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46000 2020 exam And Solution
Introduktion til vindenergi (46000)
Danmarks Tekniske Universitet
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Technical University of Denmark
Written examination, date 20 May 2020 Page 1 of 5 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 2020
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. The key specifications are given at page 5. Although the turbine is intended for offshore application, we will use it here as a land based turbine, where the tower is clamped directly into the ground.
Part 1: Rotor design and power curve for two possible diameters (40%)
We will in this part look at the rotor design and the implication of using a smaller diameter. We first consider the rotor, as specified at page 5.
1 Based on the information at page 5, calculate the power coefficient C p∗ at rated wind speed. An air density of ρair = 1. 225 kg/m 3 can be assumed.
1 We assume that the power coefficient is equal to C p∗ for all wind speeds below rated, and that the power is constant above rated wind speed. Calculate the power at a wind speed of v = 8 m/s and present a plot of the power curve from cut-in to cut-out. If 1 was not solved, a value of C p∗ = 0. 5 can be used.
1 The maximum tip speed is vtip,max = 95 m/s and is reached at rated wind speed. With this information, compute the tip speed ratio at rated wind speed. Further, under the assumption of optimum rotor operation, calculate the thrust at rated wind speed.
The assignment continues at the next page
Part 3: Design of a glass fibre tower (40%)
We here consider an alternative land based design of the tower as a glass fibre structure with the same hub height as the offshore design. The glass fibre composite is made of a resin material of Ematrix = 3. 5 GPa and fibres with Efibre = 72 GPa. The volumetric fibre fraction is Vf =60%. For simplicity we assume that the fibres are all oriented in the vertical direction along the tower. We look at an initial primitive design of uniform diameter of DTower = 8 m and a thickness of t = 10 cm.
3 Calculate the effective modulus of elasticity Ec for the glass fibre composite and next the bending stiffness EcI for the tower.
3 We consider a design thrust of T ∗ = 4 MN and the vibration that follows by a sudden emergency stop, where the thrust disappears. Calculate the deflection of the tower top at the design thrust. Calculate also the natural frequency of the turbines fore-aft tower vibration after the emergency stop. If (3) was not solved, use EcI = 1 · 1012 Nm 2.
3 For the design thrust, T ∗, explain where the largest stress occur in the tower root cross section. Calculate the stress σf in the glass fiberes. Calculate also the strain ε at the same posi- tion.
3 The tensile strength of the fibres is σmax = 3. 45 GPa. Suggest a new combination of DTower and t that makes the natural frequency fNat = 0. 17 Hz. Further calculate the largest static rotor thrust this tower can take before the glass fibres break.
Figure 1: Specifications of the IEA 15 MW reference wind turbine.
1 Assume a power coefficient of Cp = 0. 5 and calculate the rated wind speed for the alterna- tive rotor. Next apply equation (1) for the two rotors at their rated wind speed to estimate the ratio of cord length between them for a fixed value of the radius r.
We use P = 12 ρairCpARotorv 3 again, this time with the new rotor area of ARotor 2 = π/ 4 D 22 and the given power coefficient of Cp = 0. 5. We solve for v at the rated power of 15 MW and get vr 2 = 11. 59 m/s. We next write up (1) for the two rotors below each other. each at their respective rated wind speed and for one common value of r
dT 1 = 12 ρairCT 2 πr dr v 2 r 1 = B 12 ρairCL(ωr) 21 c 1 dr (3) dT 2 = 12 ρairCT 2 πr dr v 2 r 2 = B 12 ρairCL(ωr) 22 c 2 dr (4)
Due to the equality signs, the ratio of the two middle terms and the ratio of the right-most terms are equal. If we further assume that the thrust and lift coefficients are identical, we have (recall that r is the same in each equation)
v 2 r 1 v 2 r 2
= (
ωr) 21 (ωr) 22
c 1 c 2
(5)
At rated wind speed, we further have ωr = vtip,maxr/R, where R is the rotor radius. Insertion in the above equation then gives
c 2 c 1
=
(
R 2 vr 2 R 1 vr 1
) 2
= 0. 83. (6)
So, surprisingly, the smaller rotor will also need a smaller cord length. This may seem counter- intuitive, but is a result of the fixed maximum tip speed, which allows the smaller rotor to spin faster at its rated wind speed. As a bonus remark, the power curve for the new rotor is shown in figure 2 along with the original. It can be seen that the smaller rotor reaches the rated power later and produces a smaller power than the original rotor for any wind speed below rated. The larger rotor area of the original rotor will thus lead to a larger annual energy production.
Solution to part 2: Power production and turbine price (20%)
2 Calculate the annual energy production by use of the power curve from question (1). Next, calculate the capacity factor and provide a value of A that gives a capacity factor of CaF=50%.
We calculate the annual energy production by
AEP =
∫ vout
vin
pow(v) prob(v) dv × 24 × 365 = 51 GWh (7)
where the integral gives the mean power in Watt and the last factors are the number of hours per year. The probability is computed from the Weibul distribution
prob(v) =
k A
( v A
)k− 1 exp
(
−
( v A
)k) . (8)
We next calculate the capacity factor as
CaF =
AEP
Prated × 24 × 365 = 0
. 39 (9)
We repeat the above calculations for different values of A and find that CaF=0 for A = 9. 5 m/s.
2 Calculate the net present value (NPV) for the net income, for a life time of 20 years for the turbine. If the total cost of the turbine and its installation is 240 MDKK, what will the in- ternal rate of return be?
For a capacity factor of 50% we get AEP=65 GWh, which for the tariff of 0 DKK/kWh gives an annual income of 24 MDKK. With running expenses of 5 MDKK, the annual net income is thus 19 MDKK. We calculate the net present value for 20 years of operation as
NPV =
∑ 20
j=
Annual net income (1 + d)j
= 190 MDKK (10)
where d = 0. 08 is the discount rate. With an initial cost of 240 MDKK, the total NPV is -50 MDKK. The internal rate of return is the value of d where the total NPV is exactly zero. We repeat the above calculations for various d values, until the total NPV is zero. This is achived at d = 0. 050 and the internal rate of return is thereby 5%.
Solution to part 3: Design of a glass fibre tower (40%)
3 Calculate the effective modulus of elasticity Ec for the glass fibre composite and next the bending stiffness EcI for the tower.
We use the rule of mixture to achive Ec = Vf Ef + (1 − Vf )Em = 44. 6 GPa. Next, for the cross sectional moment of inertia we use the thin-wall formula
I =
π 8
tD 3 = 20 m 4 (11)
Alternatively, one can consider the tubular cross section as the difference between two circular disks and calculate I as I = 64 π (D 4 outer − D 4 inner). We get next get EIeff = 879 GNm 2.
3 We consider a design thrust of T ∗ = 4 MN and the vibration that follows by a sudden emergency stop, where the thrust disappears. Calculate the deflection of the tower top at the design thrust. Calculate also the natural frequency of the turbines fore-aft tower vibration after the emergency stop. If (3) was not solved, use EcI = 1 · 1012 Nm 2.
For a cantilever beam with a point load at the end, the deflection is u = 3 1 EI P ℓ 3. We insert the hub height of 150 m for ℓ, the given design thrust load for P and the EI value of question (3) to achieve utower top = 5. 02 m. Next, for the natual frequency of a mass-less beam with a concentrated top mass, we have
fn =
1
2 π
√
3 EI
M ℓ 3
(12)
46000 2020 exam And Solution
Kursus: Introduktion til vindenergi (46000)
Universitet: Danmarks Tekniske Universitet
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