Nominal-π derivation, losses and UPF phasor diagram

A 200 km line is a medium line, so its shunt capacitance cannot be ignored. In the nominal-Π model the total series impedance $Z = R + jX$ is lumped in the middle and $Y/2$ is placed at each end (all quantities per phase).

 IS       IL    Z = R + jX        IR
 o-->--+-----[ R ]--[ jX ]-----+-->--o
       |                       |
 VS  [Y/2] Ic1           Ic2 [Y/2]  VR
       |                       |
 o-----+-----------------------+-----o

Relation between sending and receiving quantities

Current through the receiving-end shunt: $I_{C2} = \frac{Y}{2}V_R$

Series current: $I_L = I_R + \frac{Y}{2}V_R$

$$\begin{aligned} V_S &= V_R + ZI_L = \left(1 + \frac{ZY}{2}\right)V_R + ZI_R \\ I_S &= I_L + \frac{Y}{2}V_S = Y\left(1 + \frac{ZY}{4}\right)V_R + \left(1 + \frac{ZY}{2}\right)I_R \end{aligned}$$

In matrix form:

$$\begin{bmatrix} V_S \\ I_S \end{bmatrix} = \begin{bmatrix} 1 + \frac{ZY}{2} & Z \\ Y\left(1 + \frac{ZY}{4}\right) & 1 + \frac{ZY}{2} \end{bmatrix}\begin{bmatrix} V_R \\ I_R \end{bmatrix}$$

Active and reactive power losses (3-phase)

With $Y/2 = j\omega C/2$ at each end:

$$\begin{aligned} P_{loss} &= 3|I_L|^2 R = P_S - P_R \\ Q_{loss} &= 3|I_L|^2X - 3\frac{\omega C}{2}\left(|V_S|^2 + |V_R|^2\right) = Q_S - Q_R \end{aligned}$$

The first term of $Q_{loss}$ is the reactive power absorbed by the series inductance. The second is the reactive power generated by the two shunt capacitances. A negative $Q_{loss}$ means the line supplies net reactive power, which is normal at light load.

Phasor diagram for unity p.f. load

Take $V_R$ as reference; $I_R$ is in phase with $V_R$.

1. $I_{C2}$ leads $V_R$ by 90°.
2. $I_L = I_R + I_{C2}$ leads $V_R$ slightly.
3. $V_S = V_R + I_LR + jI_LX$: $I_LR$ is parallel to $I_L$, $I_LX$ is perpendicular to $I_L$.
4. $I_{C1}$ leads $V_S$ by 90°; $I_S = I_L + I_{C1}$.

                  IS
         IC1 ^   ^         VS
              \  |  IL   / |
               \ | /    /  | IL.X
          IC2 ^ \|/    /   |
              | /   d /____|
              |/    /  IL.R
   O------------------------> VR , IR

At unity p.f. the $I_LX$ drop is almost at right angles to $V_R$, so $|V_S|$ is only slightly greater than $|V_R|$ while the load angle $\delta$ is noticeable. $I_S$ leads $I_R$ because of the two charging currents.