Nominal-T 100 km line, 20 MW at 110 kV

Given: $l = 100$ km, $z = 0.2 + j0.4\ \Omega$/km, $y = j2.5\times10^{-6}$ S/km; load 20 MW at 110 kV, 0.9 p.f. lagging; nominal-T model.

Step 1: Total parameters

$$\begin{aligned} Z &= (0.2 + j0.4)\times100 = 20 + j40\ \Omega, \quad \frac{Z}{2} = 10 + j20\ \Omega \\ Y &= j2.5\times10^{-6}\times100 = j2.5\times10^{-4}\ \text{S} \end{aligned}$$

Step 2: Receiving-end values

$$\begin{aligned} V_R &= \frac{110}{\sqrt3} = 63.51\angle 0^\circ\ \text{kV} \\ I_R &= \frac{20\times10^6}{\sqrt3\times110\times10^3\times0.9} = 116.6\angle -25.84^\circ = 104.97 - j50.84\ \text{A} \end{aligned}$$

 IS     Z/2            Z/2     IR
 o-->--[R/2 jX/2]--+--[R/2 jX/2]-->--o
                   |
 VS              [ Y ] Ic   (VC)     VR
                   |
 o-----------------+-----------------o

Step 3: Mid-point voltage

$$\begin{aligned} V_C &= V_R + I_R\frac{Z}{2} = 63{,}509 + (104.97 - j50.84)(10 + j20) \\ &= 65{,}575 + j1591 = 65.59\angle 1.39^\circ\ \text{kV} \end{aligned}$$

Step 4: Shunt current and sending-end current

$$\begin{aligned} I_C &= YV_C = j2.5\times10^{-4}(65{,}575 + j1591) = -0.40 + j16.39\ \text{A} \\ I_S &= I_R + I_C = 104.58 - j34.45 = 110.1\angle -18.23^\circ\ \text{A} \end{aligned}$$

Step 5: Sending-end voltage

$$\begin{aligned} V_S &= V_C + I_S\frac{Z}{2} = 65{,}575 + j1591 + (104.58 - j34.45)(10 + j20) \\ &= 67{,}310 + j3338 = 67.39\angle 2.84^\circ\ \text{kV/phase} \\ V_{S(L)} &= \sqrt3\times67.39 = 116.73\ \text{kV} \end{aligned}$$

Check with ABCD: $A = 1 + ZY/2 = 0.995\angle 0.14^\circ$, $B = Z(1 + ZY/4) = 44.61\angle 63.51^\circ\ \Omega$, $C = Y$; $AV_R + BI_R$ gives the same $V_S$.

Sending-end p.f. $= \cos(2.84^\circ + 18.23^\circ) = 0.9331$ lagging; $P_S = 20.77$ MW.

Answer: $I_S = 110.1\angle -18.23^\circ$ A, $V_S = 116.73$ kV (line) $= 67.39\angle 2.84^\circ$ kV per phase.