Transmission Line Modeling

In the nominal-T model of a medium transmission line, the total series impedance $Z = R + jX$ is split into two halves $Z/2$ at each end, and the total shunt admittance $Y = j\omega C$ is placed at the middle.

Circuit

 Is   Z/2        Vc        Z/2   Ir
 o---[R/2 X/2]---+---[R/2 X/2]---o
 +               |               +
 Vs            Ic| Y             Vr
 -               |               -
 o---------------+---------------o

Phasor diagram (lagging load)

Take $V_R$ as reference; $I_R$ lags it by $\phi_R$.

Currents:

        Ic (leads Vc by ~90 deg)
        ^
        |    Is = Ir + Ic
        |   /
        |  /
   O----+-+--------------> Vr (reference)
         \
          \  phi_R (lag)
           v Ir

Voltages (drops added head to tail):

                               Vs
                              /|
                             / | Is.X/2 (90 deg
                            /  |  ahead of Is)
                           /---+ Is.R/2 (along Is)
                       Vc /|
                         / | Ir.X/2 (90 deg
                        /  |  ahead of Ir)
                       /---+ Ir.R/2 (along Ir)
                      /
   O-----------------+ Vr
   Vs leads Vr by the load angle delta

Construction steps:

1. Draw $V_R$ along the reference axis; draw $I_R$ lagging by $\phi_R$.
2. Add $I_R R/2$ parallel to $I_R$ and $I_R X/2$ perpendicular (leading $I_R$ by 90°) to get $V_C$.
3. Draw $I_C$ leading $V_C$ by 90°; $I_S = I_R + I_C$.
4. Add $I_S R/2$ parallel to $I_S$ and $I_S X/2$ perpendicular to it to get $V_S$.

Derivation of ABCD constants

Voltage across the shunt branch:

$$V_C = V_R + I_R\frac{Z}{2}$$

Shunt current and sending-end current:

$$\begin{aligned} I_C &= YV_C = YV_R + \frac{YZ}{2}I_R \\ I_S &= I_R + I_C = YV_R + \left(1 + \frac{YZ}{2}\right)I_R \end{aligned}$$

Sending-end voltage:

$$\begin{aligned} V_S &= V_C + I_S\frac{Z}{2} \\ &= V_R + I_R\frac{Z}{2} + \frac{Z}{2}\left[YV_R + \left(1 + \frac{YZ}{2}\right)I_R\right] \\ &= \left(1 + \frac{YZ}{2}\right)V_R + Z\left(1 + \frac{YZ}{4}\right)I_R \end{aligned}$$

Comparing with $V_S = AV_R + BI_R$ and $I_S = CV_R + DI_R$:

Check: $AD - BC = \left(1 + \frac{YZ}{2}\right)^2 - YZ\left(1 + \frac{YZ}{4}\right) = 1$, and $A = D$ since the circuit is symmetrical.

In a long line the resistance, inductance and capacitance are distributed uniformly along the whole length. Each small element carries a different current and has a different voltage, so $V$ and $I$ are functions of distance $x$.

Starting point

Let per-unit-length series impedance be $z = r + j\omega L$ (Ω/km) and shunt admittance $y = g + j\omega C$ (S/km). Measure $x$ from the receiving end. Take an element of length $dx$ at distance $x$:

 IS      x <------ measured from receiving end
 o-->--z.dx--+--z.dx--+-- ... --z.dx--+--> IR
             |        |               |
 VS        y.dx     y.dx            y.dx    VR
             |        |               |
 o-----------+--------+-- ... --------+----o
 sending                          receiving

• Voltage rise across the element (series drop): $dV = I\,z\,dx$
• Current change across the element (shunt current): $dI = V\,y\,dx$

$$\frac{dV}{dx} = zI, \qquad \frac{dI}{dx} = yV$$

Solving

Differentiating the first equation and substituting the second:

$$\frac{d^2V}{dx^2} = z\frac{dI}{dx} = zy\,V = \gamma^2 V, \qquad \gamma = \sqrt{zy}$$

The general solution is $V = A_1 e^{\gamma x} + A_2 e^{-\gamma x}$, and from $I = \frac{1}{z}\frac{dV}{dx}$:

$$I = \frac{1}{Z_c}\left(A_1 e^{\gamma x} - A_2 e^{-\gamma x}\right), \qquad Z_c = \sqrt{z/y}$$

At $x = 0$: $V = V_R$, $I = I_R$, giving $A_1 = \frac{V_R + Z_c I_R}{2}$, $A_2 = \frac{V_R - Z_c I_R}{2}$.

$$\begin{aligned} V(x) &= \frac{V_R + Z_c I_R}{2}e^{\gamma x} + \frac{V_R - Z_c I_R}{2}e^{-\gamma x} \\ &= V_R\cosh\gamma x + Z_c I_R\sinh\gamma x \\ I(x) &= \frac{V_R}{Z_c}\sinh\gamma x + I_R\cosh\gamma x \end{aligned}$$

Why V and I differ from point to point

• $\cosh\gamma x$ and $\sinh\gamma x$ change in both magnitude and angle as $x$ changes, so $V(x)$ and $I(x)$ are different at every point.
• With $\gamma = \alpha + j\beta$, the first term $e^{\alpha x}e^{j\beta x}$ is an incident wave that grows in size and advances in phase towards the sending end; the second is a reflected wave. Their sum varies along the line.
• Only when the line is terminated in $Z_c$ ($V_R = Z_c I_R$) does the reflected wave vanish; even then the magnitude changes as $e^{\alpha x}$ and the phase as $\beta x$.
• At $x = l$ these give the sending-end values: $V_S = V_R\cosh\gamma l + Z_c I_R\sinh\gamma l$, $I_S = \frac{V_R}{Z_c}\sinh\gamma l + I_R\cosh\gamma l$.

So for a long line voltage and current are different at every point along it, unlike the short-line model, where the current stays the same along the whole line.

Both models lump the line's total series impedance $Z = R + jX$ and total shunt admittance $Y = j\omega C$. They differ only in where the shunt admittance is placed.

Nominal-Π model

Half the admittance ($Y/2$) is placed at each end.

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

$$\begin{aligned} I_{C2} &= \tfrac{Y}{2}V_R, \quad I_L = I_R + \tfrac{Y}{2}V_R \\ V_S &= V_R + Z I_L = \left(1 + \tfrac{ZY}{2}\right)V_R + Z I_R \\ I_S &= I_L + \tfrac{Y}{2}V_S = Y\left(1 + \tfrac{ZY}{4}\right)V_R + \left(1 + \tfrac{ZY}{2}\right)I_R \end{aligned}$$ $$A = D = 1 + \frac{ZY}{2}, \quad B = Z, \quad C = Y\left(1 + \frac{ZY}{4}\right)$$

Nominal-T model

The full admittance $Y$ is placed at the middle and $Z/2$ in each arm.

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

$$\begin{aligned} V_C &= V_R + \tfrac{Z}{2}I_R, \quad I_S = I_R + Y V_C = Y V_R + \left(1 + \tfrac{ZY}{2}\right)I_R \\ V_S &= V_C + \tfrac{Z}{2}I_S = \left(1 + \tfrac{ZY}{2}\right)V_R + Z\left(1 + \tfrac{ZY}{4}\right)I_R \end{aligned}$$ $$A = D = 1 + \frac{ZY}{2}, \quad B = Z\left(1 + \frac{ZY}{4}\right), \quad C = Y$$

Comparison

What can be said:

• $A$ and $D$ are identical in both, and both networks are symmetrical and reciprocal ($AD - BC = 1$).
• They are duals: the factor $(1 + ZY/4)$ multiplies $C$ in the Π model and $B$ in the T model.
• Since $ZY$ is small for medium lines, $(1 + ZY/4) \approx 1$, so both give nearly the same results; neither is exact. The exact (long-line) values lie between them.
• The Π model is preferred in load-flow studies because it adds no extra node.

Surge impedance loading (SIL) is the power delivered by a transmission line to a purely resistive load equal to its surge impedance $Z_c$ (for a lossless line, $Z_c = \sqrt{L/C}$).

$$\text{SIL} = \frac{V_L^2}{Z_c}\ \text{(MW, with } V_L \text{ in kV line-to-line)}$$

Example: a 400 kV line with $Z_c \approx 400\ \Omega$ has SIL $= 400^2/400 = 400$ MW.

Physical meaning

At SIL, the reactive power produced by the line's shunt capacitance equals the reactive power absorbed by its series inductance:

$$\omega C V^2 = \omega L I^2 \;\Rightarrow\; \frac{V}{I} = \sqrt{\frac{L}{C}} = Z_c$$

So the line neither needs nor supplies reactive power. For a lossless line the voltage magnitude is the same all along the line (flat voltage profile) and only the phase angle changes.

Significance in power systems

1. Benchmark of loading: a line loaded at SIL has a flat voltage profile and unity power factor all along it.
2. Below SIL (light load): the line generates net reactive power, so the receiving-end voltage rises (Ferranti effect). Shunt reactors are needed.
3. Above SIL (heavy load): the line absorbs reactive power, so voltage falls along the line. Shunt capacitors or series compensation are needed.
4. Loadability: practical loading of long EHV lines is expressed in multiples of SIL (e.g. a 300 km line may carry about 1.5–2 × SIL, limited by stability).
5. Raising transmission capacity: SIL increases with $V^2$ and with lower $Z_c$. This is why higher voltages, bundled conductors (which lower $L$ and raise $C$) and series capacitors are used.
6. Planning and compensation: comparing actual load with SIL tells engineers how much reactive compensation the line needs.

The long-line model (distributed parameters) is exact for any length. The short-line model keeps only the series impedance $Z$ and ignores the shunt capacitance ($A = D = 1$, $B = Z$, $C = 0$).

Long line modelled by the short-line model

A long line (above about 250 km) has a large shunt capacitance and large charging current, and the parameters are distributed. Using the short model:

• Charging current is ignored. The sending-end current is taken equal to the receiving-end current, which is wrong; at light load the real $I_S$ is mostly charging current.
• Sending-end voltage is wrong. With a lagging load the short model overestimates $V_S$ (no capacitive rise), so the voltage regulation comes out too high.
• Ferranti effect cannot be predicted. At no load the short model gives $V_R = V_S$, whereas a long line actually has $V_R > V_S$.
• Losses and efficiency are wrong, since the true current varies along the line.
• Power factor at the sending end is wrong; the reactive power generated by the line is missing.
• Reactive compensation, insulation and protection design based on these results would be unsafe.

Example: for a 300 km, 220 kV line delivering 50 MW at 0.8 lag (Z = 40 + j125 Ω, Y = j10⁻³ S), the short model gives $V_S \approx 251$ kV at p.f. 0.75 lag, whereas the Π model gives about 238 kV at p.f. 0.99 lead.

Short line modelled by the long-line model

• The long-line equations are exact for all lengths, so the results are correct.
• For a short line, $\gamma l$ is very small, so $\cosh\gamma l \approx 1$, $Z_c\sinh\gamma l \approx Z$ and $\sinh\gamma l / Z_c \approx 0$. The answers become practically the same as the short-line model.
• The only drawback is more calculation (complex hyperbolic functions) for no real gain in accuracy.

Summary

Classification of transmission line models

Based on length (for 50 Hz lines):

ABCD parameters of a short line

The shunt capacitance is neglected, so the same current flows through the whole line.

 IS = IR      R        jX        IR
 o---->----[ R ]----[ jX ]---->----o
 VS                                VR
 o---------------------------------o

$$\begin{aligned} I_S &= I_R \\ V_S &= V_R + I_R(R + jX) = V_R + Z I_R \end{aligned}$$

Comparing with $V_S = AV_R + BI_R$ and $I_S = CV_R + DI_R$:

$$A = 1, \quad B = Z = R + jX\ \Omega, \quad C = 0, \quad D = 1$$

Check: $AD - BC = 1 - 0 = 1$.

Phasor diagrams

Take $V_R$ as reference. The drop $I_R R$ is parallel to $I_R$ and $I_R X$ leads $I_R$ by 90°. $\delta$ is the angle between $V_S$ and $V_R$.

Lagging p.f. load ($I_R$ lags $V_R$ by $\phi_R$):

                               + tip of VS
                    VS     .   |
                       .       | IR.X (90 deg
                   .           |  ahead of IR)
              .  d             |
   O------------------------> VR
    \ phiR                     \ IR.R (along IR)
     \                          +
      v IR

Here $V_S > V_R$ and $V_S \approx V_R + I_R R\cos\phi_R + I_R X\sin\phi_R$, so regulation is positive and large.

Leading p.f. load ($I_R$ leads $V_R$ by $\phi_R$):

             ^ IR
            /       IR.X   + tip of VS
           /     (90 deg     \
          /    ahead of IR)   \
         /         VS  .       + IR.R
        / phiR    .           /  (along IR)
       /     . d             /
   O------------------------> VR

Here $V_S \approx V_R + I_R R\cos\phi_R - I_R X\sin\phi_R$; the $I_R X$ drop tilts backwards, so $V_S$ can be less than $V_R$ (negative regulation).

Given: $Z = 33 + j104\ \Omega$, $Y = j10^{-3}$ S (totals, so frequency is not needed), load 80 MW at 220 kV, 0.8 p.f. lagging. Per-phase values are used.

$$\begin{aligned} V_R &= \frac{220}{\sqrt3} = 127.02\angle 0^\circ\ \text{kV} \\ I_R &= \frac{80\times10^6}{\sqrt3\times220\times10^3\times0.8} = 262.4\angle -36.87^\circ = 209.95 - j157.46\ \text{A} \end{aligned}$$

(i) Nominal-Π method

$$\begin{aligned} ZY &= (33 + j104)(j10^{-3}) = -0.104 + j0.033 \\ A = D &= 1 + \frac{ZY}{2} = 0.948 + j0.0165 = 0.9481\angle 0.997^\circ \\ B &= Z = 109.1\angle 72.40^\circ\ \Omega \\ C &= Y\left(1 + \frac{ZY}{4}\right) = 9.74\times10^{-4}\angle 90.49^\circ\ \text{S} \end{aligned}$$ $$\begin{aligned} V_S &= AV_R + BI_R = (120{,}410 + j2096) + (23{,}304 + j16{,}638) \\ &= 143{,}720 + j18{,}734 = 144.93\angle 7.43^\circ\ \text{kV/phase} \\ V_{S(L)} &= \sqrt3 \times 144.93 = 251.03\ \text{kV} \\ I_S &= CV_R + DI_R = 200.58 - j22.09 = 201.8\angle -6.29^\circ\ \text{A} \end{aligned}$$

p.f. $= \cos(7.43^\circ + 6.29^\circ) = \cos 13.71^\circ = 0.9715$ lagging

$P_S = 3V_SI_S\cos\phi_S = 85.24$ MW (and $Q_S = 20.80$ Mvar)

(ii) Long-line (exact) method

$$\begin{aligned} \gamma l &= \sqrt{ZY} = 0.3303\angle 81.20^\circ = 0.05055 + j0.3264 \\ Z_c &= \sqrt{Z/Y} = 330.3\angle -8.80^\circ\ \Omega \\ A = D &= \cosh\gamma l = 0.9484 + j0.0162 = 0.9485\angle 0.98^\circ \\ B &= Z_c\sinh\gamma l = 107.2\angle 72.71^\circ\ \Omega \\ C &= \frac{\sinh\gamma l}{Z_c} = 9.828\times10^{-4}\angle 90.32^\circ\ \text{S} \end{aligned}$$ $$\begin{aligned} V_S &= AV_R + BI_R = 143{,}270 + j18{,}538 = 144.46\angle 7.37^\circ\ \text{kV/phase} \\ V_{S(L)} &= 250.23\ \text{kV} \\ I_S &= CV_R + DI_R = 200.98 - j21.11 = 202.1\angle -6.00^\circ\ \text{A} \end{aligned}$$

p.f. $= \cos(7.37^\circ + 6.00^\circ) = 0.9729$ lagging, $P_S = 85.21$ MW ($Q_S = 20.25$ Mvar)

Results and comment

• The two methods differ by only about 0.3 % in voltage, so for 280 km the nominal-Π model is reasonably accurate.
• The long-line method is exact. The error of the Π model grows with length, so the exact model should be used for lines well above 250 km.
• In both, $I_S < I_R$ because the line's charging current partly cancels the lagging load current.

Answer: Nominal-Π: $V_S = 251.03$ kV, $I_S = 201.8$ A, p.f. 0.9715 lag, $P_S = 85.24$ MW. Long line: $V_S = 250.23$ kV, $I_S = 202.1$ A, p.f. 0.9729 lag, $P_S = 85.21$ MW.

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 $I_L$:

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

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.

500 km line modelled as a medium line (nominal-Π or T)

• A 500 km line is a long line. Its parameters are distributed, and $\gamma l$ is large (about 0.5 rad at 50 Hz), so $\cosh\gamma l$ and $\sinh\gamma l$ are no longer close to $1 + ZY/2$ and $\gamma l$.
• The nominal model lumps all capacitance at one or two points. This gives errors in $A$, $B$, $C$ and hence in the sending-end voltage, current, power factor, regulation and losses. The error grows rapidly with length.
• The Ferranti rise at light load and the reactive power generated by the line are wrongly estimated, so the size of shunt reactors and the insulation level may be chosen wrongly.
• The results are therefore inaccurate and unsafe for design. The exact long-line model (or its equivalent-Π) must be used.

50 km line modelled by the long-line model

• The long-line equations are exact for any length, so the results will be correct.
• For 50 km, $\gamma l$ is very small, so $\cosh\gamma l \approx 1$, $Z_c\sinh\gamma l \approx Z$, $\sinh\gamma l/Z_c \approx Y \approx 0$. The answers become almost the same as the short-line model ($A = 1$, $B = Z$, $C = 0$).
• The only cost is unnecessary calculation with complex hyperbolic functions, for no practical gain.

Given: $l = 400$ km, $z = 0.105 + j0.3768\ \Omega$/km, $y = j2.822\times10^{-6}$ S/km, load 60 MVA at 0.9 p.f. lagging, 220 kV. A 400 km line is a long line, so the exact (distributed) model is used. Values are per phase.

Step 1: Total parameters

$$\begin{aligned} Z &= zl = 42 + j150.72 = 156.46\angle 74.43^\circ\ \Omega \\ Y &= yl = j1.1288\times10^{-3}\ \text{S} \end{aligned}$$

Step 2: Propagation constant and surge impedance

$$\begin{aligned} \gamma l &= \sqrt{ZY} = 0.4203\angle 82.21^\circ = 0.05693 + j0.41638 \\ Z_c &= \sqrt{Z/Y} = 372.3\angle -7.79^\circ\ \Omega \end{aligned}$$

Step 3: ABCD constants

$$\begin{aligned} A = D &= \cosh\gamma l = 0.91604 + j0.02304 = 0.9163\angle 1.44^\circ \\ \sinh\gamma l &= 0.05209 + j0.40511 = 0.4084\angle 82.67^\circ \\ B &= Z_c\sinh\gamma l = 152.1\angle 74.89^\circ\ \Omega \\ C &= \frac{\sinh\gamma l}{Z_c} = 1.097\times10^{-3}\angle 90.46^\circ\ \text{S} \end{aligned}$$

Step 4: Receiving-end quantities

$$\begin{aligned} V_R &= \frac{220}{\sqrt3} = 127.02\angle 0^\circ\ \text{kV} \\ I_R &= \frac{60\times10^6}{\sqrt3\times220\times10^3} = 157.46\angle -25.84^\circ = 141.71 - j68.64\ \text{A} \end{aligned}$$

$P_R = 60 \times 0.9 = 54$ MW, $Q_R = 60 \times 0.4359 = 26.15$ Mvar

Step 5: Sending-end voltage and current

$$\begin{aligned} V_S &= AV_R + BI_R = (116{,}350 + j2926) + (15{,}695 + j18{,}083) \\ &= 132{,}050 + j21{,}009 = 133.71\angle 9.04^\circ\ \text{kV/phase} \\ V_{S(L)} &= \sqrt3\times133.71 = 231.59\ \text{kV} \\ I_S &= CV_R + DI_R = (-1.11 + j139.34) + (131.40 - j59.61) \\ &= 130.28 + j79.74 = 152.7\angle 31.47^\circ\ \text{A} \end{aligned}$$

Step 6: Power factor and power

Angle between $V_S$ and $I_S$ $= 9.04^\circ - 31.47^\circ = -22.43^\circ$ (current leads).

Sending-end p.f. $= \cos 22.43^\circ = 0.9244$ leading

$$S_S = 3V_SI_S^* = 56.64 - j23.38\ \text{MVA}$$

Step 7: Losses

$$\begin{aligned} P_{loss} &= P_S - P_R = 56.64 - 54 = 2.64\ \text{MW} \\ Q_{loss} &= Q_S - Q_R = -23.38 - 26.15 = -49.53\ \text{Mvar} \end{aligned}$$

The negative reactive loss means the line's shunt capacitance generates 49.53 Mvar more than its series inductance absorbs. This is why the sending-end current leads.

Answer: $V_S = 231.59$ kV (line), $I_S = 152.7$ A, p.f. $= 0.9244$ leading, active power loss $= 2.64$ MW, reactive power loss $= -49.53$ Mvar (net 49.53 Mvar generated by the line). Efficiency $= 54/56.64 = 95.35\%$.

Given: $Z = 40 + j125 = 131.24\angle 72.26^\circ\ \Omega$, $Y = j10^{-3}$ S (totals), load 50 MW at 220 kV, 0.8 p.f. lagging. Per-phase working.

$$\begin{aligned} V_R &= \frac{220}{\sqrt3} = 127.02\angle 0^\circ\ \text{kV} \\ I_R &= \frac{50\times10^6}{\sqrt3\times220\times10^3\times0.8} = 164.02\angle -36.87^\circ = 131.22 - j98.41\ \text{A} \end{aligned}$$

$P_R = 50$ MW, $Q_R = 37.5$ Mvar.

(i) Short-line approximation ($A = D = 1$, $B = Z$, $C = 0$)

$$\begin{aligned} I_S &= I_R = 164.0\angle -36.87^\circ\ \text{A} \\ ZI_R &= (40 + j125)(131.22 - j98.41) = 17{,}550 + j12{,}466\ \text{V} \\ V_S &= V_R + ZI_R = 144{,}567 + j12{,}466 = 145.10\angle 4.93^\circ\ \text{kV/phase} \\ V_{S(L)} &= \sqrt3\times145.10 = 251.33\ \text{kV} \end{aligned}$$

p.f. $= \cos(4.93^\circ + 36.87^\circ) = \cos 41.80^\circ = 0.7455$ lagging

$$P_S = 3V_SI_S\cos\phi_S = 3\times145.10\times10^3\times164.02\times0.7455 = 53.23\ \text{MW}$$

(ii) Nominal-Π method

$$\begin{aligned} ZY &= (40 + j125)(j10^{-3}) = -0.125 + j0.04 \\ A = D &= 1 + \frac{ZY}{2} = 0.9375 + j0.02 = 0.9377\angle 1.22^\circ \\ B &= Z = 131.24\angle 72.26^\circ\ \Omega \\ C &= Y\left(1 + \frac{ZY}{4}\right) = -1.0\times10^{-5} + j9.6875\times10^{-4} = 9.688\times10^{-4}\angle 90.59^\circ\ \text{S} \end{aligned}$$ $$\begin{aligned} V_S &= AV_R + BI_R = (119{,}080 + j2540) + (17{,}550 + j12{,}466) \\ &= 136{,}630 + j15{,}006 = 137.45\angle 6.27^\circ\ \text{kV/phase} \\ V_{S(L)} &= 238.07\ \text{kV} \\ I_S &= CV_R + DI_R = (-1.27 + j123.05) + (124.98 - j89.64) \\ &= 123.71 + j33.41 = 128.1\angle 15.11^\circ\ \text{A} \end{aligned}$$

p.f. $= \cos(6.27^\circ - 15.11^\circ) = \cos 8.85^\circ = 0.9881$ leading (current leads voltage)

$$P_S = 3\times137.45\times10^3\times128.1\times0.9881 = 52.21\ \text{MW}$$

Comparison

The short-line model ignores about 123 A of charging current, so for a 300 km line it badly overestimates $V_S$ and $I_S$. The nominal-Π (or better, the exact long-line) model must be used.

Answer: (i) $V_S = 251.33$ kV, $I_S = 164.0$ A, p.f. 0.7455 lag, $P_S = 53.23$ MW. (ii) $V_S = 238.07$ kV, $I_S = 128.1$ A, p.f. 0.9881 lead, $P_S = 52.21$ MW.

Why long lines use uniformly distributed parameters

• Resistance, inductance and capacitance are not located at one point; every metre of conductor has its own series $r$, $L$ and shunt $C$ (and leakage $g$).
• In short and medium lines the total charging current is small, so lumping $C$ at one or two points gives little error.
• In a long line (above about 250 km) the charging current is large and the current changes continuously along the line: each section draws its own charging current. The voltage therefore also changes continuously.
• The line length becomes a noticeable fraction of the wavelength ($\lambda \approx 6000$ km at 50 Hz), so travelling-wave effects (phase shift, attenuation, reflection) appear. These can only be represented by distributed parameters.
• Lumped models then give wrong $V_S$, $I_S$, regulation and Ferranti rise, so an exact solution of the line's differential equations is used.

Current is different at different points

Let $z = r + j\omega L$ (Ω/km) and $y = g + j\omega C$ (S/km). Take an element $dx$ at distance $x$ from the receiving end.

 IS      x <------ measured from receiving end
 o-->--z.dx--+--z.dx--+-- ... --z.dx--+--> IR
             |        |               |
 VS        y.dx     y.dx            y.dx    VR
             |        |               |
 o-----------+--------+-- ... --------+----o
 sending                          receiving

Series drop and shunt current in the element:

$$\frac{dV}{dx} = zI, \qquad \frac{dI}{dx} = yV$$

Differentiating the second and substituting the first:

$$\frac{d^2I}{dx^2} = y\frac{dV}{dx} = yz\,I = \gamma^2 I, \qquad \gamma = \sqrt{zy} = \alpha + j\beta$$

General solution:

$$I(x) = K_1e^{\gamma x} + K_2e^{-\gamma x}$$

Then $V = \frac{1}{y}\frac{dI}{dx} = Z_c\left(K_1e^{\gamma x} - K_2e^{-\gamma x}\right)$, with $Z_c = \sqrt{z/y}$.

Boundary conditions at $x = 0$: $I = I_R$, $V = V_R$:

$$K_1 + K_2 = I_R, \quad K_1 - K_2 = \frac{V_R}{Z_c} \;\Rightarrow\; K_1 = \frac{1}{2}\left(I_R + \frac{V_R}{Z_c}\right),\; K_2 = \frac{1}{2}\left(I_R - \frac{V_R}{Z_c}\right)$$ $$\begin{aligned} I(x) &= \frac{1}{2}\left(I_R + \frac{V_R}{Z_c}\right)e^{\gamma x} + \frac{1}{2}\left(I_R - \frac{V_R}{Z_c}\right)e^{-\gamma x} \\ I(x) &= I_R\cosh\gamma x + \frac{V_R}{Z_c}\sinh\gamma x \end{aligned}$$

Interpretation

• $I(x)$ depends on $x$ through $\cosh\gamma x$ and $\sinh\gamma x$, which change in magnitude and angle along the line. So the current is different at every point.
• The term $\frac{V_R}{Z_c}\sinh\gamma x$ is the accumulated charging current; it is zero at the receiving end and largest at the sending end.
• In wave form, the current is the sum of an incident wave ($e^{\gamma x}$) and a reflected wave ($e^{-\gamma x}$), each attenuated by $e^{\alpha x}$ and phase-shifted by $\beta x$.
• At no load ($I_R = 0$), $I(x) = \frac{V_R}{Z_c}\sinh\gamma x$, which is zero at the receiving end but not at the sending end.
• At $x = l$: $I_S = \frac{V_R}{Z_c}\sinh\gamma l + I_R\cosh\gamma l$.

Given (per phase, per km): $r = 0.12\ \Omega$, $L = 1.5$ mH, $C = 2\ \mu$F, $l = 300$ km, $f = 50$ Hz (assumed), $V_S = 220$ kV.

Step 1: Line constants

$$\begin{aligned} z &= 0.12 + j(2\pi\times50\times1.5\times10^{-3}) = 0.12 + j0.4712\ \Omega/\text{km} \\ y &= j2\pi\times50\times2\times10^{-6} = j6.283\times10^{-4}\ \text{S/km} \\ Z &= zl = 36 + j141.37 = 145.9\angle 75.71^\circ\ \Omega \\ Y &= yl = j0.1885\ \text{S} \end{aligned}$$

Step 2: $\gamma l$ and $Z_c$

$$\begin{aligned} \gamma l &= \sqrt{ZY} = 5.244\angle 82.86^\circ = 0.6521 + j5.2032 \\ Z_c &= \sqrt{Z/Y} = 27.82\angle -7.14^\circ\ \Omega \end{aligned}$$

Step 3: ABCD constants (long line)

$$\begin{aligned} A = D &= \cosh\gamma l = \cosh\alpha l\cos\beta l + j\sinh\alpha l\sin\beta l \\ &= 0.5751 - j0.6167 = 0.8433\angle -47.0^\circ \\ \sinh\gamma l &= 0.3296 - j1.0762 = 1.126\angle -72.97^\circ \\ B &= Z_c\sinh\gamma l = 31.31\angle -80.12^\circ\ \Omega \\ C &= \frac{\sinh\gamma l}{Z_c} = 0.04046\angle -65.83^\circ\ \text{S} \end{aligned}$$

Step 4: Equivalent-T circuit

In the equivalent T, each series arm is $Z'/2$ and the shunt branch is $Y'$, with $A = 1 + Z'Y'/2$ and $C = Y'$:

$$\begin{aligned} Y' &= C = \frac{\sinh\gamma l}{Z_c} = 0.04046\angle -65.83^\circ\ \text{S} \\ \frac{Z'}{2} &= Z_c\tanh\frac{\gamma l}{2} = \frac{A - 1}{C} = 18.51\angle -58.73^\circ\ \Omega \end{aligned}$$

 IS     Z'/2          Z'/2     IR
 o-->--[18.51<-58.7]-+-[18.51<-58.7]-->--o
                     |
 VS            Y' = 0.04046<-65.8 S        VR
                     |
 o-------------------+-------------------o

Step 5: Receiving-end voltage (open circuit, $I_R = 0$)

$$\begin{aligned} V_R &= \frac{V_S}{A} = \frac{127.02\angle 0^\circ}{0.8433\angle -47.0^\circ} = 150.62\angle 47.0^\circ\ \text{kV/phase} \\ V_{R(L)} &= \sqrt3\times150.62 = 260.88\ \text{kV} \end{aligned}$$

The receiving voltage is higher than 220 kV (Ferranti effect).

Answer: $A = D = 0.8433\angle -47.0^\circ$, $B = 31.31\angle -80.12^\circ\ \Omega$, $C = 0.04046\angle -65.83^\circ$ S; equivalent T: $Z'/2 = 18.51\angle -58.73^\circ\ \Omega$ per arm, $Y' = 0.04046\angle -65.83^\circ$ S; no-load $V_R = 260.88$ kV.

Note: 2 μF/km is about 200 times the capacitance of a real overhead line (typically about 0.01 μF/km), which is why $\gamma l$ and the angles are unusual. With $C = 0.01\ \mu$F/km the same steps give $A = 0.9342\angle 1.02^\circ$, $B = 142.7\angle 76.04^\circ\ \Omega$, $C = 9.217\times10^{-4}\angle 90.33^\circ$ S and $V_R = 235.49$ kV.