4
a1 0.056 = Vo Ω r := Vo 20 m s = tanφ 1 a - ( ) Vo 1 a1 - ( ) Ω r := φ atan tanφ ( ) := φ 35.218 deg = Cn Cl cos φ () Cd sin φ () + := Cn 0.658 = c 8 π α sin φ () ( ) 2 1 a - ( )B Cn λ R := c 1.109 m = Fn 1 2 ρ c Vo 2 Cn := Fn 178.701 kg s 2 = Ct Cl sin φ () Cd cos φ () - := Ft 1 2 ρ c Vo 2 Ct := Ft 123.814 kg s 2 = 2o Punto r 7.5m := x r R := x 0.5 = Ω λ Vwind ( ) R := a 1 3 := R 15m := r 5m := λ 6 := Vwind 10 m s := rpm π 30s := mm m := Cl 0.8 := Cd 0.007 := α 4deg := α 0.07 = B 3 := ρ 1.225 kg := A π R 2 ( ) π r 2 ( ) - := Punto Iniciale x r R := x 0.333 = Ω λ Vwind ( ) R := Ω 38.197 rpm = a1 2 9 λ 2 x 2 :=

Mathcad - Calculation

Embed Size (px)

DESCRIPTION

Calculation routine for blade design - Wind turbines. Angle and chord lenght in function of size and wind speed.

Citation preview

Page 1: Mathcad - Calculation

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

:=

Page 2: Mathcad - Calculation

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=

Page 3: Mathcad - Calculation

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( )⋅+:=

Page 4: Mathcad - Calculation

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=