Wind

Question 1

how is wind formed

Answer 1

due to differences between atmospheric pressure, caused by uneven heating of the earths surface (by the sun)

-temperature differences
- warm air rises, cool air sinks
- creates low / high pressure areas
- air naturally moves from high to low - creating wind

Question 2

what causes wind to move between earths surfaces

Answer 2

-pressure gradient force: drives air from high to low pressure
-coriolis force; modifies direction of wind fow.
- frictions with earths surface: slows down wind near ground
- atmospheric stability: affects vertical air movement

Question 3

what are the two main configurations of wind turbibes

Answer 3

HAWT - Horizontal axis wind turbines

VAWT - Vertical axis wind turbines

Question 4

Explain a HAWT

Answer 4

Horizontal axis wind turbine

• rotor shaft is horizontal and faces wind.
• blades mounted on top of a tall tower
• requires yaw system to turn the rotor towards wind direction

Question 5

explain VAWT

Answer 5

Vertical axis wind turbine
- rotor shaft is vertical & perpendicular to ground
- main components located near ground

Question 6

a site has a wind speed of 6.3m/s at 30m as measured over a period of 8 months. z0 is 0.001m.

what wind speed would be seen by a wind turbine of 80m at the same site.

Answer 6

U(z) = U(zr) ( (ln(z/z0) / (ln(zr/z0))

ans = 6.89 m/s

Question 7

a site has an average annual mean wind speed of 8 m/s and an annual standard deviation of 5 m/s.

calculate rough estiamtes of weibull parameters k & C

Answer 7

k = (o/U)^-1.086
C = 2U/sqrt (pi)

ans = 1.67 / 9.03 m/s

Question 8

use estimates given to calculate annual energy yiled onsite.
average availability is 95%
wind speed power output
0-5 0
5-8 20
8-12 35
12-14 40
14-25 45
25+ 0

k = 1.67
C = 9.03

Answer 8

1) calculate k & C
2) P(U>v) = exp (-(v/c)^k)
p(u) P(U)*p(u)
0.25 5
0.24 8.4
0.075 3
0.121 5.45
total: 21.85 kW x avalibility
annual yeild = kW x days x hrs.

Question 9

an analysis of wind speed data has yeilfed an average speed of U(-) = 6m/s. Rayleigh wind speed distribution.

a) Estimate the number of hours per year that the wind speed will be between Ub = 10.5 and Ua = 9.5 during the year

b) Estimate the number of hours the wind speed is above 16 m/s

Answer 9

P(Ua < U < Ub) = C(Ub) - C(Ua)
C(U) = 1 - exp (-(pi/4)(U/U(-))^2)

ans = 429.24 hr / year
ans = 32.4 hr / year

Question 10

for a offshore site with (1/a) = 5 & Um = 20 & U = 40m/s.

what is the risk that this turbine will not last 10 years.

Answer 10

1/a = a
r = 1 - (p (<40))^10
p(<40) = e^-e*a(U(-)-um)

ans = 0.17

Question 11

Weibull Parameter k equation

Answer 11

(σ/U ̃ )^(−1⋅086)

Question 12

weibull parameter C equation

Answer 12

2U / sqrt (pi)

Question 13

Determine the power delivered per month by a hydroelectric system that has a flow rate of 20 kg/s and an available vertical drop (head) of 5 meters. Assume the energy conversion efficiency of the turbine is 80%.

Answer 13

P(per hour) = pgHQn
P x time

= 565.06 MW

Question 14

3. For hydro plant with:
   Power output P = 400kW
   Head h = 150m
   Efficiency n = 70%

a. Calculate flow rate at turbine height
b. Calculate nozzle diameter for selected hydro turbine

Answer 14

Q= P/pgHn

= 0.39 m^3 / s

A = pi * r^2
Q = Vjet * Ajet (formula)
Vjet = sqrt( 2 n g H) (formula)

= 0.1m

Question 15

wind turbines

what is the equaiton for risk

Answer 15

r = 1 - [p(<40)]^10

p(<40) = e^ -e^-a (u - um)

Question 16

steps to find out: power of airflow through a cylinder pipe

Answer 16

P = 1/2pA*V^3

A = 2(pi)r^2 + 2(pi)h