110 km, 110 kV line: sending end, regulation, efficiency

A 110 km line is a medium line; it is solved here with the nominal-π model (half the shunt admittance at each end).

Line constants

$$\begin{aligned} R &= 0.153\times110 = 16.83\ \Omega \\ X &= 2\pi(50)(1.21\times10^{-3})(110) = 41.81\ \Omega \\ Y &= 2\pi(50)(0.00958\times10^{-6})(110) = 3.311\times10^{-4}\ \text{S} \\ Y/2 &= 1.655\times10^{-4}\ \text{S} \end{aligned}$$

Receiving end

$$\begin{aligned} V_R &= 110\,000/\sqrt3 = 63\,508.5\angle0^\circ\ \text{V} \\ I_R &= \frac{20\times10^6}{3\times63\,508.5\times0.9} = 116.64\angle-25.84^\circ\ \text{A} \end{aligned}$$

Nominal-π steps

$$\begin{aligned} I_{C1} &= j\frac{Y}{2}V_R = j10.51\ \text{A} \\ I_L &= I_R + I_{C1} = 112.45\angle-21.02^\circ\ \text{A} \\ V_S &= V_R + I_L Z = 63\,508.5 + 5068.8\angle47.06^\circ \\ &= 67\,064.3\angle3.17^\circ\ \text{V} \\ I_{C2} &= j\frac{Y}{2}V_S = 11.10\angle93.17^\circ\ \text{A} \\ I_S &= I_L + I_{C2} = 108.38\angle-15.65^\circ\ \text{A} \end{aligned}$$

Results

• Sending-end line voltage $= \sqrt3\times67.064 = 116.16$ kV
• Sending-end current $= 108.38$ A
• Sending-end pf $= \cos(3.17^\circ + 15.65^\circ) = 0.947$ lagging

Regulation (no-load $V_R = V_S/|A|$, $A = 1 + YZ/2 = 0.9931\angle0.16^\circ$):

$$\%VR = \frac{67\,064.3/0.9931 - 63\,508.5}{63\,508.5}\times100 = \frac{67\,531.4 - 63\,508.5}{63\,508.5}\times100 = 6.33\%$$

Efficiency:

$$\begin{aligned} P_{loss} &= 3|I_L|^2R = 3(112.45)^2(16.83) = 0.638\ \text{MW} \\ \eta &= \frac{20}{20.638}\times100 = 96.91\% \end{aligned}$$

Answer: $V_S$ = 116.16 kV, $I_S$ = 108.38 A, pf 0.947 lag, regulation 6.33 %, efficiency 96.91 %. (A nominal-T solution gives almost the same: 116.13 kV, 6.30 %, 96.90 %.)