The Tower Of The Wind Turbine See Figure 6 Has A Height Of 3 ✓ Solved
The Tower Of The Wind Turbine See Figure 6 Has A Height Of 30 M And
The tower of the wind turbine (see Figure 6) has a height of 30 meters and a circular hollow cross-section with an inner diameter of 0.4 meters and an outer diameter of 1 meter. The rotor and hub mass is 0.3 × 10^3 kg. The tower is constructed of steel with a Young’s modulus of 200 GPa and a density of 7800 kg/m^3. When ignoring the tower's weight, determine the following:
- (a) The natural frequency of transverse vibration of the system.
- (b) The time response due to an initial transverse displacement x₀ = 0.1 meters.
- (c) The maximum velocity and accelerations.
When the tower’s weight is considered, use an equivalent weight of 23.57% of its actual weight, and repeat parts (a), (b), and (c).
Sample Paper For Above instruction
Introduction
Understanding the dynamic behavior of wind turbine towers is essential for their structural integrity and operational efficiency. This paper addresses the calculation of the natural frequency, transient response, and maximum velocities and accelerations of a steel wind turbine tower, considering both scenarios: ignoring and including the tower’s weight.
Parameters and Assumptions
The tower has a height (H) of 30 m, with a hollow circular cross-section characterized by an inner diameter (d_i) of 0.4 m and an outer diameter (d_o) of 1 m. The material (steel) has Young’s modulus (E) of 200 GPa and a density (ρ) of 7800 kg/m^3. The rotor and hub mass (m_r) is 300 kg.
Assuming the tower is modeled as a uniform beam with the properties given, the primary equations for vibration analysis are applied to determine the natural frequency, transient displacement, velocity, and acceleration.
Calculation of Structural Parameters
Cross-Sectional Area and Moment of Inertia
The cross-sectional area (A) of the hollow cylinder is
A = \frac{\pi}{4} (d_o^2 - d_i^2) = \frac{\pi}{4} (1^2 - 0.4^2) = 0.85\,m^2.
The second moment of area (I) for a hollow circular section:
I = \frac{\pi}{64} (d_o^4 - d_i^4) = \frac{\pi}{64} (1^4 - 0.4^4) ≈ 0.249\,m^4.
Mass per Unit Length
Density times cross-sectional area gives the mass per unit length (μ):
μ = ρ × A = 7800 × 0.85 ≈ 6,630\,kg/m.
Total mass of the tower (m_t) is:
m_t = μ × H = 6,630 × 30 ≈ 198,900\,kg.
Part A: Natural Frequency Calculation Ignoring Tower’s Weight
The transverse vibrations of the tower can be approximated as a cantilever beam with point load (mass of rotor and hub) at the top.
The fundamental natural frequency (f_n) for a cantilever with a concentrated mass at the free end:
f_n = \frac{1}{2\pi} \sqrt{\frac{3EI}{m_t H^3}} + \text{(additional correction for mass contribution)}.
However, because the tower's mass is significant, a more precise estimation uses the combined mass-spring system approach.
Using the formula:
f_n = \frac{1}{2\pi} \sqrt{\frac{k}{m_{total}}},
where k is the lateral stiffness, and m_total includes mass contributions.
The stiffness k:
k = 3EI / H^3 ≈ \frac{3 × 200×10^9 × 0.249}{30^3} ≈ 1.83×10^8\,N/m.
Total mass ignoring the tower's weight:
m_total ≈ m_r + m_t ≈ 300 + 198,900 ≈ 199,200\,kg.
Therefore:
f_n ≈ \frac{1}{2\pi} \sqrt{\frac{1.83×10^8}{199,200}} ≈ 0.272\,Hz.
Part B: Time Response Due to Initial Displacement
The initial displacement x₀ = 0.1 meters results in simple harmonic motion with the natural frequency computed above.
Displacement over time:
x(t) = x_0 \cos(2\pi f_n t).
Velocity and acceleration:
v(t) = -x_0 2\pi f_n \sin(2\pi f_n t),
a(t) = -x_0 (2\pi f_n)^2 \cos(2\pi f_n t).
Maximum velocity:
V_{max} = x_0 2\pi f_n ≈ 0.1 × 2\pi × 0.272 ≈ 0.171\,m/s.
Maximum acceleration:
a_{max} = x_0 (2\pi f_n)^2 ≈ 0.1 × (2\pi × 0.272)^2 ≈ 0.82\,m/s^2.
Part C: Maximum Velocity and Accelerations
As above, the maximum velocity and acceleration are approximately 0.171 m/s and 0.82 m/s² respectively.
Part D: Including Tower's Weight
The effective weight is increased by 23.57%. The mass of the tower’s weight component becomes:
m_{weight} = 0.2357 × m_t ≈ 0.2357 × 198,900 ≈ 46,876\,kg.
Adding this to the rotational mass:
m_{total\_weighted} = m_r + m_{weight} ≈ 300 + 46,876 ≈ 47,176\,kg.
Re-calculating the natural frequency:
f_{n,weighted} ≈ \frac{1}{2\pi} \sqrt{\frac{1.83×10^8}{47,176}} ≈ 0.632\,Hz.
Corresponding maximum velocity:
V_{max,weighted} ≈ 0.1 × 2π × 0.632 ≈ 0.396\,m/s.
Maximum acceleration:
a_{max,weighted} ≈ 0.1 × (2π × 0.632)^2 ≈ 2.44\,m/s^2.
Conclusions
The inclusion of the tower’s weight significantly increases the natural frequency, as well as the maximum velocities and accelerations. These calculations demonstrate how structural parameters and load considerations critically influence the dynamic response of wind turbine towers.
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