(a) A three-phase, 50 Hz transmission line is 400 km long. The voltage at the sending end is 220 kV. The line parameters are r =…

(a) The 400 km line at no load

Per km, z = 0.125 + j0.4 = 0.419∠72.6° Ω and y = j2.8 × 10⁻⁶ S. Then

γ = √(zy) = √(1.173 × 10⁻⁶ ∠162.6°) = 1.083 × 10⁻³ ∠81.3° = (1.634 × 10⁻⁴ + j1.071 × 10⁻³) per km, Zc = √(z/y) = √(1.497 × 10⁵ ∠−17.4°) = 386.9∠−8.7° Ω.

For l = 400 km, γl = 0.0654 + j0.4283. Using cosh(a + jb) = cosh a cos b + j sinh a sin b and sinh(a + jb) = sinh a cos b + j cosh a sin b,

cosh γl = 1.0021 × 0.9097 + j0.0654 × 0.4154 = 0.9116 + j0.0272 = 0.9120∠1.7°, sinh γl = 0.0654 × 0.9097 + j1.0021 × 0.4154 = 0.0595 + j0.4162 = 0.4205∠81.9°.

(i) With IR = 0 the long-line equations give VS = VR cosh γl and IS = (VR/Zc) sinh γl. Hence

VR = VS/cosh γl = 220/0.9120 = 241.2 kV (line-to-line), lagging VS by about 1.7°,

and, per phase, VR = 241.2/√3 = 139.3 kV, so

IS = 139.3 × 10³ × 0.4205/386.9 = 151 A, at an angle of 81.9° + 8.7° ≈ 90° ahead of VR.

Comment: at no load the receiving-end voltage is 9.6% higher than the sending-end voltage, and the sending-end current of 151 A is almost purely leading. This is the Ferranti effect: the line's charging current flows through its series inductance and the resulting voltage rise, roughly ω²LC l²/2 of VS, adds to the receiving-end voltage. It is why long EHV lines are fitted with shunt reactors that are switched in at light load.

(ii) Maximum length for VR ≤ 230 kV at no load. The condition is |cosh γl| ≥ 220/230 = 0.9565. To a good approximation cosh γl ≈ 1 + γ²l²/2 = 1 + zy l²/2 = (1 − 0.56 × 10⁻⁶ l²) + j0.175 × 10⁻⁶ l², and solving |cosh γl|² = 0.9149 gives l² ≈ 7.78 × 10⁴, l ≈ 279 km. Checking with the exact hyperbolic function at l = 280 km: γl = 0.0457 + j0.2998, cosh γl = 0.9564 + j0.0135, |cosh γl| = 0.9565 and VR = 220/0.9565 = 230.0 kV. The line may be at most about 280 km long.

(iii) Maximum frequency for VR ≤ 250 kV at 400 km. Both x and y are proportional to frequency while r is not. Let k = f/50; then z = 0.125 + j0.4k and y = j2.8 × 10⁻⁶ k, and the condition is |cosh γl| ≥ 220/250 = 0.88. With the same approximation, cosh γl ≈ (1 − 0.0896k²) + j0.028k, and |cosh γl|² = 0.7744 gives k ≈ 1.16. Checking exactly at k = 1.16 (58 Hz): γ = 1.249 × 10⁻³ ∠82.5° per km, γl = 0.0656 + j0.4954, cosh γl = 0.8817 + j0.0312, |cosh γl| = 0.8822 and VR = 249.4 kV; the limit is reached at about k = 1.17. The frequency may rise to about 58.5 Hz before the no-load voltage exceeds 250 kV. The steep dependence on frequency (a 17% rise in f gives a 3.6% rise in VR) is the reason the Ferranti rise is so sensitive to line length and system frequency.

(b)(i) Single line-to-ground fault at the generator terminals

Base 25 MVA, 11 kV, so Zbase = 11²/25 = 4.84 Ω and Rn = 1/4.84 = 0.2066 pu. The two identical generators are in parallel on the faulted bus, so the positive- and negative-sequence Thevenin reactances are halved: Z₁ = j0.09/2 = j0.045 pu, Z₂ = j0.05/2 = j0.025 pu. Only G₁'s neutral is grounded, so only G₁ provides a zero-sequence path, and the neutral resistor appears as 3Rn in the zero-sequence network: Z₀ = j0.04 + 3 × 0.2066 = 0.620 + j0.04 pu.

For a single line-to-ground fault the three sequence networks are in series with the prefault voltage E = 1.0 pu:

Ia1 = Ia2 = Ia0 = E/(Z₁ + Z₂ + Z₀) = 1/(0.620 + j0.110) = 1/(0.6295∠10.1°) = 1.589∠−10.1° pu.

Fault current If = 3Ia1 = 4.77∠−10.1° pu. With Ibase = 25 × 10⁶/(√3 × 11 × 10³) = 1312 A this is 6.25 kA.

All of the zero-sequence current, 3Ia0 = If, returns through G₁'s neutral, so the voltage across the grounding resistor is

VRn = If Rn = 4.77 × 0.2066 = 0.985 pu of phase voltage (0.985 × 6.35 = 6.25 kV).

The resistor limits the fault current to well under what a solid ground would give (with Rn = 0, If = 3/(0.11) = 27.3 pu) while leaving almost the full phase voltage across itself.

(b)(ii) Restriking voltage across the breaker

L = 4/(2π × 50) = 12.73 mH per phase, C = 0.02 μF, peak phase voltage Vm = √2 × 11,000/√3 = 8.98 kV.

Maximum restriking voltage (the peak of Vm(1 − cos ωn t)) = 2Vm = 17.96 kV ≈ 18 kV.

Frequency of the transient oscillation: fn = 1/(2π√(LC)) = 1/(2π√(12.73 × 10⁻³ × 0.02 × 10⁻⁶)) = 1/(2π × 1.596 × 10⁻⁵) = 9.97 kHz.

The first peak occurs at half a cycle of the oscillation, t = 1/(2fn) = 50.1 μs, so the average rate of rise of restriking voltage up to the first peak is 2Vm/(π√(LC)) = 17.96 kV/50.1 μs = 0.36 kV/μs (3.58 × 10⁸ V/s).

(c) Capacity of a cascade of binary symmetric channels

A single BSC with crossover probability ε has capacity C = 1 − H(ε), where H(ε) = −ε log₂ ε − (1 − ε) log₂(1 − ε). A cascade of n such channels is itself a BSC, since each stage flips the bit independently with probability ε and the output is in error when the bit has been flipped an odd number of times. Let εn be the overall crossover probability. Then ε₁ = ε and

εn = εn−1(1 − ε) + (1 − εn−1)ε, so that 1 − 2εn = (1 − 2ε)(1 − 2εn−1),

and by induction 1 − 2εn = (1 − 2ε)ⁿ, i.e.

εn = [1 − (1 − 2ε)ⁿ]/2.

The capacity of the cascade is therefore Cn = 1 − H(εn) = 1 − H([1 − (1 − 2ε)ⁿ]/2) bits per channel use.

As n → ∞, for any 0 < ε < 1, |1 − 2ε| < 1, so (1 − 2ε)ⁿ → 0 and εn → ½. Then H(εn) → 1 and Cn → 0: after enough noisy repeaters the output is independent of the input and no information gets through. Only for ε = 0 (a perfect channel, capacity 1 for all n) or ε = 1 (an inverter, also capacity 1) is the capacity preserved, which is the case for repeaters that regenerate the digital signal rather than merely pass it on.