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Model validity and nominal-π ABCD with phasor

Short answer

The long-line model is exact for any length because it solves distributed-parameter equations; the short-line model neglects shunt capacitance, so it fails for lines >80 km. Nominal-π ABCD: A=D=1+ZY/2, B=Z, C=Y(1+ZY/4). For lagging load, phasor diagram shows V_S built from V_R plus voltage drop across Z, and I_S = I_L + Ic1.

Explain why all lines can be represented by the long transmission line model whereas all the lines cannot be represented by the short transmission line model. Derive expressions of ABCD parameters of a nominal π-model of a medium transmission line. Also draw the phasor diagram for lagging power factor load.

Why the long-line model fits all lines but the short-line model does not

  • The long-line model uses distributed parameters and solves the line equations exactly: A=cosh⁡γlA = \cosh\gamma l, B=Zcsinh⁡γlB = Z_c\sinh\gamma l, C=sinh⁡γl/ZcC = \sinh\gamma l/Z_c. No approximation is made, so it is valid for any length. For a short line γl\gamma l is small, so A→1A \to 1, B→ZB \to Z, C→0C \to 0, which is exactly the short-line result.
  • The short-line model neglects shunt capacitance (C=0C = 0, IS=IRI_S = I_R). For medium and long lines the charging current is large, so this model gives wrong VSV_S, ISI_S, power factor and regulation, and cannot show the Ferranti effect. So it is valid only for short lines (up to about 80 km).

ABCD parameters of the nominal-Π model

Total series impedance ZZ in the middle; Y/2Y/2 at each end.

 IS       IL    Z = R + jX        IR
 o-->--+-----[ R ]--[ jX ]-----+-->--o
       |                       |
 VS  [Y/2] Ic1           Ic2 [Y/2]  VR
       |                       |
 o-----+-----------------------+-----o
IL=IR+Y2VRVS=VR+ZIL=(1+ZY2)VR+ZIRIS=IL+Y2VS=Y(1+ZY4)VR+(1+ZY2)IR\begin{aligned} I_L &= I_R + \frac{Y}{2}V_R \\ 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} A=D=1+ZY2,B=Z,C=Y(1+ZY4)A = D = 1 + \frac{ZY}{2}, \quad B = Z, \quad C = Y\left(1 + \frac{ZY}{4}\right)

Phasor diagram (lagging p.f. load, VRV_R reference)

  1. IRI_R lags VRV_R by ϕR\phi_R; IC2=jωC2VRI_{C2} = j\frac{\omega C}{2}V_R leads VRV_R by 90°.
  2. IL=IR+IC2I_L = I_R + I_{C2}.
  3. VS=VR+ILR+jILXV_S = V_R + I_LR + jI_LX (ILRI_LR parallel to ILI_L, ILXI_LX 90° ahead of ILI_L).
  4. IC1I_{C1} leads VSV_S by 90°; IS=IL+IC1I_S = I_L + I_{C1}.
       IC2 ^                    + tip of VS
           |         VS     .   |
           |            .       | IL.X
           |       .  d         |
   O---------------------------> VR
    \ \  phiR                   \ IL.R
     \  v IL                     +
      v IR       IS = IL + IC1 (IC1 is
                 90 deg ahead of VS)

✓ Answer checked by Claude Opus 5.5 (high effort)

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