Customer Qualifications Section
Lecture and Notes 1
The Story of Air in Motion by "Dutchy" the Windmill
Air and Its Motion a "Weighty" Subject
Does everyone know that the same air we know so well passing through my 20 foot long blades at 15 mph (6.7 meters per second) has a mass flow of just about one ton (tonne) per second? Yikes! Here is the story about this.
Confusion maybe starts with the different measuring systems, the British (no longer used in the UK) and the metric. In the U.S. a well-known fast food restaurant offers a "quarter pounder" hamburger. Does anyone know how this is described in the metric system around the world? The beef patty would weigh about one newton in metrics and so this could be called a "newtoner" hamburger. But metrics uses kilograms for most weights and so it might be called instead a "100 grammer" hamburger. It turns out that one newton is slightly more than the weight of 100 grams and so the beef patty would be slightly larger in this case, not to make fun of such a serious subject as food.
It is fascinating to know that the two major world almanacs in English published annually in the U.S., the "World Almanac" and the "New York Times Almanac of Record", do not make completely clear the difference between "mass" and "weight" in their sections on weights and measures, the "New York Times Almanac" being a little better at this (e.g., comparing weights to those found on the moon). Neither one of them, however, cover wind energy except each as one or two brief line items, hard to find, aggregated into totals of the energy contributions by all renewable energies on their pages covering the industry sources of electrical energy generation nor does either one take any time to describe the atmosphere and the air in it as being anything of substance except that it is just so much Oxygen and Nitrogen with a few more gasses thrown in. Air seems to lack anything of interest to say about it.
More on this in additional website updates later. Much kindness and good will to everyone out there in wind energy. Any comments or corrections are welcome as are answers to questions posed. "Dutchy"

Lecture and Notes 2
Wind Velocity As Rate Of Mass Flow
Meteorologists "own" standard meteorology and aviation "owns" aviation aerodynamics. It may come as a surprise that wind energy has a meteorology and an aerodynamics that it "owns" as well.
Wind velocity is of little use to a wind turbine without the density factor, the mass of the air per unit volume, included. Debate on measuring windspeed as a massflow rate is inconclusive until it is discovered that the windspeed in miles per hour is almost exactly the same value in numbers, using the density of air at standard conditions, as the pounds per second per yard squared of mass flow(!). So one mph = one #/sec yd^2 to a close approximation.
Not only this but in metrics this conveniently becomes close to *one half* of the same mph figure. This means that one mph of wind velocity is also equal to about one half kg/sec meter^2. Wind turbines have been typically rated for full power at a wind velocity of about 30 mph, which converts to about 13.4 meters per second in metrics, a number not often seen or remembered. But in mass flow units this is about 15 to 16 kgs/sec meter^2, much easier to keep in mind.
Even a further step may be taken. "One kilogram per second per meter squared" is still hard to say. So a new unit of measurement may be defined for it, something which is often done in science. For example, the electrical unit of alternating current frequency was given the unit name of "hertz" not many years ago (from the name of Heinrich Rudolph Hertz, the German electrodynamics physicist of the 1800s) and so 60 cycles per second is now 60 hertz.
Among the candidates whose names may be chosen for this unit, though wind energy is still a relatively new technology, the one that stands out is the one of Poul la Cour of Denmark, the early pioneer of modern wind energy from the late 1800s, whose name has largely gone unrecognized. I, "Dutchy" the Windmill, hereby nominate Poul la Cour for this honor and, in support of this nomination, the IntegEner-W website has been updated to provide information and photos about him from the past.
If accepted by the wind industry, wind velocity would be measured and reported in units of "laCours" instead of miles per hour or meters per second. Then also wind speed averages for geographical localities and even their *energy weighted* wind speed averages would be calculated and reported in "laCours".
A question to answer, then, is this. About how many pounds per second per square yard of wind mass flow rate are expressed by a value of 15 laCours? Then, how many miles per hour wind velocity?
More on this in additional website updates later. Much kindness and good will to everyone out there in wind energy. Any comments or corrections are welcome as are answers to questions posed. "Dutchy"

Lecture and Notes 3
Ultimate Windpower - What is really possible?
High school science classes teach all that is necessary to know in calculating how much kinetic energy is in the wind available for capture. It is quite a lot of fun to determine the energy in air that happens to be moving - like doing a magic trick. Watch closely.
A power rating of 1000 megawatts per electrical energy producing plant or site has become a rough "standard of generation" for many utilities in many nations. The appearance of wind energy on the scene naturally raises the question of whether windpower can meet this standard as well.
First, find the mass of a large volume of air such as that enclosed within a box that is visualized to cover one square mile of land and that is 1000 feet high. A height of one thousand feet is well within construction capabilities that have raised many tall buildings in large cities. At .0765 #s/foot^3 air density, the mass of atmospheric air enclosed within such a volume box comes to:
(.0765 x 5280^2 x 1000) / 32.174 = 6.6286 x 10^7 slugs of air mass (This is over one million tons of nothing but just air!)
The kinetic energy - (1/2)mv^2 - of such a mass, moving at 40 mph, is:
(1/2)(6.6286 x 10^7)(40 x (5280/3600))^2 = 1.1407 x 10^11 foot pounds
In KWHs, this is 1.1407 x 10^11 foot #s/ 2.6552 x 10^6 foot #s/KWH = 42961.3 KWHs equivalent
Now add wind turbine technology, in a theoretical sense, installed so as to cover fully a frontal area that faces the wind that is one mile long and 1000 feet high - one side of the volume described above - and generate power at the Betz limit of 59.3% efficiency. The power generated by the wind crossing a "blade swept area" of this magnitude, which is also quite close to being equal to one half of a square kilometer, is:
42961.3 KWHs x 40 per hour x .593 efficiency = 1,019,042 KWs = 1,019.042 MWs
So, conceptualized here is a 1000 MW wind power plant with a total blade swept area of only about one half of a square kilometer, all within the confines of no more than just one square mile of land or offshore surface area. Q.E.D.
This has some magic to it, though, like pulling a rabbit out of a hat.
For the foreseeable future and for most sites with energy weighted wind velocity averages of less than 40 mph (18 m/s) it is much more realistic to assume that what would be required is something closer to 6 to 8 square kilometers of total blade swept area and 100 square miles of land or offshore geographical area.
Proper averaging is important in understanding the windspeeds above. A wind energy site where the windspeed is zero for five days out of ten but 30 mph on the other five days has an energy weighted average windspeed of not just 15 mph - the linear average - but ((30^3)(5/10))^.333 = 23.8 mph. The same amount of energy would be produced at this location as that which would be produced if the windspeed were to be constant at 23.8 mph for all 10 days. Correct?
More on this in additional website updates later. Much kindness and good will to everyone out there in wind energy. Any comments or corrections are welcome as are answers to questions posed. "Dutchy"

Lecture and Notes 4 "But Air is 816 Times Less Dense Than Water"
Hydropower has become a successful source of electrical energy worldwide, though limited in availability, and wind energy can find inspiration in this fact since both work from the same physical principles and both do not require the intermediary physical process of high temperature thermal energy.
A comparison of sorts may be made between the two fluids in terms of their mass densities, since this bears heavily on what may be possible. Water has a density of about 62.427 pounds/foot^3 and air has a density at standard conditions of about .0765 pound/foot^3, which is only about one and a quarter ounces. A cubic yard of water weighs thereby about 1685 pounds or 84% of one ton while a cubic yard of air weighs nothing due to its buoyancy but has a mass that may be measured of about 2 pounds and one ounce. From these, the ratio of the mass densities of water to that of air may be determined and it comes to about 816.
So at first glance it would seem that any mechanical devices such as wind turbines that would be constructed to obtain a comparable amount of energy as that obtained from hydro should be 816 times as large. Yes, but only on a physical volume basis. Dimensions on a linear basis along one co-ordinate axis need only be the cube root of 816 or about 9.3 times as large, providing something a little more within reason. Flow channel areas would be the square root and so would need to be 816^.5 or 28.6 times as large. With dimensions in comparison such as these, wind energy surely may be seen as having potential that could accomplish comparable energy delivery at some time in future years.
Large volumes of air are surprisingly massive when their masses are calculated using nothing more than the above density of one and a quarter ounces per cubic foot. Mentioned in the last lecture of this series is the fact that one million tons of air is represented thereby as a volume that can be contained within a volume block 1000 feet high over one square mile of earth surface. In scanning the landscape over level terrain on a nice day one is looking out at and through many millions of tons of atmospheric air!!!
To re-emphasize this point, note that this is not millions of *pounds* of this substance known as "atmospheric air" but millions of *tons*.
For comparison, considering just the mass weight (and not the energy represented) of trainloads of coal - each loaded coal car about 100 tons and 100 cars per train - this comes to no fewer than 100 such railroad trains. The great ship Titanic that was lost so many years ago had a displacement of about 45,000 tons and so this represents the weight of about 23, almost two dozen, Titanics. The U.S. Navy now proudly carries the flag on some 10 aircraft carriers that are nuclear powered (an eleventh is being built), each aircraft carrier having a displacement of about 100,000 tons, all ten in total just about equal to the mass weight of one of these volume blocks of "nothing but empty" air over one square mile.
Hot air balloons are an example of the mass of air in action. These lighter-than-air craft (powered by the wind!) have balloon diameters of about 75 feet, thereby having a volume of some 250,000 cubic feet or a volume that displaces some 20,000 pounds of air, about 10 tons. Heating the air inside to only some 50 degrees above ambient provides all the lift they require.
To bring this to a conclusion and to finalize the comparison between hydro and wind power, the question may be asked, how large a lake that is 20 feet deep on average would be represented by this same one million ton volume block of air? Ans: 5 acres, 25 acres, 35 acres?
More on this in additional website updates later. Much kindness and good will to everyone out there in wind energy. Any comments or corrections are welcome as are answers to questions posed. "Dutchy"

Lecture and Notes 5 The Ghost of Bernoulli Lingers On
Quite sensible individuals even with a reasonable grasp of science are still known to "lose it" when they partake in discourse on some of the more controversial aspects of aerodynamics. It has not helped that material published in earlier years soon after the close of WW II, complete with speculations from returning military flight pilots and other early aviators, was instilled into the minds of high school students for decades thereafter and became its guiding mantra, material such as that in these two pages following copied from a late 40s textbook and rightly to be dismissed as having little value: (Basic Units in Physics by F.E. Stewart 1949, Republic Book Co., Inc., New York)
Since that time it has gradually dawned on most everyone with any curiosity about aerodynamics that this must be not quite true - even further that it must be just "baby talk" - and that airflow deflection must play an important, even a dominant, role in creating these forces, something that is given some newer treatment in presentations directed to popular understanding of flight by such Government agencies as NASA. Nevertheless, the logic is still not followed up with a solid basis for calculating aerodynamic forces in this newer way of thinking, something that was at least suggested in the older days as being just a matter of solving the Bernoulli Equation for the wing lift force. Nowadays animations on websites show the flow lines above and below airfoils along with some depictions of their angle of attack and the flow deflection at the trailing edge with much left unsaid beyond this.
What is needed to be done in supporting the weight of each aircraft is for the wings to force quite a large amount of air over their spans and above and below their passage downwards as they speed through it and do so with (a) the least amount of wing surface and wing structural weight and (b) drag on the engines propelling the airplane forward. It must be readily calculable somehow to arrive at just the right amount of air mass ( in "slugs" if in
the ordinary "pounds, feet, seconds" measurement system ) necessary to be so disturbed by this process and the wing configuration that needs to be made available in doing so. The same goes for wind turbine blades, which have even a greater imperative placed upon them to create some difference in the wind flow streamlines that approach them, moving them aside with great impact and into sharply different flow directions vectored into lower speeds over large volumes of space both near and far away from their surfaces as they rotate.
The impression that prevails in everything being said and done up to the present time is that the ghost of Bernoulli lingers on and wind turbine blades as a part of their active rotor areas are being fashioned in a way that satisfies only the slight bending of the few flow lines from computer animations that track near their surfaces as if this is all that matters. It should be added with haste that satisfying structural integrity requirements has become a well known study from past practice, something in which great comfort may be taken while not advancing the state of the art more rapidly than might be done without its continued assurance.
When the chapters of this series presented earlier are taken into consideration, it becomes more clearly understood that the great and weighty mass of air that is moving through the rotor areas should be given a high priority as the source of the energy that must be converted and that wind turbine design is a continuing process as the future unfolds that needs to further address the issue of extracting energy from this flow mass more efficiently.
A problem illustrative of what is being said here might entail the rate of air mass being impacted by the flight of a theoretical but quite realistic jet aircraft moving at 550 mph with a wing span of 200 feet and a weight of 400 tons at an elevation of 30,000 feet - air density .0286 #s/ft^3. For a wing angle of attack of just 2.0 degrees, the straight downward deflection impulse component of airflow by the wings is 550 mph x sin 2.0° = 19.2 mph = 28.2 feet per second ( termed "downwash" in aviation-speak ). The drawing below of the forces arising from this airflow deflection was taken from the
"Lift" Booklet of IntegEner-W:
The variation of Newton's Equation applicable to wing flow can be stated as - lift force in pounds = rate of air mass flow deflected downwards in slugs per second x velocity of the straight downward deflection impulse component in feet per second.
Solving this equation for the air mass flow results in 2.84 x 10^4 slugs per second which is about 3.19 x 10^7 ft^3/sec of air at this density. The cross sectional area of air thus impacted at the forward speed of 550 mph or 807 ft/sec is then close to 40,000 ft^2, which, if divided by the wing span of 200 feet, yields, of course, 200 feet, or an effective distance of 100 feet both above and below the wing surfaces. Due to conservation of mass in an approximately incompressible flow field over the pressure ranges involved, this is seen to be readily achievable and it is surprising but true that wings have such an impact so far removed from them in the vertical directions up and down.
Problem: if the aircraft drops to an altitude of 15,000 feet - air density of .0474 #s/ft^3 - and decreases air speed to 350 mph, what is the effective wing attack angle that is necessary to maintain the lift force at 400 tons, (a) about the same = 2.1 degrees, (b) about double = 4.2 degrees, or (c) nearly another degree = 2.9 degrees?
More on this in additional website updates later. Much kindness and good will to everyone out there in wind energy. Any comments or corrections are welcome as are answers to questions posed. "Dutchy"
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