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Maximum power over 20 + j30 Ω line

Short answer

Maximum power occurs when load angle equals impedance angle (56.31°). Using P_R,max = (|V_S||V_R|/|Z|) − (|V_R|²R/|Z|²) gives 13.58 MW. Efficiency is about 49.0% with receiving-end power factor 0.517 leading.

A 3-phase overhead line has a series impedance of 20 + j30 Ohm per phase. Find the maximum power that can be delivered at the receiving end keeping the receiving end voltage 31.2 kV while the sending end voltage is 33 kV. What will be the efficiency of the line under this condition?

Short line: Z=20+j30=36.06∠56.31∘ ΩZ = 20 + j30 = 36.06\angle56.31^\circ\ \Omega, ∣VS∣|V_S| = 33 kV, ∣VR∣|V_R| = 31.2 kV (line kV give three-phase MW/MVAr).

Maximum power

Maximum PRP_R occurs when δ=θ=56.31∘\delta = \theta = 56.31^\circ:

PR,max=∣VS∣∣VR∣∣Z∣−∣VR∣2R∣Z∣2=33×31.236.06−31.22×201300=28.56−14.98=13.58 MW\begin{aligned} P_{R,max} &= \frac{|V_S||V_R|}{|Z|} - \frac{|V_R|^2R}{|Z|^2} \\ &= \frac{33\times31.2}{36.06} - \frac{31.2^2\times20}{1300} \\ &= 28.56 - 14.98 = 13.58\ \text{MW} \end{aligned}

Reactive power at this condition

QR=−∣VR∣2X∣Z∣2=−31.22×301300=−22.46 MVArQ_R = -\frac{|V_R|^2X}{|Z|^2} = -\frac{31.2^2\times30}{1300} = -22.46\ \text{MVAr}

So the receiving end must supply 22.46 MVAr (pf =0.517= 0.517 leading).

Efficiency

I=13.582+22.4623×31.2=0.4857 kAPloss=3I2R=3(485.7)2(20)=14.16 MWη=13.5813.58+14.16×100=48.96%\begin{aligned} I &= \frac{\sqrt{13.58^2 + 22.46^2}}{\sqrt3\times31.2} = 0.4857\ \text{kA} \\ P_{loss} &= 3I^2R = 3(485.7)^2(20) = 14.16\ \text{MW} \\ \eta &= \frac{13.58}{13.58 + 14.16}\times100 = 48.96\% \end{aligned}

(Check: PS=∣VS∣2R∣Z∣2−∣VS∣∣VR∣∣Z∣cos⁡(θ+δ)=16.75−28.56cos⁡112.62∘=27.74P_S = \frac{|V_S|^2R}{|Z|^2} - \frac{|V_S||V_R|}{|Z|}\cos(\theta+\delta) = 16.75 - 28.56\cos112.62^\circ = 27.74 MW.)

Answer: Maximum power ≈ 13.58 MW; efficiency ≈ 49.0 % (receiving pf 0.517 leading).

✓ Answer checked by Claude Opus 5.5 (high effort)

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