Question

A 3-phase, 50-Hz overhead transmission line 100 km long with 132 kV between lines at the receiving end has the following constants :

resistance/km/phase = 0.15 Ω ; inductance/km/phase = 1.20 mH

capacitance/km/phase = 0.01 mF

Determine, using an approximate method of allowing for capacitance, the voltage, current and p.f. at the sending end when the load at the receiving end is 72 MW at 0.8 p.f. lagging. Draw vector diagram for the circuit assumed.

Answer 1

For a 100-km length of the line,

R = 0.15 × 100 = 15 Ω ; X_L = 314 × 1.2 × 10⁻³ × 100 = 37.7 Ω

XC = 106/314 × 0.01 × 100 = 3187 Ω

Using the nominal T-method, the equivalent circuit is shown in Fig. 41.33 (a)

V_R = 132/√3 = 76.23 kV = 76,230 V. Hence, V_R = 76,230 + j 0

Load current, I_R = 72 × 102/√3 × 132 × 103 × 0.8 = 394 A

∴ I_R = 394 (0.8 − j 0.6) = 315 − j 236 A ; Z_BC = (7.5 + j 18.85) W

Drop/phase over BC = I_RZ_BC = (315 − j 236) (7.5 + j 18.85) = 6802 + j 4180

V₁ = V_R + I_RZ_BC = (76,230 + j0) + ( 6802 + j 4180) = 88,030 + j 4180

I_S = I_C + I_R = (−1.31 + j 26) + (315 − j 236) = (313.7 − j 210) = 377.3 ∠− 33.9°

Drop/phase over AB = I_SZ_{A B} = (313.7 − j 210) (7.5 + j 18.85) = 6320 + j 4345

∴ V_S = V₁ + I_SZ_{A B} = (83,030 + j 4180) + (6320 + j 4345) = 89,350 + j 8525 = 89,750 ∠ 5.4°

Line value of sending-end voltage = √3 × 89,750 × 10⁻³ = 155.7 kV

Phase difference between V_S and I_S = 33.9° + 5.4° = 39.3° with current lagging as shown in Fig. 41.33 (b) cos φ_S = cos 39.3° = 0.774 (lag)