Chapter 10 Wind Energy
Problem 10.1
Using Figure 10.16 (for March), estimate the relative power provided by a wind turbine at an altitude
of: 600 m compared with one at ground level.
From Figure 10.16,
v(0 m) = 3.5 m/s
and
v(600 m) = 10 m/s.
From equation (11.4) the power available is:
3
3
P = (0.602 kg/m )Av
So the relative amount of power is proportional to the cube of the relative velocity or:
3
P(600 m)/P(0 m) = [v(0 m)/v(600 m)] =
Problem 10.2 (a)
Consider a wind farm consisting of 10-m diameter wind turbines with efficiencies of 40% on a
grid given by the average spacing shown in Figure 10.18.
2
For a wind velocity of: 6 m/s, what is the power output in: kW per km ?
2
(a) The area per turbine is: πr = ________________________ per turbine.
(b) The power per turbine (in Watts) from equation (10.4) will be
P = _____________________________________ W
From Figure 10.18 the average spacing of the wind turbines downwind will be: 10 times the rotor
diameter or: 100 m and the crosswind spacing will be: 3 times the rotor diameter or: 30 m.
2
Thus each turbine will occupy a land area of: _____________________________m .
2
1 km square of land area will, therefore, accommodate
_______________________________ turbines
for a total output of
_________________________________ MW
Problem 10.3
A wind turbine with a 40-m diameter rotor produces: 287-kW output in a: 10-m/s wind.
What is its efficiency?
From equation (10.5)
3
3
P = (0.602 kg/m )ηAv
Solving for η gives
2
2
For A = πr = _____________________________ m .
At a wind velocity of: v = 10 m/s then we solve for: η as:
Problem 10.4
A wind turbine with an efficiency of: 42% produces: 1-MW output at a wind velocity of:
13 m/s.
What is the turbine rotor diameter?
From equation (10.5)
3
3
P = (0.602 kg/m )ηAv
or
Solving for A gives
2
2
Then A = πr = πd /4. So
Problem 10.5
Wind velocities are not constant throughout the day.
The daily average power produced by a wind turbine is the power averaged over the wind velocity
for the day.
Calculate the average power for a turbine with a diameter of 20 m and an efficiency of 37% if,
during a 24-hour period, the wind velocity is
• 2 m/s
for
4 hours.
• 8 m/s
for
16 hours.
• 14 m/s
for
3 hours.
• 17 m/s
for
1 hour.
During each period the power will be given by equation (10.5) as
3
3
P = (0.602 kg/m )ηAv =
where v is in m/s and P is in W.
For the different periods the power available will be
for 4 h at 2 m/s →
___________________________________ W
for 16 h at 8 m/s →
___________________________________ W
for 3 h at 14 m/s →
___________________________________ W
for 1h at 17 m/s
___________________________________ W
The energy generated during these periods is the power times the duration as
for 4 h at 2 m/s →
___________________________________ kWh
for 16h at 8 m/s
___________________________________kWh
for 3 h at 14 m/s →
___________________________________kWh
for 1h at 17 m/s →
___________________________________kWh
Adding these gives the total energy over the 24 h period as
E = ___________________________________ kWh
The average power is, therefore,
<P> = ___________________________________ kW
Problem 10.7
2
A Darrieus rotor has an area of: 1500 m and operates with the optimal tip speed ratio.
What is its power output in a wind with a velocity of: 20 m/s?
From equation (10.5) the power is
3
3
P = (0.602 kg/m )ηAv
A = ___________________________________m2
v = 20 m/s,
estimating the efficiency of the Darrieus rotor from Figure 10.12 at about: 40% the above gives the
power output as
P = __________________________________________ MW