Green Electric Energy Lecture 19

ECE 333

Green Electric Energy Lecture 19 Wind, Electric Generators

Professor Tom Overbye Department of Electrical and Computer Engineering

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.

Squirrel Cage Rotor •

The rotor of many induction generators has copper or aluminum bars shorted together at the ends, looks like a cage

• Can be thought of as a pair of magnets spinning around a cage • Rotor current iR flows easily through the thick conductor bars Figure 6.15

Squirrel Cage Rotor • 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

Figure 6.16

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

Slip •

The difference in speed between the stator and the rotor NS − NR NR s= =1− (6.28) NS NS

• s = rotor slip – positive for a motor, negative for a generator 120 f • NS = no-load synchronous speed (rpm) N S = p • f = frequency (Hz) • p = number of poles • NR = rotor speed (rpm)

The Induction Machine as a Motor

Torque- slip curve for an induction motor, Figure 6.17

• • • •

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

The Induction Machine as a Generator

Figure 6.18. Single-phase, self-excited, induction generator

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 N R = (1 − s ) N S = [1 − ( −0.01)] ⋅ 3600 = 3636 rpm

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

Blade Efficiency vs. Windspeed

Figure 6.19

At lower windspeeds, the best efficiency is achieved at a lower rotational speed

Power Delivered vs. Windspeed

Figure 6.20

Impact of rotational speed adjustment on delivered power, assuming gear and generator efficiency is 70%

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

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

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

Vestas V80 1.8 MW

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

GE 1.5 MW and 3.6 MW DFIG Examples

GE 1.5 MW turbines are the best selling wind turbines in the US with 43% market share in 2008 Source: GE Brochure/manual

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 (DCAC) to 60 Hz for grid-connection Good for handling rapidly changing windspeeds

Figure 6.21

Example: GE 2.5 MW Turbines

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:

1 1 3 Pavg =  ρ Av  = ρ A ( v3 ) (6.29) avg 2 2   avg • This is why we can’t use average windspeed vavg to find the average power in the wind

Average Windspeed v ⋅ [ hours @ v ] ∑ miles of wind = = total hours ∑ [ hours @ v ]

(6.32)

vavg = ∑ vi ⋅ [ fraction of total hours @ vi ]

(6.32)

i

vavg

i

i

i

i

i

• vi = windspeed (mph) • The fraction of total hours at vi is also the probability that v = vi

Average Windspeed vavg = ∑ vi ⋅ [ fraction of total hours @ vi ]

(6.32)

vavg = ∑ vi ⋅ [ probability that v = vi ]

(6.33)

i

i

• •

This is the average windpseed in probabilistic terms Average value of v3 is found the same way: 3 v ( )

avg

= ∑ vi 3 ⋅ [ probability that v = vi ] i

(6.35)

Example Windspeed Site Data

Figure 6.22

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

Windspeed p.d.f.

v2

p ( v1 ≤ v ≤ v2 ) = ∫ f (v) dv

(6.36)

v1

∞

p ( 0 ≤ v ≤ ∞ ) = ∫ f (v)dv = 1

(6.37)

0

v2

hrs / yr ( v1 ≤ v ≤ v2 ) = 8760 ⋅ ∫ f (v)dv v1

(6.38)

Average Windspeed using p.d.f. •

This is similar to (6.33), but now we have a continuous function instead of discrete function vavg = ∑ vi ⋅ p ( v = vi ) (6.33) i

discrete

∞

vavg = ∫ v ⋅ f (v) dv 0

(6.39)

continuous

•

Same for the average of (v3) 3 3 v = v ( ) avg ∑ i ⋅ p ( v = vi ) (6.35) i

discrete

3 v ( )

∞

avg

= ∫ v 3 ⋅ f (v)dv 0

continuous

(6.40)

Weibull p.d.f. •

Starting point for characterizing statistics of windspeeds k-1

kv f (v ) =   ⋅ e cc

• k = shape parameter • c = scale parameter

v -  c

k

Weibull p.d.f. (6.41)

Weibull p.d.f.

k=2 looks reasonable for wind

Weibull p.d.f. for c = 8

Figure 6.24

Rayleigh p.d.f. •

This is a Weibull p.d.f. with k=2

2v f (v ) = 2 ⋅ e c

• •

v -  c

2

Rayleigh p.d.f. (6.42)

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

Rayleigh p.d.f. (Weibull with k=2)

Figure 6.25

Higher c implies higher average windspeeds

Rayleigh p.d.f. •

When using a Rayleigh p.d.f., there is a direct relationship between average windspeed v and scale parameter c ∞

vavg = v = ∫ v ⋅ f (v )dv

(6.39)

0

•

Substitute (6.42) into (6.39): k v ∞ 2v - c  vavg = v = ∫ v ⋅ 2 ⋅ e dv (6.43) c 0 π vavg = c ≅ 0.886 ⋅ c 2

(6.43)

Rayleigh p.d.f. •

From (6.43), we can solve for c in terms of v 2 π vavg =1.128v (6.44) vavg = c ≅ 0.886 ⋅ c (6.43) c = π 2

•

Then we can substitute this into the Rayleigh p.d.f − vk (6.42) for c k  2v    2v  π f (v ) = ⋅e Rayleigh p.d.f. (6.45) 2  2v    π   πv f (v ) = 2 ⋅ e 2v

πv −   4v 

2

Rayleigh p.d.f. (6.45)

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

Rayleigh Statistics – Average Power in the Wind

3 v ( )

∞

avg

= ∫ v 3 ⋅ f (v )dv 0

πv f (v ) = 2 ⋅ e 2v

(v )

∞

3

πv = ∫ v ⋅ 2 ⋅e 2v 0 3

avg

πv −   4v 

πv −   4v 

2

(6.40)

2

(6.45)

3 3 dv = c π 4

(6.46)

Rayleigh Statistics – Average Power in the Wind

(v )

∞

3

πv = ∫ v ⋅ 2 ⋅e 2v 0 3

avg

πv −   4v 

2

3 3 dv = c π 4

2 c= vavg =1.128v π

(v ) 3

avg

3 3 6 = ( vavg ) =1.91( vavg ) π

(6.46) (6.44)

(6.47)

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:

•

1 1 3 Pavg =  ρ Av  = ρ A ( v3 ) avg 2 2   avgwrite the (v ) in terms of v With Rayleigh assumptions, we can 3

avg

(6.29) avg

as in (6.47), and

the expression for average power in the wind is just

• This is an important and useful 6 1result

Pavg =

⋅ ρ A ( vavg ) π 2

3

(6.48)

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. Estimate windspeed at 50 m: α

1/7

 H 50   50  v50 = v10   = 6   =7.55 m/s  10   H10  Average power density in the wind at 50 m from (6.48):

6 1 3 Pavg / m = ⋅ (1.225) ( 7.55) = 504 W/m 2 π 2 2

(6.48)

Real Data vs. Rayleigh Statistics

Figure 6.26

This is why it is important to gather as much real wind data as possible

Wind Power Classification Scheme

Table 6.5

Wind Power Classification Scheme •

Table 6.5

http://www.windpoweringamerica.gov/pdfs/wind_maps/us_windmap.pdf

Estimates of Wind Turbine Energy

PW Power in the Wind

CP Rotor

PB

Power Extracted by Blades

ηg Gearbox & Generator

PE

Power to Electricity

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

•

Find the annual energy (kWh/yr) delivered

Assume standard air density, Rayleigh statistics, Class 1 surface, (total) efficiency is 30%

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) ln( H / z ) v = v0 (6.16) ln( H 0 / z ) ln(50 / 0.03) v = 5 m/s = 6.39 m/s ln(10 / 0.03)

• Then, the average power density in the wind at 50 m from (6.48) is 6 1 3 2 Pavg /m = ⋅ (1.225) ( 6.39 ) = 304.5 W/m2 π 2

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

π 2 Pavgthe=energy 0.3 ⋅ (delivered 304.5 inW/m ) ⋅ 4 ⋅ ( 48) = 165303 W a year is • Then, 2

Energy = 165.303 kW ⋅ 8760 hrs/yr = 1.44 × 106 kWh/yr