Musings on wind power
Jul. 30th, 2010 12:17 pmI'm making a lot of best-case assumptions here, and I'm hoping that people with the appropriate knowledge can check my figures and help me refine them.
As I understand it, energy tapped from wind power depends on the mass flow past the turbine blades and how much of the energy in that flow is taken out by the turbine. This in turn depends on the difference in wind velocities just before and just after the turbine face (assuming it to be a two-dimensional 'black box' turbine and neglecting turbulence issues, the rotational component of the flow imposed by going through a spinning propeller disc, and the influence of hub shape and diameter).
Okay, so mass flow = density times blade area x velocity. This is because velocity determines the length of the "cylinder" of air that flows through the blade face per unit time.
The kinetic energy of that air is ½m.V12, where V1 is the velocity just in front of the blade face. The kinetic energy given up to the blades is ½m.(V1 - V2)2, where V2 is the velocity after the blade face.
The wattage is of course the energy transferred to the blade face, which is energy per unit mass per second, so substitute mass flow for mass in that energy equation. If you make the assumption of perfection, V2 drops to zero.Mass Density of air is about 1.22kg/m3, blade area is πr2 or ¼πd2, V is in metres per second (but 1 metre per second is 3.6kph). Do all the math and if I'm right, it comes to:
Wattage = πd2V3 / 153
Or getting rid of the Greek letter: d2V3/48.8, say 50 to give an even figure for easy mental calculation and to allow for some inefficiences (altered from "48 for an even more optimistic picture").
with blade diameter in metres, air velocity in kilometres per hour (which is how the weather report gives it to the masses).
To put this in context, a one metre turbine (terribly small) operating in a 36kph wind, converting all that wind energy to useful shaft energy with no losses, provides 950W. Just enough to drive a very weak hairdrier in the morning and a low-wattage microwave oven in the evening. If you assume efficiencies of 85% at the blade face and the generator, that drops to 950 x 0.852 or about 710W. Not even enough to drive your hairdrier, but a compact microwave oven might just do it.
Allowing for future growth, there might be 7 million households (averaging 4 persons each) in Australia. Let's assume that a fair use for each household is 4kW. That's a 28 GW power requirement. We have not yet included public utilities, street lighting, transport, hospitals, industry etc. etc. We cannot alter the air density. Gusts much above 36kph tend to damage stuff, and are unreliable, and halving the wind velocity automatically quarters your power availability. The turbines can only be so big, and we can only build so many of them. Downtime for maintenance is required. Startup friction (when the blades start rotating again after a calm day) and ongoing losses plus back-EMF in the turbines are going to diminish efficiency. The blades are made of sophisticated lightweight alloys or synthetic plastics/composites, many of which derive from fossil fuel raw materials.
Then there is the issue of storage to match demand. Storage batteries are expensive, and require sophisticated alloys and chemical compounds which must be manufactured, which requires energy in turn. Batteries are not forever and require replacement. The only really perfect and "environmentally friendly" method of energy storage is to pump water out of the tailrace of a hydroelectric plant and back into the reservoir. This would normally be madness, but when you're tapping wind or solar energy which you essentially get for nothing, it's sort of affordable. The problem here is that good locations for wind turbines are not always good locations for hydroelectric schemes, and then there is the issue of transmission losses from one location to another. Oh, and our Green Party (which if I'm right will soon hold the balance of power in our Senate) objects to new dams. And coal. And nuclear. And restrictions on immigration...
Meanwhile, the Latrobe Valley power generators are sitting right next to seven hundred years of brown coal at current consumption, and we are a prominent exporter of both coal and uranium. And our population continues to grow.
ETA: This webpage advertises a 10kW turbine "enough for three average households" (so my guess per household was pretty close for a pulled-out-of-nowhere figure) based on a 5.5m/s wind speed. On my screen, the tower (described as 12m tall) measures nearly 12cm, and the blades about 4cm (so an eight metre disc area). Plugging these figures into my rough calculation - and remembering that this is back-of-the-envelope, order-of-magnitude, reality-check stuff with all sorts of losses and inefficiences negated - we get:
d-squared = 64.
v = 5.5 x 3.6 or 19.8kph, let's call it 20kph. Cube that, it's 8000.
8000 x 64 / 50 = 10,200 watts. When I repeated the calculation using 19.8kph, I got 9950 watts. Accurate to within 50 watts in ten thousand, or about 0.5%, and no worse than 2%.
Damn, that's close. Possibly too close. What do people think?
Further ETA: General Electric provides some other useful data. A 77m turbine with a rated speed of 14m/s (just over 50kph) and an 82.5m turbine with a rated speed of 11.5m/s are both rated at 1500kw. My equation gives 77 x 77 x 50 x 50 x 50/50 or about 15 MW, and 82.5 x 82.5 x 41 x 41 x 41/50, or 9MW, so they are either terribly inefficient or something is wrong with my equation, which worked so well at smaller scales. It's probably the latter, as they should provide about the same figure and are six to ten times over the rated power of both turbines.
As I understand it, energy tapped from wind power depends on the mass flow past the turbine blades and how much of the energy in that flow is taken out by the turbine. This in turn depends on the difference in wind velocities just before and just after the turbine face (assuming it to be a two-dimensional 'black box' turbine and neglecting turbulence issues, the rotational component of the flow imposed by going through a spinning propeller disc, and the influence of hub shape and diameter).
Okay, so mass flow = density times blade area x velocity. This is because velocity determines the length of the "cylinder" of air that flows through the blade face per unit time.
The kinetic energy of that air is ½m.V12, where V1 is the velocity just in front of the blade face. The kinetic energy given up to the blades is ½m.(V1 - V2)2, where V2 is the velocity after the blade face.
The wattage is of course the energy transferred to the blade face, which is energy per unit mass per second, so substitute mass flow for mass in that energy equation. If you make the assumption of perfection, V2 drops to zero.
Wattage = πd2V3 / 153
Or getting rid of the Greek letter: d2V3/48.8, say 50 to give an even figure for easy mental calculation and to allow for some inefficiences (altered from "48 for an even more optimistic picture").
with blade diameter in metres, air velocity in kilometres per hour (which is how the weather report gives it to the masses).
To put this in context, a one metre turbine (terribly small) operating in a 36kph wind, converting all that wind energy to useful shaft energy with no losses, provides 950W. Just enough to drive a very weak hairdrier in the morning and a low-wattage microwave oven in the evening. If you assume efficiencies of 85% at the blade face and the generator, that drops to 950 x 0.852 or about 710W. Not even enough to drive your hairdrier, but a compact microwave oven might just do it.
Allowing for future growth, there might be 7 million households (averaging 4 persons each) in Australia. Let's assume that a fair use for each household is 4kW. That's a 28 GW power requirement. We have not yet included public utilities, street lighting, transport, hospitals, industry etc. etc. We cannot alter the air density. Gusts much above 36kph tend to damage stuff, and are unreliable, and halving the wind velocity automatically quarters your power availability. The turbines can only be so big, and we can only build so many of them. Downtime for maintenance is required. Startup friction (when the blades start rotating again after a calm day) and ongoing losses plus back-EMF in the turbines are going to diminish efficiency. The blades are made of sophisticated lightweight alloys or synthetic plastics/composites, many of which derive from fossil fuel raw materials.
Then there is the issue of storage to match demand. Storage batteries are expensive, and require sophisticated alloys and chemical compounds which must be manufactured, which requires energy in turn. Batteries are not forever and require replacement. The only really perfect and "environmentally friendly" method of energy storage is to pump water out of the tailrace of a hydroelectric plant and back into the reservoir. This would normally be madness, but when you're tapping wind or solar energy which you essentially get for nothing, it's sort of affordable. The problem here is that good locations for wind turbines are not always good locations for hydroelectric schemes, and then there is the issue of transmission losses from one location to another. Oh, and our Green Party (which if I'm right will soon hold the balance of power in our Senate) objects to new dams. And coal. And nuclear. And restrictions on immigration...
Meanwhile, the Latrobe Valley power generators are sitting right next to seven hundred years of brown coal at current consumption, and we are a prominent exporter of both coal and uranium. And our population continues to grow.
ETA: This webpage advertises a 10kW turbine "enough for three average households" (so my guess per household was pretty close for a pulled-out-of-nowhere figure) based on a 5.5m/s wind speed. On my screen, the tower (described as 12m tall) measures nearly 12cm, and the blades about 4cm (so an eight metre disc area). Plugging these figures into my rough calculation - and remembering that this is back-of-the-envelope, order-of-magnitude, reality-check stuff with all sorts of losses and inefficiences negated - we get:
d-squared = 64.
v = 5.5 x 3.6 or 19.8kph, let's call it 20kph. Cube that, it's 8000.
8000 x 64 / 50 = 10,200 watts. When I repeated the calculation using 19.8kph, I got 9950 watts. Accurate to within 50 watts in ten thousand, or about 0.5%, and no worse than 2%.
Damn, that's close. Possibly too close. What do people think?
Further ETA: General Electric provides some other useful data. A 77m turbine with a rated speed of 14m/s (just over 50kph) and an 82.5m turbine with a rated speed of 11.5m/s are both rated at 1500kw. My equation gives 77 x 77 x 50 x 50 x 50/50 or about 15 MW, and 82.5 x 82.5 x 41 x 41 x 41/50, or 9MW, so they are either terribly inefficient or something is wrong with my equation, which worked so well at smaller scales. It's probably the latter, as they should provide about the same figure and are six to ten times over the rated power of both turbines.
![[personal profile]](https://www.dreamwidth.org/img/silk/identity/user.png){kind=small}