Performance Analysis

Reactive power compensation means supplying or absorbing reactive power at suitable points of the system, so that voltage stays within limits and transmission capacity is used well.

Why compensation is required

1. Voltage control: the voltage drop in a line is mainly $\Delta V \approx \frac{QX}{V}$. Heavy lagging loads lower the receiving voltage; light loads raise it (Ferranti effect). Compensation keeps voltage within ±5–10 %.
2. Line behaviour changes with load: below SIL a line generates Mvar (needs absorption by reactors); above SIL it absorbs Mvar (needs capacitive support).
3. Reduce losses: reactive current increases line current and $I^2R$ loss. Supplying Q locally reduces it.
4. Increase power transfer: for $P = \frac{V_SV_R}{X}\sin\delta$, series capacitors reduce $X$, and shunt support holds up $V$, raising the stability limit.
5. Improve power factor and release capacity of generators, transformers and lines.
6. Voltage stability: prevents voltage collapse in heavily loaded systems.

Methods

1. Shunt compensation

• Shunt capacitors: connected at load buses or substations to supply lagging Mvar during heavy load; raise voltage and p.f. Cheap but output falls with $V^2$.
• Shunt reactors: connected at ends of long EHV lines to absorb the line's charging Mvar at light load; control the Ferranti rise and switching overvoltages.

2. Series compensation

• Series capacitors in the line cancel part of the line inductance ($X_{eff} = X_L - X_C$, typically 25–70 % compensation). They increase power transfer and stability and reduce voltage drop. Risk: subsynchronous resonance.

3. Synchronous condensers

• Over-excited synchronous motors run without load supply Mvar; under-excited they absorb Mvar. Smooth, continuous control, but costly with high maintenance.

4. Static VAR compensators (SVC) and FACTS

• SVC: thyristor-controlled reactor (TCR) plus thyristor-switched or fixed capacitors give fast, variable Q.
• STATCOM (voltage-source converter), TCSC (controlled series capacitor), UPFC give faster and wider control.

5. Other means: tap-changing transformers (redistribute Q and adjust voltage) and generator excitation control.

The Ferranti effect is the rise of voltage at the receiving end of a long transmission line (or cable) above the sending-end voltage when the line is on open circuit or very light load.

Cause

At light load, the line's shunt capacitance draws a leading charging current $I_C$. This current flows through the series inductance, and the drop $jI_CX$ is in phase with $V_R$, so it adds to $V_R$ instead of subtracting.

                       IC ^ (leads VR by 90 deg)
                          |
   O---------------------->-----> VR
   O------------------->  VS      |
                     VS = VR - IC.X  (VS < VR)

Using the nominal-Π model at no load:

$$V_S = \left(1 + \frac{ZY}{2}\right)V_R \approx \left(1 - \frac{\omega^2LCl^2}{2}\right)V_R$$ $$V_R - V_S \approx \frac{\omega^2LCl^2}{2}V_R$$

So the rise grows with the square of line length and the square of frequency. It is significant for EHV lines above about 200–300 km and for cables (high C).

Effects

• Overvoltage stresses insulation of the line, transformers and equipment.
• Problems when energising long lines or after sudden load rejection.

Measures to tackle the Ferranti effect

1. Shunt reactors at the receiving end (and often both ends) of long EHV lines. They absorb the charging Mvar; most common method. They may be switched in at light load and out at heavy load.
2. Controlled reactive devices: SVC (thyristor-controlled reactors) and STATCOM absorb variable reactive power quickly.
3. Synchronous condensers run under-excited at the receiving end to absorb Mvar.
4. Generator under-excitation at the sending end and coordinated voltage control.
5. Switching out shunt capacitors and charging the line from the stronger end with lower initial voltage.
6. Keeping a minimum load on the line, or splitting very long lines with intermediate substations and reactors.
7. On-load tap changers on transformers to bring the receiving voltage back to normal.

Shunt compensation means connecting reactive-power devices (capacitors, reactors, or controlled devices) in parallel with the line or at the buses, to supply or absorb reactive power.

Why shunt compensation is necessary

• Voltage control at heavy load: large lagging loads draw reactive current, which causes a large voltage drop ($\Delta V \approx QX/V$). Shunt capacitors supply this Q locally and hold up the voltage.
• Voltage control at light load: below SIL the line's charging Mvar exceeds its inductive absorption, so voltage rises (Ferranti effect). Shunt reactors absorb this excess.
• Reduced losses and current: supplying Q near the load reduces line current, $I^2R$ loss and heating.
• Better power factor and released capacity of generators, transformers and lines.
• Higher power transfer and voltage stability: holding mid-point or receiving voltage up raises the transfer limit $P = \frac{V_SV_R}{X}\sin\delta$ and helps prevent voltage collapse.

Method 1: Shunt capacitors and shunt reactors (fixed or switched)

   Bus  ---+-------------+----- line
           |             |
          === C        (((L)))
           |             |
          ---           ---
       capacitor      reactor

• Shunt capacitor banks at substations and load centres supply lagging Mvar; switched in steps as load rises. Cheap, low loss, simple.
• Shunt reactors at the ends of long EHV lines absorb charging Mvar at light load and limit the Ferranti rise and switching overvoltages.
• Limitation: control is in steps; capacitor output falls as $V^2$, just when support is most needed.

Method 2: Static VAR compensator (SVC)

   Bus ----+-----------+-----------+
           |           |           |
         [TCR]       [TSC]       [Filter /
       thyristor-  thyristor-    fixed C]
       controlled   switched
        reactor    capacitor

• A thyristor-controlled reactor (TCR) varies the absorbed Q continuously by controlling the firing angle.
• Thyristor-switched capacitors (TSC) or fixed capacitors provide the capacitive range.
• Together they act as a fast, continuously variable susceptance, from capacitive to inductive, responding within a few cycles.
• Used for dynamic voltage control, damping oscillations and improving transient stability.

Other shunt methods: synchronous condensers (over- or under-excited synchronous machines giving smooth Q control) and STATCOM (voltage-source converter, faster than an SVC and keeps output at low voltage).

Reactive power compensation is the control of reactive power flow in a power system by adding devices that supply reactive power (capacitors, over-excited synchronous condensers, STATCOM in capacitive mode) or absorb it (reactors, under-excited condensers, SVC in inductive mode). The aim is to keep bus voltages within limits and to reduce reactive power flow through lines.

Reactive power does no useful work but is needed for magnetic and electric fields of motors, transformers and lines. If it is carried over long distances it causes voltage drop and losses. Compensation supplies it close to where it is needed.

Main types:

• Shunt compensation: capacitors or reactors connected in parallel at buses.
• Series compensation: capacitors connected in series with the line to cancel part of its reactance.
• Dynamic compensation: synchronous condensers, SVC, STATCOM.

Benefits to the system

1. Better voltage profile: voltage drop $\Delta V \approx (PR + QX)/V$ is reduced; overvoltage at light load (Ferranti effect) is controlled.
2. Lower losses: line current falls, so $I^2R$ losses in lines and transformers fall.
3. Improved power factor, avoiding low-p.f. penalties for consumers.
4. More transmission capacity: the same line and transformers can carry more active power (MW) because less of their rating is used by Mvar.
5. Higher stability limit: series capacitors reduce $X$ in $P = \frac{V_SV_R}{X}\sin\delta$, and shunt devices hold voltage up, improving steady-state and transient stability.
6. Voltage stability: prevents voltage collapse under heavy loading or contingencies.
7. Released generation capacity: generators supply less Mvar and can deliver more MW.
8. Economic saving: lower losses, deferred investment in new lines and equipment.

Given: 132 kV, 50 MVA at 0.8 p.f. lagging at the receiving end; $A = 0.98\angle 3^\circ$, $B = 110\angle 75^\circ\ \Omega$ per phase. Symmetrical line assumed, so $D = A$.

$$\begin{aligned} V_R &= \frac{132}{\sqrt3} = 76.21\angle 0^\circ\ \text{kV} \\ I_R &= \frac{50\times10^6}{\sqrt3\times132\times10^3} = 218.7\angle -36.87^\circ = 174.95 - j131.22\ \text{A} \end{aligned}$$

$P_R = 50\times0.8 = 40$ MW, $Q_R = 30$ Mvar.

(i) Sending-end voltage and power angle

$$\begin{aligned} AV_R &= 0.98\angle 3^\circ\times76{,}210 = 74{,}584 + j3909\ \text{V} \\ BI_R &= 110\angle 75^\circ\times218.7\angle -36.87^\circ = 24{,}056\angle 38.13^\circ = 18{,}923 + j14{,}854\ \text{V} \\ V_S &= AV_R + BI_R = 93{,}507 + j18{,}762 = 95.37\angle 11.35^\circ\ \text{kV/phase} \\ V_{S(L)} &= \sqrt3\times95.37 = 165.19\ \text{kV} \end{aligned}$$

Power angle $\delta = 11.35^\circ$.

(ii) Sending-end active and reactive power

Using the sending-end power equations with line values ($\beta = 75^\circ$, $\alpha = 3^\circ$, $D = A$):

$$\begin{aligned} P_S &= \frac{|D||V_S|^2}{|B|}\cos(\beta - \alpha) - \frac{|V_S||V_R|}{|B|}\cos(\beta + \delta) \\ Q_S &= \frac{|D||V_S|^2}{|B|}\sin(\beta - \alpha) - \frac{|V_S||V_R|}{|B|}\sin(\beta + \delta) \end{aligned}$$ $$\begin{aligned} \frac{|D||V_S|^2}{|B|} &= \frac{0.98\times165.19^2}{110} = 243.10\ \text{MVA} \\ \frac{|V_S||V_R|}{|B|} &= \frac{165.19\times132}{110} = 198.22\ \text{MVA} \\ P_S &= 243.10\cos 72^\circ - 198.22\cos 86.35^\circ = 75.12 - 12.63 = 62.49\ \text{MW} \\ Q_S &= 243.10\sin 72^\circ - 198.22\sin 86.35^\circ = 231.20 - 197.82 = 33.38\ \text{Mvar} \end{aligned}$$

(Check: the receiving-end form of the same equations gives back $P_R = 40$ MW and $Q_R = 30$ Mvar.)

(iii) Line losses

$$\begin{aligned} P_{loss} &= P_S - P_R = 62.49 - 40 = 22.49\ \text{MW} \\ Q_{loss} &= Q_S - Q_R = 33.38 - 30 = 3.38\ \text{Mvar} \end{aligned}$$

Answer: (i) $V_S = 165.19$ kV (line), $\delta = 11.35^\circ$; (ii) $P_S = 62.49$ MW, $Q_S = 33.38$ Mvar; (iii) losses $= 22.49$ MW and $3.38$ Mvar.

Note: the active loss is large because the 3° angle of $A$ implies a resistive (lossy) shunt branch: from $AD - BC = 1$, $C = 9.996\times10^{-4}\angle 39.1^\circ$ S. The series resistance alone ($110\cos 75^\circ = 28.5\ \Omega$) would account for only about 4.1 MW.

A transmission line has both series inductance and shunt capacitance, so it both absorbs and generates reactive power. Which one dominates depends on the load.

Why a line acts as a source or a sink

• Shunt capacitance generates reactive power: $Q_C = \omega CV^2$ per phase (per unit length). It depends on voltage only, so it is almost constant (voltage stays near rated).
• Series inductance absorbs reactive power: $Q_L = \omega LI^2$. It depends on the square of load current.

Comparing the two:

At SIL, $V/I = \sqrt{L/C} = Z_c$, so the two balance exactly.

Effects on the power system

As a source (light load):

• Receiving-end voltage rises above sending voltage (Ferranti effect); overvoltages stress insulation.
• Generators may have to absorb Mvar (under-excited operation), which reduces their stability margin.
• Shunt reactors are needed at line ends.

As a sink (heavy load):

• Voltage falls along the line and at the receiving end; voltage regulation becomes poor.
• Extra reactive current increases losses and reduces the capacity for active power.
• Generators must supply more Mvar; risk of voltage instability or collapse.
• Shunt capacitors, series capacitors, synchronous condensers or SVCs are needed.

So the voltage profile of a line changes with load, and reactive compensation must be adjustable (switched reactors and capacitors, SVC/STATCOM) to keep voltage within limits throughout the day.

A 100 km line is a medium line, so it is solved here with the nominal-T model: total series impedance split into two halves, with the total shunt admittance at the middle.

Line constants (per phase, total)

$$\begin{aligned} R &= 0.153 \times 100 = 15.3\ \Omega \\ X_L &= 2\pi (50)(1.21\times10^{-3})(100) = 38.01\ \Omega \\ Y &= 2\pi (50)(0.00958\times10^{-6})(100) = 3.010\times10^{-4}\ \text{S} \\ Z/2 &= 7.65 + j19.01\ \Omega \end{aligned}$$

Receiving-end quantities

$$\begin{aligned} V_R &= \frac{110\,000}{\sqrt3} = 63\,508.5\angle 0^\circ\ \text{V} \\ I_R &= \frac{20\times10^6}{3 \times 63\,508.5 \times 0.9} = 116.64\angle -25.84^\circ\ \text{A} \end{aligned}$$

Nominal-T solution

$$\begin{aligned} V_1 &= V_R + I_R \frac{Z}{2} = 65\,297.6\angle 1.41^\circ\ \text{V} \\ I_C &= jY V_1 = 19.65\angle 91.41^\circ\ \text{A} \\ I_S &= I_R + I_C = 109.05\angle -16.62^\circ\ \text{A} \\ V_S &= V_1 + I_S \frac{Z}{2} = 66\,754.4\angle 2.88^\circ\ \text{V} \end{aligned}$$

Sending-end line voltage $= \sqrt3 \times 66.754 = 115.62$ kV.

Regulation

At no load the receiving voltage is $V_S/|A|$, with $A = 1 + \frac{YZ}{2} = 0.9943\angle 0.13^\circ$:

$$\begin{aligned} V_{R0} &= \frac{66\,754.4}{0.9943} = 67\,138.3\ \text{V} \\ \%VR &= \frac{67\,138.3 - 63\,508.5}{63\,508.5}\times 100 = 5.72\% \end{aligned}$$

(If the no-load Ferranti rise is ignored, $(V_S - V_R)/V_R = 5.11\%$.)

Efficiency

$$\begin{aligned} P_{loss} &= 3\left(|I_R|^2 + |I_S|^2\right)\frac{R}{2} \\ &= 3\,(116.64^2 + 109.05^2)(7.65) = 0.585\ \text{MW} \\ \eta &= \frac{20}{20 + 0.585}\times 100 = 97.16\% \end{aligned}$$

Answer: Voltage regulation ≈ 5.72 % (sending end 115.62 kV), efficiency ≈ 97.16 %.

Ferranti effect

The Ferranti effect is the rise of receiving-end voltage above the sending-end voltage when a long or medium line is on no load or very light load.

• At light load, the line's shunt capacitance draws a charging current $I_C$ that leads the voltage by $90^\circ$.
• This current flows through the series inductance and produces a drop $jX_L I_C$, which points opposite to $V_R$ (since $jX_L \cdot jI = -X_L I$), so $V_S = V_R - X_L I_C$ is smaller than $V_R$.
• For a nominal-$\pi$ line at no load, $V_S = A V_R$ with $A = 1 - \omega^2 LC/2$, so $V_R = V_S/(1-\omega^2LC/2) > V_S$.
• The rise is roughly proportional to the square of the line length and is large for EHV lines (about 5–10 % at 400 km).

        jXL*Ic
   Vr --------> Vs  (Vs shorter than Vr)
   ^
   | Ic (leads Vr by 90 deg)

Remedy: shunt reactors at the receiving end, or keeping some load connected.

Skin effect

The skin effect is the tendency of alternating current to crowd towards the outer surface of a conductor instead of spreading uniformly over its cross-section.

• The inner filaments of a conductor link more flux than the outer ones, so they have higher inductive reactance. Current therefore takes the outer, lower-reactance path.
• The effective area carrying current falls, so the AC resistance is greater than the DC resistance.
• It increases with frequency, conductor diameter, permeability and conductivity. At 50 Hz it is small for normal conductors (a few percent) but is not negligible for large conductors.
• It is absent for DC.

Reduction: use stranded conductors (ACSR), hollow or tubular conductors, and steel core at the centre where little current flows anyway.

Use the power-circle equations (taking $D = A$). Let $A = |A|\angle\alpha$, $B = |B|\angle\beta$, $V_S$ leading $V_R$ by $\delta$. With line voltages in kV, powers come out directly in three-phase MW/MVAr.

Data: $|A| = 0.98$, $\alpha = 0.8^\circ$, $|B| = 45\ \Omega$, $\beta = 74^\circ$, $|V_S| = 132$ kV, $|V_R| = 134$ kV, $\delta = 10^\circ$.

Useful terms:

$$\begin{aligned} \frac{|V_S||V_R|}{|B|} &= \frac{132\times134}{45} = 393.07 \\ \frac{|A||V_R|^2}{|B|} &= \frac{0.98\times134^2}{45} = 391.04 \\ \frac{|A||V_S|^2}{|B|} &= \frac{0.98\times132^2}{45} = 379.46 \end{aligned}$$

Angles: $\beta-\delta = 64^\circ$, $\beta-\alpha = 73.2^\circ$, $\beta+\delta = 84^\circ$.

Receiving end

$$\begin{aligned} P_R &= 393.07\cos64^\circ - 391.04\cos73.2^\circ = 59.29\ \text{MW} \\ Q_R &= 393.07\sin64^\circ - 391.04\sin73.2^\circ = -21.07\ \text{MVAr} \\ \text{pf}_R &= \frac{59.29}{\sqrt{59.29^2 + 21.07^2}} = 0.942 \end{aligned}$$

$Q_R$ is negative, so reactive power flows from the receiving end into the line: the receiving-end pf is 0.942 leading. This is expected because $|V_R| > |V_S|$.

Sending end

$$\begin{aligned} P_S &= 379.46\cos73.2^\circ - 393.07\cos84^\circ = 68.59\ \text{MW} \\ Q_S &= 379.46\sin73.2^\circ - 393.07\sin84^\circ = -27.65\ \text{MVAr} \\ \text{pf}_S &= \frac{68.59}{\sqrt{68.59^2 + 27.65^2}} = 0.927\ \text{(leading)} \end{aligned}$$

Line losses

$$\begin{aligned} P_{loss} &= P_S - P_R = 68.59 - 59.29 = 9.30\ \text{MW} \\ Q_S - Q_R &= -27.65 - (-21.07) = -6.59\ \text{MVAr} \end{aligned}$$

The negative reactive difference means the line, on net, generates about 6.59 MVAr (its charging exceeds its $I^2X$ absorption).

Voltage regulation

No-load receiving voltage $= |V_S|/|A| = 132/0.98 = 134.69$ kV.

$$\%VR = \frac{134.69 - 134}{134}\times100 = 0.52\%$$

Answer: $P_R$ = 59.29 MW, $Q_R$ = −21.07 MVAr, pf 0.942 lead; $P_S$ = 68.59 MW, $Q_S$ = −27.65 MVAr, pf 0.927 lead; loss 9.30 MW; regulation 0.52 %.

Treat it as a short line: $A = 1\angle0^\circ$, $B = Z = 5 + j25 = 25.495\angle78.69^\circ\ \Omega$. With line kV and three-phase MW/MVAr, the receiving-end power circle equations are:

$$\begin{aligned} P_R &= \frac{|V_S||V_R|}{|Z|}\cos(\theta-\delta) - \frac{|V_R|^2}{|Z|}\cos\theta \\ Q_R &= \frac{|V_S||V_R|}{|Z|}\sin(\theta-\delta) - \frac{|V_R|^2}{|Z|}\sin\theta \end{aligned}$$

with $\theta = 78.69^\circ$.

Part 1: Compensation for $|V_S| = |V_R| = 33$ kV

$$\frac{33\times33}{25.495} = 42.714,\quad \frac{33^2}{25.495}\cos\theta = 8.377,\quad \frac{33^2}{25.495}\sin\theta = 41.885$$

Active power condition:

$$\begin{aligned} 15 &= 42.714\cos(78.69^\circ-\delta) - 8.377 \\ \cos(78.69^\circ-\delta) &= 0.5473 \Rightarrow \delta = 21.87^\circ \end{aligned}$$

Reactive power the line can deliver at this condition:

$$Q_R = 42.714\sin(56.82^\circ) - 41.885 = -6.14\ \text{MVAr}$$

The load needs $Q_L = 15\tan(\cos^{-1}0.8) = 11.25$ MVAr (lagging), and the line itself needs 6.14 MVAr fed into its receiving end. So the compensator must supply:

$$Q_C = 11.25 - (-6.14) = 17.39\ \text{MVAr}$$

Part 2: Extra load with $|V_R| = 28$ kV and the same $Q_C$

$$\frac{33\times28}{25.495} = 36.242,\quad \frac{28^2}{25.495}\cos\theta = 6.031,\quad \frac{28^2}{25.495}\sin\theta = 30.154$$

For a new load $P$ at 0.8 pf lag, the line must deliver $Q_R = 0.75P - 17.39$. So:

$$\begin{aligned} P &= 36.242\cos(78.69^\circ-\delta) - 6.031 \\ 0.75P - 17.39 &= 36.242\sin(78.69^\circ-\delta) - 30.154 \end{aligned}$$

Solving these together (eliminating $P$):

$$\delta = 31.33^\circ,\quad P = 18.52\ \text{MW},\quad Q_R = -3.49\ \text{MVAr}$$

Check: $0.75 \times 18.52 - 17.39 = -3.49$ MVAr.

Extra load $= 18.52 - 15 = 3.52$ MW.

Answer: Compensator capacity ≈ 17.39 MVAr; extra load ≈ 3.52 MW at 0.8 pf lag (total 18.52 MW at 28 kV).

Transmission efficiency rises with voltage because, for the same power, a higher voltage means a smaller current and hence much smaller $I^2R$ loss.

Consider a 3-phase line delivering power $P$ at line voltage $V$ and power factor $\cos\phi$. Let $R$ be the resistance per phase, $R = \rho l / a$.

Line current

$$I = \frac{P}{\sqrt3\, V\cos\phi}$$

Line loss

$$\begin{aligned} P_{loss} &= 3I^2R = 3\left(\frac{P}{\sqrt3 V\cos\phi}\right)^2 \frac{\rho l}{a} \\ &= \frac{P^2 \rho l}{V^2 \cos^2\phi\, a} \end{aligned}$$

Efficiency

$$\begin{aligned} \eta &= \frac{P}{P + P_{loss}} = \frac{1}{1 + \dfrac{P\rho l}{a V^2\cos^2\phi}} \\ &\approx 1 - \frac{P\rho l}{a\, V^2\cos^2\phi} \end{aligned}$$

For given $P$, $l$, $a$, $\rho$ and $\cos\phi$, the loss term is proportional to $1/V^2$. So as $V$ increases, the loss falls and $\eta$ increases.

Example

A line with $R = 10\ \Omega$/phase carries 10 MW at 0.8 pf.

Doubling the voltage cuts the loss to one quarter.

Related results

• For a fixed percentage loss, the conductor area needed is $a \propto 1/V^2$, so the conductor volume (and cost) falls as $1/V^2$.
• The percentage voltage drop $\approx I(R\cos\phi + X\sin\phi)/V$ also falls as $1/V^2$, so regulation improves.
• The limit is the cost of insulation, towers and switchgear, which rises with voltage; this sets the economic transmission voltage.

Start from the ABCD relation of a two-port line:

$$V_S = A V_R + B I_R$$

Take $V_R$ as reference and let

$$V_R = |V_R|\angle0^\circ,\quad V_S = |V_S|\angle\delta,\quad A = |A|\angle\alpha,\quad B = |B|\angle\beta$$

Receiving-end current

$$I_R = \frac{V_S - A V_R}{B} = \frac{|V_S|}{|B|}\angle(\delta-\beta) - \frac{|A||V_R|}{|B|}\angle(\alpha-\beta)$$

Receiving-end complex power

$$\begin{aligned} S_R &= P_R + jQ_R = V_R I_R^* \\ &= \frac{|V_S||V_R|}{|B|}\angle(\beta-\delta) - \frac{|A||V_R|^2}{|B|}\angle(\beta-\alpha) \end{aligned}$$

Separating real and imaginary parts:

$$\begin{aligned} P_R &= \frac{|V_S||V_R|}{|B|}\cos(\beta-\delta) - \frac{|A||V_R|^2}{|B|}\cos(\beta-\alpha) \\ Q_R &= \frac{|V_S||V_R|}{|B|}\sin(\beta-\delta) - \frac{|A||V_R|^2}{|B|}\sin(\beta-\alpha) \end{aligned}$$

With phase voltages these give power per phase; with line voltages they give total three-phase power.

Maximum receiving-end power

For fixed $|V_S|$, $|V_R|$ and line constants, only $\delta$ can vary. The second term is constant, so $P_R$ is maximum when $\cos(\beta-\delta) = 1$, i.e.

$$\delta = \beta$$ $$P_{R,max} = \frac{|V_S||V_R|}{|B|} - \frac{|A||V_R|^2}{|B|}\cos(\beta-\alpha)$$

At this point the reactive power at the receiving end is

$$Q_R = -\frac{|A||V_R|^2}{|B|}\sin(\beta-\alpha)$$

which is negative, so the load must supply (leading) reactive power to receive maximum active power.

For a short line ($A = 1\angle0^\circ$, $B = Z\angle\theta$):

$$P_{R,max} = \frac{|V_S||V_R|}{|Z|} - \frac{|V_R|^2 R}{|Z|^2}$$

Short line (15 km), so shunt capacitance is neglected.

Line constants

$$\begin{aligned} I &= \frac{5\times10^6}{\sqrt3 \times 11\,000 \times 0.8} = 328.04\ \text{A} \\ P_{loss} &= 0.12 \times 5 = 0.6\ \text{MW} = 3I^2R \\ R &= \frac{0.6\times10^6}{3\times 328.04^2} = 1.859\ \Omega \\ X &= 2\pi(50)(1.1\times10^{-3})(15) = 5.184\ \Omega \end{aligned}$$

(i) Sending-end voltage and regulation

$$\begin{aligned} V_R &= \frac{11\,000}{\sqrt3} = 6350.9\angle0^\circ\ \text{V} \\ I &= 328.04\angle-36.87^\circ\ \text{A} \\ V_S &= V_R + I(R + jX) \\ &= 6350.9 + 328.04\angle-36.87^\circ\,(1.859 + j5.184) \\ &= 7921.5\angle7.21^\circ\ \text{V} \end{aligned}$$

Line value $= \sqrt3 \times 7921.5 = 13.72$ kV.

$$\%VR = \frac{7921.5 - 6350.9}{6350.9}\times100 = 24.73\%$$

(ii) Load pf for zero regulation

Using the approximate drop $V_S - V_R \approx I(R\cos\phi \pm X\sin\phi)$, regulation is zero when the load is leading and

$$\begin{aligned} IR\cos\phi - IX\sin\phi &= 0 \\ \tan\phi &= \frac{R}{X} = \frac{1.859}{5.184} = 0.3585 \\ \phi &= 19.73^\circ,\quad \cos\phi = 0.941\ \text{leading} \end{aligned}$$

(Solving $|V_S| = |V_R|$ exactly with $P$ = 5 MW gives 0.890 leading.)

(iii) Capacitor at the receiving end

The 5 MW load stays at 0.8 lag (needs 3.75 MVAr). For zero regulation the net receiving-end reactive power must be

$$Q_{net} = -P\tan\phi = -5 \times 0.3585 = -1.793\ \text{MVAr}$$

Capacitor rating:

$$Q_C = 3.75 + 1.793 = 5.543\ \text{MVAr (3-phase)}$$

Capacitance per phase (star connected):

$$C = \frac{Q_C}{\omega V_L^2} = \frac{5.543\times10^6}{314.16\times 11\,000^2} = 145.8\ \mu\text{F}$$

Answer: (i) $V_S$ = 13.72 kV, regulation 24.73 %; (ii) pf ≈ 0.941 leading; (iii) 5.54 MVAr bank, ≈ 145.8 µF per phase (star).

Use the sending-end power-circle equation (taking $D = A$). With line voltages in kV, the result is in three-phase MW:

$$P_S = \frac{|A||V_S|^2}{|B|}\cos(\beta-\alpha) - \frac{|V_S||V_R|}{|B|}\cos(\beta+\delta)$$

Here $\delta$ is the angle by which $V_S$ leads $V_R$. Since $V_S = 132\angle0^\circ$ and $V_R = 130.5\angle10^\circ$,

$$\delta = 0^\circ - 10^\circ = -10^\circ$$

Data: $|A| = 0.97$, $\alpha = 0.5^\circ$, $|B| = 100\ \Omega$, $\beta = 75^\circ$.

Substitution

$$\begin{aligned} \frac{|A||V_S|^2}{|B|} &= \frac{0.97\times132^2}{100} = 169.01 \\ \frac{|V_S||V_R|}{|B|} &= \frac{132\times130.5}{100} = 172.26 \\ \beta-\alpha &= 74.5^\circ,\quad \beta+\delta = 65^\circ \end{aligned}$$ $$\begin{aligned} P_S &= 169.01\cos74.5^\circ - 172.26\cos65^\circ \\ &= 169.01(0.2672) - 172.26(0.4226) \\ &= 45.17 - 72.80 = -27.63\ \text{MW} \end{aligned}$$

Meaning

$P_S$ is negative because the receiving-end voltage leads the sending-end voltage by $10^\circ$. Active power flows from the leading to the lagging end, so about 27.6 MW actually flows out of the line at the sending end (from the "receiving" bus towards the "sending" bus).

Answer: $P_S$ ≈ −27.63 MW (27.63 MW flows into the sending-end bus).

Short line with $Z = 20.62\angle75.96^\circ = 5.00 + j20.00\ \Omega$ per phase.

Method

Take $V_R$ (line, kV) as reference. For three-phase $P + jQ$ in MW/MVAr at the receiving end:

$$V_S = V_R + \frac{Z\,(P - jQ)}{V_R}$$

Multiplying by $V_R$ and taking magnitudes:

$$|V_S|^2|V_R|^2 = (V_R^2 + RP + XQ)^2 + (XP - RQ)^2$$

Substitution

At 0.8 pf lagging, $Q = 0.75P$. With $V_R = 33$, $V_S = 46.85$, $R = 5$, $X = 20$:

$$\begin{aligned} (1089 + 5P + 15P)^2 + (20P - 3.75P)^2 &= (46.85\times33)^2 \\ (1089 + 20P)^2 + (16.25P)^2 &= 2\,390\,270 \end{aligned}$$

Expanding:

$$664.06P^2 + 43\,560P - 1\,204\,349 = 0$$

Taking the positive root:

$$P = 20.95\ \text{MW},\quad Q = 0.75P = 15.71\ \text{MVAr}$$

Check: $|33 + (5 + j20)(20.95 - j15.71)/33| = 46.85$ kV.

Sending-end power factor

$$\begin{aligned} I &= \frac{\sqrt{20.95^2 + 15.71^2}}{\sqrt3\times33} = 0.4581\ \text{kA} = 458.1\ \text{A} \\ P_{loss} &= 3I^2R = 3(458.1)^2(5) = 1.050\ \text{MW} \\ Q_{loss} &= 3I^2X = 3(458.1)^2(20) = 4.199\ \text{MVAr} \\ P_S &= 20.95 + 1.05 = 22.00\ \text{MW} \\ Q_S &= 15.71 + 4.20 = 19.91\ \text{MVAr} \\ \text{pf}_S &= \frac{22.00}{\sqrt{22.00^2 + 19.91^2}} = 0.741\ \text{lagging} \end{aligned}$$

Answer: Power output ≈ 20.95 MW at 0.8 pf lag; sending-end pf ≈ 0.741 lagging.

Use the receiving-end power-circle equations with $|V_S| = |V_R| = 275$ kV, $|A| = 0.85$, $\alpha = 5^\circ$, $|B| = 200\ \Omega$, $\beta = 75^\circ$ (line kV give three-phase MW/MVAr):

$$\begin{aligned} P_R &= \frac{|V_S||V_R|}{|B|}\cos(\beta-\delta) - \frac{|A||V_R|^2}{|B|}\cos(\beta-\alpha) \\ Q_R &= \frac{|V_S||V_R|}{|B|}\sin(\beta-\delta) - \frac{|A||V_R|^2}{|B|}\sin(\beta-\alpha) \end{aligned}$$

Constants

$$\frac{275\times275}{200} = 378.13,\qquad \frac{0.85\times275^2}{200} = 321.41,\qquad \beta-\alpha = 70^\circ$$

Unity pf condition: $Q_R = 0$

$$\begin{aligned} 378.13\sin(75^\circ-\delta) &= 321.41\sin70^\circ \\ \sin(75^\circ-\delta) &= 0.85\times0.9397 = 0.7987 \\ 75^\circ-\delta &= 53.01^\circ \Rightarrow \delta = 21.99^\circ \end{aligned}$$

Power received

$$\begin{aligned} P_R &= 378.13\cos53.01^\circ - 321.41\cos70^\circ \\ &= 227.51 - 109.93 = 117.58\ \text{MW} \end{aligned}$$

Answer: About 117.6 MW can be received at unity pf, with load angle $\delta$ ≈ 22.0°.

A 110 km line is a medium line; it is solved here with the nominal-π model (half the shunt admittance at each end).

Line constants

$$\begin{aligned} R &= 0.153\times110 = 16.83\ \Omega \\ X &= 2\pi(50)(1.21\times10^{-3})(110) = 41.81\ \Omega \\ Y &= 2\pi(50)(0.00958\times10^{-6})(110) = 3.311\times10^{-4}\ \text{S} \\ Y/2 &= 1.655\times10^{-4}\ \text{S} \end{aligned}$$

Receiving end

$$\begin{aligned} V_R &= 110\,000/\sqrt3 = 63\,508.5\angle0^\circ\ \text{V} \\ I_R &= \frac{20\times10^6}{3\times63\,508.5\times0.9} = 116.64\angle-25.84^\circ\ \text{A} \end{aligned}$$

Nominal-π steps

$$\begin{aligned} I_{C1} &= j\frac{Y}{2}V_R = j10.51\ \text{A} \\ I_L &= I_R + I_{C1} = 112.45\angle-21.02^\circ\ \text{A} \\ V_S &= V_R + I_L Z = 63\,508.5 + 5068.8\angle47.06^\circ \\ &= 67\,064.3\angle3.17^\circ\ \text{V} \\ I_{C2} &= j\frac{Y}{2}V_S = 11.10\angle93.17^\circ\ \text{A} \\ I_S &= I_L + I_{C2} = 108.38\angle-15.65^\circ\ \text{A} \end{aligned}$$

Results

• Sending-end line voltage $= \sqrt3\times67.064 = 116.16$ kV
• Sending-end current $= 108.38$ A
• Sending-end pf $= \cos(3.17^\circ + 15.65^\circ) = 0.947$ lagging

Regulation (no-load $V_R = V_S/|A|$, $A = 1 + YZ/2 = 0.9931\angle0.16^\circ$):

$$\%VR = \frac{67\,064.3/0.9931 - 63\,508.5}{63\,508.5}\times100 = \frac{67\,531.4 - 63\,508.5}{63\,508.5}\times100 = 6.33\%$$

Efficiency:

$$\begin{aligned} P_{loss} &= 3|I_L|^2R = 3(112.45)^2(16.83) = 0.638\ \text{MW} \\ \eta &= \frac{20}{20.638}\times100 = 96.91\% \end{aligned}$$

Answer: $V_S$ = 116.16 kV, $I_S$ = 108.38 A, pf 0.947 lag, regulation 6.33 %, efficiency 96.91 %. (A nominal-T solution gives almost the same: 116.13 kV, 6.30 %, 96.90 %.)

Short line, $Z = 4 + j8\ \Omega$/phase, $V_R = 33$ kV, load 20 MW at 0.995 pf leading.

Efficiency

$$\begin{aligned} I &= \frac{20\times10^6}{\sqrt3\times33\,000\times0.995} = 351.67\ \text{A} \\ P_{loss} &= 3I^2R = 3(351.67)^2(4) = 1.484\ \text{MW} \\ \eta &= \frac{20}{20 + 1.484}\times100 = 93.09\% \end{aligned}$$

Present regulation (for reference): $\phi = 5.73^\circ$ leading,

$$V_S = 19\,052.6 + 351.67\angle5.73^\circ\,(4 + j8) = 20\,384.3\angle8.29^\circ\ \text{V}$$

i.e. 35.31 kV line, regulation $= 6.99\%$. It is still positive because the pf is only slightly leading.

Making regulation zero

For a short line,

$$\%VR \approx \frac{I(R\cos\phi - X\sin\phi)}{V_R}\times 100 \quad (\text{leading pf})$$

This is zero when

$$\tan\phi = \frac{R}{X} = \frac{4}{8} = 0.5 \Rightarrow \cos\phi = 0.894\ \text{leading}$$

So regulation is made zero by making the load (or load plus compensator) take more leading current:

• Connect a shunt capacitor bank / synchronous condenser at the receiving end.
• Required net leading reactive power: $Q = P\tan\phi = 20\times0.5 = 10$ MVAr.
• The load already supplies $20\tan(5.73^\circ) = 2.01$ MVAr leading, so the capacitor needed is

$$Q_C = 10 - 2.01 = 7.99\ \text{MVAr}$$ $$C_{ph} = \frac{Q_C}{\omega V_L^2} = \frac{7.99\times10^6}{314.16\times33\,000^2} = 23.36\ \mu\text{F (star)}$$

(An exact solution of $|V_S| = |V_R|$ gives net pf 0.847 leading and $Q_C$ ≈ 10.55 MVAr.)

Answer: Efficiency ≈ 93.09 %. Zero regulation needs a net pf of about 0.894 leading, i.e. a shunt capacitor of about 8 MVAr at the receiving end.

Derivation of receiving-end complex power

For any line, $V_S = AV_R + BI_R$. Take

$$V_R = |V_R|\angle0^\circ,\ V_S = |V_S|\angle\delta,\ A = |A|\angle\alpha,\ B = |B|\angle\beta$$

Then

$$I_R = \frac{V_S - AV_R}{B} = \frac{|V_S|}{|B|}\angle(\delta-\beta) - \frac{|A||V_R|}{|B|}\angle(\alpha-\beta)$$

Receiving-end complex power (per phase; with line voltages it is three-phase):

$$\begin{aligned} S_R &= V_R I_R^* \\ &= \frac{|V_S||V_R|}{|B|}\angle(\beta-\delta) - \frac{|A||V_R|^2}{|B|}\angle(\beta-\alpha) \end{aligned}$$

In rectangular form:

$$\begin{aligned} P_R &= \frac{|V_S||V_R|}{|B|}\cos(\beta-\delta) - \frac{|A||V_R|^2}{|B|}\cos(\beta-\alpha) \\ Q_R &= \frac{|V_S||V_R|}{|B|}\sin(\beta-\delta) - \frac{|A||V_R|^2}{|B|}\sin(\beta-\alpha) \end{aligned}$$

Construction of the receiving-end power circle

Write the result as

$$\left(P_R + \frac{|A||V_R|^2}{|B|}\cos(\beta-\alpha)\right)^2 + \left(Q_R + \frac{|A||V_R|^2}{|B|}\sin(\beta-\alpha)\right)^2 = \left(\frac{|V_S||V_R|}{|B|}\right)^2$$

This is a circle in the $P$–$Q$ plane with:

• Centre $N$: $\left(-\frac{|A||V_R|^2}{|B|}\cos(\beta-\alpha),\ -\frac{|A||V_R|^2}{|B|}\sin(\beta-\alpha)\right)$, fixed for a fixed $|V_R|$.
• Radius $NM = \frac{|V_S||V_R|}{|B|}$.

Steps:

1. Draw $P$ (horizontal) and $Q$ (vertical) axes with origin $O$.
2. From $O$ draw line $ON$ of length $|A||V_R|^2/|B|$ at angle $(\beta-\alpha)$ below the negative $P$-axis; $N$ is the centre.
3. With centre $N$ draw an arc of radius $|V_S||V_R|/|B|$.
4. From $O$ draw the load line at angle $\phi_R$ above the $P$-axis (lagging); it cuts the circle at the operating point $M$. $M$ gives $P_R$ and $Q_R$.
5. Angle between $NM$ and the horizontal through $N$ is $(\beta-\delta)$, so $\delta$ can be read off.

          Q
          |        . M (P_R, Q_R)
          |     .   /
          |  .     /  load line (phi_R)
  --------O-------/------------- P
         /|     .
        / |  .   radius = |Vs||Vr|/|B|
       /  |.
      N (centre)
   angle (beta-alpha) below -P axis

The maximum $P_R$ is at the point where the radius $NM$ is horizontal ($\delta = \beta$). Circles for different $|V_S|$ are concentric about $N$.