Interconnected Power System

In an interconnected system, a mismatch of active power (generation ≠ load + losses) shows up as a change in frequency, while a mismatch of reactive power shows up as a change in bus voltage magnitude. The two effects are nearly independent ("decoupled") because of the high X/R ratio of transmission lines.

Active power mismatch and frequency

All synchronous machines in an interconnected system run at the same electrical speed. The rotor of each machine obeys the swing equation:

$$\frac{2H}{\omega_s}\frac{d^2\delta}{dt^2} = P_m - P_e$$

• If load increases suddenly ($P_e > P_m$), the extra energy is first taken from the kinetic energy stored in the rotating masses ($\tfrac{1}{2}J\omega^2$). The rotors slow down and system frequency falls.
• If load drops ($P_e < P_m$), the surplus energy accelerates the rotors and frequency rises.
• Since frequency is common to the whole network, the mismatch is shared by all units in proportion to their inertia and then their governor droop.

Frequency is restored in steps:

1. Primary control (governor): each governor senses the speed change and opens/closes the steam or water valve. Steady-state change is $\Delta f = -\dfrac{\Delta P_L}{\sum (1/R_i) + D}$, so a small offset remains.
2. Secondary control (AGC / LFC): the speed changer is adjusted until $\Delta f = 0$ and tie-line flows return to scheduled values (area control error $ACE = \Delta P_{tie} + B\,\Delta f \to 0$).

Reactive power mismatch and voltage

For a line between buses with voltages $V_1$, $V_2$ and reactance $X$ (R neglected, small $\delta$):

$$Q_{12} \approx \frac{V_1(V_1 - V_2)}{X} \quad\Rightarrow\quad \Delta V \approx \frac{Q X}{V}$$

• Reactive power flows from the higher-voltage bus to the lower-voltage bus, and the voltage drop across a line depends mainly on the Q it carries.
• If reactive demand exceeds local supply, Q must be imported over the reactance, so the voltage at that bus falls. Excess Q (e.g. a lightly loaded long line with its charging) makes the voltage rise (Ferranti effect).
• Unlike frequency, voltage is a local quantity. Reactive power cannot be sent far efficiently, so the mismatch must be fixed near where it occurs.

Control: generator excitation (AVR), shunt capacitors/reactors, synchronous condensers, SVC/STATCOM and on-load tap-changing transformers.

Why P–f and Q–V are decoupled

Since $X \gg R$:

$$P \approx \frac{V_1V_2}{X}\sin\delta,\qquad Q \approx \frac{V_1(V_1 - V_2\cos\delta)}{X}$$

P depends mainly on the angle $\delta$ (linked to rotor position and frequency); Q depends mainly on the voltage magnitudes. So $\partial P/\partial \delta$ and $\partial Q/\partial |V|$ are large, while $\partial P/\partial |V|$ and $\partial Q/\partial \delta$ are small.

An interconnected power system is one in which several generating stations and areas are tied together by transmission lines (tie-lines), so that they all run in synchronism and share load. Compared with isolated systems, it is cheaper and more reliable, and its frequency is held by keeping total generation equal to total load.

Advantages over an isolated system

1. Better reliability: if a unit or plant trips, the others pick up its load through tie-lines, so supply continues.
2. Less reserve capacity needed: spinning and standby reserve is shared by all areas, so each area keeps less idle capacity.
3. Economic operation: the cheapest units (e.g. run-of-river hydro in Nepal) are loaded first and power is sent to where it is needed (economic dispatch across areas).
4. Use of diversity: peak demands of different areas occur at different times, so the total installed capacity required is lower; load factor and plant factor improve.
5. Larger, more efficient units can be installed because the large system can absorb them.
6. Better use of energy resources: surplus hydro energy in the wet season can be exported (e.g. Nepal–India power trade) and imported in the dry season.
7. Better frequency stability: the combined inertia is large, so a load change produces only a small frequency change.
8. Easier maintenance scheduling, since units can be taken out while others supply load.

Real power–frequency balance

At steady state, total mechanical input equals total load plus losses, and frequency is constant. When load changes by $\Delta P_L$:

1. Inertia response: the deficit is supplied from the kinetic energy of all rotors, so speed and frequency fall (swing equation $\frac{2H}{\omega_s}\frac{d^2\delta}{dt^2} = P_m - P_e$).
2. Primary (governor) control: each governor with droop $R$ raises mechanical power by $\Delta P_{m,i} = -\Delta f / R_i$. Load also falls a little with frequency (damping $D$). Steady-state deviation:

$$\Delta f = \frac{-\Delta P_L}{\sum_i \dfrac{1}{R_i} + D}$$

All interconnected units share the load change in inverse proportion to their droop. 3. Secondary control (AGC): in each area an integral controller acts on the Area Control Error

$$ACE = \Delta P_{tie} + B\,\Delta f$$

and moves the speed-changer set points until $ACE = 0$. Then frequency returns to the nominal value (50 Hz in Nepal) and tie-line flows return to schedule, so each area finally supplies its own load change. 4. Tertiary control: economic dispatch resets the unit outputs for minimum cost.

 Load change -> f falls -> Governor opens valve
      ^                         |
      |                         v
   P_gen = P_load  <-  AGC resets speed changer

An interconnected power system is a network in which two or more generating stations or control areas are connected by transmission lines (tie-lines) and operate in parallel at the same frequency, sharing generation and load. A national grid such as Nepal's INPS, which is also tied to India, is an example.

Advantages over an isolated power system

1. Higher reliability: when a generator or plant fails, other plants supply the load through tie-lines.
2. Reduced reserve capacity: spinning reserve is shared, so each station needs less standby capacity.
3. Diversity of loads: peaks of different areas occur at different times, so total installed capacity is lower and load factor is better.
4. Economic operation: cheaper sources (hydro, base-load plants) are used fully and costly units only at peak; economic load dispatch across the whole system.
5. Larger, efficient units: bigger machines with lower cost per kW can be installed.
6. Exchange of surplus power: extra energy (e.g. wet-season hydro) can be exported and deficit imported.
7. Better frequency regulation: large combined inertia keeps frequency more stable for a given load change.
8. Easier maintenance: units can be shut down for overhaul without load shedding.

A short drawback note: fault levels rise and faults can spread, so better protection and coordination are needed.

An interconnected power system is a system in which several generating stations and areas are linked together by transmission lines (tie-lines) and run in synchronism, so that all generators jointly supply the total load. An isolated system is a single station feeding its own load.

Merits (advantages)

1. Reliability and continuity: a failed unit's load is taken up by other stations.
2. Lower reserve requirement: reserve capacity is pooled for the whole system.
3. Load diversity: different areas peak at different times, so less total capacity is needed; better load factor.
4. Economy: cheapest generation is used first (economic dispatch); costly peaking plants run less.
5. Large efficient units can be installed.
6. Power exchange: surplus energy can be sold and deficit bought (e.g. Nepal–India cross-border trade).
7. Better frequency stability due to large total inertia.
8. Flexible maintenance scheduling.

Demerits (limitations)

1. Higher fault level: more sources feed a fault, so circuit breakers of higher rupturing capacity are needed.
2. Cascading failures: a disturbance in one area can spread and cause a wide-area blackout if not isolated quickly.
3. Stability problems: long tie-lines and many machines make transient and dynamic stability harder to maintain; inter-area oscillations can occur.
4. Complex control and protection: load-frequency control, voltage control, tie-line scheduling and protection coordination become complex.
5. Need for coordination among utilities or countries on frequency, scheduling and tariffs.
6. High capital cost of tie-lines, substations and communication (SCADA) systems.

The bus admittance matrix $Y_{bus}$ relates the bus current injections to the bus voltages, $\mathbf{I}_{bus} = \mathbf{Y}_{bus}\mathbf{V}_{bus}$. It is formed by applying KCL at each bus of the network, with lines represented by their π-model.

Network and π-model

Consider three buses connected by lines 1-2, 1-3 and 2-3. Each line $i$-$k$ has series admittance $y_{ik} = 1/(R_{ik} + jX_{ik})$ and total shunt charging admittance $y'_{ik}$, half of which ($y'_{ik}/2$) is placed at each end.

          y12                 y23
  (1)----[===]------(2)------[===]----(3)
   |  \              |               /  |
 y'12/2 \  y13      y'12/2+y'23/2   / y'23/2
   |     \--[===]--------------/       |
  ===   (y'13/2 at buses 1 and 3)     ===
  I1 injected at 1, I2 at 2, I3 at 3

Steps of forming Ybus

Step 1: KCL at bus 1. Current injected = current in series branches + current in shunt branches:

$$\begin{aligned} I_1 &= y_{12}(V_1 - V_2) + y_{13}(V_1 - V_3) + \frac{y'_{12}}{2}V_1 + \frac{y'_{13}}{2}V_1 \\ &= \left(y_{12} + y_{13} + \frac{y'_{12}}{2} + \frac{y'_{13}}{2}\right)V_1 - y_{12}V_2 - y_{13}V_3 \end{aligned}$$

Step 2: KCL at bus 2 and bus 3 in the same way:

$$\begin{aligned} I_2 &= -y_{12}V_1 + \left(y_{12} + y_{23} + \frac{y'_{12}}{2} + \frac{y'_{23}}{2}\right)V_2 - y_{23}V_3 \\ I_3 &= -y_{13}V_1 - y_{23}V_2 + \left(y_{13} + y_{23} + \frac{y'_{13}}{2} + \frac{y'_{23}}{2}\right)V_3 \end{aligned}$$

Step 3: Write in matrix form.

$$\begin{bmatrix} I_1 \\ I_2 \\ I_3 \end{bmatrix} = \begin{bmatrix} Y_{11} & Y_{12} & Y_{13} \\ Y_{21} & Y_{22} & Y_{23} \\ Y_{31} & Y_{32} & Y_{33} \end{bmatrix} \begin{bmatrix} V_1 \\ V_2 \\ V_3 \end{bmatrix}$$

Step 4: Read off the elements, e.g. $Y_{11} = y_{12} + y_{13} + \frac{y'_{12}}{2} + \frac{y'_{13}}{2}$, $Y_{12} = Y_{21} = -y_{12}$.

Diagonal elements (self or driving-point admittance)

$$Y_{ii} = \sum_{k \ne i} y_{ik} + \sum_{k \ne i} \frac{y'_{ik}}{2}$$

$Y_{ii}$ is the sum of all admittances connected to bus $i$ (series and shunt). Physically, $Y_{ii} = I_i/V_i$ with all other buses short-circuited ($V_k = 0$).

Off-diagonal elements (mutual or transfer admittance)

$$Y_{ik} = Y_{ki} = -y_{ik}$$

It is the negative of the series admittance between buses $i$ and $k$; it is zero if no line joins them. Physically, $Y_{ik} = I_i/V_k$ with all buses except $k$ shorted.

Properties

• Symmetric ($Y_{ik} = Y_{ki}$) for networks without phase shifters.
• Sparse: in large systems most off-diagonal terms are zero.
• Easy to form directly by inspection and to modify when a line is added or removed.

Convert each voltage source with its series impedance into an equivalent current source (Norton form) at its bus, then write KCL at each bus.

Step 1: Source branches

Each source reactance is in series with its transformer, so the total series reactance to each bus is:

Equivalent injected currents $I_i = E_i\,y_{i0}$:

$$\begin{aligned} I_1 &= \frac{1\angle 0^\circ}{j0.2} = -j5\ \text{pu} \\ I_2 &= \frac{1\angle 0^\circ}{j0.25} = -j4\ \text{pu} \\ I_3 &= \frac{1\angle -36.87^\circ}{j0.2} = 5\angle -126.87^\circ = -3 - j4\ \text{pu} \end{aligned}$$

Step 2: Line admittances

$$\begin{aligned} y_{12} &= \frac{1}{0.02 + j0.04} = 10 - j20\ \text{pu} \\ y_{13} &= \frac{1}{0.01 + j0.03} = 10 - j30\ \text{pu} \\ y_{23} &= \frac{1}{0.0125 + j0.025} = 16 - j32\ \text{pu} \end{aligned}$$

Step 3: Node equations (KCL)

$$\begin{aligned} I_1 &= y_{10}V_1 + y_{12}(V_1 - V_2) + y_{13}(V_1 - V_3) \\ I_2 &= y_{20}V_2 + y_{12}(V_2 - V_1) + y_{23}(V_2 - V_3) \\ I_3 &= y_{30}V_3 + y_{13}(V_3 - V_1) + y_{23}(V_3 - V_2) \end{aligned}$$

Collecting terms:

$$\begin{aligned} -j5 &= (y_{10} + y_{12} + y_{13})V_1 - y_{12}V_2 - y_{13}V_3 \\ -j4 &= -y_{12}V_1 + (y_{20} + y_{12} + y_{23})V_2 - y_{23}V_3 \\ -3 - j4 &= -y_{13}V_1 - y_{23}V_2 + (y_{30} + y_{13} + y_{23})V_3 \end{aligned}$$

Step 4: Ybus elements

$$\begin{aligned} Y_{11} &= -j5 + (10 - j20) + (10 - j30) = 20 - j55 \\ Y_{22} &= -j4 + (10 - j20) + (16 - j32) = 26 - j56 \\ Y_{33} &= -j5 + (10 - j30) + (16 - j32) = 26 - j67 \\ Y_{12} &= Y_{21} = -10 + j20 \\ Y_{13} &= Y_{31} = -10 + j30 \\ Y_{23} &= Y_{32} = -16 + j32 \end{aligned}$$

Result

$$\begin{bmatrix} -j5 \\ -j4 \\ -3 - j4 \end{bmatrix} = \begin{bmatrix} 20 - j55 & -10 + j20 & -10 + j30 \\ -10 + j20 & 26 - j56 & -16 + j32 \\ -10 + j30 & -16 + j32 & 26 - j67 \end{bmatrix} \begin{bmatrix} V_1 \\ V_2 \\ V_3 \end{bmatrix}$$

Answer: $Y_{bus}$ is the 3×3 matrix above (pu); it is symmetric, and its diagonal terms include the source-branch admittances $-j5$, $-j4$, $-j5$.

With the nominal π model, each line's total series reactance is $x \times \text{length}$ and its total shunt susceptance $B = b \times \text{length}$ is split as $B/2$ at each end. Resistance is not given, so it is taken as zero.

Given: $x = 0.001$ pu/km, $b = 0.0016$ pu/km.

Step 1: Line parameters

Step 2: Diagonal elements

$Y_{ii}$ = sum of series admittances at bus $i$ + sum of half-line-charging susceptances at bus $i$:

$$\begin{aligned} Y_{11} &= (-j10 - j5) + j(0.08 + 0.16) = -j14.76 \\ Y_{22} &= (-j10 - j2.5) + j(0.08 + 0.32) = -j12.10 \\ Y_{33} &= (-j5 - j2.5) + j(0.16 + 0.32) = -j7.02 \end{aligned}$$

Step 3: Off-diagonal elements

$$\begin{aligned} Y_{12} &= Y_{21} = -y_{12} = j10 \\ Y_{13} &= Y_{31} = -y_{13} = j5 \\ Y_{23} &= Y_{32} = -y_{23} = j2.5 \end{aligned}$$

Result

$$Y_{bus} = \begin{bmatrix} -j14.76 & j10 & j5 \\ j10 & -j12.10 & j2.5 \\ j5 & j2.5 & -j7.02 \end{bmatrix}\ \text{pu}$$

Answer: $Y_{11} = -j14.76$, $Y_{22} = -j12.10$, $Y_{33} = -j7.02$, $Y_{12} = j10$, $Y_{13} = j5$, $Y_{23} = j2.5$ pu. The shunt charging only reduces the magnitude of the diagonal terms; off-diagonal terms are unaffected.

An interconnected power system links many generating stations and areas by tie-lines so that they operate in synchronism. Its two main control loops are the load-frequency (P–f) loop and the reactive power–voltage (Q–V) loop.

Advantages

• Higher reliability; a failed unit's load is picked up by others.
• Less reserve capacity, as reserve is shared.
• Load diversity gives lower installed capacity and better load factor.
• Economic dispatch of the cheapest units; power exchange between areas.
• Larger combined inertia gives better frequency regulation.

Disadvantages

• Higher fault levels, so higher-rated breakers are needed.
• A fault can cascade and black out a large area.
• Stability and protection coordination become more complex.
• Costly tie-lines and communication/SCADA systems.

Control schematic

           Speed changer (AGC set point)
                   |
   f sensor --> Governor --> Valve --> Turbine
     ^                                   |  Pm
     |                                   v
     +-----------  Generator <-----------+
     |               |  ^
  Bus voltage V      |  | Field current
     |               v  |
     +--> Comparator --> AVR --> Exciter
          (V_ref)

Load / frequency control (P–f loop)

1. A load increase makes $P_e > P_m$. The deficit is taken from rotor kinetic energy, so speed and frequency fall.
2. The speed governor senses $\Delta f$ and opens the steam/water valve, increasing $P_m$. With droop $R$, the steady-state change is $\Delta f = -\Delta P_L / (\sum 1/R + D)$; a small error remains.
3. Automatic Generation Control (AGC) adds an integral of the area control error $ACE = \Delta P_{tie} + B\Delta f$ to the speed-changer set point, bringing $\Delta f$ and tie-line deviation to zero.

Reactive power / voltage control (Q–V loop)

1. The terminal voltage is measured, rectified and compared with $V_{ref}$.
2. The error is amplified by the AVR and drives the exciter, which changes the field current $I_f$ and hence the internal emf $E_f$.
3. Higher $E_f$ makes the generator supply more Q and raises the bus voltage; lower $E_f$ absorbs Q and lowers it.
4. In the network, voltage is also held by shunt capacitors/reactors, synchronous condensers, SVC/STATCOM and tap-changing transformers.

Because $X \gg R$, P depends mainly on $\delta$ and Q mainly on $|V|$, so the two loops can be designed and operated independently.

Interconnection means tying several generating stations and control areas together by transmission lines so that they work as one synchronous system and jointly supply the total load.

Importance of interconnection

1. Reliability: if a generator or plant trips, others supply its load through tie-lines.
2. Reduced reserve: reserve capacity is shared, so less idle plant is needed.
3. Load diversity: different peak times mean lower total installed capacity and higher load factor.
4. Economy: economic dispatch uses the cheapest sources first; surplus energy can be exchanged (e.g. Nepal exporting wet-season hydro power to India).
5. Large efficient units become practical.
6. Frequency stability: large total inertia keeps frequency deviations small.
7. Maintenance can be planned without load shedding.

Real power – frequency balance

At constant frequency, total generation equals total demand plus losses: $\sum P_G = \sum P_D + P_{loss}$.

• A sudden load increase is first met from the kinetic energy of all rotating machines. By the swing equation $\frac{2H}{\omega_s}\frac{d^2\delta}{dt^2} = P_m - P_e$, they decelerate and frequency falls.
• Primary control: speed governors increase turbine input. With droop $R$ and load damping $D$, the steady-state deviation is $\Delta f = -\Delta P_L/(\sum 1/R_i + D)$. All connected units share the change.
• Secondary control (AGC): an integral controller acting on $ACE = \Delta P_{tie} + B\,\Delta f$ adjusts the speed changers until frequency returns to 50 Hz and tie-line flows return to schedule.

So, real power mismatch → frequency change, and frequency is a system-wide quantity.

Reactive power – voltage balance

For a line of reactance $X$:

$$Q_{12} \approx \frac{V_1(V_1 - V_2\cos\delta)}{X} \approx \frac{V_1\,\Delta V}{X}$$

• The voltage difference between two buses depends mainly on the reactive power flowing between them.
• If reactive demand at a bus exceeds local supply, Q must flow in over the line reactance, and the bus voltage falls. Surplus Q (light load, line charging) makes the voltage rise.
• Voltage is a local quantity; Q cannot be transmitted far because of high $I^2X$ losses, so it is balanced locally.
• Means of control: generator excitation with AVR, shunt capacitors and reactors, synchronous condensers, SVC/STATCOM, and on-load tap changers.

So, reactive power mismatch → voltage change.

Role of interconnection

Interconnection ties generating stations and areas together by tie-lines so they run in synchronism and share load. Its roles are:

• Reliability: loss of a unit is covered by other plants.
• Reserve sharing: less spinning reserve per area.
• Economy: cheapest generation is used first; surplus power can be exchanged between areas or countries.
• Load diversity: lower total installed capacity.
• Frequency stability: large combined inertia limits frequency swings.

Control of reactive power

Bus voltage depends mainly on reactive power ($\Delta V \approx QX/V$), so Q is controlled to keep voltage within limits (about ±5%). Methods:

1. Generator excitation (AVR): raising field current increases $E_f$, so the generator supplies Q (over-excited); lowering it absorbs Q (under-excited), within the capability-curve limits.
2. Shunt capacitors: supply Q at heavy load, raising voltage; switched in steps.
3. Shunt reactors: absorb Q at light load (e.g. on long EHV lines) to limit the Ferranti rise.
4. Synchronous condensers: unloaded synchronous motors that give or absorb Q smoothly by excitation control.
5. Static VAR compensators (SVC) and STATCOM: thyristor or converter-based devices with fast, continuous Q control.
6. Tap-changing transformers (OLTC): redistribute Q flow and adjust secondary voltage.
7. Series capacitors: cancel part of line reactance, reducing the reactive drop $IX$.

In an interconnected system all generators run at one common frequency. Frequency stays constant only while total generation = total load + losses. Real power/frequency control (load-frequency control, LFC) keeps this balance.

What happens on a load change

1. Inertial response (first seconds): when load rises by $\Delta P_L$, the deficit is supplied from the kinetic energy of all rotors. From the swing equation $\frac{2H}{\omega_s}\frac{d^2\delta}{dt^2} = P_m - P_e$, they slow down and frequency falls.
2. Primary control (governor): each speed governor senses $\Delta f$ and opens the turbine valve/gate. With speed regulation (droop) $R$:

$$\Delta P_{m,i} = -\frac{\Delta f}{R_i},\qquad \Delta f_{ss} = \frac{-\Delta P_L}{\sum_i \dfrac{1}{R_i} + D}$$

The load change is shared by all units of all areas, in inverse proportion to their droop. A steady-state frequency error remains, and tie-line flows change from schedule. 3. Secondary control (AGC): each area computes its Area Control Error

$$ACE_i = \Delta P_{tie,i} + B_i\,\Delta f$$

An integral controller changes the speed-changer set points until $ACE_i = 0$. Frequency returns to nominal and each area again supplies its own load, with tie flows back at schedule. 4. Tertiary control: economic dispatch sets the new outputs at minimum cost.

 dPL -> [Power system] -> df
           ^               |
           |  dPm          v
      [Turbine] <- [Governor 1/R] <- df
           ^
     speed changer <- integral(ACE)

Example: in a two-area system, if area 1 load increases, at first both areas raise generation and power flows from area 2 to area 1 over the tie-line; AGC in area 1 then raises its own generation so the tie-line flow returns to schedule.

Bus voltage magnitude in a power system is governed by the reactive power balance: Q supplied at or near a bus must equal Q consumed by loads and line reactances. A deficit lowers the voltage, a surplus raises it.

Why Q controls V

For a line of reactance $X$ with $X \gg R$:

$$\Delta V = V_1 - V_2 \approx \frac{Q_{12}X}{V_1}$$

So the voltage drop depends mainly on the reactive power carried. Q flows from the higher-voltage bus to the lower. Since $I^2X$ losses are large, Q cannot be sent over long distances; it must be balanced locally.

Sources and sinks of Q

• Sources: over-excited generators, shunt capacitors, line charging (lightly loaded lines), synchronous condensers.
• Sinks: induction motors, transformers and line reactances at heavy load, shunt reactors, under-excited generators.

How balance is maintained

1. AVR on generators: compares terminal voltage with the set value and changes field current; more excitation → more Q output → voltage rises.
2. Shunt capacitors switched in at heavy load; shunt reactors switched in at light load.
3. Synchronous condensers and SVC/STATCOM: continuous, fast Q adjustment.
4. On-load tap changers on transformers adjust the voltage ratio.
5. Series capacitors reduce net line reactance and hence the reactive drop.

Voltage is normally held within about ±5% of nominal; if Q support is not enough, voltage may fall progressively (voltage collapse).

Consequences

Real power on frequency: if load exceeds generation, the deficit is drawn from the kinetic energy of rotating machines, which slow down, so frequency falls; surplus generation makes frequency rise. Frequency is common to the whole interconnection. Large deviations cause under-frequency load shedding, damage to turbine blades, and slow motors.

Reactive power on voltage: if reactive demand exceeds local supply, Q flows in over the line reactance and causes a drop $\Delta V \approx QX/V$, so bus voltage falls; excess Q (light load, line charging) makes it rise. Voltage is a local quantity. Low voltage overheats motors and can lead to voltage collapse; high voltage stresses insulation.

Basic control mechanisms

 P-f loop:  df -> Governor -> Valve -> Turbine -> Pm
                    ^ AGC (integral of ACE)
 Q-V loop:  dV -> AVR -> Exciter -> Field -> Ef -> Q

• P–f: speed governor (primary, with droop) and Automatic Generation Control acting on $ACE = \Delta P_{tie} + B\Delta f$ (secondary) to bring frequency back to 50 Hz.
• Q–V: generator AVR/excitation control, shunt capacitors and reactors, synchronous condensers, SVC/STATCOM and on-load tap changers.

Role of reactive power injection or consumption at a bus

Bus voltage depends mainly on the net reactive power at that bus, since $\Delta V \approx QX/V$ for a line with $X \gg R$.

• Injecting Q at a bus (generator over-excited, capacitor, condenser) reduces the reactive current drawn over the lines, so the voltage rises.
• Consuming Q at a bus (inductive load, reactor, under-excited generator) increases the drop across line reactance, so the voltage falls.

So Q injection/absorption is the main tool to hold the bus voltage at its specified value; this is why a PV bus has its Q left free (within limits) to keep $|V|$ fixed.

Generator on an infinite bus

For a cylindrical-rotor generator with internal emf $E_f$, terminal voltage $V$ (fixed by the infinite bus), synchronous reactance $X_s$ and load angle $\delta$:

$$P = \frac{E_fV}{X_s}\sin\delta,\qquad Q = \frac{E_fV\cos\delta - V^2}{X_s}$$

With P fixed by the turbine, Q is set by the excitation $E_f$:

• $E_f\cos\delta > V$ (over-excited): Q > 0, generator supplies Q (lagging pf).
• $E_f\cos\delta < V$ (under-excited): Q < 0, generator absorbs Q (leading pf).

Maximum reactive power limit

The upper limit is set by field (rotor) heating. The maximum field current gives $E_{f,max}$, so

$$Q_{max} = \frac{V}{X_s}\sqrt{E_{f,max}^2 - \left(\frac{PX_s}{V}\right)^2} - \frac{V^2}{X_s}$$

On the P–Q capability chart this is a circle of radius $VE_{f,max}/X_s$ centred at $(0, -V^2/X_s)$. At high P, the armature current limit (circle of radius $VI_{a,max}$ about the origin) may set the bound instead.

Minimum reactive power limit

When the generator is under-excited:

1. Steady-state stability limit: as $E_f$ is reduced, $\delta$ increases for the same P; at $\delta = 90^\circ$ the machine loses synchronism. A practical margin (e.g. $\delta \le 70^\circ$) defines the limit.
2. Stator end-region heating in under-excited operation, which further restricts absorption.
3. A minimum excitation limit set by the exciter.

        Q (supplied)
        ^    field-heating limit
        |  .---.
        | /     \  armature-current limit
  ------+--------+-----> P
        | \     /  stability / end-heating limit
        |  '---'
        v Q (absorbed)

So in load-flow studies a PV bus has $Q_{min} \le Q_G \le Q_{max}$; if a limit is hit, Q is fixed at that limit and the bus is treated as a PQ bus.

A synchronous generator on an infinite bus has its terminal voltage $V_t$ and frequency fixed by the bus. With constant mechanical input (constant P), it can still be made to supply or absorb any reactive power within limits by changing its field excitation.

Equations

With synchronous reactance $X_s$ (R neglected), internal emf $E_f$ and load angle $\delta$:

$$E_f = V_t + jX_sI_a,\qquad P = \frac{E_fV_t}{X_s}\sin\delta,\qquad Q = \frac{E_fV_t\cos\delta - V_t^2}{X_s}$$

• P depends on the turbine input; since $V_t$ and $X_s$ are fixed, $E_f\sin\delta$ = constant. So as $E_f$ changes, the tip of $E_f$ moves along a line parallel to $V_t$.
• $I_a\cos\phi$ (active current) also remains constant.
• Q depends on $E_f\cos\delta - V_t$, which changes with excitation.

Three cases

     locus of Ef tip (P constant)
   --------x-------x--------x-------
          /       /        /
     Ef1 /   Ef2 /    Ef3 /     Ef1 < Ef2 < Ef3
        /       /        /
       O--------------------> Vt
   Ef1: under-excited (absorbs Q)
   Ef3: over-excited (supplies Q)

Explanation

• Increasing field current raises $E_f$; $\delta$ decreases and $I_a$ swings to lagging. The generator sends Q into the bus.
• Decreasing field current lowers $E_f$; $\delta$ increases and $I_a$ becomes leading. The generator draws Q from the bus.
• Terminal voltage remains constant because the infinite bus holds it.

Limits

Maximum Q is set by field heating (maximum $E_f$) and armature current; minimum Q (maximum absorption) by the steady-state stability limit ($\delta$ near 90°) and stator end heating. This is how generators are used for voltage/Q support in a real grid.

An Automatic Voltage Regulator (AVR) holds the generator terminal (bus) voltage at a set value by automatically adjusting the field excitation, which changes the generator's internal emf and hence its reactive power output.

Equivalent circuit of the synchronous generator

Neglecting armature resistance, the generator is an emf $E_f$ behind synchronous reactance $X_s$:

      jXs       Ia
  +--[~~~~]----->----+---- Vt (bus)
  |                  |
 (Ef)               Load
  |                  |
  +------------------+

$$V_t = E_f - jX_sI_a$$

Taking $V_t$ as reference and $I_a = I_p - jI_q$ (active and reactive components):

$$E_f = V_t + X_sI_q + jX_sI_p$$

For small $\delta$, $|E_f| \approx V_t + X_sI_q$, so

$$V_t \approx |E_f| - X_sI_q = |E_f| - \frac{X_sQ}{V_t}$$

Principle:

• Terminal voltage falls when reactive load $Q$ (lagging current $I_q$) increases, because of the drop $X_sI_q$.
• To keep $V_t$ constant, $|E_f|$ must be raised. Since $E_f = 4.44fN\Phi$ and $\Phi$ depends on field current $I_f$, this is done by increasing $I_f$.
• Active load mainly changes the angle $\delta$, not $|V_t|$ (term $jX_sI_p$ is at 90°), so voltage control is essentially reactive-power control.

AVR loop

Vref -->(+)--> Amplifier --> Exciter --> Gen field
         ^-                                  |
         |                                   v
       Rectifier <-- PT <-------- Terminal voltage Vt
       & filter
       (+ stabilising feedback)

1. Sensing: a potential transformer measures $V_t$; it is rectified and filtered.
2. Comparison: the measured value is compared with $V_{ref}$; error $e = V_{ref} - |V_t|$.
3. Amplification: the error is amplified (amplifier gain $K_A$, time constant $T_A$).
4. Exciter: the amplified signal changes the exciter output voltage, hence field voltage and field current $I_f$.
5. Generator: $E_f$ changes, Q output changes, and $V_t$ is driven back to $V_{ref}$.
6. A stabilising (rate) feedback loop damps oscillations and improves response.

Operation

• Load Q rises → $V_t$ falls → positive error → more $I_f$ → higher $E_f$ → generator supplies more Q → $V_t$ restored.
• Load Q falls (or line charging high) → $V_t$ rises → negative error → less $I_f$ → generator absorbs Q → $V_t$ restored.

The AVR acts within the generator capability limits (field heating limit for Q supply, under-excitation/stability limit for Q absorption). Being fast (time constants of the order of a second or less), it also improves transient stability by raising excitation during faults.

Consider a cylindrical-rotor synchronous generator with internal emf $E_f\angle\delta$, terminal voltage $V_t\angle 0^\circ$, synchronous reactance $X_s$ and negligible armature resistance.

Derivation

Armature current:

$$I_a = \frac{E_f\angle\delta - V_t\angle 0}{jX_s}$$

Complex power delivered at the terminal:

$$\begin{aligned} S &= V_tI_a^* = V_t\left[\frac{E_f\angle -\delta - V_t}{-jX_s}\right] = \frac{jV_t\left(E_f\cos\delta - jE_f\sin\delta - V_t\right)}{X_s} \\ &= \frac{V_tE_f\sin\delta}{X_s} + j\,\frac{V_tE_f\cos\delta - V_t^2}{X_s} \end{aligned}$$

So

$$P = \frac{E_fV_t}{X_s}\sin\delta,\qquad Q = \frac{V_t(E_f\cos\delta - V_t)}{X_s}$$

Showing δ is due to real power alone

Write $I_a = I_a\cos\phi - jI_a\sin\phi$ (lagging). Then

$$E_f = V_t + jX_sI_a = (V_t + X_sI_a\sin\phi) + j\,X_sI_a\cos\phi$$ $$\tan\delta = \frac{X_sI_a\cos\phi}{V_t + X_sI_a\sin\phi}$$

• The imaginary (quadrature) part of $E_f$, which creates the angle $\delta$, is $X_sI_a\cos\phi = X_sP/V_t$, i.e. proportional only to real power.
• The reactive current $I_a\sin\phi$ adds only to the in-phase part of $E_f$, changing its magnitude, not producing any angle by itself.

Two limiting cases make this clear:

1. P = 0 (pure reactive load, $\cos\phi = 0$): $E_f\sin\delta = 0$, so $\delta = 0$ however large Q is; $E_f$ and $V_t$ are in phase and differ only in magnitude.
2. Q = 0 (unity pf): $E_f = V_t + jX_sI_a$, and $\delta = \tan^{-1}(X_sI_a/V_t)$, a non-zero angle set by P.

Hence the power angle $\delta$ between $E_f$ and $V_t$ exists only because real power is delivered ($\delta = 0 \Leftrightarrow P = 0$), while Q affects mainly the magnitude of $E_f$. This is the basis of P–δ (P–f) and Q–V decoupling.

The bus admittance matrix relates bus injected currents to bus voltages, $\mathbf{I} = \mathbf{Y}_{bus}\mathbf{V}$. Its elements are found from KCL at each bus.

3-bus network

Buses 1, 2 and 3 are connected by lines 1-2, 2-3 and 1-3. Each line is represented by a π model: series admittance $y_{ik}$ and half of the line-charging admittance $y'_{ik}/2 = jB_{ik}/2$ at each end. $I_1, I_2, I_3$ are injected currents.

      I1             I2             I3
      |      y12     |      y23     |
     (1)----[===]---(2)----[===]---(3)
      |              |              |
    y10            y20            y30
   (shunt)        (shunt)        (shunt)
      |______________|______________|
             reference (ground)
   line 1-3 (y13) joins buses 1 and 3
   y10 = y'12/2 + y'13/2, etc.

KCL at each bus

$$\begin{aligned} I_1 &= y_{10}V_1 + y_{12}(V_1 - V_2) + y_{13}(V_1 - V_3) \\ I_2 &= y_{20}V_2 + y_{12}(V_2 - V_1) + y_{23}(V_2 - V_3) \\ I_3 &= y_{30}V_3 + y_{13}(V_3 - V_1) + y_{23}(V_3 - V_2) \end{aligned}$$

Rearranging,

$$\begin{bmatrix} I_1 \\ I_2 \\ I_3 \end{bmatrix} = \begin{bmatrix} y_{10} + y_{12} + y_{13} & -y_{12} & -y_{13} \\ -y_{12} & y_{20} + y_{12} + y_{23} & -y_{23} \\ -y_{13} & -y_{23} & y_{30} + y_{13} + y_{23} \end{bmatrix} \begin{bmatrix} V_1 \\ V_2 \\ V_3 \end{bmatrix}$$

General expressions

$$Y_{ii} = y_{i0} + \sum_{k \ne i} y_{ik},\qquad Y_{ik} = -y_{ik}\ (i \ne k)$$

so that for an $n$-bus system $I_i = \sum_{k=1}^{n} Y_{ik}V_k$.

• $Y_{ii}$ (self/driving-point admittance) = sum of all admittances terminating at bus $i$.
• $Y_{ik}$ (mutual/transfer admittance) = negative of the series admittance between $i$ and $k$; zero if not connected.

Effect of line shunt parameters

With line charging, $y_{i0} = \sum_k \dfrac{jB_{ik}}{2}$ (plus any shunt capacitor/reactor at bus $i$).

1. Only diagonal elements change. Each $Y_{ii}$ gains the term $+j\sum B_{ik}/2$. Off-diagonal elements $Y_{ik} = -y_{ik}$ are unchanged, because shunt branches connect a bus to ground, not to another bus.
2. Since series admittance is inductive (negative imaginary part) and charging is capacitive (positive), the shunt term reduces the magnitude of the imaginary part of $Y_{ii}$.
3. Without shunts, each row of $Y_{bus}$ sums to zero (matrix is singular, no reference). Shunt elements give a path to ground, making $Y_{bus}$ non-singular so $Z_{bus} = Y_{bus}^{-1}$ exists.
4. In load flow, the shunt terms add reactive generation, raising voltages at light load; neglecting them is acceptable only for short lines.

The bus admittance matrix $Y_{bus}$ is an $n \times n$ matrix that relates the currents injected at the $n$ buses of a power network to the bus voltages measured from the reference (ground):

$$\mathbf{I}_{bus} = \mathbf{Y}_{bus}\,\mathbf{V}_{bus},\qquad I_i = \sum_{k=1}^{n} Y_{ik}V_k$$

For a 3-bus system:

$$\begin{bmatrix} I_1 \\ I_2 \\ I_3 \end{bmatrix} = \begin{bmatrix} Y_{11} & Y_{12} & Y_{13} \\ Y_{21} & Y_{22} & Y_{23} \\ Y_{31} & Y_{32} & Y_{33} \end{bmatrix} \begin{bmatrix} V_1 \\ V_2 \\ V_3 \end{bmatrix}$$

Diagonal elements: self (driving-point) admittance

$$Y_{ii} = y_{i0} + \sum_{k \ne i} y_{ik} = \left.\frac{I_i}{V_i}\right|_{V_k = 0,\ k \ne i}$$

$Y_{ii}$ is the sum of all admittances connected to bus $i$, including series line admittances and shunt elements (half line charging, capacitors, reactors). It equals the current injected at bus $i$ per unit voltage at bus $i$ when all other buses are shorted to ground.

Off-diagonal elements: mutual (transfer) admittance

$$Y_{ik} = -y_{ik} = \left.\frac{I_i}{V_k}\right|_{V_j = 0,\ j \ne k}$$

$Y_{ik}$ is the negative of the series admittance between buses $i$ and $k$. It is the current injected at bus $i$ per unit voltage applied at bus $k$ with all other buses shorted. If buses $i$ and $k$ are not directly connected, $Y_{ik} = 0$.

Properties

• Symmetric: $Y_{ik} = Y_{ki}$ (no phase-shifting transformers).
• Sparse: most off-diagonal elements are zero in large networks.
• Diagonally dominant; formed easily by inspection.
• Used directly in Gauss–Seidel and Newton–Raphson load flow.

Bus admittance and bus impedance matrices

• Bus admittance matrix $Y_{bus}$: relates injected bus currents to bus voltages, $\mathbf{I} = \mathbf{Y}_{bus}\mathbf{V}$. It is sparse and formed by inspection.
• Bus impedance matrix $Z_{bus}$: the inverse, $\mathbf{V} = \mathbf{Z}_{bus}\mathbf{I}$, with $Z_{bus} = Y_{bus}^{-1}$. It is a full matrix, formed by the building algorithm or inversion, and used mainly in fault analysis.

Formulation for a 4-bus system

Assume lines 1-2, 1-3, 2-3, 2-4 and 3-4 with series admittances $y_{ik}$, and shunt admittances $y_{i0}$ at each bus (half line charging).

   (1)------y12------(2)
    |                 |  \
   y13              y23   y24
    |                 |     \
   (3)----------------+     (4)
    \_____________y34_______/
   (3)-(2) is y23; no line 1-4
   each bus also has y_i0 to ground

KCL at each bus:

$$\begin{aligned} I_1 &= y_{10}V_1 + y_{12}(V_1 - V_2) + y_{13}(V_1 - V_3) \\ I_2 &= y_{20}V_2 + y_{12}(V_2 - V_1) + y_{23}(V_2 - V_3) + y_{24}(V_2 - V_4) \\ I_3 &= y_{30}V_3 + y_{13}(V_3 - V_1) + y_{23}(V_3 - V_2) + y_{34}(V_3 - V_4) \\ I_4 &= y_{40}V_4 + y_{24}(V_4 - V_2) + y_{34}(V_4 - V_3) \end{aligned}$$

In matrix form:

$$\begin{bmatrix} I_1 \\ I_2 \\ I_3 \\ I_4 \end{bmatrix} = \begin{bmatrix} Y_{11} & -y_{12} & -y_{13} & 0 \\ -y_{12} & Y_{22} & -y_{23} & -y_{24} \\ -y_{13} & -y_{23} & Y_{33} & -y_{34} \\ 0 & -y_{24} & -y_{34} & Y_{44} \end{bmatrix} \begin{bmatrix} V_1 \\ V_2 \\ V_3 \\ V_4 \end{bmatrix}$$

with $Y_{11} = y_{10} + y_{12} + y_{13}$, $Y_{22} = y_{20} + y_{12} + y_{23} + y_{24}$, $Y_{33} = y_{30} + y_{13} + y_{23} + y_{34}$, $Y_{44} = y_{40} + y_{24} + y_{34}$.

General equations of elements

$$Y_{ii} = y_{i0} + \sum_{k \ne i} y_{ik},\qquad Y_{ik} = Y_{ki} = -y_{ik}$$

Properties of Ybus

1. Symmetric about the leading diagonal (no phase shifters).
2. Sparse: $Y_{ik} = 0$ when no line joins $i$ and $k$ (here $Y_{14} = 0$).
3. Diagonally dominant: $|Y_{ii}| \ge \sum_{k\ne i} |Y_{ik}|$.
4. Non-singular only when shunt elements to ground exist.
5. Easy to form by inspection and to modify for adding or removing a line.

In load-flow analysis each bus has four quantities: $P$, $Q$, $|V|$ and $\delta$. Two are specified and two are found by the solution. Buses are classified according to which two are known.

Bus classification

• Slack bus: only one in the system. Its P and Q make up the difference between scheduled generation and load + losses (losses are unknown before the solution). Its voltage angle serves as reference.
• PV bus: real power output is set by the turbine and voltage magnitude by the AVR. If the computed Q exceeds its limits, Q is fixed at the limit and the bus becomes a PQ bus.
• PQ bus: net injection $P = P_G - P_D$, $Q = Q_G - Q_D$ is known; voltage is found.

Ybus for a four-bus system

Take a four-bus network with lines 1-2, 1-3, 2-3, 2-4 and 3-4. Each line has series admittance $y_{ik}$ and shunt $y'_{ik}/2$ at each end; total shunt at bus $i$ is $y_{i0}$.

Applying KCL:

$$\begin{aligned} I_1 &= y_{10}V_1 + y_{12}(V_1 - V_2) + y_{13}(V_1 - V_3) \\ I_2 &= y_{20}V_2 + y_{12}(V_2 - V_1) + y_{23}(V_2 - V_3) + y_{24}(V_2 - V_4) \\ I_3 &= y_{30}V_3 + y_{13}(V_3 - V_1) + y_{23}(V_3 - V_2) + y_{34}(V_3 - V_4) \\ I_4 &= y_{40}V_4 + y_{24}(V_4 - V_2) + y_{34}(V_4 - V_3) \end{aligned}$$ $$\mathbf{Y}_{bus} = \begin{bmatrix} y_{10} + y_{12} + y_{13} & -y_{12} & -y_{13} & 0 \\ -y_{12} & y_{20} + y_{12} + y_{23} + y_{24} & -y_{23} & -y_{24} \\ -y_{13} & -y_{23} & y_{30} + y_{13} + y_{23} + y_{34} & -y_{34} \\ 0 & -y_{24} & -y_{34} & y_{40} + y_{24} + y_{34} \end{bmatrix}$$

General rule: $Y_{ii} = y_{i0} + \sum_{k\ne i} y_{ik}$ and $Y_{ik} = -y_{ik}$ ($= 0$ if no line, e.g. $Y_{14}$).

These elements are then used in the power-flow equations

$$P_i - jQ_i = V_i^*\sum_{k=1}^{4} Y_{ik}V_k$$

which are solved for the unknowns listed in the table above.