Environmental Engineering Reference
In-Depth Information
Figure 1.8 Distribution of wind speeds across the United States (U.S. Department of Energy).
A turbine such as the one shown in Figure 1.9 with radius R
¼
63.5m, and
1.2 kg/m
3
for air at 20
C yields
assuming v
¼
8m/s, taking
r¼
5
2
1
28
3
P
¼ gp
63
:
:
=
2
¼ g
3
:
89 MW
:
The best ef
ciency in practice is about 0.4, giving 1.56MW/turbine at the assumed
8m/s, which is a favorable value, shown in the dark areas of the windmap, Figure 1.8,
typically in the United States in a band running from Texas to Minnesota.
It is quite easy to show that the maximum ef
ciency is about 0.59 (Betzs law)
(http://c21.phas.ubc.ca/article/wind-turbines-betz-law-explained.) by realizing that
the speed v
0
behind the turbine is reduced, and the average speed is v
av
1
/
2
(v
v
0
).
¼
þ
Thus, the corrected formula is
v
0
2
Þ¼p
R
2
v
2
P
ð
R
r
v
av
ð
Þ=
2
:
ð
1
:
12
Þ
This formula provides a maximum power at most 0.59 of the unperturbed power
P
0
(R)
¼p
R
2
r
v
3
/2. This corresponds to v
v/3, so one can see why the wind turbines
are not longitudinally arranged because the exit air velocity is quite reduced.
Consider an array of such turbines, spaced by 10 R. Then the power per unit
ground area delivered by the array of the designated turbines at 8m/s is 1.56MW/
(635m)
2
¼
3.86W/m
2
. A rough comparison with solar cells is that an average solar
power at earth is 205W/m
2
with an expected ef
ciency around 0.15, thus 30.75W/m
2
.
The possibility exists of having both solar cells and wind turbines in the same area,
plausible if the area is not cultivated. Questions of the installation costs are deferred,
but the starting estimate of $1/(peak installed watt) generally is useful.
We can ask how large a windfarm is needed to generate 500 GW, approximately
the electricity used in the United States? If we take 3.86W/m
2
, the answer is
¼
