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Mid-line reactor for Vr = Vs at no load

Short answer

For a lossless line, the required reactor at midpoint is XL = (Zc/2) cot(βl/2). With Zc=316.23 Ω and βl=28.46°, XL = 623.5 Ω per phase (≈257 MVAr at 400 kV).

A 400 kV, 500 km, 50 Hz HV transmission line has r = 0.031 Ω/km, l = 1 mH/km and C = 10 nF/km. Calculate the value of XL of a reactor required at the middle of the line as shown in figure below so that the receiving end voltage at no load could be controlled as Vr = Vs. [Figure: line from Vs to Vr with a single shunt reactor jXL connected from the midpoint of the line to ground]

Treat the line as lossless for the reactor calculation (as usual; rr has a negligible effect, checked below). Split it into two equal halves, each with electrical length θ=βl/2\theta = \beta l/2, with the reactor jXLjX_L at the midpoint.

Line constants:

β=ωLC=314.161×10−3×10×10−9=9.9346×10−4 rad/kmβl=9.9346×10−4×500=0.4967 rad=28.46∘θ=βl/2=0.2484 rad=14.23∘Zc=L/C=10−3/10−8=316.23 Ω\begin{aligned} \beta &= \omega\sqrt{LC} = 314.16\sqrt{1\times10^{-3} \times 10\times10^{-9}} = 9.9346\times10^{-4}\ \text{rad/km} \\ \beta l &= 9.9346\times10^{-4} \times 500 = 0.4967\ \text{rad} = 28.46^\circ \\ \theta &= \beta l/2 = 0.2484\ \text{rad} = 14.23^\circ \\ Z_c &= \sqrt{L/C} = \sqrt{10^{-3}/10^{-8}} = 316.23\ \Omega \end{aligned}

Without the reactor: Vr/Vs=1/cos⁡28.46∘=1.137V_r/V_s = 1/\cos 28.46^\circ = 1.137 pu.

Derivation

Second half (midpoint M to the open receiving end, Ir=0I_r = 0):

Vm=Vrcos⁡θ,Im2=jVrZcsin⁡θV_m = V_r\cos\theta, \qquad I_{m2} = j\frac{V_r}{Z_c}\sin\theta

At M: the current leaving the first half feeds the reactor and the second half:

Im1=VmjXL+Im2=−jVrcos⁡θXL+jVrsin⁡θZcI_{m1} = \frac{V_m}{jX_L} + I_{m2} = -j\frac{V_r\cos\theta}{X_L} + j\frac{V_r\sin\theta}{Z_c}

First half:

Vs=Vmcos⁡θ+jZcsin⁡θ Im1=Vr[cos⁡2θ−sin⁡2θ+ZcXLsin⁡θcos⁡θ]=Vr[cos⁡βl+Zc2XLsin⁡βl]\begin{aligned} V_s &= V_m\cos\theta + jZ_c\sin\theta\,I_{m1} \\ &= V_r\left[\cos^2\theta - \sin^2\theta + \frac{Z_c}{X_L}\sin\theta\cos\theta\right] \\ &= V_r\left[\cos\beta l + \frac{Z_c}{2X_L}\sin\beta l\right] \end{aligned}

Condition Vr=VsV_r = V_s

cos⁡βl+Zc2XLsin⁡βl=1  ⇒  XL=Zcsin⁡βl2(1−cos⁡βl)=Zc2cot⁡βl2\cos\beta l + \frac{Z_c}{2X_L}\sin\beta l = 1 \;\Rightarrow\; X_L = \frac{Z_c\sin\beta l}{2(1 - \cos\beta l)} = \frac{Z_c}{2}\cot\frac{\beta l}{2} XL=316.232×cot⁡(14.23∘)=158.11×3.943=623.5 Ω\begin{aligned} X_L &= \frac{316.23}{2} \times \cot(14.23^\circ) \\ &= 158.11 \times 3.943 = 623.5\ \Omega \end{aligned}

Check: cos⁡28.46∘+316.232×623.5sin⁡28.46∘=0.8792+0.2536×0.4766=1.000\cos 28.46^\circ + \dfrac{316.23}{2 \times 623.5}\sin 28.46^\circ = 0.8792 + 0.2536 \times 0.4766 = 1.000 ✓. Including r=0.031 Ωr = 0.031\ \Omega/km, the same reactor gives ∣Vs/Vr∣=1.00002|V_s/V_r| = 1.00002, so neglecting rr is justified.

Reactor rating (3-phase, 400 kV): Q=4002/623.5=256.6Q = 400^2/623.5 = 256.6 MVAr.

 Vs o------ 250 km ------+------ 250 km ------o Vr
                         |
                        jXL = 623.5 ohm
                         |
 ------------------------+--------------------- gnd

Answer: XL≈X_L \approx 623.5 Ω per phase (about 257 MVAr) at the midpoint.

✓ Answer checked by Claude Opus 5.5 (high effort)

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