Question

A 3-phase, 50-Hz transmission line, 100 km long delivers 20 MW at 0.9 p.f. lagging and at 110 kV. The resistance and reactance of the line per phase per km are 0.2 Ω and 0.4 Ω respectively while the capacitive admittance is 2.5 × 10⁻⁶ S per km. Calculate (a) the voltage and current at the sending end and (b) the efficiency of transmission. Use the nominal T-method.

Answer 1

Resistance for 100 km = 0.2 × 100 = 20 Ω

Reactance for 100 km = 0.4 × 100 = 40 Ω

Capacitive admittance for 100 km = 2.5 × 10⁻⁶ × 100 = 2.5 × 10⁻⁴ S

Let us take the receiving-end voltage E_R as reference vector.

E_R = 110/√3 = 63.5 kV ; IR = 20 × 106/√3 × 110 × 103 × 0.9 = 116.6 A

cos φ_R = 0.9 ; sin φ_R = 0.435 (from tables)

With reference to Fig.we have

E_R = (63.5 + j0) kV ; I_R = 116.6(0.9 − j0. 435) = 105 − j50.7 ; Z_BC = (10 + j20)

Voltage drop between points B and C is

V_BC = I_RZ_BC = (105 − j50.7) (10 + j20) = (2,064 + j1,593) V

E₁ = E_R + I_RZ_BC = (63,500 + 2,064 + j1,593) = 65,564 + j1,593

I_C = E₁Y = (65,564 + j1,593) × j2.5 × 10⁻⁴ = (−0.4 + j16.4) A

I_S = I_R + I_C = (105 − j50.7) + (−0.4 + j16.4) = (104.6 − j34.3) = 110.1 ∠18º 9

Drop between points A and B is

V_AB = I_SZ_{A B} = (104.6 − j34.3) (10 + j20) = 1,732 + j1,749

E_S = E₁ + V_{A B} = (65,564 + j1,593) + (1,732 + j1,749) = 67,296 + j3,342 = 67,380 ∠2º 51 

(a) Sending-end voltage (line value) is 67,380 × √3 = 116,700 V = 116.7 kV Sending-end current = 110.1 A

Note. Phase difference between E_S and I_S is (−18º9 − 2º51) = − 21º with current lagging. Hence, p.f. at sending-end is cos φ _S = cos 21º = 0.934 (lag).

(b) Copper loss for three phases between points B and C is

= 3 × 116.6² × 10 = 0.408 MW

Copper loss for three phases between points A and B is = 3 × 110.1² × 10 = 0.363 MW

Total Cu loss for 100 km of line length = 0.408 + 0.363 = 0.771 MW 

Transmission η = 20 × 100/20.771 = 96.27 %

Note. Line losses could also be computed by finding the line input which = √3 E_S.I_S cos φ_S.