Question

A 3-phase transmission line, 100 km long has following constants: resistance per km per phase = 0.28 Ω; inductive reactance per km per phase = 0.63 Ω. Capacitive susceptance per km per phase = 4 × 10⁻⁶ siemens. If the load at the receiving end is 75 MVA at 0.8 p.f. lagging with 132 kV between lines calculate sending-end voltage, current and p.f. Use nominal-π-method.

Answer 1

For a line of length 100 km,

resistance/phase = 0.28 × 100 = 28 Ω ;

inductive reactance/phase = 0.63 × 100 = 63 Ω

Capacitive susceptance/phase = 4 × 10⁻⁶ × 100 = 4 × 10⁻⁴ S

Capacitive susceptance at each end = 2 × 10⁻⁴ S

V_R = 132 × 10³ / √3 = 76,230 V ; V_R = (76,230 + j 0) ; I_R = 75 × 10⁶ / √3 × 132 × 10³ × 0.8 = 410

I_R = 410 (0.8 − j 0.6) = 328 − j 246

Y_C1 = j 2 × 10⁻⁴ S ; I_C1 = V_R . Y_C1

= 76,230 × j 2 × 10⁻⁴ = j 15.25 A

I_L = I_R + I_C1 = 328 − j 231 A

Drop per conductor = I_L . Z_L = (328 − j 231) (28 + j 63) = 23,737 + j 14,196

V_S = V_R + I_L Z_L = 76,230 + 23,737 + j 14,196 = 99,967 + j 14,196 = 100,970 ∠8.1º

Line value of sending-end voltage = 100,970 × 10⁻³ × √3 = 174.9 kW

I_C = Y_S Y_C2 = 100,970 ∠8.1º × 2 × 10⁻⁴ ∠90º = 20.19 ∠98.1º = −2.84 + j 20

I_S = I_C ² + I_L = (−2.84 + j 20) + (328 − j 231) = 325.2 − j 211 = 388 ∠−33º 

Angle between V_S and I_S = 8.1º + 33º = 41.4º ; cos φ = 0.877 (lag)