4. Competencias para la oralización civil
4.1 ESTIGMAS Y ESTEREOTIPOS FRENTE A ARQUETIPOS IDEALES
c) The non-dimensional inflow can be written as:
( )
Thus, the equation can be written as:( ) ( ) ( ) ( )
The ratio of the angles at 15.24 m/s (50 ft/s), designated with a subscript of f, to the lead lag angle at 7.62 m/s (25 ft/s), designated with a subscript of i, can be written, after some manipulation, as:
( ) ( ( ) ( ) )
The radial velocity is 50 rpm = 5.24 rad/s.
The axial induction factor at 7.62 m/s (25 ft/s), ai, is 0.30.
The axial induction factor at 15.24 m/s (50 ft/s), a , is 0.25. f
The radius is 4.57 m = 15 feet.
Substituting, one gets:
( ) ( ( ( ) ) ( ( )( )( ) ) )
Thus, this two fold increase in wind speed increases the steady state lead lag angle 5.6 times.
4.11 Using the lead–lag equations described in Problem 4.10:
a) Write the matrix equation for lead–lag motion if all of the terms are ignored except those that have to do with steady winds and vertical wind shear.
b) Solve the equations and find the expression for the lead–lag angle with steady winds and vertical shear. What is the effect of vertical wind shear on each term of the lead–lag angle in the absence of yaw rate, gravity, and crosswind?
SOLUTION
a) The equation for the lead lag motion, assuming zero yaw rate, gravity, and crosswind, is:
The steady state lead lag angle can be found using Cramer's Rule:
( )
The determinant of the top matrix is:
( )
The determinant of the bottom matrix is:
( )
(
2−1) ( [
0− 2−1)( )
2]
= 2(
2−1)
2This constant blade bending in the lead lag direction is the result of steady wind and shows no effect of wind shear in the absence of yaw rate, gravity, and crosswind.
Cosine Term, ζ 1c
The coefficient of the cosine term can also be found using Cramer's Rule:
( )
The determinant of the top matrix is:
( ) ( )
The determinant of the bottom matrix is the same as before:
( )
Thus, the perturbation introduced on top of the steady state term, in the absence of yaw rate, gravity, and crosswind, is negative when the cosine of the azimuth is positive (in the lower half of the rotor disk) where there is less wind. The lead lag bending angle is increased in the upper half of the rotor disk.
Sine Term, ζ 1s
The coefficient of the sine term can also be found using Cramer's Rule:
( )
The determinant of the top matrix is:
( )
Thus, in the absence of yaw rate, gravity, and crosswind, wind shear only affects the cosine term in the lead lag angle equation.
4.12 Consider the data file, Flap1.txt. This data represents 10 minutes of blade bending moment data taken from an operating wind turbine. The turbine has a nominal operating speed of 60 rpm. The data was sampled at 40 Hz. Find the psd of the data. What are the predominant frequencies? What do you suspect that these frequencies might be caused by?
SOLUTION
Using the MiniCodes, the psd the range of interest is shown below. The peak frequencies are 1.07 Hz and 4.307 Hz. The spike at 1.07 is presumably due to rotation. The other one may be due to a blade vibration mode.
psd
-20 0 20 40 60 80 100 120 140 160 180 200
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10
psd
4.13 Consider the following turbine. The tower is 50 m high, has diameter of 2 m at the top and 3.36 m at the bottom. The mass of the nacelle and rotor is 20,400 kg. The wall thickness is 0.01384 m. Assume that the density of steel is 7,700 kg/m3 and that its modulus of elasticity is 160 GPa. The problem considers a number of approaches (successively improving) to estimate the natural frequency (1st mode) of the turbine.
a) Assume that the tower is not tapered, but has a constant diameter of 2 m. What is the natural frequency using the simple equation, but ignoring the mass of the rotor nacelle assembly (RNA) on the top of the tower.
b) Repeat (a) using the Euler method. Do this directly or with the MiniCodes.
c) Repeat (a) using the Myklestad method. Use the MiniCodes.
d) Repeat (a), taking into account the weight of the RNA the top, using the simple equation.
e) Repeat (d), using the Myklestad method. This will require some approximations to account for the weight of the RNA.
f) Repeat (e), but take into account the actual taper of the tower.
SOLUTION
Parts of this problem can be solved with the help of the MiniCodes. An Excel spreadsheet is also useful for solving the simpler parts and for preparing the input files for the MiniCodes. See, for example, V47.xls.
a)
( ) ( )( )
( )( )
(
E)
HzL m m f EI
nacelle tower
736 . 50 0 248 , 33 23 . 0
0426 . 0 9 160 3 2
1 23
. 0
3 2
1
3
0 3 = =
= +
π π
b) See MiniCodes screenshot: ω=4.51rad /s, fo = 0.72 Hz
c) This is for straight tower with no weight, using the Myklestad method. The answer is almost the same. Ten identical sections works well enough:
Input: tower straight_2.csv Use:
0 0.086 0.043 0.1 0.086 0.043 0.2 0.086 0.043 0.3 0.086 0.043 0.4 0.086 0.043 0.5 0.086 0.043 0.6 0.086 0.043 0.7 0.086 0.043 0.8 0.086 0.043 0.9 0.086 0.043 1 0.086 0.043
d)
( ) ( )( )
( )( )
(
E)
HzL m m f EI
nacelle tower
384 . 50 0 400 , 20 248 , 33 23 . 0
0426 . 0 9 160 3 2
1 23
. 0
3 2
1
3
0 3 =
= +
= +
π π
e) We used 20 sections. The top one was heavier to include effect of weight. Note that method uses averages, so we needed to introduce an extra section.
Use:
0 1.146 0.0426 0.05 1.146 0.0426 0.0501 0.0864 0.0426 0.1 0.0864 0.0426 0.15 0.0864 0.0426 0.2 0.0864 0.0426 0.25 0.0864 0.0426 0.3 0.0864 0.0426 0.35 0.0864 0.0426 0.4 0.0864 0.0426 0.45 0.0864 0.0426 0.5 0.0864 0.0426 0.55 0.0864 0.0426 0.6 0.0864 0.0426 0.65 0.0864 0.0426 0.7 0.0864 0.0426 0.75 0.0864 0.0426 0.8 0.0864 0.0426 0.85 0.0864 0.0426 0.9 0.0864 0.0426 0.95 0.0864 0.0426
1 0.0864 0.0426
f) This is whole thing, much wider at the base, much more steel. The RNA is only approximated here; could have used average area of top 2 sections to find factor to include effect of weight:
For inputs we used:
0 1.1461 0.0426 0.05 1.1461 0.0519 0.0501 0.0896 0.0519 0.1 0.0928 0.0612 0.15 0.0961 0.0704 0.2 0.0993 0.0797 0.25 0.1026 0.089 0.3 0.1058 0.0983 0.35 0.1091 0.1076 0.4 0.1123 0.1169 0.45 0.1156 0.1262 0.5 0.1188 0.1354 0.55 0.122 0.1447 0.6 0.1253 0.154 0.65 0.1285 0.1633 0.7 0.1318 0.1726 0.75 0.135 0.1819 0.8 0.1383 0.1912 0.85 0.1415 0.2004 0.9 0.1448 0.2097 0.95 0.148 0.219
1 0.1512 0.2283
Summary
Tower RNA Method Frequency, Hz
Straight No Simple 0.736
Straight No Euler 0.72
Straight No Myklestad 0.74
Straight Yes Simple 0.38
Straight Yes Myklestad 0.40
Tapered Yes Myklestad 0.78
Conclusion:
Adding weight to top reduces stiffness; making the tower wider at the base increases stiffness.