Correction Exercice 4 - Rendement d`une éolienne

DV=Seve=Sv =Ssvs
ve> v > vs
Ss> S > Se
1
2ρve2+P0=1 2ρv2+P+
1
2ρv2+P−=1 2ρvs2+P0
−→
Ffluide→helice =P+S−→ ux−P−S−→ ux
−→
Fhelice→f luide =−(P+−P−)S−→ ux
S∗
t S
x
p∗(t) = p(t) et p∗(t+dt) = p(t+dt)−dmve+dmvs
dm =ρDVdt dt
S
p(t+dt) = p(t)

p∗(t+dt)−p∗(t) = ρDV(vs−ve)dt
D−→ p∗
Dt =ρDV(vs−ve)−→
ux
S∗
D−→ p∗
Dt =−→
Fhelice→f luide +−→ Fpression→f luide
P0
D−→ p∗
Dt =−(P+−P−)S−→
ux
D−→ p∗
Dt
ρDV(vs−ve) = −(P+−P−)S
P+P−
DV=Sv
v=ve+vs
2
v ve> v > vs
S∗
dE∗
c=Phelice→f luide dt =1
2dm(v2
s−v2 e)
Pfluide→helice =1
2ρv2
eDVÃ1−v2
s
v2
e!
DV=Sv =Sve+vs
2
Pfluide→helice =1
2ρv2
eSve
1
2Ã1 + vs ve!Ã1−v2
s
v2
e!
f(x) = 1
2(1 + x)(1 −x2)
1
2ρv2
eSve
f
x= 1

f f0(x) = 1
2¡1−2x−3x2¢
x x2+2
3x−1 3= 0
x=−1 ou 1
3
x f Ã1
3!=16 27 > f(0) = 1 2
xmax =1
3et Pfluide→helicemax =8 27 ρSv3
e
ρair = 1,4−3S=πd2
4Pfluide→helicemax = 2,1
Ss
S=v vs
=1
2
ve+vs
vs
=1
2Ã1 x+ 1!
x=xmax =1
3
Ss
S= 2
ds=√2d'112
ds
Ffluide→helice =1
2ρv2
e¡1−x2 max¢S= 3,1.105
∼17
vmax =Rω = 40 ×17 ×2π/60 = 71 −1=
256 −1
340 −1
100 −1
15 −1
15 −1