Green Electric Energy Lecture 19
Transcript of Green Electric Energy Lecture 19
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Lecture 19Wind, Electric Generators
Professor Tom OverbyeDepartment of Electrical and
Computer Engineering
ECE 333 Green Electric Energy
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Announcements
• Start reading Chapter 6.
• Homework 8 is 6.3, 6.5, 6.8, 6.14; due on Tuesday Nov 10.
• Wind Farm field trip will be on Thursday from 8 am to 4 pm – turn in forms by today to sign up.
• Exam 2 is Thursday November 19 in class.
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Squirrel Cage Rotor
• The rotor of many induction generators has copper or aluminum bars shorted together at the ends, looks like a cage
Figure 6.15
• Can be thought of as a pair of magnets spinning around a cage
• Rotor current iR flows easily through the thick conductor bars
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Squirrel Cage Rotor
Figure 6.16
• Instead of thinking of a rotating stator field, you can think of a stationary stator field and the rotor moving counterclockwise
• The conductor experiences a clockwise force
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The Inductance Machine as a Motor
• The rotating magnetic field in the stator causes the rotor to spin in the same direction
• As rotor approaches synchronous speed of the rotating magnetic field, the relative motion becomes less and less
• If the rotor could move at synchronous speed, there would be no relative motion, no current, and no force to keep the rotor going
• Thus, an induction machine as a motor always spins somewhat slower than synchronous speed
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Slip
• The difference in speed between the stator and the rotor
1 (6.28)S R R
S S
N N Ns
N N
• s = rotor slip – positive for a motor, negative for a generator
• NS = no-load synchronous speed (rpm)
• f = frequency (Hz) • p = number of poles
• NR = rotor speed (rpm)
120S
fN
p
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The Induction Machine as a Motor
• As load on motor increases, rotor slows down
• When rotor slows down, slip increases
• “Breakdown torque” increasing slip no longer satisfies the load and rotor stops
• Braking- rotor is forced to operate in the opposite direction to the stator field
Torque- slip curve for an induction motor, Figure 6.17
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The Induction Machine as a Generator
• The stator requires excitation current– from the grid if it is grid-connected or– by incorporating external capacitors
• Windspeed forces generator shaft to exceed synchronous speed
Figure 6.18. Single-phase, self-excited, induction generator
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The Induction Machine as a Generator
• Slip is negative because the rotor spins faster than synchronous speed
• Slip is normally less than 1% for grid-connected generator
• Typical rotor speed
(1 ) [1 ( 0.01)] 3600 3636 rpmR SN s N
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Speed Control
• Necessary to be able to shed wind in high-speed winds
• Rotor efficiency changes for different Tip-Speed Ratios (TSR), and TSR is a function of windspeed
• To maintain a constant TSR, blade speed should change as windspeed changes
• A challenge is to design machines that can accommodate variable rotor speed and fixed generator speed
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Blade Efficiency vs. Windspeed
Figure 6.19
At lower windspeeds, the best efficiency is achieved at a lower rotational speed
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Power Delivered vs. Windspeed
Figure 6.20
Impact of rotational speed adjustment on delivered power, assuming gear and generator efficiency is 70%
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Pole-Changing Induction Generators
• Being able to change the number of poles allows you to change operating speeds
• A 2 pole, 60 Hz, 3600 rpm generator can switch to 4 poles and 1800 rpm
• Can do this by switching external connections to the stator and no change is needed in the rotor
• Common approach for 2-3 speed appliance motors like those in washing machines and exhaust fans
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Variable-Slip Induction Generators
• Purposely add variable resistance to the rotor
• External adjustable resistors - this can mean using a wound rotor with slip rings and brushes which requires more maintanance
• Mount resistors and control electronics on the rotor and use an optical fiber link to send the rotor a signal for how much resistance to provide
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Variable Slip Example: Vestas V80 1.8 MW
• The Vestas V80 1.8 MW turbine is an example in which an induction generator is operated with variable rotor resistance (opti-slip).
• Adjusting the rotor resistance changes the torque-speed curve
• Operates between 9 and 19 rpm
Source: Vestas V80 brochure
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Vestas V80 1.8 MW
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Doubly-Fed Induction Generators
• Another common approach is to use what is called a doubly-fed induction generator in which there is an electrical connection between the rotor and supply electrical system using an ac-ac converter
• This allows operation over a wide-range of speed, for example 30% with the GE 1.5 MW and 3.6 MW machines
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GE 1.5 MW and 3.6 MW DFIG Examples
Source: GE Brochure/manual
GE 1.5 MW turbines are the best selling wind turbines in the US with 43% market share in 2008
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Indirect Grid Connection Systems
• Wind turbine is allowed to spin at any speed
• Variable frequency AC from the generator goes through a rectifier (AC-DC) and an inverter (DC-AC) to 60 Hz for grid-connection
• Good for handling rapidly changing windspeeds
Figure 6.21
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Example: GE 2.5 MW Turbines
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Average Power in the Wind
• How much energy can we expect from a wind turbine?
• To figure out average power in the wind, we need to know the average value of the cube of velocity:
• This is why we can’t use average windspeed vavg to find the average power in the wind
3 31 1 (6.29)
2 2avg avgavg
P Av A v
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Average Windspeed
hours@miles of wind
(6.32)total hours hours@
i ii
avgi
i
v vv
v
• vi = windspeed (mph)
• The fraction of total hours at vi is also the probability that v = vi
fraction of total hours@ (6.32)avg i ii
v v v
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Average Windspeed
• This is the average windpseed in probabilistic terms
• Average value of v3 is found the same way:
fraction of total hours@ (6.32)avg i ii
v v v
probability that = (6.33)avg i ii
v v v v
3 3 probability that = (6.35)i iavgi
v v v v
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Example Windspeed Site Data
Figure 6.22
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Wind Probability Density Functions
Windspeed probability density function (p.d.f) – between 0 and 1, area under the curve is equal to 1
Figure 6.23
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Windspeed p.d.f.
• f(v) = windspeed p.d.f.
• Probability that wind is between two windspeeds:
• # of hours/year that the wind is between two windspeeds:
2
1
1 2 ( ) (6.36)v
v
p v v v f v dv
0
0 ( ) = 1 (6.37)p v f v dv
2
1
1 2/ 8760 ( ) (6.38)v
v
hrs yr v v v f v dv
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Average Windspeed using p.d.f.
• This is similar to (6.33), but now we have a continuous function instead of discrete function
• Same for the average of (v3)
0
( ) (6.39)avgv v f v dv
= (6.33)avg i i
i
v v p v v
3 3 = (6.35)i iavgi
v v p v v
3 3
0
( ) (6.40)avg
v v f v dv
discrete
continuous
continuous
discrete
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Weibull p.d.f.
• Starting point for characterizing statistics of windspeeds
k-1-
( ) e Weibull p.d.f. (6.41)
kv
ck vf v
c c
• k = shape parameter • c = scale parameter
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Weibull p.d.f.
Figure 6.24
k=2 looks reasonable
for wind
Weibull p.d.f. for c = 8
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Rayleigh p.d.f.
• This is a Weibull p.d.f. with k=2
• Typical starting point when little is known about the wind at a particular site
• Fairly realistic for a wind turbine site – winds are mostly pretty strong but there are also some periods of low wind and high wind
2
-
2
2( ) e Rayleigh p.d.f. (6.42)
v
cvf v
c
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Rayleigh p.d.f. (Weibull with k=2)
Figure 6.25
Higher c implies higher average windspeeds
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Rayleigh p.d.f.
• When using a Rayleigh p.d.f., there is a direct relationship between average windspeed v and scale parameter c
• Substitute (6.42) into (6.39):
0
( ) (6.39)avgv v v f v dv
-
20
2e (6.43)
kv
cavg
vv v v dv
c
0.886 (6.43)2avgv c c
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Rayleigh p.d.f.
• From (6.43), we can solve for c in terms of v
• Then we can substitute this into the Rayleigh p.d.f (6.42) for c
0.886 (6.43)2avgv c c
2
=1.128 (6.44)avgc v v
2
2
2( ) e Rayleigh p.d.f. (6.45)
2
k
kv
v
vf v
v
2
42
( ) e Rayleigh p.d.f. (6.45)2
v
vvf v
v
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Rayleigh Statistics – Average Power in the Wind
• Can use Rayleigh statistics when all you know is the average windspeed
• Anemometer – Spins at a rate proportional to windspeed– Has a revolution counter that indicates “miles” of wind that pass– Dividing “miles” of wind by elapsed hours gives the average
windspeed (miles/hour)– “Wind odometer”– About $200 each– Easy to use
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Rayleigh Statistics – Average Power in the Wind
• Assume the wind speed distribution is a Rayleigh distribution
• To find average power in the wind, we need (v3)avg
• From (6.40) and the Rayleigh p.d.f. (6.45):
• Then for a Rayleigh distribution we have
3 3
0
( ) (6.40)avg
v v f v dv
2
42
( ) e (6.45)2
v
vvf v
v
2
3 3 342
0
3e = c (6.46)
2 4
v
v
avg
vv v dv
v
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Rayleigh Statistics – Average Power in the Wind
• This is (v3)avg in terms of c, but we can use (6.44) to write c in terms of vavg
• Then we have (v3)avg in terms of vavg :
2
3 3 342
0
3e = c (6.46)
2 4
v
v
avg
vv v dv
v
2=1.128 (6.44)avgc v v
3 33 6=1.91 (6.47)avg avgavg
v v v
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Rayleigh Statistics – Average Power in the Wind
• To figure out average power in the wind, we need to know the average value of the cube of velocity:
• With Rayleigh assumptions, we can write the (v3)avg in terms of vavg as in (6.47), and the expression for average power in the wind is just
• This is an important and useful result
3 31 1 (6.29)
2 2avg avgavg
P Av A v
36 1 (6.48)
2avg avgP A v
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Example 6.10 – Average Power in the Wind
Estimate average power density in the wind at 50 m when the windspeed at 10 m has vavg = 6m/s. Assume Rayleigh statistics, α=1/7, and ρ=1.225 kg/m3.
1/7
5050 10
10
506 =7.55 m/s
10
Hv v
H
Estimate windspeed at 50 m:
Average power density in the wind at 50 m from (6.48):
32 26 1/ (1.225) 7.55 = 504 W/m (6.48)
2avgP m
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Real Data vs. Rayleigh Statistics
This is why it is important to gather as much real wind data as possible
Figure 6.26
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Wind Power Classification Scheme
Table 6.5
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Wind Power Classification Scheme
• Table 6.5
http://www.windpoweringamerica.gov/pdfs/wind_maps/us_windmap.pdf
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• Not all of the power in the wind is retained - the rotor spills high-speed winds and low-speed winds are too slow to overcome losses
• Depends on rotor, gearbox, generator, tower, controls, terrain, and the wind
• Overall conversion efficiency (Cp·ηg) is around 30%
Estimates of Wind Turbine Energy
WPBP EP
Power in the Wind
Power Extracted by Blades
Power to Electricity
PCRotor Gearbox &
Generator
g
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Ex. 6.11 – Annual Energy from a Wind Turbine
• NEG Micon 750/48 (750 kW and 48 m rotor)
• Tower is 50 m
• In the same area, vavg is 5m/s at 10 m
• Assume standard air density, Rayleigh statistics, Class 1 surface, (total) efficiency is 30%
• Find the annual energy (kWh/yr) delivered
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Ex. 6.11 Annual Energy from a Wind Turbine
• We need to use (6.16) to find v at 50 m, where z for roughness Class 1 is 0.03 m (from Table 6.4)
• Then, the average power density in the wind at 50 m from (6.48) is
00
ln( / ) (6.16)
ln( / )
H zv v
H z
ln(50 / 0.03)5 m/s 6.39 m/s
ln(10 / 0.03)v
32 26 1/m (1.225) 6.39 = 304.5 W/m
2avgP
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Ex. 6.11 Annual Energy from a Wind Turbine
• The rotor diameter is 48 m and the total efficiency is 30%, so the average power from the wind turbine is
• Then, the energy delivered in a year is
220.3 304.5 W/m 48 = 165303 W 4avgP
6Energy 165.303 kW 8760 hrs/yr = 1.44 10 kWh/yr