Advanced Power Systems

1
Advanced Power Systems

• ECE 0909.402-01, 0909.504-01
• Lecture 7 Wind Generation
• 21 March 2005
• Dr. Peter Mark Jansson PP PE
• Associate Professor Electrical and Computer
  Engineering

2
admin announcements

• Mid-Term at 600 today in auditorium
• Final Project Selection due with HW 5 today
• Graduate section individual project
• Undergraduate section up to 2 per team

3
Final Project (30)

• NOTE Peer Review Assessment will be part of
  presentation grade

4
See revised class schedule

• Posted on Web
• Final Presentation Dates
• 18 April, 25 April, 2 May, 9 May

5
New homework

• HW 6 due next Monday 28 Mar
• now posted on web
• 6.1, 6.2, 6.3, 6.4, 6.5, 6.7, 6.8, 6.12, 6.15

6
Aims of Todays Lecture

• Part One complete summary of ch. 6 concepts
• Wind-turbine Generators
• break at 600 p.m.
• Part Two mid-term covers through todays concepts

7
Wind Turbines

• Wind energy is proportional to V3, Why?

8
Simple Rule of Thumb

• Annual Energy (kWh) 1.64 D2 V3
• D rotor diameter, meters
• V annual average wind speed (m/sec)

9
Wind Turbine types

• VAWT
• HAWT downwind
• HAWT upwind, requires yaw control

10
Power in the Wind
11
Why using average wind speed is a poor
approximation of site potential

• Lets use our new equation to consider the
  following
• What is energy in 200 hours of 7m/s wind?
• Compare with energy in 100 hours of 4m/s and 100
  hours at 10m/s
• (NOTE both average 7m/s)?

12
Power in the Wind
13
LM 1

• How much power is there in 500 hours of 6m/s wind
  (per m2)?

14
Impact of Tower Height
15
? and Z

• Tables 6.3 6.4
• Page 320

16
Example

• If you have anemometer data from a 10-m tower (6
  m/s ave.) across a surface of crops, hedges, and
  shrubs and want to estimate how much higher the
  wind is at 40 meters, how do you do it?

17
solution
18
LM 2

• What is the average wind speed at 50 meters if it
  is 7m/s at an anemometer sited 10 meters above
  the ground in a small village?

19
Max theoretical rotor efficiency

• Max theoretical is called Betz efficiency
• For typical turbines this is 59
• Under ideal conditions todays turbines can
  achieve 80 of the max theoretical
• So many turbines range between 45-50

20
Rayleigh Probability Density Function

• With wind we typically do not have a normal
  distribution.
• A Weibull or Rayleigh p.d.f. is much more common,
  we will learn the Rayleigh a form of Wiebull when
  k2. (see below)

21
Rayleigh Probability Density Function

• With the assumption of a Rayleigh distribution we
  have a very powerful analytical tool.
• To calculate the number of hours in a given year
  we will experience wind above or below a certain
  level the process is quite simple

22
Example

• If we have a site where the wind speed (average)
  is 7 m/s and we assume a Rayleigh distribution,
  how many hours per year will the wind speed be
  less than 4 m/s (the cut in velocity for an NEG
  micon 1000/54 wind turbine)?

23
LM 3

• If we have a site where the wind speed (average)
  is 6 m/s and we assume a Rayleigh distribution,
  how many hours per year will the wind speed be
  less than 3 m/s?

24
Solution

• How many hours per year will the wind speed be
  less than 4 m/s?

25
Solution

• How many hours per year will the wind speed be
  less than 4 m/s?
• 0.2262 x 8760 1982 hours

26
Using this and a WT Power Spec

• Combining these two we can predict effectively
  the power generated at any given site

27
Rayleigh PDF and Power Curve

• 1) build a spreadsheet
• 2) calculate hours at each speed (see below).
• 3) use manufacturers data for turbine generation
  at each wind speed
• 4) estimate annual generation

28
LM 4

• What are the hours per year we will have wind
  speeds at 5, 6, 7 and 8 m/s if the average wind
  speed is 7 m/s?

29
Problem 6.15

• Excellent illustration

30
New homework

• HW 6 due next Monday 28 Mar
• now posted on web
• 6.1, 6.2, 6.3, 6.4, 6.5, 6.7, 6.8, 6.12, 6.15