EP MINTEGIA 2009/02/25 (ebazpena)

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λ λ λ (J 1 ) 1 (J 1 ) 2 (J 1 ) 3 (I C ) 1 = 30A n 1 n 2 1 = 15 n 1 n 2 2 = 30 n 1 n 2 3 = 30 I SARE (I C ) 2 = 90A (I C ) 3 = 60A Δ Δ λ

description

 

Transcript of EP MINTEGIA 2009/02/25 (ebazpena)

Page 1: EP MINTEGIA 2009/02/25 (ebazpena)

λ  λ  λ 

(J1)1  (J1)2  (J1)3 

(IC)1= 30A 

n1n2

1

=15

n1n2

2

= 30

n1n2

3

= 30

ISARE 

(IC)2= 90A  (IC)3= 60A 

Δ  Δ λ 

Page 2: EP MINTEGIA 2009/02/25 (ebazpena)

n2n1

1

=115

n2n1

2

=130

⇒n2n1

2

=12⋅n2n1

1

n2n1

3

=130

⇒n2n1

3

=12⋅n2n1

1

(V13)1

(V13)3 30o

(V1)2

(IC)1= 30A

(IC)2= 90A (IC)2 = 3(IC)1

(IC)3= 60A (IC)3 = 2(IC)1

30o

Page 3: EP MINTEGIA 2009/02/25 (ebazpena)

λ

λ

(J1)1

(IC)1 = 30A

n1n2

1

=15

30 60 90 120 150 180 210 240 270 300 330 360

(VL)1

(Is1)1

(V13)1 (V23)1 (V21)1 (V31)1 (V32)1 (V12)1

(J1)1 = (Ip1)1

(IC)1

-(IC)1

n2n1

1

⋅ IC( )1 = x

−n2n1

1

⋅ IC( )1 = −x

(V32)1

Page 4: EP MINTEGIA 2009/02/25 (ebazpena)

30 60 90 120 150 180 210 240 270 300 330 360

(VL)2

(Is1)2

(V2)2 (V3)2 (V1)2

(Ip1)2

(IC)2

23

n2n1

2

⋅ IC( )2 =23

⋅12n2n1

1

⋅ 3 ⋅ IC( )1[ ] =

n2n1

1

⋅ IC( )1

= x

−13

n2n1

2

⋅ IC( )2 = −13

⋅12n2n1

1

⋅ 3 ⋅ IC( )1[ ] = −

12⋅n2n1

1

⋅ IC( )1 = −12x

λ

(J1)2

n1n2

2

= 30

(IC)2 = 90A

Δ 

Page 5: EP MINTEGIA 2009/02/25 (ebazpena)

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(VL)2

(Is3)2

(V2)2 (V3)2 (V1)2

(Ip3)2

(IC)2

x

−12x

λ

(J1)2

n1n2

2

= 30

(IC)2 = 90A

Δ 

Page 6: EP MINTEGIA 2009/02/25 (ebazpena)

Δ

λ

(J1)2

n1n2

2

= 30

(Ip3)2

(Ip1)2

(J1)2 = (Ip1)2 - (Ip3)2

30 60 90 120 150 180 210 240 270 300 330 360 €

−12x

x

−12x

x

−32x

32x

(IC)2 = 90A

Page 7: EP MINTEGIA 2009/02/25 (ebazpena)

30 60 90 120 150 180 210 240 270 300 330 360

(VL)3

(Is1)3

(V13)3 (V23)3 (V21)3 (V31)3 (V32)3 (V12)3

(IC)3

-(IC)3

n2n1

3

⋅ IC( )3 =12⋅n2n1

1

⋅ 2 ⋅ IC( )1[ ] =

n2n1

1

⋅ IC( )1 = x

−n2n1

3

⋅ IC( )3 = −xΔ

λ

(J1)3

n1n2

3

= 30

(IC)3 = 60A

(Ip1)3

Page 8: EP MINTEGIA 2009/02/25 (ebazpena)

30 60 90 120 150 180 210 240 270 300 330 360

(VL)3

(Is3)3 (IC)3

-(IC)3

n2n1

3

⋅ IC( )3 = x

−n2n1

3

⋅ IC( )3 = −xΔ

λ

(J1)3

n1n2

3

= 30

(IC)3 = 60A

(Ip3)3

(V13)3 (V23)3 (V21)3 (V31)3 (V32)3 (V12)3

Page 9: EP MINTEGIA 2009/02/25 (ebazpena)

x

−x

Δ

λ

(J1)3

n1n2

3

= 30

(IC)3 = 60A

(Ip3)3

x

−x

(Ip1)3

(J1)3 = (Ip1)3 - (Ip3)3

x€

2x

−x

−2x

30 60 90 120 150 180 210 240 270 300 330 360

Page 10: EP MINTEGIA 2009/02/25 (ebazpena)

(J1)3

x€

2x

−x

−2x30 60 90 120 150 180 210 240 270 300 330 360

(J1)2

(J1)1

x

−x

ISARE

30 60 90 120 150 180 210 240 270 300 330 360

−2x

−3x

5x

−4x

4x

ISARE =12π

⋅ 2 ⋅ π6⋅52x

2

+ 2 ⋅ π6⋅72x

2

+ 2 ⋅ π3⋅92x

2

+ 2 ⋅ π6⋅ 2x( )2 + 2 ⋅ π

6⋅ x( )2

= x ⋅ 12π

⋅ 2 ⋅ π6⋅254

+ 2 ⋅ π6⋅494

+ 2 ⋅ π3⋅814

+ 2 ⋅ π6⋅ 4 + 2 ⋅ π

6

= x ⋅ 12⋅5024

+9824

+16212

+86

+26

= x ⋅ 1

2⋅51224

= x ⋅ 32

3

ISARE = x ⋅ 323

=n2n1

1

⋅ IC( )1

⋅323

=115⋅ 30

⋅353

= 6.53A

−32x

32x

−x

3x

2x

x

−5x