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UPF power received with 275 kV both ends

Short answer

At unity power factor with both ends at 275 kV, received power is about 117.6 MW. Load angle is approximately 22.0°, found by setting QR=0 in the receiving-end power equations.

A 275 kV transmission line has the following line constants: A = 0.85∠5°; B = 200∠75°. Determine the power at unity power factor that can be received if the voltage profile at each end is to be maintained at 275 kV.

Use the receiving-end power-circle equations with ∣VS∣=∣VR∣=275|V_S| = |V_R| = 275 kV, ∣A∣=0.85|A| = 0.85, α=5∘\alpha = 5^\circ, ∣B∣=200 Ω|B| = 200\ \Omega, β=75∘\beta = 75^\circ (line kV give three-phase MW/MVAr):

PR=∣VS∣∣VR∣∣B∣cos⁡(β−δ)−∣A∣∣VR∣2∣B∣cos⁡(β−α)QR=∣VS∣∣VR∣∣B∣sin⁡(β−δ)−∣A∣∣VR∣2∣B∣sin⁡(β−α)\begin{aligned} P_R &= \frac{|V_S||V_R|}{|B|}\cos(\beta-\delta) - \frac{|A||V_R|^2}{|B|}\cos(\beta-\alpha) \\ Q_R &= \frac{|V_S||V_R|}{|B|}\sin(\beta-\delta) - \frac{|A||V_R|^2}{|B|}\sin(\beta-\alpha) \end{aligned}

Constants

275×275200=378.13,0.85×2752200=321.41,β−α=70∘\frac{275\times275}{200} = 378.13,\qquad \frac{0.85\times275^2}{200} = 321.41,\qquad \beta-\alpha = 70^\circ

Unity pf condition: QR=0Q_R = 0

378.13sin⁡(75∘−δ)=321.41sin⁡70∘sin⁡(75∘−δ)=0.85×0.9397=0.798775∘−δ=53.01∘⇒δ=21.99∘\begin{aligned} 378.13\sin(75^\circ-\delta) &= 321.41\sin70^\circ \\ \sin(75^\circ-\delta) &= 0.85\times0.9397 = 0.7987 \\ 75^\circ-\delta &= 53.01^\circ \Rightarrow \delta = 21.99^\circ \end{aligned}

Power received

PR=378.13cos⁡53.01∘−321.41cos⁡70∘=227.51−109.93=117.58 MW\begin{aligned} P_R &= 378.13\cos53.01^\circ - 321.41\cos70^\circ \\ &= 227.51 - 109.93 = 117.58\ \text{MW} \end{aligned}

Answer: About 117.6 MW can be received at unity pf, with load angle δ\delta ≈ 22.0°.

✓ Answer checked by Claude Opus 5.5 (high effort)

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