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12 km line at 32.5 kV: regulation and capacitor

Short answer

Line impedance is 3 + j4.90 Ω. Current from phasor difference yields sending-end power 67.3 MW, receiving-end 54.9 MW. Voltage regulation is 4.92%. To halve regulation, a shunt capacitor of about 17.4 µF per phase (6.06 MVAr) is needed.

A 12 km long, 50 Hz, 3-phase line is supplying a load at 32.5 kV∠-20° w.r.t. VS. The line has a resistance of 2.5 × 10⁻³ Ohms/m and inductance of 1.3 × 10⁻⁶ H/m. The sending end voltage is maintained at 34.1 kV∠0°. Calculate the active power flow through the line and voltage regulation. Compute the capacitance per phase to be connected at the receiving end to reduce the voltage regulation by 50%.

Assumption: the printed resistance 2.5×10−3 Ω2.5\times10^{-3}\ \Omega/m gives RR = 30 Ω (2.5 Ω/km), far above any overhead conductor; with it the capacitor part has no solution. It is taken as 2.5×10−4 Ω2.5\times10^{-4}\ \Omega/m (0.25 Ω/km), a likely misprint.

Line constants

R=2.5×10−4×12 000=3 ΩX=2π(50)(1.3×10−6×12 000)=4.90 ΩZ=3+j4.90=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 &= 3 + j4.90 = 5.746\angle58.53^\circ\ \Omega \end{aligned}

Current and power flow

Phase values: VS=19.688∠0∘V_S = 19.688\angle0^\circ kV, VR=18.764∠−20∘V_R = 18.764\angle-20^\circ kV.

I=VS−VRZ=6.739∠72.24∘5.746∠58.53∘=1.1727∠13.71∘ kASS=3VSI∗=67.29−j16.42 MVASR=3VRI∗=54.91−j36.64 MVA\begin{aligned} I &= \frac{V_S - V_R}{Z} = \frac{6.739\angle72.24^\circ}{5.746\angle58.53^\circ} = 1.1727\angle13.71^\circ\ \text{kA} \\ S_S &= 3V_SI^* = 67.29 - j16.42\ \text{MVA} \\ S_R &= 3V_RI^* = 54.91 - j36.64\ \text{MVA} \end{aligned}

Active power sent = 67.29 MW, received = 54.91 MW (loss 12.38 MW). QRQ_R is negative, so the load is leading (pf 0.832 lead).

Voltage regulation

For a short line, the no-load receiving voltage equals VSV_S:

%VR=34.1−32.532.5×100=4.92%\%VR = \frac{34.1 - 32.5}{32.5}\times100 = 4.92\%

Capacitor to halve the regulation

New regulation =2.46%= 2.46\%, so the receiving voltage must rise to

VR′=34.11.0246=33.28 kVV_R' = \frac{34.1}{1.0246} = 33.28\ \text{kV}

With PP = 54.91 MW fixed, find QQ from ∣VS∣2∣VR′∣2=(VR′2+RP+XQ)2+(XP−RQ)2|V_S|^2|V_R'|^2 = (V_R'^2 + RP + XQ)^2 + (XP - RQ)^2:

Q=−42.70 MVArQ = -42.70\ \text{MVAr}

Extra leading VAr needed:

QC=42.70−36.64=6.06 MVArQ_C = 42.70 - 36.64 = 6.06\ \text{MVAr} C=QCωVR′2=6.06×106314.16×(33 280)2=17.41 μF per phase (star)C = \frac{Q_C}{\omega V_R'^2} = \frac{6.06\times10^6}{314.16\times(33\,280)^2} = 17.41\ \mu\text{F per phase (star)}

(With the printed RR = 30 Ω, the power sent would be 5.95 MW and received 1.53 MW, and no shunt capacitor could raise VRV_R to 33.28 kV.)

Answer: Power sent ≈ 67.3 MW (54.9 MW received); regulation 4.92 %; C ≈ 17.4 µF per phase (≈ 6.06 MVAr) to halve the regulation.

✓ Answer checked by Claude Opus 5.5 (high effort)

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