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150 km line with mid-point shunt capacitor

Short answer

With a shunt capacitor of susceptance 4×10⁻⁶ S at the midpoint, the transmission efficiency is approximately 96.7%.

A 150 km long, 50 Hz, 132 kV, 3-phase overhead line with series impedance of 0.1 + j0.3 Ω/m and shunt admittance of 3 × 10⁻⁶∠90° S/km is feeding a load of 35 MW at 0.9 p.f. lagging. Determine the transmission line efficiency when a shunt capacitor of susceptance 4 × 10⁻⁶∠90° S is connected at the middle of the line.

Assumptions: the series impedance is taken as 0.1+j0.3 Ω0.1 + j0.3\ \Omega/km (Ω/m is a misprint). The load is at 132 kV. The line is modelled as two 75 km nominal-π sections with the capacitor at the junction.

Section constants (75 km each)

Zh=(0.1+j0.3)×75=7.5+j22.5 ΩYh=j3×10−6×75=j2.25×10−4 S,Yh/2=j1.125×10−4 S\begin{aligned} Z_h &= (0.1 + j0.3)\times75 = 7.5 + j22.5\ \Omega \\ Y_h &= j3\times10^{-6}\times75 = j2.25\times10^{-4}\ \text{S},\quad Y_h/2 = j1.125\times10^{-4}\ \text{S} \end{aligned}

At the middle bus, the two half-section admittances and the capacitor add: Ymid=j(2.25×10−4+4×10−6)=j2.29×10−4Y_{mid} = j(2.25\times10^{-4} + 4\times10^{-6}) = j2.29\times10^{-4} S.

Vs    Zh          Vm          Zh      Vr
o---/\/\/---+-----+-----+---/\/\/---o
|           |     |     |           |
Yh/2      Yh/2   Ysh   Yh/2       Yh/2
|           |     |     |           |
o-----------+-----+-----+-----------o

Receiving-end section

VR=132/3=76 210∠0∘ VIR=35×1063×132 000×0.9=170.09∠−25.84∘ AI1=IR+jYh2VR=170.09∠−25.84∘+j8.57=166.54∠−23.19∘ AVm=VR+I1Zh=78 889∠2.15∘ V\begin{aligned} V_R &= 132/\sqrt3 = 76\,210\angle0^\circ\ \text{V} \\ I_R &= \frac{35\times10^6}{\sqrt3\times132\,000\times0.9} = 170.09\angle-25.84^\circ\ \text{A} \\ I_1 &= I_R + j\frac{Y_h}{2}V_R = 170.09\angle-25.84^\circ + j8.57 = 166.54\angle-23.19^\circ\ \text{A} \\ V_m &= V_R + I_1Z_h = 78\,889\angle2.15^\circ\ \text{V} \end{aligned}

Middle bus and sending section

Imid=YmidVm=18.07∠92.15∘ AI2=I1+Imid=159.64∠−17.32∘ AVS=Vm+I2Zh=81 270∠4.25∘ VIS=I2+jYh2VS=156.51∠−14.20∘ A\begin{aligned} I_{mid} &= Y_{mid}V_m = 18.07\angle92.15^\circ\ \text{A} \\ I_2 &= I_1 + I_{mid} = 159.64\angle-17.32^\circ\ \text{A} \\ V_S &= V_m + I_2Z_h = 81\,270\angle4.25^\circ\ \text{V} \\ I_S &= I_2 + j\frac{Y_h}{2}V_S = 156.51\angle-14.20^\circ\ \text{A} \end{aligned}

Sending-end line voltage =140.76= 140.76 kV.

Efficiency

SS=3VSIS∗=36.20+j12.08 MVAη=3536.20×100=96.69%\begin{aligned} S_S &= 3V_SI_S^* = 36.20 + j12.08\ \text{MVA} \\ \eta &= \frac{35}{36.20}\times100 = 96.69\% \end{aligned}

(Check: 3(∣I1∣2+∣I2∣2)(7.5)=1.203(|I_1|^2 + |I_2|^2)(7.5) = 1.20 MW loss.)

A 4×10−64\times10^{-6} S capacitor supplies only about V2B≈0.07V^2B \approx 0.07 MVAr, so it hardly changes the result (without it, η\eta = 96.68 %).

Answer: Transmission efficiency ≈ 96.7 %.

✓ Answer checked by Claude Opus 5.5 (high effort)

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