Rotor Blades 6 - Example on Rotor Blade Calculation
This is a modal window.
The media could not be loaded, either because the server or network failed or because the format is not supported.
Formal Metadata
| Title |
| |
| Title of Series | ||
| Number of Parts | 8 | |
| Author | ||
| Contributors | ||
| License | CC Attribution - ShareAlike 4.0 International: You are free to use, adapt and copy, distribute and transmit the work or content in adapted or unchanged form for any legal purpose as long as the work is attributed to the author in the manner specified by the author or licensor and the work or content is shared also in adapted form only under the conditions of this | |
| Identifiers | 10.5446/65290 (DOI) | |
| Publisher | 014nnvj65 (ROR) | |
| Release Date | ||
| Language | ||
| Other Version |
Content Metadata
| Subject Area | |||||
| Genre | |||||
| Abstract |
| ||||
| Keywords |
|
00:00
DiagramComputer animation
00:29
Lecture/Conference
01:11
Lecture/Conference
03:06
Lecture/Conference
03:44
Computer animation
05:18
Lecture/ConferenceDiagram
07:54
Computer animation
08:14
Computer animation
Transcript: English(auto-generated)
00:13
Hello and welcome. In this video I want to show with the help of an example how we can dimension a rotor blade using the blade element theory. For the dimensioning I have chosen
00:26
an example wind energy plant which is the Anacon E48. This plant has a nominal power of 800 kilowatts, it has a rotor radius of 24 meters, a nominal wind speed of 12 meters
00:43
per second and the number of rotor blades is 3. The tip speed velocity is 78 meters per second. And I assume that the lift force coefficient CA is 1.2 and the drag force coefficient is 0.02.
01:02
In case of all calculations the rotor is divided into blade elements with the thickness dr of 1 meter. First of all we calculate the optimal pitch angle tether for each blade element. Therefore we first calculate the axial velocity VA. This velocity has the same number for each
01:25
blade element and is two-third of the wind velocity thus in our case it is 8 meters per second. And afterwards the tip speed ratio is calculated. This is the tip speed divided by the wind speed and in our example it is 6.5. After that some calculations for the element
01:49
at the tip with a radius of 24 meters are carried out. For this element the resulting velocity VR is calculated in the beginning and this is the square root of U to the
02:03
power of 2 and VA to the power of 2. So in our case it is 78.41 meters per second. And afterwards we calculate the angle delta thus the sum of angle of attack and pitch
02:21
angle. And this angle is calculated by the r cross cosine of U divided by VR. The result is 0.1 in radian measure thus 5.86 degrees. For this blade element we have
02:42
to assume a value for the angle theta which is our constructive parameter. We assume that this angle at this blade element is 0 degrees and after that we can calculate the angle of attack alpha. The angle alpha is the total angle delta minus the pitch angle
03:01
theta thus it is also 5.86 degrees. And afterwards the same calculation is carried out for the next blade element which is one meter closer to the rotor blade center and thus has a radius of 23 meters. In the beginning we calculate the circumferential
03:22
velocity U according to the equation tip speed ratio multiplied with the radius of the blade element and divided by the total rotor radius multiplied with the wind velocity. And in our case we get a result of 74.75 meters per second.
03:44
After that we calculate the resulting velocity VR. It is calculated by the square root of U to the power of 2 and VA to the power of 2 and is 75.18 meters per second in our example.
04:01
Afterwards the sum of the angle of attack and the pitch angle is calculated. It is again the r cross cosiness of U divided by VR. In our case this leads to 0.11 in radian measure thus 6.11 degrees. The requirement in order to reach equal flow compared
04:27
to the previous element is that the angle of attack alpha is the same compared to the element before. So in our example it has to be 5.86 degrees again.
04:42
The pitch angle theta, the constructive parameter can be calculated again by subtracting the angle of attack alpha from the total angle delta. And thus for this blade element the pitch angle theta is 0.25 degrees. The same calculation has to be carried out for
05:04
all further elements to the center of the rotor blade. For all further elements the assumption that the angle of attack alpha is the same at all blade elements has to be valid. After that I will calculate the optimal blade
05:22
depth dB for our example. In the beginning the calculation is carried out at the blade tip at a rotor radius of 24 meters. All the values of the equation are given so the optimal blade depth at this element is 0.88 meters. And after that this calculation
05:45
is carried out for the next blade element at a radius of 23 meters. And when inserting the given values we reach an optimal blade depth of 0.91 meters for this element.
06:03
The same calculation has to be carried out for all further blade elements until the center of the rotor. In this graph here you can see the optimal blade depth Tb and the pitch angle theta depending on the blade radius R. The blade
06:20
depth is shown on the left y axis and the pitch angle is shown on the right y axis. On the x axis the radius 0 corresponds to the center of the rotor blade and the radius 24 meters corresponds to the blade tip. It is visible that the blade depth and the
06:43
rotor radius. After that I want to show an example where I calculate the relevant forces at each blade element. In the beginning we calculate the lift force F A. First of all we calculate the lift force at the blade tip thus at
07:01
the radius R of 24 meters. And the lift force is calculated according to the here shown formula as explained in a previous video. All values are given or we have calculated them in a previous step. So it is important that the corresponding values for this blade
07:22
element are taken for the blade depth Tb and the velocity Vr. With these values the result for the lift force is 3.85 kilonewton for this blade element with a thickness of 1 meter. The same calculation must be carried out for all further elements to the
07:45
center of the blade. Afterwards the drag force is calculated. This is also calculated at the blade tip in the beginning. And when the given and already calculated values are taken into account the result of the drag force at this blade element
08:03
is 0.06 kilonewton. And the same calculation has to be carried out for all further elements until the center of the blades. Afterwards the tangential force F T is calculated. In the beginning this force is also calculated for the first element at the rotor tip at
08:25
a radius of 24 meters. And all values for this blade element are given from the previous calculations. We get a result of 0.33 kilonewton for this blade element. And this calculation is again carried out for all further elements until the center
08:45
of the blade. And finally we calculate the axial velocity F ax. This velocity is first calculated for the blade element at the blade tip having a radius of 24 meters. And we know this formula from a previous video. The values are given
09:08
from previous calculations. Thus the axial velocity for this blade element is 3.83 kilonewton. And this calculation again is carried out for all further elements until
09:22
the center of the blade. In this picture the calculated forces are shown depending on the rotor radius. A rotor radius of 0 at the X axis corresponds to the center of the rotor blade. And the radius of 24 meters corresponds to the blade
09:42
tip. You can see that the lift force and the axial force have the highest values. And the drag force has the lowest value as wanted. The lift force and the axial force increase with increasing rotor radius. And in the next video we will see why most
10:03
wind turbines have just three rotor blades. Thank you very much.
Recommendations
Series of 8 media