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Wind Energy Exercises Week 6
Introduktion til vindenergi (46000)
Danmarks Tekniske Universitet
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- A fiber-polymer composite spar cap of a wind turbine rotor blade has a fatigue strength coefficient of C=500 MPa and an exponent of m=10 at R=-1. During a blade certification test, the cap is subject to a constant equivalent stress amplitude of σa=150 MPa.Calculate the lifetime of the cap.
Forhåndsvisning af tekst
46000 Introduction to Wind Energy
Exercises: Wind turbine technology
What are the pros and cons of a down wind turbine relatively to an upwind turbine?
What is the reason for giving the H-rotor a spiral shape?
Summarize the advantages of vertical-axis wind turbines.
Find a small wind turbine for private use. Characterise its technology. What effect does it offer? Height? Aerodynamic work principle.
A Savonious rotor is placed in a constant wind of 8 m/s. The drag coefficient for a hollow half-cylinder is 2 if the wind blows into it and 1 if the wind blows towards its outward side. These drag coefficients are based on the frontal area. Assume a height of 5m and a total diameter of 5m. If no generator is mounted, what is the equilibrium rate of rotation? What power would you expect it to yield?
Why can’t the monopile substructure be installed at all depths?
What is the average depth of Anhold wind farm? What foundation type was used?
Show that
ω 6 ω 1
=
r 1 r 2
r 3 r 4
r 5 r 6
(1)
for configuration (d) on the gear ratio slide.
In a four-gear straight-line gear train (see figure 9 on the gear ratio slide), the smallest gear is 2 inch in diameter and the largest gear has a diameter of 12 inches. If the mechanism is used for speed reduction, how much is the gear ratio between the input and output?
In a gear train similar to that in figure 9(d) on the gear ratio slide, used as a speed reducer, the first gear has 15 teeth and the second gear has 2 sets of teeth. One has 30 teeth and the other has 14. The third gear has two sets of teeth, one with 32 teeth and the other with 20 teetch. The last gear has 44 teach. Determine at what rpm the input shaft should rotate such that the output shaft has a speed of 179 rpm.
In a planetary gear suppose Ns = 44 and Nr = 100. If the sun is connected to the shaft of an engine whose rpm is 450 rpm, what is the speed of the arm?
In problem (11) suppose that the arm is the input and the sun is the output. Find the sun gear rpm.
If the sun gear in problem 12 is subject to 200 kNm torque, how much torque must the engine provide?
Explain what double bearing and torque support means for a drive train.
What are the pros and cons of direct drive?
Explain the drive train concepts of table 6 in the text book in depth to your site mate. In depth.
The depth is 14 m. Monopiles were used.
- Show that
ω 6 ω 1
=
r 1 r 2
r 3 r 4
r 5 r 6
(5)
for configuration (d) on the gear ratio slide. ... Proof not shown here.
- In a four-gear straight-line gear train (see figure 9 on gear ratio slide), the smallest gear is 2 inch in diameter and the largest gear has a diameter of 12 inches. If the mecha- nism is used for speed reduction, how much is the gear ratio between the input and output? We have
n 4 n 1
=
N 1
N 4
=
D 1
D 4
=
2. 5
12 .6 = 0
. 1984 ≈ 1 / 5 (6)
The gear ratio is thus approximately 1/5.
- In a gear train similar to that in figure 9(d) of the gear ratio slide, used as a speed reducer, the first gear has 15 teeth and the second gear has 2 sets of teeth. One has 30 teeth and the other has 14. The third gear has two sets of teeth, one with 32 teeth and the other with 20 teetch. The last gear has 44 teach. Determine at what rpm the input shaft should rotate such that the output shaft has a speed of 179 rpm. The gear ratio is calculated by equation (9) of the text book. We obtain
n 6 n 1
=
N 1
N 2
N 3
N 4
N 5
N 6
= 15
30
14
32
20
44 = 0
. 0994 (7)
hereby the input shaft must rotate with 179 rpm/0=1800 rpm.
- In a planetary gear suppose Ns = 44 and Nr = 100. If the sun is connected to the shaft of an engine whose rpm is 450 rpm, what is the speed of the arm? We use the formula
ωA = ωs
Ns Nr + Ns
= 450 rpm
44
100 + 44 = 137
.5 rpm (8)
In problem (11) suppose that the arm is the input and the sun is the output. Find the sun gear rpm. The ratio is the inverse of the above result. Hereby ωs = 450 rpm 144/44 = 1473 rpm.
If the sun gear in problem (12) is subject to 200 kNm torque, how much torque must the engine provide? The mechanical power is conserved (losses neglected). Therefore τA = (ωs/ωA) 200 kNm=144/44 200 kNm=655 kNm.
14–16: no solution given here.
Wind Energy Exercises Week 6
Kursus: Introduktion til vindenergi (46000)
Universitet: Danmarks Tekniske Universitet
