Announcements

Announcements • • • •

Start reading Chapter 6. Homework 7 is due now. Homework 8 is 6.3, 6.5, 6.8, 6.14; due on Tuesday Nov 10. Wind Farm field trip will be on Nov 5 from about 8am to 4pm – turn in forms to sign up.

Maximum Rotor Efficiency • Two extreme cases, and neither makes sense• Downwind velocity is zero – turbine extracted all of the •

power Downwind velocity is the same as the upwind velocity – turbine extracted no power

• Albert Betz 1919 - There must be some ideal slowing of the wind so that the turbine extracts the maximum power

Maximum Rotor Efficiency • •

Constraint on the ability of a wind turbine to convert kinetic energy in the wind into mechanical power Think about wind passing though a turbine- it slows down and the pressure is reduced so it expands

Figure 6.9

Power Extracted by The Blades 1 Pb = m&( v 2 − vd 2 ) 2

• • •

ṁ = mass flow rate of air within stream tube v = upwind undisturbed windspeed vd = downwind windspeed

•

From the difference in kinetic energy between upwind and downwind air flows and (6.2)

(6.18)

Determining Mass Flow Rate • Easiest to determine at the plane of the rotor because we know the •

cross sectional area A Then, the mass flow rate from (6.3) is

m&= ρ Avb

(6.19)

• Assume the velocity through the rotor vb is the average of upwind velocity v and downwind velocity vd:

v + vd vb = 2

 v + vd  m&= ρ A   2  

Power Extracted by the Blades •

Then (6.18) becomes

1  v + vd  2 2 Pb = ρ A  v − v ( d )  2  2  • Define vd λ= v

•

(6.20)

(6.21)

Rewrite (6.20) as

1  v + λv  2 2 2 Pb = ρ A  v − λ v )  ( 2  2 

(6.22)

Power Extracted by the Blades 1  v + λv  2 2 2 Pb = ρ A  v − λ v )  ( 2  2 

(6.22)

3 2 3 3 3 3 v λ v λ v λ v  v + λv  2 2 2 +   (v −λ v ) = 2 2 2 2  2  v3 ( 1 + λ ) -λ 2 ( 1 + λ )  = 2 v3 ( 1 + λ ) ( 1 − λ 2 )  =  2 1 3 1 Pb = ρ Av ⋅ ( 1 + λ ) ( 1 − λ 2 )  (6.22) 2 2

PW = Power in the wind

CP = Rotor efficiency

Maximum Rotor Efficiency •

Find the speed windspeed ratio λ which maximizes the rotor efficiency, CP

•

From the previous slide 2 3 1 1 λ λ λ CP = ( 1 + λ ) ( 1 − λ 2 )  = - + 2 2 2 2 2 Set the derivative of rotor efficiency to zero and solve for λ:

∂CP =-2λ + 1 − 3λ 2 = 0 ∂λ ∂CP =3λ 2 + 2λ − 1 = 0 ∂λ ∂CP = ( 3λ − 1) ( λ + 1) = 0 ∂λ

1 λ= 3

maximizes rotor efficiency

Maximum Rotor Efficiency •

Plug the optimal value for λ back into CP to find the maximum rotor efficiency: 1  1   1 C P =  1 +   1 − 2 2  3   3

  16 = 59.3%  =   27

(6.26)

• The maximum efficiency of 59.3% occurs when air is slowed to 1/3 of its upstream rate • Called the “Betz efficiency” or “Betz’ law”

Maximum Rotor Efficiency

Rotor efficiency CP vs. windspeed ratio λ

Figure 6.10

Tip-Speed Ratio (TSR) • •

Efficiency is a function of how fast the rotor turns Tip-Speed Ratio (TSR) is the speed of the outer tip of the blade divided by windspeed

Rotor tip speed rpm × π D Tip-Speed-Ratio (TSR) = = (6.27) Wind speed 60v

• D = rotor diameter (m) • v = upwind undisturbed windspeed (m/s) • rpm = rotor speed, (revolutions/min)

Tip-Speed Ratio (TSR) • •

TSR for various rotor types Rotors with fewer blades reach their maximum efficiency at higher tip-speed ratios

Figure 6.11

Example 6.7 • 40-m wind turbine, three-blades, 600 kW, windspeed is 14 m/s, air density is 1.225 kg/m3

a. Find the rpm of the rotor if it operates at a TSR of 4.0 b. Find the tip speed of the rotor c. What gear ratio is needed to match the rotor speed to the generator speed if the generator must turn at 1800 rpm? d. What is the efficiency of the wind turbine under these conditions?

Example 6.7 a. Find the rpm of the rotor if it operates at a TSR of 4.0 Rewriting (6.27),

Tip-Speed-Ratio (TSR) ⋅ 60v rpm = πD 4.0 ⋅ 60sec/min ⋅14m/s rpm = = 26.7 rev/min We can also express this as seconds per revolution: π ⋅ 40m/rev 26.7 rev/min rpm = = 0.445 rev/sec or 2.24 sec/rev 60 sec/min

Example 6.7 b. Tip speed rpm × π D From (6.27): Rotor tip speed= 60 sec/min Rotor tip speed = (rev/sec) × π D Rotor tip speed = 0.445 rev/sec ⋅ π 40 m/rev = 55.92 m/s

c. Gear Ratio Generator rpm 1800 Gear Ratio = = = 67.4 Rotor rpm 26.7

Example 6.7 d. Efficiency of the complete wind turbine (blades, gear box, generator) under these conditions From (6.4): 1 1 π 3 2 PW = ρ Av = ( 1.225)  ⋅ 40  143 = 2112 kW 2 2 4  Overall efficiency: 600 kW η= = 28.4% 2112 kW

Synchronous Machines • • • • •

Spin at a rotational speed determined by the number of poles and by the frequency The magnetic field is created on their rotors Create the magnetic field by running DC through windings around the core A gear box is needed between the blades and the generator 2 complications – need to provide DC, need to have slip rings on the rotor shaft and brushes

Asynchronous Induction Machines • • • • • •

Do not turn at a fixed speed Acts as a motor during start up as well as a generator Do not require exciter, brushes, and slip rings The magnetic field is created on the stator instead of the rotor Less expensive, require less maintanence Most wind turbines are induction machines

Rotating Magnetic Field • Imagine coils in the stator of this 3-phase generator • Positive current iA flows from A to A’ • Magnetic fields from positive currents are shown by the bold arrows

Figure 6.13

Rotating Magnetic Field • Three-phase currents are flowing in the stator • At ωt = 0, iA is at the maximum positive value and iB=iC are both negative

Resultant magnetic flux points vertically down

Figure 6.14 (a)

Rotating Magnetic Field • Three-phase currents are flowing in the stator • At ωt = π/3, iC is at the maximum negative value and iA=iB are both positive

Resultant magnetic flux moves clockwise by 60 degrees

Figure 6.14 (b)

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

Wind Farms • The study in Figure 6.28 considered square arrays, but square • • •

arrays don’t make much sense Rectangular arrays with only a few long rows are better Recommended spacing is 3-5 rotor diameters between towers in a row and 5-9 diameters between rows Offsetting or staggering the rows is common