Nominal-π phasor, sending end, regulation, efficiency

In the nominal-Π model of a medium line, the total series impedance $Z = R + jX$ is in the middle and half the shunt admittance, $Y/2 = j\omega C/2$, is connected at each end.

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

Phasor diagram (lagging load, $V_R$ as reference)

Construction:

1. Draw $V_R$ along the reference axis and $I_R$ lagging by $\phi_R$.
2. $I_{C2} = j\frac{\omega C}{2}V_R$ leads $V_R$ by 90°.
3. $I_L = I_R + I_{C2}$.
4. From the tip of $V_R$ add $I_L R$ (parallel to $I_L$) and $I_L X$ (90° ahead of $I_L$) to get $V_S$.
5. $I_{C1} = j\frac{\omega C}{2}V_S$ leads $V_S$ by 90°; $I_S = I_L + I_{C1}$.

          IC2 ^           IL.X  /VS
              |              | /
              |              |/ IL.R
   O-----------------------> VR
    \ \  phiR
     \  \ IL     (IS = IL + IC1,
      v   v       IC1 is 90 deg ahead of VS)
      IR

Sending-end voltage

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

Sending-end current

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

So $A = D = 1 + ZY/2$, $B = Z$, $C = Y(1 + ZY/4)$.

Voltage regulation

At no load ($I_R = 0$), $V_{R(NL)} = V_S/A$. Hence

$$\%\,\text{VR} = \frac{|V_S|/|A| - |V_R|}{|V_R|}\times 100 = \frac{\left|V_S\right| - \left|1 + \frac{ZY}{2}\right|\left|V_R\right|}{\left|1 + \frac{ZY}{2}\right|\left|V_R\right|}\times100$$

Note: $|V_S|/|A|$ is used, not $|V_S|$, because the shunt capacitance raises the no-load receiving voltage (Ferranti effect).

Efficiency

Only the series resistance carries loss, and it carries current $I_L$:

$$\begin{aligned} P_{loss} &= 3|I_L|^2 R \\ \eta &= \frac{P_R}{P_R + 3|I_L|^2R}\times100 = \frac{3|V_R||I_R|\cos\phi_R}{3|V_S||I_S|\cos\phi_S}\times100 \end{aligned}$$