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 = [tex]\sqrt{\frac{z}{y} }[/tex] = [tex]\sqrt{\frac{0.03 i + j 0.35}{j4.4*10^-6 } }[/tex]
= [tex]\sqrt{79837.128< 4.899^o}[/tex] = 282.6 < -2.45°
hence Zc = 282.34 - j 12.08 Ω
B) Calculate gl
gl = [tex]\sqrt{zy} * d[/tex]
d = 500
z = 0.03 i + j 0.35
y = j4.4*10^-6 S/km
gl = [tex]\sqrt{0.03 i + j 0.35* j4.4*10^-6} * 500[/tex]
= [tex]\sqrt{1.5456*10^{-6} < 175.1^0} * 500[/tex]
= 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) = [tex]\frac{e^{gl} + e^{-gl} }{2}[/tex]
sin h (gl) = [tex]\frac{e^{gl}-e^{-gl} }{2}[/tex]
