Aufgabe 1 a) Umformung des Netzes in Stern-ESB:

a)

Umformung des Netzes in Stern-ESB:

( + ) Ω

=

⋅ Ω

= e ^° 0,3021 j0,1409 3

Z_N,Y 1 ⁱ²⁵

V 9 , 3 230

V U_N,Y = 400 =

Zerlegung der Verbraucherimpedanz in Real- und Imaginärteil:

Ω +

=

⋅ Ω

=50 e ^° 40,96 j28,68 Z_V,Y ⁱ³⁵

A 589 , 82 4

, 28 j 26 , 41

V 9 , 230 Z

Z I U

Y , N Y , V

Y , N

1 =

Ω

= +

= +

A 649 , 2 3 I

I_N,Str = _L1 =

V 4 , 229 Z

I

U_V,Str = ₁⋅ _V,Y =

b)

Blindleistungskompensation, d. h. die Admittanz von Verbraucher und Kondensator muss reell werden:

mS 5 , 11 Z j

j 1 Y

Y , V Y

,

C =−







 ℑ

−

=

F 5 , Hz 36 50 2

mS 5 , 11 f 2

C_Y Y^C,Y = µ

⋅

= π

⋅

= π

c)

Umwandlung der mit L2 und L3 verbundenen Elemente in eine Ersatzspannungsquelle ergibt:

2 Z Z Z

2 ; U 2

U

U_E U^{V W U} _E ^N,Y+ ^V,Y

=

− + =

=

( ) ( )

°

⋅ −

Ω = +

+ Ω +

⋅

= ⋅ +

= − ^j34,9

E Y , N

E U K

1 6,86A e

68 , 28 j 96 , 40 1409

, 0 j 3021 , 0 5 , 1

V 9 , 230 5 , 1 Z

Z U I U

Erreger- wicklung α

Erreger- wicklung α

α 2π π

H_δ,max

α 2π π

H_δ,max

m 50 kA m 10 2

A 1 100 2

I

H w^{f f} ₃ =

⋅

= ⋅ δ

⋅

= ⋅ ₋

δ

c)

Vs 314 m

10 m 50

10 A Am 50

10 Vs 4 A H A

B ₀ ^{7 3 4 2}

P = ⋅ =µ ⋅ ⋅ = π⋅ ⋅ ⋅ ⋅ ⋅ = µ

Φ _{δ δ δ δ} ^{− −}

a)

Ω

=

⋅

= 12,1

S u U X

N 2 N k Y ,

T (rein induktiv)

5 , 5 20 / 110

ü= =

b)

( )

(

)

MVA 30 j 40 U

I S

1

*

1 − = −

=

=

( + + ) = ⋅ ^°

=

⋅ +

= _{1N 1 T,Y} ^j6.3

2´ U I jX 63,3 5,7 j7,6 kV 69,4kV e

U

kV 6 , 12

´ ü U U₂ = 1⋅ ₂ =

kV 9 , 21 3 U

U_2LL = ₂ ⋅ =

c)

kA 5 , X 31

´ ü U I

Y , T

2

K = ⋅ =

X_T,Y begrenzt den Kurzschlussstrom auf den etwa 20-fachen Bemessungsstrom.

Sternschaltung

Nm min 219

960 2

min 60 s kW 22 n

2

M P ₁

N N

N =

⋅ π

⋅

⋅ =

= π ₋

b)

(

)

A 44

I_1N = ⋅ − − ² = − ⇒ −

{ } ^{81,1A 4,05cm}

Nm 219

Nm A 480

0 , M 37

I M I

N Kipp N

1

WKipp =ℜ ⋅ = ⋅ = ⇒

Zeichnerische Lösung: Einzeichnen I_1N, Kreis mit I_WKipp: Schnittpunkt mit neg. imaginärer Achse ergibt Mittelpunkt des Kreises; s. Blatt

Rechnerische Lösung:

Kreismittelpunkt I_M:

{ }I j I { }I j102,1A

I_M =ℑ _1N − ⋅ ²_WKipp −ℜ _1N ² =− ; _{I0 =IM +jIWKipp =−j21A}

{^I } ^j183,2A

j I

I_∞ = _M− ℑ _WKipp =−

c)

Kipp Kipp Kipp

´

s s s s

2 M

M

+

= ⇒ _{s s s 0}

M M

s²−2 ^´Kipp ⋅ ⋅ _Kipp + ²_Kipp =

% 16 , 16 s 03 , 4 s M 1

M M

s M _{N N}

N Kipp

´ N

Kipp

´

Kipp ⋅ = ⋅ =

















 −





 + 

= und

% 83 , 6 s 422 , 0 s M 1

M M

s M _{Kipp Kipp}

2

c Kipp

´ c

Kipp

´

c ⋅ = ⋅ =

















 −







− 

=

⇒ n=931min⁻¹

a)

p = f_N/n_N = 2

Ω

=

⋅ ⋅

= 5,2

K 1 3 I X U

N C N 1

MW 351 I

U 3

S_N = ⋅ _N⋅ _N = kV 39 X I_N⋅ ₁ =

(U I X sin ) (I X cos ) 49 kV

U_PN = _NY + _N⋅ ₁⋅ ϕ_N ² + _N⋅ ₁⋅ ϕ_N ² = (Sternspannung)

b)

A 2138 j U 3 I Q

N ) b

b =

= ⋅

kV 1 , 11 X I_b⋅ ₁ =

{ }I X 26,7 kV U

U_Pb = _NY+ℑ _b ⋅ ₁=

A 109 U I

I U _fN

PN Pb

fb = ⋅ =

c)

Leistung der Turbine erhöhen; Polradspannung erhöhen.

W P 714

P

N mechanisch , N el ,

N =

= η

A 88 , U 3 cos I P

N N N

N

aN =

⋅ ϕ

⋅

= η

Ω

− =

=

+ 14,3

I P R P

R ₂

N N el , N f a

Ω

=

=

+ 35,6

I X Q

X ₂

N Nl f a

b)

(R R ) (X X ) ^{6,0 A}

I U

2 f a 2 f a

N

k =

+ + +

=

Nm 59 , n 1 2 M P

N N

N =

⋅

= π

Nm 8 , 3 I M

M I_{2 N}

N 2 K

K = ⋅ =

c)

A 18 , 2 M I

I M _N

N

=

⋅

=

(^{X X} ) ^I (^{R R} ) ^{I 129V}

U

U_i,N= ²_N− _a+ _f ²⋅ ²_N − _a+ _f ⋅ _N = V

9 , I 12

I n U n U

N N N , i

i = ⋅ ⋅ =

( )

(U R R I) (X X ) I 88,8V U= _i+ _a+ _f ⋅ ²+ _a+ _f ²⋅ ² =