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Nominal-T relations, losses and Ferranti condition

Short answer

In the nominal-T model, A = D = 1 + ZY/2, B = Z(1 + ZY/4), C = Y. Active loss = 3(|I_S|² + |I_R|²)R/2, reactive loss includes line inductive and capacitive parts. Ferranti effect occurs at light load when |1 + ZY/2| < 1, making receiving voltage exceed sending voltage due to capacitance.

Derive the expressions relating sending end current and voltage to the receiving end voltage and current in a medium length transmission line by nominal-T model. Express the active and reactive losses in the line and the condition for voltage at the receiving end being higher than at the sending end.

In the nominal-T model, the total shunt admittance Y=jωCY = j\omega C is at the middle of the line and Z/2=(R+jX)/2Z/2 = (R + jX)/2 is in each arm (per phase).

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

Sending-end voltage and current

VC=VR+Z2IRIS=IR+YVC=YVR+(1+ZY2)IRVS=VC+Z2IS=(1+ZY2)VR+Z(1+ZY4)IR\begin{aligned} V_C &= V_R + \frac{Z}{2}I_R \\ I_S &= I_R + YV_C = YV_R + \left(1 + \frac{ZY}{2}\right)I_R \\ V_S &= V_C + \frac{Z}{2}I_S = \left(1 + \frac{ZY}{2}\right)V_R + Z\left(1 + \frac{ZY}{4}\right)I_R \end{aligned}

So A=D=1+ZY2A = D = 1 + \frac{ZY}{2}, B=Z(1+ZY4)B = Z\left(1 + \frac{ZY}{4}\right), C=YC = Y.

Active and reactive losses (3-phase)

Current ISI_S flows in the sending half and IRI_R in the receiving half:

Ploss=3(∣IS∣2+∣IR∣2)R2Qloss=3(∣IS∣2+∣IR∣2)X2−3ωC∣VC∣2\begin{aligned} P_{loss} &= 3\left(|I_S|^2 + |I_R|^2\right)\frac{R}{2} \\ Q_{loss} &= 3\left(|I_S|^2 + |I_R|^2\right)\frac{X}{2} - 3\omega C|V_C|^2 \end{aligned}

The last term is the reactive power generated by the line capacitance. Equivalently, Ploss=PS−PRP_{loss} = P_S - P_R and Qloss=QS−QRQ_{loss} = Q_S - Q_R.

Condition for receiving voltage higher than sending voltage

At no load or light load (IR≈0I_R \approx 0):

VS=(1+ZY2)VR  ⇒  VR=VS1+ZY2V_S = \left(1 + \frac{ZY}{2}\right)V_R \;\Rightarrow\; V_R = \frac{V_S}{1 + \frac{ZY}{2}}

With Z=R+jXZ = R + jX and Y=jωCY = j\omega C:

1+ZY2=(1−ωCX2)+jωCR21 + \frac{ZY}{2} = \left(1 - \frac{\omega CX}{2}\right) + j\frac{\omega CR}{2}

Neglecting the small RR term, ∣VR∣>∣VS∣|V_R| > |V_S| when

∣1+ZY2∣<1⇔0<ωCX2<2\left|1 + \frac{ZY}{2}\right| < 1 \quad\Leftrightarrow\quad 0 < \frac{\omega CX}{2} < 2

which is always true for a practical line (and RR small). So at light load the receiving voltage rises above the sending voltage; this is the Ferranti effect. The rise is

∣VR∣−∣VS∣≈ωCX2∣VS∣=ω2LC l22∣VS∣|V_R| - |V_S| \approx \frac{\omega CX}{2}|V_S| = \frac{\omega^2LC\,l^2}{2}|V_S|

so it grows with the square of line length. With load, VR>VSV_R > V_S happens when the capacitive charging current outweighs the drop of the load current, i.e. when the load is below about SIL or is leading.

✓ Answer checked by Claude Opus 5.5 (high effort)

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