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Voltage 150 km along a 400 km line

Short answer

The point 150 km from the sending end is 250 km from the receiving end. Using the long-line equation, the voltage at that point is 428.6 kV line-to-line (247.4∠10.97° kV per phase relative to V_R).

A 50 Hz, 400 km long, 400 kV, 3-phase overhead transmission line delivering a power of 400 MW at 0.95 power factor lagging has a series resistance of 3.2 × 10⁻⁵ Ohm/m, inductance of 1.066 × 10⁻⁶ H/m and shunt capacitance of 1.09 × 10⁻¹¹ F/m per phase. Determine the voltage at 150 km from the sending end toward the receiving end.

Given: 50 Hz, l=400l = 400 km, 400 kV, 400 MW at 0.95 p.f. lagging; r=3.2×10−5 Ωr = 3.2\times10^{-5}\ \Omega/m =0.032 Ω= 0.032\ \Omega/km, L=1.066×10−6L = 1.066\times10^{-6} H/m =1.066= 1.066 mH/km, C=1.09×10−11C = 1.09\times10^{-11} F/m =0.0109 μ= 0.0109\ \muF/km. Receiving voltage 400 kV.

The point 150 km from the sending end is x=400−150=250x = 400 - 150 = 250 km from the receiving end.

Step 1: Line constants

z=0.032+j2π×50×1.066×10−3=0.032+j0.3349 Ω/kmy=j2π×50×1.09×10−8=j3.424×10−6 S/kmγ=zy=1.073×10−3∠87.27∘=5.11×10−5+j1.0721×10−3 /kmZc=z/y=313.4∠−2.73∘ Ω\begin{aligned} z &= 0.032 + j2\pi\times50\times1.066\times10^{-3} = 0.032 + j0.3349\ \Omega/\text{km} \\ y &= j2\pi\times50\times1.09\times10^{-8} = j3.424\times10^{-6}\ \text{S/km} \\ \gamma &= \sqrt{zy} = 1.073\times10^{-3}\angle 87.27^\circ = 5.11\times10^{-5} + j1.0721\times10^{-3}\ /\text{km} \\ Z_c &= \sqrt{z/y} = 313.4\angle -2.73^\circ\ \Omega \end{aligned}

Step 2: Receiving-end values

VR=4003=230.94∠0∘ kVIR=400×1063×400×103×0.95=607.7∠−18.19∘=577.35−j189.77 A\begin{aligned} V_R &= \frac{400}{\sqrt3} = 230.94\angle 0^\circ\ \text{kV} \\ I_R &= \frac{400\times10^6}{\sqrt3\times400\times10^3\times0.95} = 607.7\angle -18.19^\circ = 577.35 - j189.77\ \text{A} \end{aligned}

Step 3: Hyperbolic functions at x = 250 km

γx=0.01278+j0.26803cosh⁡γx=0.96437+j0.00338=0.9644∠0.20∘Zcsinh⁡γx=83.10∠84.61∘=7.81+j82.74 Ω\begin{aligned} \gamma x &= 0.01278 + j0.26803 \\ \cosh\gamma x &= 0.96437 + j0.00338 = 0.9644\angle 0.20^\circ \\ Z_c\sinh\gamma x &= 83.10\angle 84.61^\circ = 7.81 + j82.74\ \Omega \end{aligned}

Step 4: Voltage at the point

Vx=VRcosh⁡γx+ZcIRsinh⁡γx=230.94×0.9644∠0.20∘+83.10∠84.61∘×0.6077∠−18.19∘ kV=247.4∠10.97∘ kV/phaseVx(L)=3×247.4=428.58 kV\begin{aligned} V_x &= V_R\cosh\gamma x + Z_cI_R\sinh\gamma x \\ &= 230.94\times0.9644\angle 0.20^\circ + 83.10\angle 84.61^\circ\times0.6077\angle -18.19^\circ\ \text{kV} \\ &= 247.4\angle 10.97^\circ\ \text{kV/phase} \\ V_{x(L)} &= \sqrt3\times247.4 = 428.58\ \text{kV} \end{aligned}

For comparison, at the sending end (x=400x = 400 km) the same formula gives VS=438.13V_S = 438.13 kV (line), so the voltage falls steadily from 438.1 kV to 400 kV along the line.

Answer: Voltage 150 km from the sending end ≈428.6\approx 428.6 kV (line-to-line), i.e. 247.4∠10.97∘247.4\angle 10.97^\circ kV per phase relative to VRV_R.

✓ Answer checked by Claude Opus 5.5 (high effort)

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