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Wind Energy Exercises Week 12
Introduktion til vindenergi (46000)
Danmarks Tekniske Universitet
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46000 Introduction to Wind Energy
Exercise: Electricity generation
What type of generator does the DTU 10 MW turbine have?
Plot ω against wind speed. Which control regimes can you identify?
Plot the tip speed ratio, λ, against wind speed
Plot the power coefficient against wind speed. Does CP = 4a(1 − a) 2 hold?
Plot torque (aerodynamic and generator) as function of wind speed
Plot torque as function of ω
Solution
What type of generator does the turbine have? The turbine is equipped with a medium-speed generator, with a maximum rotational speed of 480 rpm. It is connected to the wind turbine rotor through a multiple-stage gearbox with a gear box ratio of 50. The efficiency of the generator is 94 %.
Plot ω against wind speed. Which control regimes can you identify?
Figure 1: Rotational speed ω as function of wind speed.
The plot is shown in figure 1. From cut-in at 4 m/s to 7 m/s the turbine is operated with a con- stant rotational speed of 6 rpm (regime 1). From 7 m/s to 11 m/s (rated wind speed) the turbine operates at a fixed tip speed ratio and thus with linearly increasing rotational speed (regime 2). Above rated speed, the rotational speed is fixed (regime 3). The findings are further documented in figure 2 which shows the tip speed of the rotor blade, the electrical power (obtained as 0. times the mechanical power) and the tip speed ratio λ.
Plot the tip speed ratio, λ, against wind speed The plot is included in figure 2. First, in regime 1, λ drops due to the fixed rotational speed. Next, in regime 2, λ is constant. Finally, in regime 3, λ decreases to keep a fixed rotational speed.
Plot the power coefficient against wind speed. Does CP = 4a(1 − a) 2 hold? Figure 3 shows CP plotted against wind speed. CP has been found as CP = P/(0. 5 ρAv 3 ) where P is the mechanical power. The power coefficient increases in regime 1 where ω is con- stant. This is because the optimum λ is gradually reached here. In regime 2 where λ is kept constant at its optimum value, CP has a constant value of 0 which is less than the Betz limit of 0. The Betz limit, however, can only be reached with an infinite number of blades and
no losses. It is thus an idealized limit. In regime 3, CP decreases, as the power potential of the wind is not used fully due to the generator rating.
Figure 3: Power coefficient CP and estimated value based on induction factor and 1D momentum theory.
To check if CP = 4a(1 − a) 2 holds, CT is read from table 3 in the reference turbine report. The induction factor a is found from 1D momentum theory CT = 4a(1 − a). This second-order equation has two roots for each value of CT. Here, from inspection, the smallest is the relevant one. Next, a test value of CP = 4a(1 − a) 2 can be calculated. The test value is included in figure 3. First, we see that the test value is larger than the tabulated value. This is due to the as- sumption of negligible energy losses in the 1D momentum theory. Therefore CP = 4a(1 − a) 2 does not hold.
- Plot torque (aerodynamic and generator) as function of wind speed The aerodynamic torque is found from Pmech = τaeroωaero. Next as
Paero = τaeroωaero = Pgen = τgenωgen = τgenNgearωaero (1)
we have τgen = τaero/Ngear. Note that the generator losses are not included here as we consider the input to the generator before conversion to electricity. The torque values are plotted in figure 4, upper and middle panel. The curves only differ by the gearing factor. The torque increases through regime 1 and 2 and is kept constant in regime 3. In regime 2, λ is constant. Therefore ω is proportional to the wind speed v and CP is constant. More-over, as P = τ ω = 1/ 2 ρACP v 3 , the torque τ must be proportional to v 2 in regime 2. This can be seen in the figure for v > 7 m/s.
- Plot torque as function of ω The plot is shown in the lower panel of figure 4. In regime 1, ω is constant at 6 rpm and τ increases as the wind speed increases. In regime 2 we found in question (5) that ω ∼ v and τ ∼ v 2 (here ∼ means ’is proportional to’). Therefore τ ∼ ω 2
Figure 4: Torque as function of wind speed and rotor speed. Upper: Aerodynamic torque versus wind speed. Middle: Generator torque versus wind speed. Lower: Aerodynamic torque versus rotor speed.
Wind Energy Exercises Week 12
Kursus: Introduktion til vindenergi (46000)
Universitet: Danmarks Tekniske Universitet
