4.(a)(i) The configuration of a 400 KV 3 phase line is shown in figure. The radius of each sub-conductor is 2 cm. Calculate the charging mega volt-amperes if line is operating at 50 Hz and has a length of 300 km. (10 marks)

4.(a)(ii) Calculate the m

Question

4.(a)(i) The configuration of a 400 KV 3 phase line is shown in figure. The radius of each sub-conductor is 2 cm. Calculate the charging mega volt-amperes if line is operating at 50 Hz and has a length of 300 km. (10 marks)

4.(a)(ii) Calculate the most economical overall diameter of insulation of a cable to be operated at 400 KV, 3 phase power system if maximum stress is limited to 100 KV/cm. (10 marks)

4.(b) Derive the conditions of balance of an Anderson's bridge and also draw the phasor diagram of the bridge under balanced condition. Determine the unknown quantities in terms of known parameters and comment on easy convergence of balance of the bridge. (20 marks)

4.(c) The approximate magnitude plot, obtained experimentally, of a nonminimum phase system is shown in figure. Calculate the phase in degrees at w = 3 rad/sec. (10 marks)

UPSC Answer Check · Accepted Answer

4.(a)(i) For the two-conductor bundle, the equivalent capacitance radius is
Ds = √(r d) = √(0.02 m × 0.5 m) = 0.1 m.
The phase-centre distances are Dab = 10 m, Dbc = 10 m, Dca = 20 m, so
GMD = ∛(10 × 10 × 20) = 12.599 m.
For a long 3-phase line, neglecting earth, the capacitance per phase to neutral is
C = 2π ε₀ / ln(GMD/Ds) per metre.
This formula is valid for a long line with earth neglected and for RMS voltage; the bundle spacing is small compared with phase spacing.
Here ln(12.599/0.1) = ln(125.992) = 4.836, so
C = 2π(8.854×10⁻¹² F/m)/4.836 = 1.150×10⁻¹¹ F/m.
For 300 km, Cph = 1.150×10⁻¹¹ × 3.00×10⁵ = 3.45×10⁻⁶ F.
At 400 kV line-to-line and 50 Hz, ω = 100π rad/s, and total charging reactive power is
Q = 3Vph²ωCph = VLL²ωCph.
Thus Q = (4.00×10⁵ V)² × 100π × 3.45×10⁻⁶ F = 1.73×10⁸ var = 173 MVAR.

4.(a)(ii) For a single-core cable, radial electric stress at radius x is
E(x) = V/(x ln(b/a)),
where a is conductor radius, b is outer insulation radius, and V is phase voltage. The maximum stress is at the conductor:
g = V/(a ln(b/a)).
Using the full limit g, b = a exp(V/(g a)). Let y = V/(g a); then b = (V/g)e^y/y.
Minimizing b gives db/dy = 0, so y = 1. The second derivative is positive at y = 1, so this is a minimum. Hence a = V/g, b = eV/g, and the most economical overall insulation diameter is
D = 2b = 2eV/g.
For 3-phase 400 kV, V = 400/√3 kV; g = 100 kV/cm. Therefore
D = 2e(400/√3)/100 cm = (8e/√3) cm = 12.54 cm.
D = 12.5 cm.

4.(b) Let the bridge source be A-C and detector B-D. The arms are
Z₁ = R + jωL (unknown), Z₂ = R₂, Z₃ = 1/(jωC₁), Z₄ = R₃ + 1/(jωC₂).
All elements are linear, so the bridge balance theorem applies. With A at voltage V and C at 0,
V_B = V Z₂/(Z₁ + Z₂), V_D = V Z₃/(Z₃ + Z₄).
Balance requires V_B = V_D, so
Z₂(Z₃ + Z₄) = Z₃(Z₁ + Z₂), hence Z₁Z₃ = Z₂Z₄.
Substituting,
(R + jωL)/(jωC₁) = R₂(R₃ + 1/(jωC₂)).
Multiplying by jωC₁,
R + jωL = R₂C₁/C₂ + jωR₂R₃C₁.
Equating real and imaginary parts gives
R = R₂C₁/C₂ and L = R₂R₃C₁.
The known parameters are R₂, R₃, C₁ and C₂.

Phasor diagram under balance: take I₂, the current in the R₃-C₂-C₁ branch, as horizontal reference. Since VBC = VDC, the drop I₁R₂ is vertical downward; hence I₁ is 90° behind I₂. Draw I₂R₃ to the right and I₂/(ωC₂) downward to form VAD; draw ωL I₁ to the right and I₁R downward to form VAB. These coincide. Add I₁R₂, or equivalently I₂/(ωC₁), downward to get the common VAC. The detector is between B and D; because VAB and VAD are equal phasors from the same node A, B and D are at the same potential. The balance equations are independent of ω, and R₃ adjusts L while C₂ adjusts R independently; hence convergence is easy, especially for low inductance.

4.(c) The asymptote has a break at 2 rad/s and a break at 20 rad/s; the downward slope between them is -20 dB/dec, so it corresponds to a pole at 2 and a zero at 20. Nonminimum phase means the zero is right-half-plane, so take
G(s) = (1 - s/20)/(1 + s/2).
The gain is chosen so that the low-frequency magnitude is 0 dB, as shown.
The phase is
φ = -tan⁻¹(ω/2) - tan⁻¹(ω/20).
At ω = 3 rad/s,
φ = -tan⁻¹(3/2) - tan⁻¹(3/20) = -56.31° - 8.53° = -64.8°.

4.(a)(i) The configuration of a 400 KV 3 phase line is shown in figure. The radius of each sub-conductor is 2 cm. Calculate the…

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