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200 mile line: Zc, propagation constant, SIL

Short answer

Characteristic impedance Zc = 406.4 angle -5.48° Ω; propagation constant gamma = 2.075×10⁻³ angle 84.52° per mile; attenuation constant alpha = 1.981×10⁻⁴ Np/mi; phase constant beta = 2.065×10⁻³ rad/mi; surge impedance loading SIL = 119.1 MW (using |Zc| from sqrt(z/y)).

A 200 mile long, 3-phase, 220 kV, 60 Hz overhead transmission line has series impedance of 0.8431∠79.04° Ω/mi and shunt admittance of 5.105 × 10⁻⁶∠90° S/mi. Determine the characteristic impedance, propagation constant, phase shift constant, attenuation constant and SIL of the line.

Given: l=200l = 200 mi, 220 kV, 60 Hz, z=0.8431∠79.04∘ Ωz = 0.8431\angle 79.04^\circ\ \Omega/mi, y=5.105×10−6∠90∘y = 5.105\times10^{-6}\angle 90^\circ S/mi.

Characteristic impedance

Zc=zy=0.8431∠79.04∘5.105×10−6∠90∘=165,152∠−10.96∘=406.4∠−5.48∘ Ω=404.5−j38.8 Ω\begin{aligned} Z_c &= \sqrt{\frac{z}{y}} = \sqrt{\frac{0.8431\angle 79.04^\circ}{5.105\times10^{-6}\angle 90^\circ}} = \sqrt{165{,}152\angle -10.96^\circ} \\ &= 406.4\angle -5.48^\circ\ \Omega = 404.5 - j38.8\ \Omega \end{aligned}

Propagation constant

γ=zy=0.8431×5.105×10−6∠169.04∘=2.075×10−3∠84.52∘ per mile=1.981×10−4+j2.065×10−3 per mile\begin{aligned} \gamma &= \sqrt{zy} = \sqrt{0.8431\times5.105\times10^{-6}\angle 169.04^\circ} \\ &= 2.075\times10^{-3}\angle 84.52^\circ\ \text{per mile} \\ &= 1.981\times10^{-4} + j2.065\times10^{-3}\ \text{per mile} \end{aligned}

For the whole line: γl=0.4149∠84.52∘=0.0396+j0.4130\gamma l = 0.4149\angle 84.52^\circ = 0.0396 + j0.4130.

Attenuation and phase-shift constants

α=Re(γ)=1.981×10−4 Np/mi(αl=0.0396 Np)β=Im(γ)=2.065×10−3 rad/mi(βl=0.4130 rad=23.66∘)\begin{aligned} \alpha &= \text{Re}(\gamma) = 1.981\times10^{-4}\ \text{Np/mi}\quad(\alpha l = 0.0396\ \text{Np}) \\ \beta &= \text{Im}(\gamma) = 2.065\times10^{-3}\ \text{rad/mi}\quad(\beta l = 0.4130\ \text{rad} = 23.66^\circ) \end{aligned}

Surge impedance loading

SIL=VL2∣Zc∣=(220)2406.4=119.1 MW\text{SIL} = \frac{V_L^2}{|Z_c|} = \frac{(220)^2}{406.4} = 119.1\ \text{MW}

(Strictly, SIL is defined with the lossless surge impedance L/C=x/b=0.8277/5.105×10−6=402.7 Ω\sqrt{L/C} = \sqrt{x/b} = \sqrt{0.8277/5.105\times10^{-6}} = 402.7\ \Omega, which gives 120.2 MW. The difference is under 1%.)

Answer: Zc=406.4∠−5.48∘ ΩZ_c = 406.4\angle -5.48^\circ\ \Omega, γ=2.075×10−3∠84.52∘\gamma = 2.075\times10^{-3}\angle 84.52^\circ /mi, β=2.065×10−3\beta = 2.065\times10^{-3} rad/mi, α=1.981×10−4\alpha = 1.981\times10^{-4} Np/mi, SIL =119.1= 119.1 MW (120.2 MW with the lossless ZcZ_c).

✓ Answer checked by Claude Opus 5.5 (high effort)

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