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Chapter 4 1. A 15-m/sec wind at 101.3 kPa and 20 C enters enters a two-bladed wind turbine with a diameter diameter of 15 m. m. Calculate the the following: (a) The power of the incoming wind. (b) The theoretical maximum power that could be extracted. (c) A reasonable attainable turbine power. (d) The speed in RPM required for part (c). (e) The torque for part (c). R := 287.⋅
J
P := 101300⋅ Pa
kg⋅K
2
D := 15.0⋅ m V := 15⋅
m sec
T := 293.0 ⋅ K 2
A := 0.25⋅ π ⋅ D
A = 176. 176.71 715 5⋅ m 3
CpBetz := 0.5926
P
ρ :=
ρ =
R⋅T
1.205 ⋅
kg 3
m
Capture area
Power wind := 0.5⋅ ρ ⋅ A⋅ V
Power wind = 359.233 359.233 ⋅ kW
Power max := CpBetz⋅ Power wind
Power max = 212.881 212.881 ⋅ kW
Cpreasonable := 0.45 Power reasonable := Cpreasonable ⋅ Power wind Power reasonable
=
161.655 ⋅ kW
The maximum power coefficient for a two-bladed wind turbine occurs at an advance ratio of 11 (see Figure 4.6). The rotational speed then becomes
Ω :=
ω :=
11
Ω V ⋅ 0.5⋅ D
ω=
22
1
ω=
s
Rev := 2⋅ π
τ :=
Power reasonable
ω
τ = 1.7 1.744 ×
22⋅ Hz
RPM := 2 5 ft ⋅ lb
10
2
s
Rev min
ω=
τ = 5.4 5.42 ×
3
210.085 ⋅ RPM
10 ⋅ ft⋅ lbf
2. Compute the power coefficients for a Vestas V52-850-kW (Table 4.3) wind turbine at wind speeds of 10 m/s and 15 m/s. What are the advance ratios for these wind speeds? How do the power coefficients compare with the expected (Figure 4.6)? Table 4.3 Vestas V52-850 kW specifications Rated capacity (kW) Cut-in speed (m/sec) Cut-out speed (m/sec) Rated wind speed (m/sec) Rotor diameter (m) Swept area (m2) Rotor speed (rpm)
850 4 25 16 52 2124 14-31.4
Compute the density R := 287.⋅
J kg⋅K
P := 101300⋅ Pa 2
D := 52.0⋅ m
A := 0.25⋅ π ⋅ D
T := 303.0 ⋅ K
ρ := 3
2
A = 2.124 × 10 ⋅ m
P
ρ =
R⋅ T
1.165 ⋅
kg
Capture area
V = 10 m/sec V := 10⋅
m
3
3
Power wind := 0.5⋅ ρ ⋅ A ⋅ V
Power wind = 1.237 × 10 ⋅ kW
Power := 500⋅ kW
Read from Figure 4.24
Cp :=
Ω :=
Read from Figure 4.6
sec
10
ω := Ω ⋅
V 0.5⋅ D
ω=
3.846 ⋅ Hz
rpm :=
ω⋅
Power Power wind
60⋅ sec min⋅ 2 ⋅ π
rpm
Cp
=
=
0.404
36.728⋅
1 min
V = 15 m/sec V := 15⋅
m
3
3
Power wind := 0.5⋅ ρ ⋅ A ⋅ V
Power wind = 4.175 × 10 ⋅ kW
Power := 840⋅ kW
Read from Figure 4.24
Cp :=
Ω :=
Read from Figure 4.6
sec
10
ω := Ω ⋅
V 0.5⋅ D
ω=
5.769 ⋅ Hz
rpm :=
ω⋅
Power Power wind
60⋅ sec min⋅ 2 ⋅ π
rpm
3
m
Cp
=
=
0.201
55.092⋅
1 min
At 10 m/sec, the power coefficient of 0.404 is reasonable for operation in the maximum power coefficient range. At 15 m/sec, the power coefficient is 0.201, indicating operation in the constant power range. These values are reasonable. At an advance ratio of 10, the rotor speeds are in excess of those specified in Table 4.3; however, the advance ratio value of 10 from Figure 4.6 is for a two-bladed device not a three-bladed wind turbine, so some difference is expected.
3. A General Electric 1.5se wind turbine is used for this exercise. Information on the GE 1.5se is provided in Table 4.2. (a) Estimate the power coefficient at the rated speed and at 10 m/sec. (b) Explain the importance of part (a). (c) Estimate the kWh production of a GE 1.5se device in a wind distribution with c = 10 m/sec and k = 2. Show plots similar to those examined in the chapter.
ρ :=
1.225 ⋅
kg
mph :=
density
3
m
mi
2
D := 70.5⋅ m
kW := 1000⋅ watt define kW
define mph
hr 3
A := 0.25⋅ π ⋅ D
2
A = 3.904 × 10 ⋅ m
Capture area
V = 10 m/sec V := 10⋅
m sec
3
Power wind := 0.5⋅ ρ ⋅ A ⋅ V
Power wind = 2.391
3
×
10 ⋅ kW
×
10 ⋅ kW
Power := 1000⋅ kW Read from Figure 4.22 Cp :=
Power
Cp
Power wind
=
0.418
Cpnom := Cp
V = 13 m/sec (rated s peed) V := 13⋅
m sec
3
Power wind := 0.5⋅ ρ ⋅ A ⋅ V
Power wind = 5.253
3
Power := 1500⋅ kW Read from Figure 4.22 Cp :=
c := 10⋅
Power
Cp
Power wind m
0.286
k := 2.0
sec
h ( v , k , c) :=
=
k ⎛ v ⎞ c
⋅
⎝ c ⎠
specify values of the scale parameter and shape parameter
− ⎛ ⎞ v
k− 1
c ⋅ e ⎝ ⎠ 3
PowerDen( V) := 0.5⋅ ρ ⋅ V
k
Weibull distribution function definition. Power density.
Generate plots for specified parameters:
v := 0 ⋅
m sec
, 0.1⋅
m sec
.. 50⋅
m sec
Wind speed range
800
c e s / m r e p r a e y r e p s r u o H
700 600 500 400 300 200 100 0
0
5
10
15
20
25
30
35
40
45
50
Wind speed, m/sec 0.00125
c e s / m r e p r a e y r e p s r u o H
V := 7 ⋅
0.001 7.5 .10
4
5 .10
4
2.5 .10
4
m sec
Given d
0
h ( V , k , c) = 0
2.5 .10
4
d V
5 .10
4
Vmode := Find( V)
7.5 .10
4
Vmode = 7.071⋅
0.001 0.00125
0
5
10
15
20
25
30
35
40
45
m sec
50
Wind speed, m/sec Vmode := Vmode
PowerDen Vmode = 216.551⋅
(
)
watt 2
m m
∞⋅
⌠ ⎮ ⎮ Vmean := ⎮ ⎮ ⌡0⋅
sec
c
⎝ c ⎠
Vmean = 8.862 ⋅
m sec
PowerDen Vmean = 426.325 ⋅
(
)
watt 2
m
m sec
3
∞⋅
⌠ ⎮ ⎮ Vrmc := ⎮ ⎮ ⌡0⋅
k
⎛ v ⎞ k− 1 − k ⎛ v ⎞ c ⋅ ⋅ e ⎝ ⎠ ⋅ v d v
m sec
k ⎛ v ⎞ c m sec
⋅
⎝ c ⎠
k− 1
− ⎛ ⎞ v
k
c 3 ⋅ e ⎝ ⎠ ⋅ v d v
Vrmc = 10.995⋅
m sec
PowerDen Vrmc = 814.221 ⋅
(
)
watt 2
m
Power( v ) := 0.5⋅ ρ ⋅ h ( v , k , c) ⋅ v
3
Power density available with Cp = 1.0.
The energy available for a power coefficient of 0.5. 1000⋅
⌠ ⎮ ⎮ Energy := ⎮ ⎮ ⌡0⋅
) m * m ( / ) s / m ( / ) r y / h W k ( n i y t i s n e d y g r e n E
m
⎡
sec
0.25⋅ ρ ⋅ ⎢
k ⎛ v ⎞
⋅
⎣ c ⎝ c ⎠
k− 1
− ⎛ ⎞
⋅e
v
⎤
k
⎝ c ⎠ ⎥ ⋅ 8760⋅ hr ⋅ v3 d v
⎦
yr
Energy
=
hr
3
3.566 × 10 ⋅ kW⋅
yr⋅m
m sec 750 700 650 600 550 500 450 400 350 300 250 200 150 100 50 0
0
5
10
15
20
25
30
Wind speed in m/sec Power density Powerinmax := Vcutin := 4 ⋅
m
1500⋅ kW A
Powerin max = 0.384 ⋅
kW
Generator input
2
m
Cut-in speed
sec
PowerDen( v ) := 0.5⋅ Cpnom⋅ ρ ⋅ v
3
Vcutout := 25⋅
m
Cut-off speed
sec
Power density available with specified Cp.
2
10
) m * m ( / W k n i y t i s n e d r e w o P
9 8 7 6 5 4 3 2 1 0
0
5
10
15
20
25
30
Wind speed in m/sec PowerDenCon( v) :=
0⋅
kW 2
if v
< Vcutin
m
Powerinmax if PowerDen( v ) 0⋅
kW 2
if v
> Powerinmax
Piece-wise continuous function defined to implement the cut-in, cut-out, and rated power constraints.
> Vcutout
m
PowerDen( v) otherwise
Power density when control scheme is applied.
) m * m ( / W k n i y t i s n e d r e w o P
8 7 6 5 4 3 2 1 0
0
5
10
15
20
Wind speed in m/sec
25
30
CpV( v ) :=
PowerDenCon( v) ⋅Cpnom
Power coefficient as a function of speed.
PowerDen( v)
Energy( v ) := PowerDen( v) h ( v , k , c) ⋅ 8760⋅ hr
Energy density available with nominal Cp.
EnergyCon( v) := PowerDenCon( v) ⋅ h ( v , k , c) ⋅ 8760⋅ hr
Energy density with controls.
0.5 0.4
) V ( p C
0.3 0.2 0.1 0
0
5
10
15
20
25
30
Wind speed in m/sec
) m * m ( / ) s / m ( / ) r y / h W k ( n i y t i s n e d y g r e n E
400 350 300 250 200 150 100 50 0
0
5
10
15
20
25
30
Wind speed in m/sec
35
40
45
50
1000⋅
⌠ ⎮ EnergyCon := ⎮ ⎮ ⌡0⋅
CaptureRatio :=
sec
EnergyCon( v) ⋅
1 yr
d v
EnergyCon
=
3
1.614 × 10 ⋅ kW⋅
Energy extracted per
hr yr⋅m
m
2
sec
1000⋅
⌠ ⎮ Energymax := ⎮ ⎮ ⌡0⋅
m
m sec
Energy( v) ⋅
1 yr
d v
Energymax = 2.983
m sec
EnergyCon Energymax
CaptureRatio = 0.541
×
3
10 ⋅ kW⋅
Energy extracted per
hr yr⋅m
year per m2 for system with controls.
2
year per m2 with no controls.
4. A 27-mph wind at 14.65 psia and 70 F enters a wind turbine with a 1000-ft 2 crosssectional area. Calculate (a) the power of the incoming wind. (b) the theoretical maximum power that could be extracted. (c) a reasonable attainable turbine power. (d) the torque for part (c). Rg := 53.35 ⋅
ft⋅ lbf
P := 14.65 ⋅
lb⋅R
2
A := 1000⋅ ft
lbf
T := 530⋅ R
2
in
mph :=
Capture area 3
V := 27⋅ mph
Power wind := 0.5⋅ ρ ⋅ A ⋅ V
CpBetz := 0.5926
ρ :=
P
ρ =
Rg⋅T
0.075
lb 3
ft
mi hr
Power wind = 97.621 ⋅ kW Power wind = 130.911⋅ hp
Power max := CpBetz⋅ Power wind
Power max = 57.85⋅ kW
Cpreasonable := 0.45 Power reasonable := Cpreasonable ⋅ Power wind Power reasonable
=
43.929⋅ kW
The maximum power coefficient for a two-bladed wind turbine occurs at an advance ratio of 11 (see Figure 4.6). The rotational speed then becomes
Ω :=
ω :=
11
Ω V ⋅ 0.5⋅ D
ω =
3.767
1
ω=
s
Rev := 2⋅ π
τ :=
Power reasonable
ω
3.767⋅ Hz
RPM :=
min
2
τ = 2.768 ×
5 ft ⋅ lb
10
2
s
ω=
Rev
τ = 8.602 ×
3
10 ⋅ ft⋅ lbf
35.968⋅ RPM
5. Consider two cases: (1) a constant wind velocity twice the mean wind velocity and operating half the time, and (2) a constant wind velocity three times the same mean velocity operating one third of the time. At all other times the wind velocity is zero. Determine for each case the ratio of the total energy available from the wind to the total wind energy available at the mean velocity continuously. The energy available is the power available, Eq. (4-2), times the hours of operation. If a constant wind speed is a Factor time the mean but is only operated 1/Factor of the time, then the energy ratio, energy available for case described divided by the energy available for continuous operation at the mean speed is
1 Energy RatioFactor =
2
8760 ⋅ hr ⎞ (Factor ⋅V )3 ⋅ ⎛ ⎜ ⎟ ⎝ Factor ⎠ = Factor 2 1 3 ρ AV ⋅ 8760 ⋅ hr
ρ A
2
2
EnergyRatio2 := 2
EnergyRatio2
=
4
2
EnergyRatio3 := 3
EnergyRatio3
=
9
This problem really illustrates the importance of wind speed in wind turbine energy extraction.
6. A 15-ft diameter wind turbine operates in 25 ft/sec wind at 1 atm and 60 F. The turbine is used to pump 60 F water from a 30-ft deep well. How much water in cfs can be pumped if the overall efficiency of the wind-turbine-pump system is 0.25? Estimate the power that can be delivere to the water. Rg := 53.35 ⋅
ft⋅ lbf
D := 15⋅ ft ft V := 25⋅ sec
lb⋅R
P := 14.7⋅
lbf
T := 520⋅ R
2
ρ :=
in 2
A := 0.25⋅ π ⋅ D
2
A = 176.715 ft 3
Power wind := 0.5⋅ ρ ⋅ A ⋅ V
P
ρ =
Rg⋅ T
0.076
lb ft
3
Capture area
Power wind = 4.439 ⋅ kW
Cpreasonable := 0.45 Power extracted := Cpreasonable⋅ Power wind
Power wind = 5.953⋅ hp
Power extracted = 1.998⋅ kW
If the overall system efficiency is 0.25, the power imparted to the fluid is Power fluid := 0.25⋅ Power extracted
Power fluid = 0.499⋅ kW
Power fluid = 0.67⋅ hp
The power to the water is Q∆P = QρH.
ρ water :=
62.2⋅
lb 3
ft Q :=
Power fluid
ρ water ⋅ H
H := 30⋅ ft⋅
lbf lb 3
Q = 88.598gpm
Q = 0.197 ⋅
ft
sec
7. A wind turbine-generator is designed to attain the full-load capacity with a wind velocity of 48 km/h. The rotor diameter is 50 m. If the power coefficient is 0.48 and generator efficiency is 0.85, calculate the rated output for 1 atm and 22 C. Compute the density R := 287.⋅
J kg⋅K
P := 101300⋅ Pa 2
T := 295.0 ⋅ K
D := 50.0⋅ m
A := 0.25⋅ π ⋅ D
=
km V := 48.0⋅ hr
Power wind := 0.5⋅ ρ ⋅ A ⋅ V
A
3
P
ρ := 3
2
1.963 × 10 ⋅ m
ρ =
R⋅T
3
Power wind = 2.784 × 10 ⋅ kW 3
Power extracted = 1.336
×
Power output := 0.85⋅ Power extracted
Power output = 1.136
10 ⋅ kW
×
10 ⋅ kW 3
kg 3
m
Capture area
Power extracted := 0.48⋅ Power wind
1.196 ⋅
8. The U. S. Department of Energy constructed a Darrieus vertical axis wind turbine in Sandia, NM. The machine was 60-ft tall, 30 ft in diameter, and swept an area of 1200 ft2. Estimate the power this device can produce at a wind speed of 20 mph. Rg := 53.35 ⋅
ft⋅ lbf lb⋅R
2
A := 1200⋅ ft
V := 20⋅ mph
P := 14.7⋅
lbf
T := 530⋅ R
2
in
mph :=
Capture area 3
Power wind := 0.5⋅ ρ ⋅ A ⋅ V
CpBetz := 0.5926
ρ :=
P Rg⋅T
ρ =
0.075
lb 3
ft
mi hr
Power wind = 47.775 ⋅ kW Power wind = 64.067⋅ hp
Power max := CpBetz⋅ Power wind
Power max = 28.311⋅ kW
For the Darrieus turbine, Figure 4.6 suggest a reasonable power coefficieint to be 0.45. Cpreasonable := 0.45 Power reasonable := Cpreasonable ⋅ Power wind Power reasonable
=
21.499⋅ kW
The maximum power coefficient for a two-bladed wind turbine occurs at an advance ratio of 6 (see Figure 4.6).
9. An early NASA/DOE wind turbine consisted of a 125-ft diameter, two-bladed, horizontal-axis rotor. Maximum power was achieved at a wind speed of 19 mph. For these conditions estimate: (a) The power generated in kW. (b) The rotor speed in RPM. (c) The velocity downstream of the rotor (Vo). Rg := 53.35 ⋅
ft⋅ lbf
P := 14.7⋅
lb⋅R
lbf
T := 530⋅ R
2
ρ :=
in 2
ρ =
Rg⋅T
A := 0.25⋅ π ⋅ D
V := 19⋅ mph
Power wind := 0.5⋅ ρ ⋅ A ⋅ V
A = 1.227 × 10 ft Capture area 3
0.075
lb 3
ft
4 2
D := 125⋅ ft
CpBetz := 0.5926
P
mph :=
mi hr
Power wind = 418.89 ⋅ kW Power wind = 561.741 ⋅ hp
Power max := CpBetz⋅ Power wind
Power max = 248.234⋅ kW
Cpreasonable := 0.45 Power reasonable := Cpreasonable ⋅ Power wind Power reasonable
=
188.501 ⋅ kW
The maximum power coefficient for a two-bladed wind turbine occurs at an advance ratio of 11 (see Figure 4.6). The rotational speed then becomes
Ω :=
ω :=
11
Ω V ⋅ 0.5⋅ D
ω =
1
4.905
ω=
s
Rev := 2⋅ π mdot :=
ρ A ⋅ ⋅V
mdot
4.905⋅ Hz
RPM := 4 lb
=
2.56 × 10
Rev
ω=
min
46.835⋅ RPM
s
The power extracted at the turbine must be equal to the rate of change of kinetic energy at the turbine. At the turbine the Betz analysis shows that the velocity is 2/3 times the freestream, so that the mass flow rate is 2/3 mdot. Application of the energy equation, Eq. (3-1), yields
Power = mdotturbine⋅
2
Vout :=
V
Vout :=
V
2
−
−
⎛ V
2
in
⎝
2
−
Vout 2
2⋅ Power reasonable mdot⋅ 0.66667 2 ⋅ Power max mdot⋅ 0.66667
⎞
2
⎠
=
2 3
mdot⋅
⎛ V
2
in
⎝
Vout = 15.887
Vout = 9.289
−
2 ft s
ft s
Vout 2
⎞
2
⎠ Vout
=
10.832⋅ mph
Vout
=
6.333 ⋅ mph
For the case of maximum power extraction, the exit velocity is exactly 1/3 of the freestream velocity--in agreement with the Betz analysis.
10. A wind turbine with a rotor diameter of 40 m possesses a power coefficient of 0.30 3 in an 8 m/s wind. The density is 1.2 kg/m . The turbine is to be used in a wind farm that is to serve a community of 100,000 (average family size of 4). Each house will require 3 kW. The wind farm will have a turbine spacing of 2.4 rotor diameters perpendicular to the wind and 8 rotor diameters parallel to the wind. The wind farm will have ten turbines perpendicular to the wind. (a) Estimate the power production from one turbine. (b) How many turbines will be required in the wind farm for the community? (c) Estimate the dimensions of the wind farm. (d) How many acres will be required for the wind farm? (e) If the average house is on a 0.25-acre lot, how large will the wind farm be in comparison to the community? (f) What does this problem infer about wind power feasibility in an urban setting?
ρ :=
1.2⋅
kg
D := 40⋅ m
3
2
3
A := 0.25⋅ π ⋅ D
2
A = 1.257 × 10 ⋅ m Capture area
m V := 8
m
3
Power wind := 0.5⋅ ρ ⋅ A ⋅ V
sec
Cp := 0.30
Power wind = 386.039 ⋅ kW
Power turbine := Cp⋅ Power wind
Power total := Number :=
100000 4
⋅ 3 ⋅ kW
Power total Power turbine
Power turbine 4
Number
=
Number := 648
647.603
4
Afootprint = 3.072
×
Atotal := Number⋅ Afootprint
Atotal = 1.991
7
Ratio :=
100000 4
Atotal Ahouses
⋅ 0.25 ⋅ acre
115.812 ⋅ kW
Power total = 7.5 × 10 ⋅ kW
Afootprint := 2.4⋅ D⋅ 8 ⋅ D
Ahouses :=
=
Power wind = 517.687⋅ hp
×
Number required
2
10 ⋅ m 2
10 ⋅ m
Atotal = 4.919
×
3
10 ⋅ acre
3
Ahouses = 6 .25 × 10 ⋅ acre
Ratio
=
0.787
Thus, the wind farm will occupy nearly as much area as the houses. Such an arrangement in an urban setting is land intensive. Moreover, because of the nature of wind, wind energy cannot be considered as a base load so that backup will be required.
11. What is the power coefficient for the Vestas V52-850 kW wind turbine at rated conditions? R := 287.⋅
J kg⋅K
P := 101300⋅ Pa
T := 303.0 ⋅ K
ρ :=
P
ρ =
R⋅T
2
V := 16⋅
m sec
A := 0.25⋅ π ⋅ D
Power := 850⋅ kW
3
3
Power Power wind
Cp
Capture area 3
Power wind := 0.5⋅ ρ ⋅ A ⋅ V Cp :=
2
A = 2.124 × 10 ⋅ m
Power wind = 5.067 × 10 ⋅ kW
=
0.168
kg 3
m
Read from Table 4.3 D := 52.0⋅ m
1.165 ⋅
12. What is the power coefficient for the Bergey Excel wind turbine at rated conditions? R := 287.⋅
J kg⋅K
P := 101300⋅ Pa
T := 303.0 ⋅ K
ρ :=
P R⋅T
ρ =
Read from Table 4.4 2
2
D := 6.7⋅ m
A := 0.25⋅ π ⋅ D
m V := 13.8⋅ sec
Power wind := 0.5⋅ ρ ⋅ A ⋅ V
Power := 10⋅ kW
A = 35.257⋅ m 3
Cp :=
Power Power wind
Cp
Capture area
Power wind = 53.967⋅ kW
=
0.185
1.165 ⋅
kg 3
m
13. The Utopia Wind Turbine Company advertises that their two-bladed, 20-m diameter wind turbine/generator will produce 600 kW in a 15 m/s wind. The air density is 1.18 kg/m3. Do you believe their claim? Explain. R := 287.⋅
J kg⋅K
2
D := 20.0⋅ m V := 15⋅
m sec
P := 101300⋅ Pa
A := 0.25⋅ π ⋅ D
CpBetz := 0.5926
T := 293.0 ⋅ K
ρ := 2
A = 314.159 ⋅ m 3
P R⋅ T
ρ =
1.205 ⋅
kg 3
m
Capture area
Power wind := 0.5⋅ ρ ⋅ A⋅ V
Power wind = 638.636⋅ kW
Power max := CpBetz⋅ Power wind
Power max = 378.456⋅ kW
The Betz limit for these conditions is 378 kW. Hence, a stated value of 600 kW exceeds the Betz limit. The claim is not true.
14. Repeat Example 4.3 with characteristics of wind turbine from a manufacturer’s web site. If available, use a wind distribution consistent with your location. The solution basis for this problem is presented in the solution to Problem 4.3. Parameters from the manufacturer's data must be inserted into the work sheet. The determination of the Weibull parameters depends on the local weather conditions.
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