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12 km 33 kV line: losses, zero-regulation capacitor

Short answer

With corrected R=3 Ω, X=4.90 Ω, line current is 1.173 kA. Active loss is 12.39 MW, reactive loss 20.24 MVAr. For zero regulation, a capacitor of about 22.3 µF per phase (8.1 MVAr) is required.

A 12 km long 3-phase line is supplying a load at 33 kV. The line has a resistance of 2.5 × 10⁻³ Ohms/m and inductance of 1.3 × 10⁻⁶ H/m. If the sending end voltage is maintained at 34 kV∠20°, compute the active and reactive power losses in the line. Compute the capacitance per phase that needs to be connected at the end of the line to make the voltage regulation zero.

Assumption: the printed 2.5×10−3 Ω2.5\times10^{-3}\ \Omega/m gives RR = 30 Ω for 12 km (2.5 Ω/km), unrealistic and larger than XX; with it zero regulation is impossible by any shunt capacitor. It is taken as 2.5×10−4 Ω2.5\times10^{-4}\ \Omega/m (likely misprint).

Line constants

R=2.5×10−4×12 000=3 ΩX=2π(50)(1.3×10−6×12 000)=4.90 ΩZ=5.746∠58.53∘ Ω\begin{aligned} R &= 2.5\times10^{-4}\times12\,000 = 3\ \Omega \\ X &= 2\pi(50)(1.3\times10^{-6}\times12\,000) = 4.90\ \Omega \\ Z &= 5.746\angle58.53^\circ\ \Omega \end{aligned}

Line current

Phase values: VS=19.630∠20∘V_S = 19.630\angle20^\circ kV, VR=19.053∠0∘V_R = 19.053\angle0^\circ kV.

VS−VR=6.741∠95.16∘ kVI=6.741∠95.16∘5.746∠58.53∘=1.1732∠36.63∘ kA\begin{aligned} V_S - V_R &= 6.741\angle95.16^\circ\ \text{kV} \\ I &= \frac{6.741\angle95.16^\circ}{5.746\angle58.53^\circ} = 1.1732\angle36.63^\circ\ \text{kA} \end{aligned}

Losses

Ploss=3I2R=3(1173.2)2(3)=12.39 MWQloss=3I2X=3(1173.2)2(4.90)=20.24 MVAr\begin{aligned} P_{loss} &= 3I^2R = 3(1173.2)^2(3) = 12.39\ \text{MW} \\ Q_{loss} &= 3I^2X = 3(1173.2)^2(4.90) = 20.24\ \text{MVAr} \end{aligned}

(Check: SS=66.20−j19.78S_S = 66.20 - j19.78 MVA, SR=53.81−j40.01S_R = 53.81 - j40.01 MVA; differences 12.39 MW and 20.24 MVAr.)

Capacitance for zero regulation

Zero regulation means ∣VR∣=∣VS∣=34|V_R| = |V_S| = 34 kV at full load. Keep the load PP = 53.81 MW and find the net QQ:

344=(342+3P+4.90Q)2+(4.90P−3Q)2⇒Q=−48.12 MVAr34^4 = (34^2 + 3P + 4.90Q)^2 + (4.90P - 3Q)^2 \Rightarrow Q = -48.12\ \text{MVAr}

The load already supplies 40.01 MVAr (leading), so the capacitor must supply

QC=48.12−40.01=8.11 MVArQ_C = 48.12 - 40.01 = 8.11\ \text{MVAr} C=QCωVL2=8.11×106314.16×34 0002=22.34 μF per phase (star)C = \frac{Q_C}{\omega V_L^2} = \frac{8.11\times10^6}{314.16\times34\,000^2} = 22.34\ \mu\text{F per phase (star)}

(With the printed RR = 30 Ω the losses would be 4.43 MW and 0.72 MVAr.)

Answer: Losses ≈ 12.39 MW and 20.24 MVAr; C ≈ 22.3 µF per phase (≈ 8.1 MVAr) for zero regulation.

✓ Answer checked by Claude Opus 5.5 (high effort)

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