you can afford it.
So what does all this theory do for us? It does tell us how and what to optimize, but can it tell us how much power to expect on a monthly basis from our wind generator? No. It does not.
For a good estimate what we would need is a monthly profile of wind speed for our location that shows the expected number of hours of each wind speed per month. We could then use the equations above to figure the power output for each wind speed and time duration and sum the numbers to get a monthly total. This would still just be an estimate and the sort of survey we need likely does not exist for our site.
So, a somewhat simpler approach using just average wind speed has been used. This approach makes use of a "capacity factor" derived from the average wind speed and the wind speed it which the generator produces its rated power. The factor so derived is then simply multiplied by the rated power of the generator and the number of hours in the month to get total energy produced.
Here is a part of such a table.
Avg Wind Capacity Factor (Cf) Speed Vr=16 Vr=20 Vr=24 Vr=28 8 .2 .1 10 .35 .2 .1 12 .47 .3 .2 .12 14 .41 .3 .2
Where Vr is the wind speed where the generator produces its rated power.
Returning to our example above, if our 960 Watt output at 20 MPH wind is the rating from the manufacture, then our Vr is 20 and our rated power is 960. What then is the total expected monthly power output if our average wind speed is 12 MPH?
From the table, our Cf is .3, so our monthly total is .3 * 960 * 720 = 207.36 KWH.
Notice from the table that an increase in average wind speed of just 3 to 4 MPH will double the expected monthly power output. Remember that tower? How much tower is required to get a 3 MPH speed increase. That depends on what the average is to start with. Our equation only gives us a multiplier.
Speed @ 5 ft Height Speed @ Height 8 10 9.2 25 11 50 12.7 12 10 13.8 25 16.6 50 19 14 10 16.1 25 19.3 50 22.2
OK, deep breath, what does that mean. If the wind speed measured 5 feet above ground is 8 MPH, then the wind at 10 feet is 9.2, at 25 feet is 11, and at 50 feet is 12.7. Since we are looking for a 3 MPH improvement to double our power output, we can see that in a region that gets an average wind speed of 8 MPH, a 50 foot tower will gives us the doubling we're looking for.
Higher wind speeds are better. A 25 foot tower allows us to double our output if the average wind speed is 12 MPH. Likewise for 14 MPH. A tower is probably a good investment.
One last rule of thumb, the crudest of all, is to take the generator power output at the average wind speed and multiply it by 1440 to get expected monthly power production in KW-H.
For example, a generator that produces 500 watts at the site's average wind speed of 12 MPH would (by this R.O.T.) deliver 720 KW-H.
This is crude, but it is quick, and all of this is estimation anyway. One number has about as much chance of being right as another.
Since wind speed is soooo crucial to a successful wind plant, we should (1) get all the wind measurements we can for our site and (2) consider the tower as carefully as the generator. Actual on-site measurements of wind speed would be a good idea and as well as placement where the wind is least obstructed by buildings, trees, etc.
There are some other things to consider as well in addition to the generator and tower. There are two inconvenient characteristics of wind power. First, the power output of the generator is highly variable. Since the majority of the power comes from the minority of the wind, the generator must be "over designed" for the average. Most generators don't even generate any power below 8 MPH -- it isn't worth it. The more important number is the top speed where generation must shutdown for structural reasons. One good storm will produce more in a few hours than the rest of the month combined so the machine must continue to generate in these high wind speeds.
Second, there can be very long periods of little or no wind followed by short bursts of more power than you can use. This requires a much larger energy store (battery) than would be required for a motor-driven generator or solar cell system. Since the wind can't be "shut off" like a motor can, it is reasonably certain that some of the power production of a wind generator will occur when the batteries are full, which means the wind power will be lost. This will impact pay-back period.
This article first appeared in the March 22, 2000 issue of OffgridOnlie, our free weekly newsletter. |