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46000 2021 exam And Solution
Introduktion til vindenergi (46000)
Danmarks Tekniske Universitet
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Technical University of Denmark
Written examination, date 21 May 2021 Page 1 of 6 pages
Course name: Introduction to Wind Energy
Course number: 46000
Aids allowed: All aid
Exam duration: 4 hours
Weighting: As indicated on the individual parts
46000 Introduction to Wind Energy
Exam
Spring 2021
We consider here the recent 15 MW IEA reference wind turbine, which has been designed by NREL and DTU. It is a 3-bladed up-wind turbine with direct drive generator.
1 We now look at the cord distribution for the 4-bladed turbine and compare it to the cord distribution of the 3-bladed turbine. We know that we can compute the cord distribution for an optimum turbine with the following relation:
c =
(
8 πra sin 2 φ
)
F
B(1 − a) [CL cos φ + CD sin φ]
(2)
In the previous equation, F is the Prandtl’s tip loss factor:
F = 2
π
arccos
{
e−f
}
(3)
f =
B
2
(R − r) r sin φ
(4)
where R = D/ 2 is the turbine radius, and r is a coordinate varying from 0 to R (from the center to the tip of the blade). Plot the ratio between the cord length c of the 3-bladed rotor and that of the 4-bladed rotor as a function of the coordinate r. You can assume that the tip speed ratio, the induction and CL, CD are identical between the two rotors.
Part 2: Power production and NPV computations (30%)
2 We now assume that the turbines are installed in a location which is characterized by a roughness length of z 0 = 0[m]. Using a u∗ = 0[m/s], compute the mean velocity at hub height U ̄hub assuming neutral stability.
2 Now assume that the wind conditions at hub height are represented by a Weibull distri- bution where k = 2[−] and A = ̄Uhub from question 2. Plot the resulting Weibull distribution as a function of the wind velocity. Calculate the an- nual energy production for the two above mentioned turbines, the one with B = 3 and the one with B = 4. If questions 1 and 1 were not solved, you can use CP∗ 4 B = 0. 48 and P 4 B = 15[M W ] to solve this exercise. If 2 was not solved, a value of A = 9[m/s] can be used.
2 Now we will check whether the extra energy production guaranteed by the extra blade compensates the extra cost. For each turbine, consider a fixed cost for the tower, nacelle and installation of 170 MDKK. Furthermore, each blade costs 10 MDKK. The electricity provider pays a tariff of 0 .37[DKK/kW h]. The discount rate is 7 .5%. The annual running expenses can be considered fixed, and are 5 MDKK. Compute the NPV for the two turbines for a 20 years life time. For this particular case, does it pay off to have 4 rather than 3 blades?
The assignment continues at the next page
Part 3: Design of the blade for the 3-bladed wind turbine (30%)
3 One of the turbine blades is now stored at the port, waiting to be shipped offshore where it will be mounted on a turbine, see Figure 1. The blade is resting on two stands, and is only sub- ject to its own distributed weight. It can be assumed here that the blade mass of mb = 65[ton] is uniformly distributed along the blade length of Lb = 117[m]. Compute the maximum mo- ment Mrest for this configuration.
Figure 1: Schematization of the blade resting at the port.
3 The design moment is taken as Mdesign = 3 · Mrest. The supporting structure of the blade can be approximated by a sandwich structure, with height H and with a skin material area of Askin. The fiberglass composite has a Eskin = 55[GP a]. Compute the minimum area of material that is needed to keep the maximum deformation due to the design load below εskin = 0[−]. To find H as a function of the blade length, use H = 0. 02 Lb, where Lb is the length of the blade.
3 Assume that the whole weight of the blade is due to the fiberglass composite in the sup- port structure. Neglect any other contribution. The density of the glass fibers in the composite is ρf = 2500[kg/m 3 ], and the density of the resin composing the matrix of the fiberglass mate- rial is ρm = 1100[kg/m 3 ].
Compute the volume fraction of the fibers Vf. Next, if the stiffness of the glass fibers is Ef = 90 .0[GP a], what is the stiffness Em of the matrix material? Assume that all the glass fibers are oriented axially (0 degrees).
Solution to part 1: Rotor design and power curve for two possible diame-
ters (40%)
1 We assume an air density of ρair = 1. 225 kg/m 3. For the three bladed rotor, we assume that the power coefficient is equal to CP∗ 3 B = 0[−] for all wind speeds below rated, and that the power is constant the above rated wind speed. Present a plot of the power curve from cut-in to cut-out.
We use
P 3 B = 12 ρairCp∗ 3 BARotorU 3
and insert the rated wind speed of Ur = 10. 59 m/s. The rated power is 15 MW.
Figure 3: Power curve of the three bladed rotor
1 Equation (5), from Wilson et al. (1976), describes the maximum power coefficient CP,max[−] as a function of the tip speed ratio λ[−].
CP,max =
(
16
27
)
λ
[
λ + 1
.32 +
( λ− 8 20
) 2
B 23
]− 1
−
(0)λ 2 CL CD
(
λ + 21 B
) (5)
In the above equation, λ is the tip speed ratio, B is the number of blades, CL is the lift coeffi- cient and CD is the drag coefficient. Find for what ratio of CL/CD the formula predicts a maximum power coefficient equal to CP∗ 3 B= 0[−]. Plot CP,max vs. λ for this value of CL/CD.
Implementing the formula, one can find that for CL/CD = 51. 7 , λ = 9., B = 3 we obtain a CP∗ 3 B= 0[−]. The curve is plotted in Figure 4.
Figure 4: CP vs. λ for CL/CD = 51. 7
1 We use Equation (5) to compute the new optimum CP∗ 4 B, assuming the same tip speed ratio λ and the same CL/CD from question 1. The rated wind speed of the 4-bladed wind turbine is the same as for the 3-bladed wind turbine. What is the new rated power P 4 B of the 4-bladed wind turbine? Present a plot of the power curve from cut-in to cut-out of the 3-bladed wind turbine and the 4-bladed tubine on top of each other. You can assume that the CP below rated wind speed is equal to CP∗ 4 B, and that the power is equal to P 4 B above rated. Briefly discuss the differences between the two curves.
Using the same formula in the previous point, we obtain a CP∗ 4 B = 0[−]. The new rated power is computed via
P 4 B = 1
2
ρairC∗P 4 BU rated 3 ARotor = 15[M W ]
Figure 5: Power curve of the three bladed vs. four bladed rotor
The curves are very similar.
1 We now look at the cord distribution for the 4-bladed turbine and compare it to the cord
induction factors will change between the two designs, as the optimization of the rotor is de- pendent on the number of blades.
Part 2: Power production and NPV computations (30%)
2 We now assume that the turbines are installed in a location which is characterized by a roughness length of z 0 = 0[m]. Using a u∗ = 0[m/s], compute the mean velocity at hub height U ̄hub assuming neutral stability.)
The formula for the log-profile is
U (z) =
u∗ κ
(
ln
z z 0
)
By substituting u∗ = 0[m/s], z 0 = 0[m] we obtain U ̄hub = 10[m/s].
2 Now assume that the wind conditions at hub height are represented by a Weibull distri- bution where k = 2[−] and A = ̄Uhub from question 2. Plot the resulting Weibull distribution as a function of the wind velocity. Calculate the an- nual energy production for the two above mentioned turbines, the one with B = 3 and the one with B = 4. If questions 1 and 1 were not solved, you can use CP∗ 4 B = 0. 48 and P 4 B = 15[M W ] to solve this exercise. If 2 was not solved, a value of A = 9[m/s] can be used.
A = 10[m/s]
The plot of the Weibull distribution is in Figure 7.
Figure 7: Plot of the Weibull distribution with A = 10[m/s] and k = 2[−]
2 Now we will check whether the extra energy production guaranteed by the extra blade compensates the extra cost. For each turbine, consider a fixed cost for the tower, nacelle and
installation of 170 MDKK. Furthermore, each blade costs 10 MDKK. The electricity provider pays a tariff of 0 .37[DKK/kW h]. The discount rate is 7 .5%. The annual running expenses can be considered fixed, and are 5 MDKK. Compute the NPV for the two turbines for a 20 years life time. For this particular case, does it pay off to have 4 rather than 3 blades?
We get the two power curves as a function of the wind velocity from question 1.
AEPB=3 =
∫ u=25[m/s]
u=4[m/s]
p(u)P 3 B (u) · 24 · 365 = 74[GW h]
AEPB=4 =
∫ u=25[m/s]
u=4[m/s]
p(u)P 4 B (u) · 24 · 365 = 75[GW h]
For the three-bladed wind turbine, the yearly income is
Annual Net Income(B = 3) = 74[GW h]· 0. 37 · 10 − 3 [DKK/W h]−5[M DKK] = 22[M DKK]
The NPV calculation is done with a d = 0. 075
NPV(B = 3) =
∑ 20
j=
Annual net income (1 + d)j
= 230. 72 MDKK
Subtracting the initial expenses of 200 MDKK, we are left with 30 MDKK. For the four-bladed wind turbine, the yearly income is
Annual Net Income(B = 4) = 75[GW h]· 0. 37 · 10 − 3 [DKK/W h]−5[M DKK] = 23[M DKK]
The NPV calculation is done with a d = 0. 075
NPV(B = 4) =
∑ 20
j=
Annual net income (1 + d)j
= 234. 43 MDKK
Subtracting the initial expenses of 210 MDKK (one extra blade), we are left with 24 MDKK. So, in this particular case, it does not pay off to add one blade to the wind turbine. The extra income due to the extra power production does not justify the extra expense.
Part 3: Design of the blade for the 3-bladed wind turbine (40%)
3 One of the turbine blades is now stored at the port, waiting to be shipped offshore where it will be mounted on a turbine, see Figure 1. The blade is resting on two stands, and is only sub- ject to its own distributed weight. It can be assumed here that the blade mass of mb = 65[ton] is uniformly distributed along the blade length of Lb = 117[m]. Compute the maximum moment Mrest for this configuration. With g = 9[m/s 2 ] we obtain
Mrest = 1 8
mbgLb = 9[M N m]
That can be reworked into
Em =
Eskin − Vf Ef 1 − Vf
= 5[GP a]
46000 2021 exam And Solution
Kursus: Introduktion til vindenergi (46000)
Universitet: Danmarks Tekniske Universitet
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