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

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

Step-by-step 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{(z)/(y) }` =
`\sqrt{(0.03 i + j 0.35)/(j4.4*10^-6 ) }`

=
`√(79837.128< 4.899^o)` = 282.6 < -2.45°

hence Zc = 282.34 - j 12.08 Ω

B) Calculate gl

gl =
`√(zy) * d`

d = 500

z = 0.03 i + j 0.35

y = j4.4*10^-6 S/km

gl =
`√(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) =
`(e^(gl) + e^(-gl) )/(2)`

sin h (gl) =
`(e^(gl)-e^(-gl) )/(2)`