Question

A 500-km, 500-kV, 60-Hz, uncompensated three-phase line has a positivesequence series impedance. z = 5 0.03 1 + j 0.35 V/km and a positive sequence shunt admittance y = j4.4*10^26 S/km. Calculate:(a) Zc, (b) (gl), (c) the exact ABCD parameters for this line.

Answer 1

Answer:

A) 282.34 - j 12.08 Ω

B) 0.0266 + j 0.621 / unit

C)

A = 0.812 < 1.09° per unit

B =  164.6 < 85.42°Ω  

C =  2.061 * 10^-3 < 90.32° s

D =  0.812 < 1.09° per unit

Explanation:

Given data :

Z ( impedance ) = 0.03 i  + j 0.35 Ω/km

positive sequence shunt admittance ( Y ) = j4.4*10^-6 S/km

A) calculate Zc

Zc = $\sqrt{\frac{z}{y} }$  =  $\sqrt{\frac{0.03 i + j 0.35}{j4.4*10^-6 } }$    

    = $\sqrt{79837.128< 4.899^o}$   =  282.6 < -2.45°

hence Zc = 282.34 - j 12.08 Ω

B) Calculate  gl

gl = $\sqrt{zy} * d$  

 d = 500

 z = 0.03 i  + j 0.35

 y = j4.4*10^-6 S/km

gl =  $\sqrt{0.03 i + j 0.35* j4.4*10^-6} * 500$

   = $\sqrt{1.5456*10^{-6} < 175.1^0} * 500$

   = 0.622 < 87.55 °

gl = 0.0266 + j 0.621 / unit

C) exact ABCD parameters for this line

A = cos h (gl) . per unit  =  0.812 < 1.09° per unit ( as calculated )

B = Zc sin h (gl) Ω  = 164.6 < 85.42°Ω  ( as calculated )

C = 1/Zc  sin h (gl) s  =  2.061 * 10^-3 < 90.32° s ( as calculated )

D = cos h (gl) . per unit = 0.812 < 1.09° per unit ( as calculated )

where :  cos h (gl)  = $\frac{e^{gl} + e^{-gl} }{2}$

             sin h (gl) = $\frac{e^{gl}-e^{-gl} }{2}$