ETME 491
Wind Energy Engineering
Fall 2015
Test #1
1. The annual energy production estimates are provided on page (4) of the brochure, for
average wind speeds of 5.0, 5.5, 6.0, 6.5, 7.0, and 7.5 m/s.
Create a well-labeled table showing (a) the ideal annual energy output using the Carlin
method (ref: equation 2.82), and (b) turbine capacity factor at each of these six wind
speeds. Assume 100% availability.
Solution Method:
(a) 10 pts. Ideal annual energy output (per Carlin) uses Pave = ρ* (.6667D)2*Uave3
to create a table with a row for each of the six wind speeds. The rotor diameter has a fixed
value of D = 24.4 m, and the standard atmospheric density ρ = 1.225 kg/m3 is also a fixed
value. Then multiply each of these instantaneous power values by 8760 hrs/year to
find annual energy production values. (Just like in the text’s example…)
Wind Speeds,
m/s
Energy per
manufacturer
MWHr/yr
Power per Carlin W
Ideal Annual
Energy Output
(kWHr)
5
40517.55556
196000
354933.7867
5.5
53928.86644
240000
472416.8701
6
70014.336
284000
613325.5834
6.5
89017.06956
325000
779789.5293
7
111180.1724
364000
973938.3106
7.5
136746.75
399000
1197901.53
(b) 10 pts. Capacity Factor is defined as actual output (not ideal output!) divided by
the output possible if the turbine runs at rated power over the same time period – in
this case, that is one year. The actual energy output was given by the manufacturer,
for the same average wind speed regimes.
Energy per
manufacturer
MWHr/yr
196000
240000
284000
325000
364000
399000
Per the given specs, the power rating is
actually 95 KW (despite the 100kw name)
In a full year this turbine nameplate
rating would be 95KW*8760 hrs/year
832200
832200
832200
832200
832200
832200
Capacity Factor
0.23552
0.288392
0.341264
0.390531
0.437395
0.479452
2. The turbine is available for purchase with tower (hub) heights of 37m, 29m, or 22m. The
rated wind speed for this turbine is 12 m/s. If the turbine experiences this 12 m/s wind
speed at the tallest hub height of 37m, what wind speeds are predicted at each of the
other two hub heights? Display your results in a well-labeled table, where you
(a) Obtain your power law exponent using the “Counihan method” (ref: equation
2.38), with the surface roughness zo value obtained from Table 2.2 for fallow
field terrain.
(b) Obtain your power law exponent using the “Justus method” (ref: equation
2.37.)
20 pts.
Solution Method:
(a) Counihan method – For a “Fallow field” surface the table gives z0 = 30mm (= 0.030
meters.) Power law exponent using equation 2.38 gives α=0.1309.
(b)Justus method uses wind speed only, equation 2.37 gives α=0.171.
Table below summarizes values
Hub Height
(m)
Wind Speed
(m/s)
37
29
22
12
12
12
Wind
Speed
Counihan
(m/s)
Wind
Speed
Justus
(m/s)
12
11.623
11.210
12
11.510
10.979
3. Published power output at the rated 12 m/s speed is predicted at 94.7 kW. From the
Power Curve provided, approximately what power output could be expected at the wind
speeds you found for the 29m and 22m heights?
10 pts.
Solution Method:
Look up on power curve as stated, the detail of the graph is not great but the values
(determined using either of the above methods) are approximately as shown below:
Wind
Hub Height:
Speed:
Predicted
(m)
(m/s)
Power: (kW)
37
12
94.5
29
11.5
92.5
22
11
90
4. The energy production numbers on page (4) of the brochure are based on Rayleigh
Wind Distribution assumptions, and standard air density of 1.225 kg/m3. Plot both the
Rayleigh Probability Density Function and Rayleigh Cumulative Distribution Function
graphs for a 7.5 m/s average wind speed.
20 pts.
Solution Method: Use definitions of Raleigh PDF and CDF as we did in class:
Rayleigh probability density function, p(U) = (π/2)*(U/Uave2)*exp[-(π/4)*(U/ Uave)2]
Rayleigh cumulative density function, F(U) = 1-exp[-(π/4)*(U/ Uave)2]
Wind Speed
(m/s)
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
Rayleigh PDF
0
0.027524301
0.05279148
0.073850084
0.089302576
0.098452469
0.101330183
0.098604094
0.091405975
0.081113502
0.069134535
0.056729701
0.044895228
0.034311663
0.025350369
0.018121065
0.012540851
0.008407151
0.005461899
0.003440127
0.002101255
0.001245015
0.000715752
0.000399329
0.00021625
0.000113685
Rayleigh CDF
0
0.013858628
0.054292753
0.118032437
0.200116262
0.294528488
0.394923268
0.495315714
0.590638052
0.677095931
0.75230461
0.815226074
0.865957719
0.905436989
0.935124737
0.956717202
0.971917881
0.98228167
0.989128385
0.993513003
0.996235809
0.997875885
0.998834365
0.999377949
0.999677175
0.999837075
Rayleigh PDF
0.12
0.1
Probability
0.08
0.06
Rayleigh PDF
0.04
0.02
0
0
5
10
15
20
25
30
Wind Speed (m/s)
Rayleigh CDF
1.2
Cumulative Density
1
0.8
0.6
Rayleigh CDF
0.4
0.2
0
0
5
10
15
Wind Speed (m/s)
20
25
30
5. Based on the 7.5 m/s average wind speed probability results of problem 4, what is the
probability that the wind speed will exceed the cutout speed of the turbine?
10 pts.
Solution Method:
p(U > 25) = 1- F(25) = 1- .999837
0.000163
6. The blade design uses an airfoil with lift coefficient Cl= 1.4450, and drag coefficient Cd=
0.014. If the turbine rotates at 40 rpm at a 12 m/s wind speed, solve for the maximum
achievable power coefficient Cpmax. (ref: equation 3.145.)
20 pts.
Solution Method: Recall I modified the Cd value from 0.0014 to a more realistic
Cd = 0.014. . . Using this value:
Omega @ 40 RPM = 2*pi*N/60 = 4.186 radians/sec
Radius = D/2 = 12.2 m
Tip speed is therefore 12.2m*4.186 rad/sec = 51.07 m/s, and tip speed ratio λ = 4.256
Coeff of lift
Cl
1.445
Coeff of
drag
Cd
0.014
Lift to drag
ratio
Cl/Cd
103.2142857
B is simply the number of blades, so B = 3 for this turbine.
The referenced equation using this value results in Cpmax = 0.491