Almost correct and maybe the math gives correlation to energy withdrawn from the wind. The correct idea is that the energy is proportional to the area of wind blades rotational area (disc) and the reduction of wind speed. That is the true reason it gives energy. It easy to understand that the mass the wind turbine handles id the area of the disc multiplied with incoming wind speed. m=A*v(wind speed)*roo (The density of air). m^2 * m/s * kg/m^3 = kg. F=ma a = deceleration of the air mass handled. Some of the energy is lost because of the rotor efficency due to friction which is mainly released as air vortex. At these speeds the temperature change is minimal, except maybe at the blade tip where local air velocity is approaching to speeds where air is compressible, close to the speed of sound. Power in wats, W = F/dt.
Parts of this lesson are somewhat helpful - but only if you already know how it all works. That's another way of saying that this basically shows us why most students coming out of college know almost nothing. Backing up and starting from the beginning, let's take the obvious language problem. I had to listem to him say things like "Air Foil Data" up to five times then catch it written on screen in order to realize he was saying something other than "Awfa". There's no point in trying to pass on knowledge, especailly intuitive knowlgedge to someone who speaks a different lanugage than the instructor. (And I comprehend foreign dialects better than most.) Also, mathematics does no good without a base of intuitive understanding of what it's based upon. So, the best you can say about this is -- it's nearly, but not totally, useless.
It seems strange for you to be criticising a lecturer from a Danish University for their use of English. I could understand every single word without any issue. Perhaps you don't comprehend foreign dialects better than most?
Don't expect to watch a youtube video on wind turbine forces and understand everything if you don't have a basic knowledge of mathematics and physical phenomena beforehand. This can be easily complemented with independent study, and there's lots of books you can get to do so. Also, as a non-native english speaker, I understood every word he has spoken; maybe you're wrong about "comprehending foreign dialects better than most"?
Betz limit has been smashed and debunked.wind pushes what it touches it can't push on the gaps between the blades.there is no kinetic energy in the wind there is force Mv squared kinetic energy is the energy of consistent work from a consistent force regards Graham Flowers MEng
I think with out the forces analyzed on the wind turbine model i wont be able to understand how those forces acting on a wind blade. Really thankful for your amazing illustration ..
I had this thought yesterday that palm trees were the outside of a sugar cane plant that turn sea water into sugar Is that possible for turbines Is it capable of freezing tge water back to bon threatening heat levels of total melt down I guess either way is a mekt down considering hot hot cold is
sir, we really loved your demonstrations. you show the imaginary velocity in front of us and we feel the concepts. thank you so much for your valuable time. great effort sir and team
In the flow equation 6:21, there is a V and a v (upper-case and lower-case letter), does it mean something a different? or is it just a typing error? Thank you in advance, really useful :)
Ok , if you were to design at least a 50% more efficient wind generator, now that you have your equations, what would it look like now? Thanks for the vid.
In all of these velocity triangle explanations, the rotor rotation is drawn with an arrow opposing the tangential component of the lift force vector. The omega*r term is pointing to the left in all the drawings, and yet the lift vector's tangential component is to the right...which way will the rotor turn? It should be in the direction of the net tangential vector, not, opposed to it like it's drawn. I'm confused on nomenclature.
9:12 in the lift/drag formulas you should be multiplying by the surface area of the wing, not the chord length. Otherwise your answer is just in units of kg/s^2.
Hi this is a wind energy problem .I would be very grateful if you could help me out in this problem."Given a wind speed of 8𝑚 /𝑠 and a blade tip speed ratio of 3. Calculate the blade chord angle at radius ratios 𝑅′ of 0.2, 0.6 & 1 for 1.An angle of attack of 0 degree 2.An angle of attack of 16 degree"
7:02 says lift is perp chordline, 8:34 says drag is perp lift, Vrel and drag act on same line, chord line and Vrel are not parallel, mistake? idk if 8:34 is an unstated assumption or if either timestamp is an error. thanks otherwise
@@DTUWindEnergy I would like to know the answer from @tahmin, but I can intuit he is thinking about installing the generator down in the floor, but not taking into account the huge torque, so the torsion stress on that "shaft" to connect it from the top to the bottom, so the cost will be huge as well), I just assume he is thinking about that. But I would like to know what he/she means. Best regards,
Hello Mr.Bredmose thank you for this amazing explanation. I want to point out that at the beginning, in the velocity triangle , you call a' the axial induction instead it is tangential/swirl induction factor if i think I am right. Thank you.
At 4:36, why is density not taken into account in the formula of Cp? I am working with tidal turbines, so I would like to know the formula of Cp by taking into account the effect of water density
Dear Satish. Thank you for getting in touch. I have forwarded your question and will get back to you as soon as I have an answer. Best regards Betina B. L. Winther DTU Wind Energy Communications
Dear Satish The density enters through the power expression P=1/2 rho A Cp V^3. The power coefficient, Cp, however, is dimensionless - for example the Betz limit has Cp=16/27. Best regards Henrik Bredmose, DTU Wind Energy
Thank you very much for your response. As for as I know the formula of Lift is L = Cl*1/2*rho*c*width*Vrel^2. At 9:46, in the formula of lift and drag, the width of the blade section is not included. Have you considered unit width here or is that any other concept?
Wonderful Video......and thanks a lot for the last method you used to illustrate the real aerofoil by projecting it on a piece of paper...... However I am afraid that the line that you have drawn is not the chord line, it can be only the chord line in case a symmetrical aerofoil and it seems that you have drawn a chambered aerofoil......But again really appreciated !
I think the line he had drawn is chord line, because the definition of chord line is the line from the leading edge to the trailing edge. Chamber line is the locus of midway between upper and lower surface. Please correct me if I am wrong.
Thanks for your comments! I agree that the chord line is defined as the straight line that connects the leading and trailing edge of the airfoil. The line on the slides is thus the chord line. Best regards Henrik Bredmose, DTU Wind Energy
@@DTUWindEnergy Typical: No distinction made between "chord line" in the classic sense of usage and "zero lift line" which is what you actually mean. Only in fully symmetrical airfoil profiles are they the same. Someone who has a shallow understanding is invariably confused by this.
My math skills aren't quite what yours are, so I have to ask some dumb questions. Let's suppose, the rotor is fixed pitch, such that I cannot feather them out knife edge to the wind, to minimize hub load. My small turbine has a tail vane, which always faces the rotor into the wind. Is the axial hub load less or more, if stopped completely, than if allowed to free-wheel? Is the hub load at the maximum when loaded to the optimum blade tip to wind speed ratio? Last question: From the calculations you presented, may I conclude, that the torque goes up on the square of wind speed, while the RPM goes up proportionately to the wind speed, thus, doubling the wind speed gives 8X the power? If that's true, high wind speeds could over run a permanent magnet alternator, causing it to lose control of the rotor, allowing it to basically free wheel. The torque of a permanent magnet alternator is limited by the field strength of the magnets.
I simply loved it when you superimposed the paper on the blade to explain your point. Wish the world had more teachers like you!
me too,first time I understood that diagram
Excellent video, would be nice to have more such educational videos for beginners. For example explaining twist of the blade and so on...
Classic ....love it
Almost correct and maybe the math gives correlation to energy withdrawn from the wind. The correct idea is that the energy is proportional to the area of wind blades rotational area (disc) and the reduction of wind speed. That is the true reason it gives energy. It easy to understand that the mass the wind turbine handles id the area of the disc multiplied with incoming wind speed. m=A*v(wind speed)*roo (The density of air). m^2 * m/s * kg/m^3 = kg. F=ma a = deceleration of the air mass handled. Some of the energy is lost because of the rotor efficency due to friction which is mainly released as air vortex. At these speeds the temperature change is minimal, except maybe at the blade tip where local air velocity is approaching to speeds where air is compressible, close to the speed of sound. Power in wats, W = F/dt.
not understand until 8:07
Parts of this lesson are somewhat helpful - but only if you already know how it all works. That's another way of saying that this basically shows us why most students coming out of college know almost nothing. Backing up and starting from the beginning, let's take the obvious language problem. I had to listem to him say things like "Air Foil Data" up to five times then catch it written on screen in order to realize he was saying something other than "Awfa". There's no point in trying to pass on knowledge, especailly intuitive knowlgedge to someone who speaks a different lanugage than the instructor. (And I comprehend foreign dialects better than most.) Also, mathematics does no good without a base of intuitive understanding of what it's based upon. So, the best you can say about this is -- it's nearly, but not totally, useless.
Excellent comment. The instructor needs to develop his English before giving speeches on such complicated topics.
It seems strange for you to be criticising a lecturer from a Danish University for their use of English. I could understand every single word without any issue. Perhaps you don't comprehend foreign dialects better than most?
Don't expect to watch a youtube video on wind turbine forces and understand everything if you don't have a basic knowledge of mathematics and physical phenomena beforehand. This can be easily complemented with independent study, and there's lots of books you can get to do so. Also, as a non-native english speaker, I understood every word he has spoken; maybe you're wrong about "comprehending foreign dialects better than most"?
At least they put all the equations on screen.
can you speak louder
The explanation is perfect however the sound quality is awful.
Betz limit has been smashed and debunked.wind pushes what it touches it can't push on the gaps between the blades.there is no kinetic energy in the wind there is force Mv squared kinetic energy is the energy of consistent work from a consistent force regards Graham Flowers MEng
I think with out the forces analyzed on the wind turbine model i wont be able to understand how those forces acting on a wind blade. Really thankful for your amazing illustration ..
I had this thought yesterday that palm trees were the outside of a sugar cane plant that turn sea water into sugar
Is that possible for turbines
Is it capable of freezing tge water back to bon threatening heat levels of total melt down
I guess either way is a mekt down considering hot hot cold is
What is the difference between a catapult an oil rig or wind turbine?
Design? Same winds different names different resource?
sir, we really loved your demonstrations. you show the imaginary velocity in front of us and we feel the concepts. thank you so much for your valuable time. great effort sir and team
In the flow equation 6:21, there is a V and a v (upper-case and lower-case letter), does it mean something a different? or is it just a typing error? Thank you in advance, really useful :)
Haha youre right
I have taken one of his class and he is the best professor I have ever met.
Ok , if you were to design at least a 50% more efficient wind generator, now that you have your equations, what would it look like now? Thanks for the vid.
In all of these velocity triangle explanations, the rotor rotation is drawn with an arrow opposing the tangential component of the lift force vector. The omega*r term is pointing to the left in all the drawings, and yet the lift vector's tangential component is to the right...which way will the rotor turn? It should be in the direction of the net tangential vector, not, opposed to it like it's drawn. I'm confused on nomenclature.
9:12 in the lift/drag formulas you should be multiplying by the surface area of the wing, not the chord length. Otherwise your answer is just in units of kg/s^2.
Why don't we consider and calculate aerodynamics for 5 blades instead of 3?
Is it possible to have a negative pitch angle in the low wind speed region? what effect does this create on the WT performance?
Hi this is a wind energy problem .I would be very grateful if you could help me out in this problem."Given a wind speed of 8𝑚 /𝑠 and a blade tip speed ratio of 3. Calculate the blade chord angle at radius ratios 𝑅′ of 0.2, 0.6 & 1 for 1.An angle of attack of 0 degree 2.An angle of attack of 16 degree"
I got the concept, awesome idea to show the forces on the blade, thanks❤
can you please comment below and explain properly what is the definition of axial induction factor and angular induction factor.
Sir I have a good idea to catch the wind and turn it into power
Where can I submit my idea
7:02 says lift is perp chordline, 8:34 says drag is perp lift, Vrel and drag act on same line, chord line and Vrel are not parallel, mistake? idk if 8:34 is an unstated assumption or if either timestamp is an error. thanks otherwise
Very good explanation and video. Thank you.
I cant understand why people use the worst design in wind turbine?
Why generator is in the head?
Dear Tahmim.
Thank you for your comment. Would you care to elaborate?
Best regards
Betina Brink Laursen
DTU Wind Energy Communications
@@DTUWindEnergy I would like to know the answer from @tahmin, but I can intuit he is thinking about installing the generator down in the floor, but not taking into account the huge torque, so the torsion stress on that "shaft" to connect it from the top to the bottom, so the cost will be huge as well), I just assume he is thinking about that. But I would like to know what he/she means. Best regards,
GREAT
I like his energy not to mention he explained it very well with a nice demonstration!
I will let him know. Thank you for your reply
Thank you sooo much for this video. Soo simple and easy to understand 🔥🔥
That's it. I am going to put DTU on my list for grad school.
Hello Mr.Bredmose thank you for this amazing explanation. I want to point out that at the beginning, in the velocity triangle , you call a' the axial induction instead it is tangential/swirl induction factor if i think I am right. Thank you.
chek wind turbine news 2018 heppolt wind
Are there places to get that model
Could you please show the formula for chord length?
Thank you sir! you just saved my day
Very nice! I like it!
Hello!
I want do to MS in wind energy topic. Please allow me
Hi. Please contact us by sending an email to communication@windenergy.dtu.dk.
Kind regards
Charlotte, DTU Wind Energy
At 4:36, why is density not taken into account in the formula of Cp? I am working with tidal turbines, so I would like to know the formula of Cp by taking into account the effect of water density
Dear Satish.
Thank you for getting in touch. I have forwarded your question and will get back to you as soon as I have an answer.
Best regards
Betina B. L. Winther
DTU Wind Energy Communications
Dear Satish
The density enters through the power expression P=1/2 rho A Cp V^3. The power coefficient, Cp, however, is dimensionless - for example the Betz limit has Cp=16/27.
Best regards
Henrik Bredmose, DTU Wind Energy
Thank you very much for your response.
As for as I know the formula of Lift is L = Cl*1/2*rho*c*width*Vrel^2. At 9:46, in the formula of lift and drag, the width of the blade section is not included. Have you considered unit width here or is that any other concept?
Yes, the formula is pr width - so force per width. You would then integrate the resulting forces along the blade.
Best regards
Henrik
Thank you very much for your response. The video was very useful.
yes.
I wonder what happen if you electronically charge the the blades?:)
bro i had to watch this like 20 times and i still didnt know what u were saying
Wonderful Video......and thanks a lot for the last method you used to illustrate the real aerofoil by projecting it on a piece of paper......
However I am afraid that the line that you have drawn is not the chord line, it can be only the chord line in case a symmetrical aerofoil and it seems that you have drawn a chambered aerofoil......But again really appreciated !
I think the line he had drawn is chord line, because the definition of chord line is the line from the leading edge to the trailing edge. Chamber line is the locus of midway between upper and lower surface. Please correct me if I am wrong.
Thanks for your comments! I agree that the chord line is defined as the straight line that connects the leading and trailing edge of the airfoil. The line on the slides is thus the chord line.
Best regards
Henrik Bredmose, DTU Wind Energy
@@DTUWindEnergy Typical: No distinction made between "chord line" in the classic sense of usage and "zero lift line" which is what you actually mean. Only in fully symmetrical airfoil profiles are they the same. Someone who has a shallow understanding is invariably confused by this.
My math skills aren't quite what yours are, so I have to ask some dumb questions. Let's suppose, the rotor is fixed pitch, such that I cannot feather them out knife edge to the wind, to minimize hub load. My small turbine has a tail vane, which always faces the rotor into the wind. Is the axial hub load less or more, if stopped completely, than if allowed to free-wheel? Is the hub load at the maximum when loaded to the optimum blade tip to wind speed ratio? Last question: From the calculations you presented, may I conclude, that the torque goes up on the square of wind speed, while the RPM goes up proportionately to the wind speed, thus, doubling the wind speed gives 8X the power? If that's true, high wind speeds could over run a permanent magnet alternator, causing it to lose control of the rotor, allowing it to basically free wheel. The torque of a permanent magnet alternator is limited by the field strength of the magnets.
WHAT AN AMAZING VIDEO! I've been trying to understand the triangle of forces (lift and drag) for days now, thank you!
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