Proteccion de Sistemas de Potencia 2

CURSO 

PROTECCION DE SISTEMAS  ELECTRICOS  DE POTENCIA CAPITULO 2  PROFESOR: ING. BERNARDIN BERNARDINO O ROJAS  VERA

FASORES

FASORES

FASORES

FASORES

    φ φ φ φ φ φ φ φ φ  φ

φ  φ φ φ

 ω ω  φ  ω φ φ ω φ

ω  φ  φ ω

ω   ω

  φ φ φ φ

  

! "  # $

v ( t ) = V m c o s ( ω t+ θ ) v ( t ) = V m s in in (ω t + θ )





Vm ∠ θ°

V m ∠ (θ − 9 0 ) °

 ω ω ω ω

  $ Problema 9.16 Utilizando fasores, calcular

a) 3 cos(20t + 10) 10) – 5 cos(20t-30) cos(20t-30) b) 40 sen(50t) sen(50t) + 30 cos(50t-45) c) 20 sen 400t + 10 cos(400t+60) – 5 sen(400t-20) (a)

3∠ 10° − 5∠-30° = 2.954 + j0.5209 – 4.33 + j2.5 = -1.376 + j3.021 = 3.32∠ 114.49° Therefore, 3 cos(20t + 10°) – 5 cos(20t – 30°) = 3.32 cos(20t + 114.49°)

(b)

4∠ -90° + 3∠-45° = -j40 + 21.21 – j21.21 = 21.21 – j61.21 = 64.78∠ -70.89° Ther herefore, 40 sin( in(50t) + 30 30 cos( os(50t – 45°) = 64.78 cos(50t – 70.89°)

(c)

Using sinα = cos(α − 90°), 20∠ -90° + 10∠60° − 5∠-110° = -j20 + 5 + j8.66 + 1.7101 1.7101 + j4.699 = 6.7101 – j6.641 = 9.44∠-44.7° Ther herefore, 20 sin sin((400t) + 10 10 cos( os(400t + 60°) – 5 sin(400t – 20°)



Element

Time domain

R

frequency domain

    =     

  = 

L C

   =

Z1

Impedance

V= I

  

V = ω    I

 

%

        

V=

%  ω    

I

Admitance

Z=

Y=

Z = ω   

Y=

Z=

%

%   %  

ω  



Y = ω    

 ω    

Z2 Zeq

Zeq

ZN

 = % + & +    

   = .

Z1

% % %

+

% &

+

%     

Z2



ZN

 = % + & +