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Calculation routine for blade design - Wind turbines. Angle and chord lenght in function of size and wind speed.
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a1 0.056=
Vo Ω r⋅:=
Vo 20m
s=
tanφ1 a−( ) Vo⋅
1 a1−( ) Ω⋅ r⋅:= φ atan tanφ( ):= φ 35.218 deg=
Cn Cl cos φ( )⋅ Cd sin φ( )⋅+:= Cn 0.658=
c8 π⋅ α⋅ sin φ( )( )2
⋅ 1 a−( ) B⋅ Cn⋅ λ⋅
R⋅:= c 1.109m=
Fn1
2ρ⋅ c⋅ Vo
2⋅ Cn⋅:= Fn 178.701
kg
s2
=
Ct Cl sin φ( )⋅ Cd cos φ( )⋅−:=
Ft1
2ρ⋅ c⋅ Vo
2⋅ Ct⋅:= Ft 123.814
kg
s2
=
2o Punto r 7.5m:=
xr
R:= x 0.5=
Ωλ Vwind⋅( )
R:=
a1
3:= R 15m:= r 5m:= λ 6:= Vwind 10
m
s:= rpm
π
30s:= m³ m m⋅ m⋅:=
Cl 0.8:= Cd 0.007:= α 4deg:= α 0.07= B 3:= ρ 1.225kg
m³:=
A π R2
⋅( ) π r2
⋅( )−:=Punto Iniciale
xr
R:= x 0.333=
Ωλ Vwind⋅( )
R:=
Ω 38.197 rpm=
a12
9 λ2
⋅ x2
⋅
:=
φ 34.061 deg=φ atan tanφ( ):=tanφ1 a−( ) Vo⋅
1 a1−( ) Ω⋅ r⋅:=
Vo 40m
s=
Vo Ω r⋅:=
a1 0.014=a12
9 λ2
⋅ x2
⋅
:=
Ω 38.197 rpm=
Ωλ Vwind⋅( )
R:=
x 0.667=xr
R:=
r 10m:=3o Punto
Ft 258.257kg
s2
=Ft1
2ρ⋅ c⋅ Vo
2⋅ Ct⋅:=
Ct Cl sin φ( )⋅ Cd cos φ( )⋅−:=
Fn 385.008kg
s2
=Fn1
2ρ⋅ c⋅ Vo
2⋅ Cn⋅:=
c 1.051m=c8 π⋅ α⋅ sin φ( )( )2
⋅ 1 a−( ) B⋅ Cn⋅ λ⋅
R⋅:=
Cn 0.664=Cn Cl cos φ( )⋅ Cd sin φ( )⋅+:=
φ 34.354 deg=φ atan tanφ( ):=tanφ1 a−( ) Vo⋅
1 a1−( ) Ω⋅ r⋅:=
Vo 30m
s=
Vo Ω r⋅:=
a1 0.025=a12
9 λ2
⋅ x2
⋅
:=
Ω 38.197 rpm=
Ft 987.831kg
s2
=Ft1
2ρ⋅ c⋅ Vo
2⋅ Ct⋅:=
Ct Cl sin φ( )⋅ Cd cos φ( )⋅−:=
Fn 1.501 103
×kg
s2
=Fn1
2ρ⋅ c⋅ Vo
2⋅ Cn⋅:=
c 1.019m=c8 π⋅ α⋅ sin φ( )( )2
⋅ 1 a−( ) B⋅ Cn⋅ λ⋅
R⋅:=
Cn 0.668=Cn Cl cos φ( )⋅ Cd sin φ( )⋅+:=
φ 33.854 deg=φ atan tanφ( ):=tanφ1 a−( ) Vo⋅
1 a1−( ) Ω⋅ r⋅:=
Vo 60m
s=
Vo Ω r⋅:=
a1 6.173 103−
×=a12
9 λ2
⋅ x2
⋅
:=
Ω 38.197 rpm=
Ωλ Vwind⋅( )
R:=
x 1=xr
R:=
r 15m:=4o Punto
Ft 447.268kg
s2
=Ft1
2ρ⋅ c⋅ Vo
2⋅ Ct⋅:=
Ct Cl sin φ( )⋅ Cd cos φ( )⋅−:=
Fn 674.224kg
s2
=Fn1
2ρ⋅ c⋅ Vo
2⋅ Cn⋅:=
c 1.032m=c8 π⋅ α⋅ sin φ( )( )2
⋅ 1 a−( ) B⋅ Cn⋅ λ⋅
R⋅:=
Cn 0.667=Cn Cl cos φ( )⋅ Cd sin φ( )⋅+:=
Torque kW 1000W:=
Ct 4 a⋅ 1 a−( )⋅:=
A 628.319 m2
=
T Ct1
2⋅ ρ⋅ A⋅ Vwind
2⋅:= T 3.421 10
4× N=
Cp 4 a⋅ 1 a−( )2
⋅:=
P Cp1
2⋅ ρ⋅ A⋅ Vwind( )
3⋅:= P 228.056 kW=