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46000 2022 exam And Solution
Introduktion til vindenergi (46000)
Danmarks Tekniske Universitet
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Forhåndsvisning af tekst
Technical University of Denmark
Written examination, date 18 May 2022 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 2022
We consider here a design for a 2 bladed offshore wind turbine, with a rated power of P 0 = 12 .2[M W ], a diameter of D = 225[m] and a hub height of hhub = 150[m]. The rota- tional speed at rated power is ωr = 1[rad/s], which is achieved for a rated wind speed of Vr = 10[m/s]. The cut-in and cut-out speeds are respectively 3[m/s] and 25[m/s].
Note: please report your results in SI units if possible.
Part 2: Dimensioning of the tower: natural frequency (30%)
We here look at the tower dimensioning and at its influence on the natural frequency of the turbine structure.
2 The top mass of the turbine (sum of hub, blades and nacelle mass) is mrna = 9. 52 e5[kg]. The tower is made up of a steel (E = 2. 11 e11[P a]) tubular cylinder with constant outer diame- ter of Dtower = 9[m] and constant thickness of ttower = 0[m].
Assuming that the turbine can be modeled as a cantilever beam of length L = 150[m] with a top mass of mrna, compute the natural frequency ωnat of the structure. (Second moment of area of tubular section: I = π 4 (r 42 − r 41 ) where r 2 is the outer radius, and r 1 is the inner radius.)
Due to the load imbalance analyzed in Part 1, the rotor will generate a time-varying aerody- namic force on the structure with frequency 2 ωr. If the tower frequency ωnat is close to 2 ωr, we expect the tower natural frequency to be excited by this aerodynamic force. Are the two frequencies close here?
2 The engineers designing the turbine would like to increase the natural frequency of the structure to ωnat,new = 3[rad/s] by modifying the thickness of the turbine tower. What is the thickness ttower,new that allows to achieve ωnat,new = 3[rad/s]?
2 Assume that the rotor thrust at the rated wind speed can be computed with a thrust co- efficient CT [−] as from optimum rotor 1D momentum theory. Compute the rated thrust on the rotor Tr[N ] and the moment at the tower base Mr[N m]. What is the maximum stress in the material, using the new configuration of the tower?
If the previous question was not solved, use ttower,new = 0[m].
Part 3: Economy and AEP of original turbine vs. turbine with leading edge
erosion (30%)
In this part, we analyze the annual energy production and the associated economy of the two bladed rotor in two different working conditions.
3 In this question, we plot the power curve of the turbine. First, compute the power coef- ficient at rated CP,r.
Now plot the power curve for the turbine as from the following relation
P =
1
2
ρCP,rArotorV 3 , V < Vr
P 0 , V ≥ Vr
(1)
Next, calculate the AEP of this turbine for a wind climate described by a Weibull distribution with parameters A = 7[m/s] and k = 2[−].
3 After some years of operation, due to the high maximum tip speed velocity Vtip = ωrR, the wind turbine operator starts to observe leading edge erosion. To avoid further erosion prob- lems, he has the possibility to reduce the maximum tip speed to Vtip,red = 0. 9 Vtip, at the price of a reduced power output of the turbine for some velocity range. Plot the power curve associated with the reduced tip speed operation, which is described by the following relations
P =
1
2
ρCP,rArotorV 3 , V < Vs
P (Vs) +
P 0 − P (Vs) Vr,red − Vs
(V − Vs), Vs ≤ V < Vr,red
P 0 , V ≥ Vr,red
(2)
In the above equation, the new rated wind speed Vr,red associated with the reduced tip speed is computed by assuming that
V tip 2 Vr = V tip,red 2 Vr,red (3)
while
Vs = Vr
Vtip,red Vtip
(4)
Next, re-calculate the AEP with the new power curve for the same climate of the previous ques- tion. If the previous question was not solved, use CP,r = 0[−].
Figure 1: Q1.
Solution to part 1: Load on the wind turbine in shear flow (40%)
1 The turbine is operating in a neutral boundary layer, which is described by a log profile with a roughness length of z 0 = 0[m]. What is the friction velocity u∗ that allows to achieve the Vr at hub height? Plot the velocity profile between 1 and 300 meters height using a log profile is the wind velocity Vtop at the highest point reached by the blade tip ztop and the wind velocity Vbot at the lowest point zbot?
Due to the shear, there will be a larger speed above hub height than below. See the plot in Fig. 1. It’s easy to invert the log profile formula to find Eq. (5)
u∗ =
Vrκ log zhub/z 0
= 0[m/s] (5)
By substituting htop = hhub + 0. 5 D and hbot = hhub − 0. 5 D into Eq. (6)
V =
u∗ κ
log
z z 0
(6)
we obtain V 0 ,top = 11[m/s] and V 0 ,bot = 9[m/s]. 1 Consider the position where both blades are vertical, one pointing upwards and one down- wards. Sketch the velocity triangles at ztop and zbot, using the two local incoming wind velocities Vtop and Vbot and assuming a = 1/ 3. Use some scale in both axes to indicate the vector lengths.
Next, compute the relative velocity Vrel and the flow angle φ at the two locations when the turbine is operating at the rated conditions.
If the previous question was not solved, use Vbot = 9[m/s] and Vtop = 11[m/s].
Figure 2: Velocity trianges for Q1.
We observe in Fig. 1 the velocity triangles. They are not so different from each other, since the main contribution to the relative velocity is from the rotationally induced velocity. The flow angles are quite low, due to the large tip speed ratio, being φbot = 2[deg] and φtop = 3 .118[deg]. They were found by using the simple relation in Eq. (7), where, from optimum rotor design, it was compute that a′ = 0 when a = 1/ 3.
φ = arctan
ωR V (1 − a)
(7)
In the above equation, V is either the top or bottom velocity. By using Pythagora’s theorem on the velocity triange, we find that Vrel,top = 138[m/s] and Vrel,bot = 138[m/s].
1 Assume that the chord length at the blade tip is c = 1[m]. Assuming an air density of ρair = 1[kg/m 3 ], a lift coefficient of Cl = 1[−] and a suitable drag coefficient Cd, compute the sectional lift L′[N/m] and sectional drag D′[N/m] for the top and bottom location.
Next, combine the two vectors L′ and D′ and find the normal force pn[N/m] at the ztop and zbot locations.
If the previous question was not solved, use Vrel,top = 130[m/s] and Vrel,bot = 129[m/s], φtop = 4[deg] and φbot = 2[deg]
Now it becomes quite easy to compute the lift, as we can assume a lift coefficient and we alreay have the relative velocities. This can be done via (8)
L[N/m] = 1 2
ρcV rel 2 (8)
Plugging in the values for Vrel,top and Vrel,bot, we obtain Ltop = 17284[N/m] and Lbot = 17269 .310[N/m]. If we compute the Reynolds number at the tip we obtain
Re =
cVrel ν
≈ 13 · 106 (9)
.
For this range of Reynolds numbers, one can easily get a CL/CD = 100, so here we assume CD = 0. 01. This leads to a Dtop = 172[N/m] and Dbot = 172[N/m].
frequencies close here?)
The area moment of inertial of the tower is estimated via
Itow =
π 64 (
D tower 4 − (Dtower − 2 ttower) 4 ) = 30[m 4 ] (12)
Note: same as the provided formula, I have taken 214 outside of the parentheses. This can be further used to find the natural angular frequency of the tower ωtower
ωtower =
√
3 EsteelItower MRN Ah 3 hub
= 2[rad/s] (13)
This is exactly twice the rated RPM of the turbine, which means that we expect a large amount of resonance and therefore tower excitation. We need to increase the natural frequency to avoid this bad design.
2 The engineers designing the turbine would like to increase the natural frequency of the structure to ωnat,new = 3[rad/s] by modifying the thickness of the turbine tower. What is the thickness ttower,new that allows to achieve ωnat,new = 3[rad/s]?
By playing with the value of the thickness, one can see that the conditions is achieved with ttower,new = 0[m].
2 Assume that the rotor thrust at the rated wind speed can be computed with a thrust co- efficient CT [−] as from optimum rotor 1D momentum theory. Compute the rated thrust on the rotor Tr[N ] and the moment at the tower base Mr[N m]. What is the maximum stress in the material, using the new configuration of the tower?
If the previous question was not solved, use ttower,new = 0[m].
If we consider a CT = 8/ 9 as from 1D momentum theory, we can easily compute the rated thrust as
Tr = 1 2
ρair
πD 2 4
V r 2 = 2[M N ] (14)
. By multiplying the rated thrust by the hub height, one obtains the tower-base moment at the rated wind speed as Mr = Tr · hhub = Mr = 371[M N m]. Using
σmax =
M
I z max = 33[M P a] (15)
Part 3: Economy and AEP of original turbine vs. turbine with leading edge
erosion (30%)
3 In this question, we plot the power curve of the turbine. First, compute the power coefficient at rated CP,r.
Now plot the power curve for the turbine as from the following relation
P =
1
2
ρCP,rArotorV 3 , V < Vr
P 0 , V ≥ Vr
(16)
Next, calculate the AEP of this turbine for a wind climate described by a Weibull distribution with parameters A = 7[m/s] and k = 2[−]. The power curve of the turbine is plotted in red in Fig. 3, while in blue we observe the Weibull- distributed hourly mean speed data.
Figure 3: power curve and probability distribution of the wind.
The rated power coefficient can be computed by the well-known relation
Pr = 1 2
ρair
πD 2 4
V r 3 Cp,r (17)
Inverting the relation with Pr = 12[M W ], we find Cp,r = 0[−]. Now, the AEP is nothing but the integral of the power curve times the probability.
AEP =
∫
P (V )p(V )dV · 8760[kW h/yr] (18)
, where P [W ] is the turbine power and p[−] is the Weibull probability, using A = 7[m/s] and k = 2. 26. The resulting AEP is AEPbase = 33[GW h]. 3 After some years of operation, due to the high maximum tip speed velocity Vtip = ωrR,
(Q3), and the turbine survives for the 10 remaining years as planned.
Assuming a fixed electricity price of 0 .25[DKK/kW h], a fixed yearly running cost of 1 e6[DKK], an initial expense of 70 e6[DKK] and a discount rate of 6%, calculate the net present value (NPV) for scenarios A and B for the whole investment. Which scenario would you advice? Which is the internal rate of return for each scenario? If the previous question was not solved, use AEPQ 3. 1 = 40[GW h], AEPQ 3. 2 = 38[GW h]. The two turbines earn the same amount of money in the first 10 years. From the 11th, you have to consider a different revenue for both. For scenario A
N P V =
∑ 17
j=
AEPbase · elprice − OP EX (1 + d)j
− CAP EX (23)
while for scenario B
N P V =
∑ 10
j=
AEPbase · elprice − OP EX (1 + d)j
+
∑ 20
j=
AEPred · elprice − OP EX (1 + d)j
− CAP EX (24)
When performing the summation with the present values for the two scenarios, you obtain N P Vbase = 4[M DKK] and N P Vred = 10[M DKK]. Despite the lower yearly en- ergy capture, it’s more beneficial for the project to have 3 more years of operation.
When computing the interest rates that zero out the present value, you find the IRR. They are for this exercise IRRbase = 7[%] and IRRred = 8[%].
46000 2022 exam And Solution
Kursus: Introduktion til vindenergi (46000)
Universitet: Danmarks Tekniske Universitet
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