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Equivalent-T of 300 km long line

Short answer

Per-phase total Z=145.9∠75.71° Ω, Y=j0.1885 S, γl=5.244∠82.86°, Zc=27.82∠-7.14° Ω. ABCD: A=D=0.8433∠-47.0°, B=31.31∠-80.12° Ω, C=0.04046∠-65.83° S. Equivalent-T: Z'/2=18.51∠-58.73° Ω, Y'=0.04046∠-65.83° S. With VS=220 kV, open-circuit receiving voltage VR(L)=260.88 kV.

A 220 kV, three phase transmission line is 300 km long. The line has resistance of 0.12 ohm per phase per km, line inductance of 1.5 mH per phase per km and shunt capacitance of 2 μF per phase per km. Calculate ABCD parameters of the line with long line model in equivalent T-model. If the line is excited by 220 kV from the sending end, calculate the receiving voltage.

Given (per phase, per km): r=0.12 Ωr = 0.12\ \Omega, L=1.5L = 1.5 mH, C=2 μC = 2\ \muF, l=300l = 300 km, f=50f = 50 Hz (assumed), VS=220V_S = 220 kV.

Step 1: Line constants

z=0.12+j(2π×50×1.5×10−3)=0.12+j0.4712 Ω/kmy=j2π×50×2×10−6=j6.283×10−4 S/kmZ=zl=36+j141.37=145.9∠75.71∘ ΩY=yl=j0.1885 S\begin{aligned} z &= 0.12 + j(2\pi\times50\times1.5\times10^{-3}) = 0.12 + j0.4712\ \Omega/\text{km} \\ y &= j2\pi\times50\times2\times10^{-6} = j6.283\times10^{-4}\ \text{S/km} \\ Z &= zl = 36 + j141.37 = 145.9\angle 75.71^\circ\ \Omega \\ Y &= yl = j0.1885\ \text{S} \end{aligned}

Step 2: γl\gamma l and ZcZ_c

γl=ZY=5.244∠82.86∘=0.6521+j5.2032Zc=Z/Y=27.82∠−7.14∘ Ω\begin{aligned} \gamma l &= \sqrt{ZY} = 5.244\angle 82.86^\circ = 0.6521 + j5.2032 \\ Z_c &= \sqrt{Z/Y} = 27.82\angle -7.14^\circ\ \Omega \end{aligned}

Step 3: ABCD constants (long line)

A=D=cosh⁡γl=cosh⁡αlcos⁡βl+jsinh⁡αlsin⁡βl=0.5751−j0.6167=0.8433∠−47.0∘sinh⁡γl=0.3296−j1.0762=1.126∠−72.97∘B=Zcsinh⁡γl=31.31∠−80.12∘ ΩC=sinh⁡γlZc=0.04046∠−65.83∘ S\begin{aligned} A = D &= \cosh\gamma l = \cosh\alpha l\cos\beta l + j\sinh\alpha l\sin\beta l \\ &= 0.5751 - j0.6167 = 0.8433\angle -47.0^\circ \\ \sinh\gamma l &= 0.3296 - j1.0762 = 1.126\angle -72.97^\circ \\ B &= Z_c\sinh\gamma l = 31.31\angle -80.12^\circ\ \Omega \\ C &= \frac{\sinh\gamma l}{Z_c} = 0.04046\angle -65.83^\circ\ \text{S} \end{aligned}

Step 4: Equivalent-T circuit

In the equivalent T, each series arm is Z′/2Z'/2 and the shunt branch is Y′Y', with A=1+Z′Y′/2A = 1 + Z'Y'/2 and C=Y′C = Y':

Y′=C=sinh⁡γlZc=0.04046∠−65.83∘ SZ′2=Zctanh⁡γl2=A−1C=18.51∠−58.73∘ Ω\begin{aligned} Y' &= C = \frac{\sinh\gamma l}{Z_c} = 0.04046\angle -65.83^\circ\ \text{S} \\ \frac{Z'}{2} &= Z_c\tanh\frac{\gamma l}{2} = \frac{A - 1}{C} = 18.51\angle -58.73^\circ\ \Omega \end{aligned} 
 IS     Z'/2          Z'/2     IR
 o-->--[18.51<-58.7]-+-[18.51<-58.7]-->--o
                     |
 VS            Y' = 0.04046<-65.8 S        VR
                     |
 o-------------------+-------------------o

Step 5: Receiving-end voltage (open circuit, IR=0I_R = 0)

VR=VSA=127.02∠0∘0.8433∠−47.0∘=150.62∠47.0∘ kV/phaseVR(L)=3×150.62=260.88 kV\begin{aligned} V_R &= \frac{V_S}{A} = \frac{127.02\angle 0^\circ}{0.8433\angle -47.0^\circ} = 150.62\angle 47.0^\circ\ \text{kV/phase} \\ V_{R(L)} &= \sqrt3\times150.62 = 260.88\ \text{kV} \end{aligned}

The receiving voltage is higher than 220 kV (Ferranti effect).

Answer: A=D=0.8433∠−47.0∘A = D = 0.8433\angle -47.0^\circ, B=31.31∠−80.12∘ ΩB = 31.31\angle -80.12^\circ\ \Omega, C=0.04046∠−65.83∘C = 0.04046\angle -65.83^\circ S; equivalent T: Z′/2=18.51∠−58.73∘ ΩZ'/2 = 18.51\angle -58.73^\circ\ \Omega per arm, Y′=0.04046∠−65.83∘Y' = 0.04046\angle -65.83^\circ S; no-load VR=260.88V_R = 260.88 kV.

Note: 2 μF/km is about 200 times the capacitance of a real overhead line (typically about 0.01 μF/km), which is why γl\gamma l and the angles are unusual. With C=0.01 μC = 0.01\ \muF/km the same steps give A=0.9342∠1.02∘A = 0.9342\angle 1.02^\circ, B=142.7∠76.04∘ ΩB = 142.7\angle 76.04^\circ\ \Omega, C=9.217×10−4∠90.33∘C = 9.217\times10^{-4}\angle 90.33^\circ S and VR=235.49V_R = 235.49 kV.

✓ Answer checked by Claude Opus 5.5 (high effort)

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