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Reducing Ferranti effect mathematically

Short answer

Ferranti effect is reduced by shunt reactors, which absorb charging current. With a reactor XL at the receiving end, Vr/Vs=1/[cos βl+(Zc/XL) sin βl]; selecting XL=Zc cot(βl/2) makes Vr=Vs.

How the Ferranti effect can be reduced in over voltage issues? Explain mathematically.

The Ferranti effect is the rise of receiving-end voltage above the sending-end voltage on a long line at no load or light load. The line's charging current flows through its series inductance and raises the voltage along the line. It is a main cause of temporary overvoltage.

Mathematical explanation

For a lossless line of length ll, phase constant β=ωLC\beta = \omega\sqrt{LC} and surge impedance Zc=L/CZ_c = \sqrt{L/C}:

Vs=Vrcos⁡βl+jZcIrsin⁡βlV_s = V_r\cos\beta l + jZ_c I_r\sin\beta l

At no load Ir=0I_r = 0:

VrVs=1cos⁡βl\frac{V_r}{V_s} = \frac{1}{\cos\beta l}

At 50 Hz, β≈6∘\beta \approx 6^\circ per 100 km. For example, a 400 km line with L=1L = 1 mH/km and C=10C = 10 nF/km gives βl=0.397\beta l = 0.397 rad and Vr=1.085 VsV_r = 1.085\,V_s. A 1500 km line (βl=90∘\beta l = 90^\circ) gives an infinite voltage.

Reduction by shunt reactor compensation

Connect a shunt reactor jXLjX_L at the receiving end. At no load, the reactor current is

Ir=VrjXLI_r = \frac{V_r}{jX_L}

Substituting:

Vs=Vrcos⁡βl+jZcsin⁡βl⋅VrjXLVrVs=1cos⁡βl+ZcXLsin⁡βl\begin{aligned} V_s &= V_r\cos\beta l + jZ_c\sin\beta l \cdot \frac{V_r}{jX_L} \\ \frac{V_r}{V_s} &= \frac{1}{\cos\beta l + \dfrac{Z_c}{X_L}\sin\beta l} \end{aligned}

The extra positive term ZcXLsin⁡βl\dfrac{Z_c}{X_L}\sin\beta l in the denominator reduces VrV_r. For Vr=VsV_r = V_s:

cos⁡βl+ZcXLsin⁡βl=1  ⇒  XL=Zcsin⁡βl1−cos⁡βl=Zccot⁡βl2\cos\beta l + \frac{Z_c}{X_L}\sin\beta l = 1 \;\Rightarrow\; X_L = \frac{Z_c\sin\beta l}{1 - \cos\beta l} = Z_c\cot\frac{\beta l}{2}

Physically, the reactor draws a lagging current that cancels part of the line's leading charging current. A smaller net current flows through the series inductance, so the voltage rise is smaller.

Other methods

  1. Shunt reactors at the ends and at intermediate stations; switched or controlled reactors for varying load.
  2. Static VAR compensators (SVC) and STATCOMs, which absorb reactive power under light load.
  3. Series capacitors are not used for this. They reduce the effective electrical length for power transfer but do not cure the no-load rise.
  4. Sectionalising long lines with intermediate substations, so that βl\beta l of each section is small.
  5. Generator under-excitation (absorbing VArs), within stability limits.
  6. Proper switching sequence: energise the line from the stronger end and connect the reactors before or with the line.

✓ Answer checked by Claude Opus 5.5 (high effort)

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