Development of Modern Power System - A Brief Historical Preview

Introduction of Modern Power System

Series Parameters of Transmission Lines

Line Resistance

Inductance of a Straight Conductor

Internal Inductance

External Inductance

Inductance of a Single-phase Line

Inductance of Three - Phase Lines with Symmetrical Spacing

Inductance of Three - Phase Lines with Asymmetrical Spacing

Transposed Line

Composite Conductors

Bundled Conductors

Shunt Parameters of Transmission Lines

Capacitance of a Straight Conductor

Capacitance of a Single - Phase Line

Capacitance of a Three-Phase Transposed Line

Effect of Earth on the Calculation Capacitance

Synchronous Machine Model

Transformer Model

Balanced Operation of a Three-Phase Circuit

Per Unit Representation

Development of Modern Power System - A Brief Historical Preview

The development of the modern day electrical energy system took a few centuries. Prior to 1800, scientists like William Gilbert, C. A. de Coulomb, Luigi Galvani, Benjamin Franklin, Alessandro Volta etc. worked on electric and magnetic field principles. However, none of them had any application in mind. They also probably did not realize that their work will lead to such an exciting engineering innovation. They were just motivated by the intellectual curiosity.

Between 1800 and 1810 commercial gas companies were formed - first in Europe and then in North America. Around the same time with the research efforts of scientists like Sir Humphrey Davy, Andre Ampere, George Ohm and Karl Gauss the exciting possibilities of the use of electrical energy started to dawn upon the scientific community.

In England, Michael Faraday worked on his induction principle between 1821 and 1831. The modern world owes a lot to this genius. Faraday subsequently used his induction principle to build a machine to generate voltage. Around the same time American engineer Joseph Henry also worked independently on the induction principle and applied his work on electromagnets and telegraphs.

For about three decades between 1840 and 1870 engineers like Charles Wheatstone, Alfred Varley, Siemens brothers Werner and Carl etc. built primitive generators using the induction principle. It was also observed around the same time that when current carrying carbon electrodes were drawn apart, brilliant electric arcs were formed. The commercialization of arc lighting took place in the decade of 1870s. The arc lamps were used in lighthouses and streets and rarely indoor due to high intensity of these lights. Gas was still used for domestic lighting. It was also used for street lighting in many cities.

From early 1800 it was noted that a current carrying conductor could be heated to the point of incandescent. Therefore the idea of using this principle was very tempting and attracted attention. However the incandescent materials burnt very quickly to be of any use. To prevent them from burning they were fitted inside either vacuum globes or globes filled with inert gas. In October 1879 Thomas Alva Edison lighted a glass bulb with a carbonized cotton thread filament in a vacuum enclosed space. This was the first electric bulb that glowed for 44 hours before burning out. Edison himself improved the design of the lamp later and also proposed a new generator design.

The Pearl Street power station in New York City was established in 1882 to sell electric energy for incandescent lighting. The system was direct current three-wire, 220/110 V and supplied Edison lamps for a total power requirement of 30 kW.

The only objective of the early power companies was illumination. However we can easily visualize that this would have resulted in the under utilization of resources. The lighting load peaks in the evening and by midnight it reduces drastically. It was then obvious to the power companies that an elaborate and expensive set up would lay idle for a major amount of time. This provided incentive enough to improve upon the design of electric motors to make them commercially viable. The motors became popular very quickly and were used in many applications. With this the electric energy era really and truly started.

However with the increase in load large voltage and unacceptable drops were experienced, especially at points that were located far away from the generating stations due to poor voltage regulation capabilities of the existing dc networks. One approach was to transmit power at higher voltages while consuming it at lower voltages. This led to the development of the alternating current.

In 1890s the newly formed Westinghouse Company experimented with the new form of electricity, the alternating current. This was called alternating current since the current changed direction in synchronism with the generator rotation. Westinghouse Company was lucky to have Serbian engineer Nicola Tesla with them. He not only invented polyphase induction motor but also conceived the entire polyphase electrical power system. He however had to face severe objection from Edison and his General Electric Company who were the proponents of dc. The ensuing battle between ac and dc was won by ac due to the following factors:

Transformers could boost ac voltage for transmission and could step it down for distribution.

The construction of ac generators was simpler.

The construction of ac motors was simpler. Moreover they were more robust and cheaper than the dc motors even though not very sophisticated.

With the advent of ac technology the electric power could reach more and more people. Also size of the generators started increasing and transmission level voltages started increasing. The modern day system contains hundreds of generators and thousands of buses and is a large interconnected network.

Modern electric power systems have three separate components - generation, transmission and distribution. Electric power is generated at the power generating stations by synchronous alternators that are usually driven either by steam or hydro turbines. Most of the power generation takes place at generating stations that may contain more than one such alternator-turbine combination. Depending upon the type of fuel used, the generating stations are categorized as thermal, hydro, nuclear etc. Many of these generating stations are remotely located. Hence the electric power generated at any such station has to be transmitted over a long distance to load centers that are usually cities or towns. This is called the power transmission. In fact power transmission towers and transmission lines are very common sights in rural areas.

Modern day power systems are complicated networks with hundreds of generating stations and load centers being interconnected through power transmission lines. Electric power is generated at a frequency of either 50 Hz or 60 Hz.

In an interconnected ac power system, the rated generation frequency of all units must be the same. In India the frequency is 50 Hz.

The basic structure of a power system is shown in Fig. 1.1.

Fig. 1.1 A typical power system.

It contains a generating plant, a transmission system, a subtransmission system and a distribution system. These subsystems are interconnected through transformers T1 , T2 and T3 . Let us consider some typical voltage levels to understand the funtioning of the power system. The electric power is generated at a thermal plant with a typical voltage of 22 kV (voltage levels are usually specified line-to-line). This is boosted up to levels like 400 kV through transformer T1 for power transmission. Transformer T2 steps this voltage down to 66 kV to supply power through the subtransmission line to industrial loads that require bulk power at a higher voltage. Most of the major industrial customers have their own transformers to step down the 66 kV supply to their desired levels. The motivation for these voltage changes is to minimize transmission line cost for a given power level. Distribution systems are designed to operate for much lower power levels and are supplied with medium level voltages.

The power distribution network starts with transformer T3, which steps down the voltage from 66 kV to 11 kV. The distribution system contains loads that are either commercial type (like office buildings, huge apartment complexes, hotels etc) or residential (domestic) type. Usually the commercial customers are supplied power at a voltage level of 11 kV whereas the domestic consumers get power supply at 400-440 V. Note that the above figures are given for line-to-line voltages. Since domestic customers get single-phase supplies, they usually receive 230-250 V at their inlet points. While a domestic customer with a low power consumption gets a single-phase supply, both industrial and commercial consumers get three-phase supplies not only because their consumption is high but also because many of them use three-phase motors. For example, the use of induction motor is very common amongst industrial customers who run pumps, compressors, rolling mills etc.

The main components of a power system are generators, transformers and transmission lines.

In this module we shall discuss the models of these components that will be used subsequently in power system studies.

Section I: Series Parameters of Transmission Lines

Line Resistance

Inductance of a Straight Conductor

Inductance of a Single-phase Line

Inductance of Three-Phase Lines with Symmetrical Spacing

Inductance of Three-Phase Lines with Asymmetrical Spacing

Transposed Line

Composite Conductors

Bundled Conductors

Overhead transmission lines and transmission towers are a common sight in rural India. The transmission towers are usually made of steel and are solidly erected with a concrete base. The three-phase conductors are supported by the towers through insulators. The conductors are usually made of aluminum or its alloys. Aluminum is preferred over copper as an aluminum conductor is lighter in weight and cheaper in cost than copper conductor of the same resistance.

The conductors are not straight wires but strands of wire twisted together to form a single conductor to give it higher tensile strength. One of the most common conductors is aluminum conductor, steel reinforced (ACSR). The cross sectional view of such a conductor is shown in Fig. 1.2. The central core is formed with strands of steel while two layers of aluminum strands are put in the outer layer. The other type of conductors that are in use are all aluminum conductor (AAC), all aluminum alloy conductor (AAAC), aluminum conductor, alloy reinforced (ACAR).

Fig. 1.2 Cross sectional view of an ACSR conductor.

Line Resistance

It is very well known that the dc resistance of a wire is given by

where ρ is the resistivity of the wire in Ω - m, l is the length in meter and A is the cross sectional area in m2 . Unfortunately however the resistance of an overhead conductor is not the same as that given by the above expression. When alternating current flows through a conductor, the current density is not uniform over the entire cross section but is somewhat higher at the surface. This is called the skin effect and this makes the ac resistance a little more than the dc resistance. Moreover in a stranded conductor, the length of each strand is more that the length of the composite conductor. This also increases the value of the resistance from that calculated in (1.1).

Finally the temperature also affects the resistivity of conductors. However the temperature rise in metallic conductors is almost linear in the practical range of operation and is given by

where R1 and R2 are resistances at temperatures t1 and t2 respectively and T is a constant that depends on the conductor material and its conductivity. Since the resistance of a conductor cannot be determined accurately, it is best to determine it from the data supplied by the manufacturer

Inductance of a Straight Conductor

From the knowledge of high school physics we know that a current carrying conductor produces a magnetic field around it. The magnetic flux lines are concentric circles with their direction specified by Maxwell's right hand thumb rule ( i.e., if the thumb of the right hand points towards the flow of

current then the fingers of the fisted hand point towards the flux lines ). The sinusoidal variation in the current produces a sinusoidal variation in the flux. The relation between the inductance, flux linkage and the phasor current is then expressed as

where L is the inductance in Henry, λ is the flux linkage in Weber-turns and I is the phasor current in Ampere.

A. Internal Inductance

Consider a straight round (cylindrical) conductor, the cross-section of which is shown in Fig. 1.3. The conductor has a radius of r and carries a current I . Ampere's law states that the magnetomotive force (mmf) in ampere-turns around a closed path is equal to the net current in amperes enclosed by the path. We then get the following expression

where H is the magnetic field intensity in At/m, s is the distance along the path in meter and I is the current in ampere.

Let us denote the field intensity at a distance x from the center of the conductor by Hx. It is to be noted that Hxis constant at all points that are at a distance x from the center of the conductor.

Therefore Hx is constant over the concentric circular path with a radius of x and is tangent to it. Denoting the current enclosed by Ix we can then write

Fig. 1.3 Cross section of a round conductor.

If we now assume that the current density is uniform over the entire conductor, we can write

Substituting (1.6) in (1.5) we get

Assuming a relative permeability of 1, the flux density at a distance of x from the center of the conductor is given by

where µ0 is the permeability of the free space and is given by 4π X 10-7 H/m.

The flux inside (or outside) the conductor is in the circumferential direction . The two directions that are perpendicular to the flux are radial and axial . Let us consider an elementary area that has a dimension of dx m along the radial direction and 1 m along the axial direction. Therefore the area perpendicular to the flux at all angular positions is dx X 1 m2 . Let the flux along the circular strip be denoted by dφ x and this is given by

Note that the entire conductor cross section does not enclose the above flux. The ratio of the cross sectional area inside the circle of radius x to the total cross section of the conductor can be thought about as fractional turn that links the flux dφ x. Therefore the flux linkage is

Integrating (1.10) over the range of x , i.e., from 0 to r , we get the internal flux linkage as

Wbt/m

Then from (1.3) we get the internal inductance per unit length as

H/m

For μ≠1, μr H/m

It is interesting to note that the internal inductance is independent of the conductor radius.

B. External Inductance

Let us consider an isolated straight conductor as shown in Fig. 1.4. The conductor carries a current I . Assume that the tubular element at a distance x from the center of the conductor has a field intensity Hx . Since the circle with a radius of x encloses the entire current, the mmf around the element is given by

and hence the flux density at a radius x becomes

Fig 1.4 A Conductor with two external points

The entire current I is linked by the flux at any point outside the conductor. Since the distance x is greater than the radius of the conductor, the flux linkage dλx is equal to the flux dφx. Therefore for 1 m length of the conductor we get

The external flux linkage between any two points D1 and D2, external to the conductor is

Wbt/m

From (1.3) we can then write the inductance between any two points outside the conductor as

H/m

For μ≠1, H/m where μr =relative permeability

Inductance of a Single-phase Line

Consider two solid round conductors with radii of r1 and r2 as shown in Fig. 1.5. One conductor is the return circuit for the other. This implies that if the current in conductor 1 is I then the current in conductor 2 is -I . First let us consider conductor 1. The current flowing in the conductor will set up flux lines. However, the flux beyond a distance D + r2 from the center of the conductor links a net current of zero and therefore does not contribute to the flux linkage of the circuit. Also at a distance

(1.13)

less than D - r2 from the center of conductor 1 the current flowing through this conductor links the flux. Moreover since D >> r2 we can make the following approximations

Fig. 1.5 A single-phase line with two conductors.

Therefore from (1.12) and (1.17) we can specify the inductance of conductor 1 due to internal and external flux as

We can rearrange L1 given in (1.18) as follows

Substituting r1 = r1 e

1/4 in the above expression we get

The radius r1 can be assumed to be that of a fictitious conductor that has no internal flux but with the same inductance as that of a conductor with radius r1 .

In a similar way the inductance due current in the conductor 2 is given by

Therefore the inductance of the complete circuit is

If we assume r1 = r2 = r , then the total inductance becomes

where r = re1/4.

H/m (1.18)

Inductance of Three-Phase Lines with Symmetrical Spacing

Consider the three-phase line shown in Fig. 1.6. Each of the conductors has a radius of r and their centers form an equilateral triangle with a distance D between them. Assuming that the currents are balanced, we have

Consider a point P external to the conductors. The distance of the point from the phases a, b and c are denoted by Dpa, Dpband Dpcrespectively.

Fig. 1.6 Three-phase symmetrically spaced conductors and an external point P.

Let us assume that the flux linked by the conductor of phase-a due to a current Ia includes the internal flux linkages but excludes the flux linkages beyond the point P . Then from (1.18) we get

The flux linkage with the conductor of phase-a due to the current Ib , excluding all flux beyond the point P , is given by (1.17) as

Similarly the flux due to the current Ic is

Therefore the total flux in the phase-a conductor is

The above expression can be expanded as

From (1.22) we get

Substituting the above expression in (1.27) we get

(1.27)

Now if we move the point P far away, then we can approximate Dpa Dpb Dpc.. Therefore their logarithmic ratios will vanish and we can write (1.28) as

Hence the inductance of phase-a is given as

Note that due to symmetry, the inductances of phases b and c will be the same as that of phase-a given above, i.e., Lb= Lc = L a .

Inductance of Three-Phase Lines with Asymmetrical Spacing

It is rather difficult to maintain symmetrical spacing as shown in Fig. 1.6 while constructing a transmission line. With asymmetrical spacing between the phases, the voltage drop due to line inductance will be unbalanced even when the line currents are balanced. Consider the three-phase asymmetrically spaced line shown in Fig. 1.7 in which the radius of each conductor is assumed to be r . The distances between the phases are denoted by Dab, Dbcand Dca. We then get the following flux linkages for the three phases

Fig. 1.7 Three-phase asymmetrically spaced line.

Let us define the following operator

Note that for the above operator the following relations hold

Let as assume that the current are balanced. We can then write

Substituting the above two expressions in (1.31) to (1.33) we get the inductance of the three phases as

(1.31)

(1.32)

(1.33)

(1.36)

It can be seen that the inductances contain imaginary terms. The imaginary terms will vanish only when Dab= Dbc= Dca. In that case the inductance will be same as given by (1.30).

Transposed Line

The inductances that are given in (1.36) to (1.38) are undesirable as they result in an unbalanced circuit configuration. One way of restoring the balanced nature of the circuit is to exchange the positions of the conductors at regular intervals. This is called transposition of line and is shown in Fig.1.8. In this each segment of the line is divided into three equal sub-segments. The conductors of each of the phases a, b and c are exchanged after every sub-segment such that each of them is placed in each of the three positions once in the entire segment. For example, the conductor of the phase-a occupies positions in the sequence 1, 2 and 3 in the three sub-segments while that of the phase-b occupies 2, 3 and 1. The transmission line consists of several such segments.

Fig. 1.8 A segment of a transposed line.

In a transposed line, each phase takes all the three positions. The per phase inductance is the average value of the three inductances calculated in (1.36) to (1.38). We therefore have

This implies

From (1.35) we have a + a2 = - 1. Substituting this in the above equation we get

The above equation can be simplified as

Defining the geometric mean distance ( GMD ) as

equation (1.41) can be rewritten as

Notice that (1.43) is of the same form as (1.30) for symmetrically spaced conductors. Comparing these two equations we can conclude that GMD can be construed as the equivalent conductor spacing. The GMD is the cube root of the product of conductor spacings.

(1.39)

(1.37)

(1.38)

Composite Conductors

So far we have considered only solid round conductors. However as mentioned at the beginning of Section 1.1, stranded conductors are used in practical transmission line. We must therefore modify the equations derived above to accommodate stranded conductors. Consider the two groups of conductors shown in Fig. 1.9. Of these two groups conductor x contains n identical strands of radius rx while conductor y contains m identical strands of radius ry . Conductor x carries a current I the return path of which is through conductor y . Therefore the current through conductor y is - I .

Fig. 1.9 Single-phase line with two composite conductors.

Since the strands in a conductor are identical, the current will be divided equally among the strands. Therefore the current through the strands of conductor x is I / n and through the strands of conductor y is -I/m . The total flux linkage of strand a is given by

We can write (1.44) as

The inductance of the strand a is then given by

In a similar way the inductances of the other conductors are also obtained. For example,

The average inductance of any one of the strands in the group of conductor x is then

Conductor x is composed of n strands that are electrically parallel. Even though the inductance of the different strand is different, the average inductance of all of them is the same as Lav, x . Assuming that the average inductance given above is the inductance of n parallel strands, the total inductance of the conductor x is

(1.49)

Substituting the values of La , Lbetc. in the above equation we get

where the geometric mean distance ( GMD ) and the geometric mean radius ( GMR ) are given respectively by

The inductance of the conductor y can also be similarly obtained. The geometric mean radius GMRy will be different for this conductor. However the geometric mean distance will remain the same.

Bundled Conductors

So far we have discussed three-phase systems that have only one conductor per phase. However for extra high voltage lines corona causes a large problem if the conductor has only one conductor per phase. Corona occurs when the surface potential gradient of a conductor exceeds the dielectric strength of the surrounding air. This causes ionization of the area near the conductor. Corona produces power loss. It also causes interference with communication channels. Corona manifests itself with a hissing sound and ozone discharge. Since most long distance power lines in India are either 220 kV or 400 kV, avoidance of the occurrence of corona is desirable.

The high voltage surface gradient is reduced considerably by having two or more conductors per phase in close proximity. This is called conductor bundling . The conductors are bundled in groups of two, three or four as shown in Fig. 1.10. The conductors of a bundle are separated at regular intervals with spacer dampers that prevent clashing of the conductors and prevent them from swaying in the wind. They also connect the conductors in parallel.

The geometric mean radius (GMR) of two-conductor bundle is given by

where Ds is the GMR of conductor. The GMR for three-conductor and four-conductor bundles are given respectively by

The inductance of the bundled conductor is then given by where n=2,3 ....

where the geometric mean distance is calculated assuming that the center of a round conductor is the same as that of the center of the bundle.

(1.51)

(1.52)

(1.53)

(1.54)

(1.55)

Fig. 1.10 Bundled conductors: (a) 2-conductor, (b) 3-conductor and (c) 4-conductor bundles

Section II: Shunt Parameters Of Transmission Lines

Capacitance of a Straight Conductor

Capacitance of a Single-Phase Line

Capacitance of a Three-Phase Transposed Line

Effect of Earth on the Calculation of Capacitance

Capacitance in a transmission line results due to the potential difference between the conductors. The conductors get charged in the same way as the parallel plates of a capacitor. Capacitance between two parallel conductors depends on the size and the spacing between the conductors. Usually the capacitance is neglected for the transmission lines that are less than 50 miles (80 km) long. However the capacitance becomes significant for longer lines with higher voltage. In this section we shall derive the line capacitance of different line configuration.

Capacitance of a Straight Conductor

Consider the round conductor shown in Fig. 1.11. The conductor has a radius of r and carries a charge of q coulombs. The capacitance C is the ratio of charge q of the conductor to the impressed voltage, i.e.,

The charge on the conductor gives rise to an electric field with radial flux lines where the total electric flux is equal to the charge on the conductor. By Gauss's law, the electric flux density at a cylinder of radius x when the conductor has a length of 1 m is

C/m2

The electric filed intensity is defined as the ratio of electric flux density to the permittivity of the medium. Therefore

V/m

Fig. 1.11 Cylindrical conductor with radial flux lines.

Now consider the long straight conductor of Fig. 1.12 that is carrying a positive charge q C/m. Let two points P1 and P2 be located at distances D1 and D2 respectively from the center of the conductor. The conductor is an equipotential surface in which we can assume that the uniformly distributed charge is concentrated at the center of the conductor. The potential difference V12 between the points P1 and P2 is the work done in moving a unit of charge from P2 to P1 . Therefore the voltage drop between the two points can be computed by integrating the field intensity over a radial path between the equipotential surfaces, i.e.,

Fig. 1.12 Path of integration between two points external to a round straight conductor.

Capacitance of a Single-Phase Line

Consider the single-phase line consisting of two round conductors as shown in Fig. 1.5. The separation between the conductors is D . Let us assume that conductor 1 carries a charge of q1 C/m while conductor 2 carries a charge q2 C/m. The presence of the second conductor and the ground will disturb field of the first conductor. However we assume that the distance of separation between the conductors is much larger compared to the radius of the conductor and the height of the conductor is much larger than D for the ground to disturb the flux. Therefore the distortion is small and the charge is uniformly distributed on the surface of the conductor.

Assuming that the conductor 1 alone has the charge q1 , the voltage between the conductors is

Similarly if the conductor 2 alone has the charge q2 , the voltage between the conductors is

V (1.60)

The above equation implies that

From the principle of superposition we can write

For a single-phase line let us assume that q1 (= -q2 ) is equal to q . We therefore have

Assuming r1 = r2= r3, we can rewrite (1.64) as

Therefore from (1.57) the capacitance between the conductors is given by

The above equation gives the capacitance between two conductors. For the purpose of transmission line modeling, the capacitance is defined between the conductor and neutral. This is shown in Fig. 1.13. Therefore the value of the capacitance is given from Fig. 1.13 as

Fig. 1.13 (a) Capacitance between two conductors and (b) equivalent capacitance to ground.

Capacitance of a Three-Phase Transposed Line

Consider the three-phase transposed line shown in Fig. 1.14. In this the charges on conductors of phases a, b and c are qa, qband qc espectively. Since the system is assumed to be balanced we have

Fig. 1.14 Charge on a three-phase transposed line.

Using superposition, the voltage Vab for the first, second and third sections of the transposition are given respectively as

Then the average value of the voltage is

This implies

The GMD of the conductors is given in (1.42). We can therefore write

Similarly the voltage Vac is given as

Adding (1.74) and (1.75) and using (1.68) we get

For a set of balanced three-phase voltages

Therefore we can write

Combining(1.76 and (1.77) we get

Therefore the capacitance to neutral is given by

V

(1.73)

V (1.78)

For bundled conductor

where

Effect of Earth on the Calculation of Capacitance

Earth affects the calculation of capacitance of three-phase lines as its presence alters the electric field lines. Usually the height of the conductors placed on transmission towers is much larger than the spacing between the conductors. Therefore the effect of earth can be neglected for capacitance calculations, especially when balanced steady state operation of the power system is considered. However for unbalanced operation when the sum of the three line currents is not zero, the effect of earth needs to be considered.

Section III Synchronous Machine Model

The schematic diagram of a synchronous generator is shown in Fig. 1.15. This contains three stator windings that are spatially distributed. It is assumed that the windings are wye-connected. The winding currents are denoted by ia , ib and ic. The rotor contains the field winding the current through which is denoted by if . The field winding is aligned with the so-called direct ( d ) axis. We also define a quadrature ( q ) axis that leads the d -axis by 90°. The angle between the d-axis and the a-phase of the stator winding is denoted by θd.

Fig. 1.15 Schematic diagram of a synchronous generator.

Let the self-inductance of the stator windings be denoted by Laa, Lbb, Lcc such that

and the mutual inductance between the windings be denoted as

F/m (1.79)

(1.80)

The mutual inductances between the field coil and the stator windings vary as a function of θd and are given by

The self-inductance of the field coil is denoted by Lff.

The flux linkage equations are then given by

For balanced operation we have

Hence the flux linkage equations for the stator windings (1.85) to (1.87) can be modified as

For steady state operation we can assume

Also assuming that the rotor rotates at synchronous speed ωs we obtain the following two equations

where θd0 is the initial position of the field winding with respect to the phase-a of the stator winding at time t = 0. The mutual inductance of the field winding with all the three stator windings will vary as a function of θd, i.e.,

(1.81)

(1.82)

(1.83)

(1.84)

(1.85)

(1.86)

(1.87)

(1.88)

(1.89)

(1.90)

(1.91)

(1.92)

(1.93)

(1.94)

(1.95)

(1.96)

(1.97)

Substituting (1.92), (1.94), (1.95), (1.96) and (1.97) in (1.89) to (1.91) we get

Since we assume balanced operation, we need to treat only one phase. Let the armature resistance of the generator be R . The generator terminal voltage is given by

where the negative sign is used for generating mode of operation in which the current leaves the terminal. Substituting (1.98) in (1.101) we get

The last term of (1.102) is the internal emf ea that is given by

where the rms magnitude Ei is proportional to the field current

Since θd0 is the position of the d -axis at time t = 0, we define the position of the q -axis at that instant as

Therefore (1.94) can be rewritten as

Substituting (1.105) in (1.103) we get

Hence (1.102) can be written as

The equivalent circuit is shown in Fig. 1.16. Let the current ia lag the internal emf ea by θa . The stator currents are then

(1.98)

(1.99)

(1.100)

(1.105)

(1.107)

(1.108)

(1.109)

(1.110)

(1.111)

Fig. 1.16 Three-phase equivalent circuit of a synchronous generator.

The single-phase equivalent circuit is shown in Fig. 1.17. The phase angle θabetween eaand ia is rather difficult to measure under load as ea is the no load voltage. To avoid this, we define the phase angle between νa and ia to be θ . We assume that ea leads νaby δ . Therefore we can write Then the voltages and currents shown in Fig. 1.17 are given as Equations (1.113) to (1.115) imply that The synchronous impedance is then defined as The terminal voltage equation is then

Fig. 1.17 Single-phase equivalent circuit of a synchronous generator.

(1.112)

(1.113)

(1.114)

(1.115)

(1.116)

(1.117)

(1.118)

Section IV: Transformer Model

The equivalent circuit of a single-phase transformer is shown in Fig. 1.18. In this the primary voltage and currents are denoted by V1 and V2 respectively. The current entering the primary terminals is I1. The core loss component is represented by Rcwhile the magnetizing reactance is denoted by Xm . The leakage inductance of the transformer is denoted by Xeqand Req is transformer winding resistance. It is to be noted that all the quantities are referred to the primary side. The turns ratio of the transformer is given by N1 : N2 .

The impedance of the shunt branch is much larger compared to that of the series branch. Therefore we neglect Rcand Xm. Again of the series parameters, Reqis much smaller than Xeq. We can therefore neglect the series impedance. Therefore the transformer can be represented by the leakage reactance Xeq. The single-phase transformer equivalent circuit, when referred to the primary side, is as shown is Fig. 1.19 (a). The equivalent circuit, when referred to the secondary side, is shown in Fig. 1.19 (b) where a = N1 / N2 .

Fig. 1.18 Equivalent circuit of a single-phase transformer

Fig. 1.19 Simplified equivalent circuit of a single-phase transformer: (a) when referred to the primary side and (b) when referred to the secondary side.

Section V: Balanced Operation Of a Three-Phase Citcuit

In the language of Power Systems, a three-phase circuit is said to be balanced if the following conditions are true.

If all the sources and loads are y-connected.

There is no mutual inductance between the phases.

All neutrals are at the same potential.

As a consequence of the points (2) and (3) above, all phases are decoupled.

All network variables are balanced sets in the same sequence as the sources.

Consider the three-phase circuit shown in Fig. 1.20 that contains three balanced sources Ea , Eband Ec along with three balanced source impedances, each of value Zs. The sources supply two balanced loads - one wye-connected with impedance of Zy and the other Δ-connected with impedance of ZΔ . Since this is a balanced network, the sum of the currents at the neutrals N (or n ) is zero. Therefore the neutral are at the same potential. Transforming the Δ-connected load to an equivalent y, we get the per phase equivalent circuit as shown in Fig. 1.21. In this fashion an entire power system can be converted into its per phase equivalent. The line diagram showing a per phase equivalent circuit is called a single-line diagram.

Section VI: Per Unit Representation

In a power system different power equipment with different voltage and power levels are connected together through various step up or step down transformers. However the presence of various voltage and power levels causes problem in finding out the currents (or voltages) at different points in the network. To alleviate this problem, all the system quantities are converted into a uniform normalized platform. This is called the per unit system . In a per unit system each system variable or quantity is normalized with respect to its own base value. The units of these normalized values are per unit (abbreviated as pu) and not Volt, Ampere or Ohm. The base quantities chosen are:

VA base ( Pbase ): This is the three-phase apparent power (Volt-Ampere) base that is

common to the entire circuit.

Voltage Base ( Vbase ): This is the line-to-line base voltage. This quantity is not uniform for the entire circuit but gets changed by the turns ratio of the transformer.

Fig. 1.20 Three balanced sources supplying two balanced load through balanced source

impedances.

Fig. 1.21 Per phase equivalent circuit of the network of Fig. 1.20.

Based on the above two quantities the current and impedance bases can be defined as

Assume that an impedance Z is defined as Z1 per unit in a base impedance of Zbase _ old . Then we have

The impedance now has to be represented in a new base value denoted as Z base_new . Therefore

From (1.120) Z2 can be defined in terms of old and new values of VA base and voltage base as

(1.119)

(1.120)

(1.121)

(1.122)

(1.123)

Let us consider the circuit shown in Fig. 1.19 (a) which contains the equivalent circuit of a transformer. Let the transformer rating be

500 MVA, 220/22 kV with a leakage reactance of 10%.

The VA base of the transformer is 500 MVA and the voltage bases in the primary and secondary side are 200 kV and 22 kV respectively. Therefore the impedance bases of these two sides are

Ω and W

where the subscripts 1 and 2 refer to the primary (high tension) and secondary (low tension) sides respectively. Assume that the leakage reactance is referred to the primary side. Then for 10%, i.e., 0.1 per unit leakage reactance we have

Ω

The above reactance when referred to the secondary side is

Ω

Hence the per unit impedance in the secondary side is 0.0968/0.968 = 0.1. Therefore we see that the per unit leakage reactance is the same for both sides of the transformer and, as a consequence, the transformer can be represented by only its leakage reactance. The equivalent circuit of the transformer is then as shown in Fig. 1.22. Since this diagram only shows the reactance (or impedance) of the circuit, this is called the reactance (or impedance ) diagram .

Fig. 1.22 Per unit equivalent circuit of a transformer.

Example 1.2:

Consider the 50 Hz power system the single-line diagram of which is shown in Fig. 1.23. The system contains three generators, three transformers and three transmission lines. The system ratings are

Generator G1 200 MVA, 20 kV, Xd = 15%

Generator G2 300 MVA, 18 kV, Xd = 20%

Generator G3 300 MVA, 20 kV, Xd = 20%

Transformer T 1 300 MVA, 220Y/22 kV, Xd = 10%

Transformer T 2 Three single-phase units each rated 100 MVA, 130Y/25 kV, X = 10%

Transformer T 3 300 MVA, 220/22 kV, X = 10%

The transmission line reactances are as indicated in the figure. We have to draw the reactance diagram choosing the Generator 3 circuit as the base.

Fig. 1.23 Single-line diagram of the power system of Example 1.2.

As we have chosen the circuit of Generator3 as the base, the base MVA for the circuit is 300. The high voltage side of transformer T2 is connected wye. Therefore its ratedline to line voltage is √3 X 130 = 225 kV. Since the low voltage side is connected in D , its line to line voltage is 25 kV. The base voltages are chosen as discussed below.

Since the base voltage of G3 is 20 kV, the base voltage between T3 and bus 1 will be 20 X 10 = 200 kV. Also as there is no transformer connected in bus 1, the base voltage of 200 kV must be chosen for both the lines that are connected to either side of bus 1. Then the base voltage for the circuit of G1 will also be 20 kV. Finally since the turns ratio of T2 is 9 (= 225 ÷ 25), the base voltage in the G 2 side is 200 ÷ 9 = 22.22 kV. The base voltages are also indicated in Fig. 1.23.

Once the base voltages for the various parts of the circuit are known, the per unit values for the various reactances of the circuit are calculated according to (1.123) for a base MVA of 300. These are listed below.

Generator G1

Generator G2

Generator G3

Transformer T 1

Transformer T 2

Transformer T 3

The base impedance of the transmission line is

Ω

Therefore the per unit values of the line impedances are

pu and pu

The impedance diagram is shown in Fig. 1.24.

Fig. 1.24 The impedance diagram of the system of Fig. 1.23.

Closure

This completes our discussion on the modeling of power system components. In the subsequent portion of this course we shall use these models to construct a power system and use the per unit notation and the impedance diagram to represent the system.

Fig. 1.24 The impedance diagram of the system of Fig. 1.23.

CHAPTER 2

Overview

As we have discussed earlier in Chapter 1 that the transmission line parameters include series resistance and inductance and shunt capacitance. In this chapter we shall discuss the various models of the line. The line models are classified by their length. These classifications are

Short line approximation for lines that are less than 80 km long.

Medium line approximation for lines whose lengths are between 80 km to 250 km.

Long line model for lines that are longer than 250 km.

These models will be discussed in this chapter. However before that let us introduce the ABCD parameters that are used for relating the sending end voltage and current to the receiving end voltage and currents.

Section I: ABCD Parameters

Consider the power system shown in Fig. 2.1. In this the sending and receiving end voltages are denoted by VS and VR respectively. Also the currents IS and IR are entering and leaving the network respectively. The sending end voltage and current are then defined in terms of the ABCD parameters as

From (2.1) we see that

This implies that A is the ratio of sending end voltage to the open circuit receiving end voltage. This quantity is dimension less. Similarly,

i.e., B , given in Ohm, is the ratio of sending end voltage and short circuit receiving end current. In a similar way we can also define

The parameter D is dimension less.

Fig. 2.1 Two port representation of a transmission network.

Section II: Short Line Approximation

(2.1)

(2.2)

(2.3)

Ω

(2.4)

mho

(2.5)

(2.6)

The shunt capacitance for a short line is almost negligible. The series impedance is assumed to be lumped as shown in Fig. 2.2. If the impedance per km for an l km long line is z0 = r + jx , then the total impedance of the line is Z = R + jX = lr + jlx . The sending end voltage and current for this approximation are given by

Therefore the ABCD parameters are given by

Fig. 2.2 Short transmission line representation.

2.3 Medium Line Approximation

Normal -π Representation

Normal- T Representation

Medium transmission lines are modeled with lumped shunt admittance. There are two different representations - nominal- π and nominal-T depending on the nature of the network. These two are discussed here one by one.

Nominal- π Representation

In this representation the lumped series impedance is placed in the middle while the shunt admittance is divided into two equal parts and placed at the two ends. The nominal- π representation is shown in Fig. 2.3. This representation is used for load flow studies, as we shall see later. Also a long transmission line can be modeled as an equivalent π -network for load flow studies.

Fig. 2.3 Nominal- p representation.

Let us define three currents I1 , I2 and I3 as indicated in Fig. 2.3. Applying KCL at nodes M and N we get

(2.7)

(2.8)

(2.9)

(2.10)

Substituting (2.11) in (2.10) we get

Therefore from (2.11) and (2.12) we get the following ABCD parameters of the nominal- p representation

Nominal- T Representation

In this representation the shunt admittance is placed in the middle and the series impedance is divided into two equal parts and these parts are placed on either side of the shunt admittance. The nominal-T representation is shown in Fig. 2.4. Let us denote the midpoint voltage as VM . Then the application of KCL at the midpoint results in

Fig. 2.4 Nominal-T representation.

Rearranging the above equation can be written as

Now the receiving end current is given by

Substituting the value of VM from (2.16) in (2.17) and rearranging we get

(2.11)

(2.12)

(2.13)

(2.14)

(2.16)

(2.17)

Furthermore the sending end current is

Then substituting the value of VM from (2.16) in (2.19) and solving

Then the ABCD parameters of the T-network are

Section IV: Long Line Model

For accurate modeling of the transmission line we must not assume that the parameters are lumped but are distributed throughout line. The single-line diagram of a long transmission line is shown in Fig. 2.5. The length of the line is l . Let us consider a small strip Δx that is at a distance x from the receiving end. The voltage and current at the end of the strip are V and I respectively and the beginning of the strip are V + ΔV and I + Δ I respectively. The voltage drop across the strip is then ΔV . Since the length of the strip is Δx , the series impedance and shunt admittance are z Δx and y Δx . It is to be noted here that the total impedance and admittance of the line are

Fig. 2.5 Long transmission line representation. From the circuit of Fig. 2.5 we see that

Again as x 0, from (2.25) we get

(2.24)

(2.25)

(2.26)

(2.18)

(2.19)

(2.20)

(2.21)

(2.22)

Now for the current through the strip, applying KCL we get

The second term of the above equation is the product of two small quantities and therefore can be neglected. For x 0 we then have

Taking derivative with respect to x of both sides of (2.26) we get

Substitution of (2.28) in the above equation results

The roots of the above equation are located at ±√( yz ). Hence the solution of (2.29) is of the form

Taking derivative of (2.30) with respect to x we get

Combining (2.26) with (2.31) we have

Let us define the following two quantities

Then (2.30) and (2.32) can be written in terms of the characteristic impedance and propagation constant as

Let us assume that x = 0. Then V = VR and I = IR . From (2.35) and (2.36) we then get

Solving (2.37) and (2.38) we get the following values for A1 and A2 .

(2.27)

(2.28)

(2.29)

(2.30)

(2.31)

(2.32)

(2.33)

(2.34)

(2.35)

(2.36)

(2.37)

(2.38)

Also note that for x = l we have V = Vs and I = IS . Therefore replacing x by l and substituting the values of A1 and A2 in (2.35) and (2.36) we get

Noting that

We can rewrite (2.39) and (2.40) as

The ABCD parameters of the long transmission line can then be written as

(2.39)

(2.40)

(2.41)

(2.42)

(2.43)

(2.44)

(2.45)

Let's do an Example

Example 2.1

Consider a 500 km long line for which the per kilometer line impedance and admittance are given respectively by z = 0.1 + j 0.5145 W and y = j 3.1734 x 10 -6 mho. Therefore

and

We shall now use the following two formulas for evaluating the hyperbolic forms

Application of the above two equations results in the following values

Therefore from (2.43) to (2.45) the ABCD parameters of the system can be written as

Equivalent- π Representation of a Long Line

The π -equivalent of a long transmission line is shown in Fig. 2.6. In this the series impedance is denoted by Z while the shunt admittance is denoted by Y From (2.13) to (2.15) the ABCD parameters are defined as

Fig. 2.6 Equivalent p representation of a long transmission line.

Comparing (2.44) with (2.47) we can write

where Z = zl is the total impedance of the line. Again comparing (2.43) with (2.46) we get

Rearranging (2.50) we get

where Y = yl is the total admittance of the line. Note that for small values of l , sinh γl = γl and tanh ( γl /2) = γl /2. Therefore from (2.49) we get Z = Z and from (2.51) we get Y = Y . This implies that when the length of the line is small, the nominal- p representation with lumped parameters is fairly accurate. However the lumped parameter representation becomes erroneous as the length of the line increases.

Example 2.2

Consider the transmission line given in Example 2.1. The equivalent system parameters for both lumped and distributed parameter representation are given in Table 2.1 for three different line lengths. It can be seen that the error between the parameters increases as the line length increases.

Table 2.1 Variation in equivalent parameters as the line length changes

Length of the Line (km)

Lumped Parameters Distributed Parameters

Z(Ω) Y(mho)

100 52.41 < 79o 3.17 X 10-4 < 90o 52.27< 79o 3.17 X 10-4 < 89.98o

(2.46)

(2.47)

(2.48)

Ω (2.49)

(2.50)

(2.51)

250 131.032 < 79o 7.93 X 10-4 < 90o 128.81< 79.2o

8.0 X 10-4 < 89.9o

500 262.064 < 79o 1.58 X 10-3 < 90o 244.61<79.8o 1.64 X 10-3 < 89.6o

Characterization of a Long Lossless Line

For a lossless line, the line resistance is assumed to be zero. The characteristic impedance then becomes a

pure real number and it is often referred to as the surge impedance . The propagation constant becomes a

pure imaginary number. Defining the propagation constant as γ = jβ and replacing l by x we can rewrite (2.41)

and (2.42) as

The term surge

impedance loading or SIL is often used to indicate the nominal capacity of the line. The surge impedance is

the ratio of voltage and current at any point along an infinitely long line. The term SIL or natural power is a

measure of

power delivered

by a transmission

line when

terminated by surge impedance and is given by

where V0 is the rated voltage of the line.

At SIL ZC = VR / IR and hence from equations (2.52) and (2.53) we get

This implies that as the distance x changes, the magnitudes of the voltage and current in the above equations

do not change. The voltage then has a flat profile all along the line. Also as ZC is real, V and I are in phase with

each other all through out the line. The phase angle difference between the sending end voltage and the

receiving end voltage is then θ = β l. This is shown in Fig. 2.7.

Fig. 2.7 Voltage-current relationship in naturally loaded line.

Voltage and Current Characteristics of an SMIB System

For the analysis presented below we assume that the magnitudes of the voltages at the two ends are the same. The sending and receiving voltages are given by

(2.52)

(2.53)

(2.54)

(2.55)

(2.56)

where δ is angle between the sources and is usually called the load angle . As the total length of

the line is l , we replace x by l to obtain the sending end voltage from (2.39) as

Solving the above equation we get

Substituting (2.59) in (2.52), the voltage equation at a point in the transmission line that is at a distance x from the receiving end is obtained as

In a similar way the current at that point is given by

When the system is unloaded, the receiving end current is zero ( IR = 0). Therefore we can rewrite (2.58) as

Substituting the above equation in (2.52) and (2.53) we get the voltage and current for the unloaded system as

and (2.57)

(2.58)

(2.59)

(2.60)

(2.61)

(2.62)

(2.63)

Example 2.4

Consider the system given in Example 2.3. It is assumed that the system is unloaded with VS = VR = 1< 0° per unit. The voltage and current profiles for the unloaded system is shown in Fig. 2.9. The maximum voltage of 1.2457 per unit occurs at the receiving end while the maximum current of 0.7428 per unit is at the sending end. The current falls monotonically from the sending end and voltage rises monotonically to the receiving end. This rise in voltage under unloaded or lightly loaded condition is called Ferranti effect .

Fig. 2.9 Voltage and current profile over an unloaded transmission line.

Mid Point Voltage and Current of Loaded Lines

The mid point voltage of a transmission line is of significance for the reactive compensation of transmission lines. To obtain an expression of the mid point voltage, let us assume that the line is loaded (i.e., the load angle d is not equal to zero). At the mid point of the line we have x = l /2 such that βx = q /2. Let us denote the midpoint voltage by VM . Let us also assume that the line is symmetric, i.e., VS = VR = V. We can then rewrite equation (2.60) to obtain

Again noting that

We obtain the following expression of the mid point voltage

The mid point current is similarly given by

The phase angle of the mid point voltage is half the load angle always. Also the mid point voltage and current are in phase, i.e., the power factor at this point is unity. The variation in the magnitude of voltage with changes in load angle is maximum at the mid point. The voltage at this point decreases with the increase in δ . Also as the power through a lossless line is constant through out its length and the mid point power factor is unity, the mid point current increases with an increase in δ

Example 2.5

Consider the transmission line discussed in Example 2.4. Assuming the magnitudes of both sending and receiving end voltages to be 1.0 per unit, we can compute the magnitude of the mid point voltage as the load angle ( δ ) changes. This is given in Table 2.2. The variation in voltage with δ is shown in Fig. 2.10.

Table 2.2 Changes in the mid-point voltage magnitude with load angle

δ in degree VM in per unit

20 1.0373

25 1.0283

30 1.0174

(2.65)

(2.66)

Fig. 2.10 Variation in voltage profile for a loaded line

Power in a Lossless Line

The power flow through a lossless line can be given by the mid point voltage and current equations given in (2.66) and (2.67). Since the power factor at this point is unity, real power over the line is given by

If V = V0 , the rated voltage, we can rewrite the above expression in terms of the natural power as

For a short transmission line we have

where X is the total reactance of the line. Equation (2.71) then can be modified to obtain the well known power transfer relation for the short line approximation as

In general it is not necessary for the magnitudes of the sending and receiving end voltages to be same. The power transfer relation given in (2.72) will not be valid in that case. To derive a general expression for power transfer, we assume

If the receiving end real and reactive powers are denoted by PR and QR respectively, we can write from (2.52)

(2.71)

(2.72)

(2.73)

(2.74)

and

Equating real and imaginary parts of the above equation we get

and

Rearranging (2.76) we get the power flow equation for a losslees line as

To derive expressions for the reactive powers, we rearrange (2.75) to obtain the reactive power delivered to the receiving end as

Again from equation (2.61) we can write

The sending end apparent power is then given by

Equating the imaginary parts of the above equation we get the following expression for the reactive generated by the source

The reactive power absorbed by the line is then

It is important to note that if the magnitude of the voltage at the two ends is equal, i.e., VS = VR = V, the reactive powers at the two ends become negative of each other, i.e., QS = QR . The net reactive power absorbed by the line then becomes twice the sending end reactive power, i.e., QL = 2QS . Furthermore, since cosθ ≈ 1 for small values of θ, the reactive powers at the two ends for a short transmission line are given by

(2.75)

(2.76)

(2.77)

(2.78)

(2.79)

(2.80)

(2.81)

The reactive power absorbed by the line under this condition is given by

Example 2.6

Consider a short, lossless transmission line with a line reactance of 0.5 per unit. We assume that the magnitudes of both sending and receiving end voltages to be 1.0 per unit. The real power transfer over the line and reactive power consumed by the line are shown in Fig. 2.11. The maximum real power is 2.0 per unit and it occurs for δ = 90° . Also the maximum reactive power consumed by the line occurs at δ = 180 ° and it has a value of 8 per unit.

Fig. 2.11 Real power flow and reactive power consumed by a transmission line.

(2.82)

Chapter 3: Network Admittance and Impedance Matrices

As we have seen in Chapter 1 that a power system network can be converted into an equivalent impedance diagram. This diagram forms the basis of power flow (or load flow) studies and short circuit analysis. In this chapter we shall discuss the formation of bus admittance matrix (also known as Ybus matrix) and bus impedance matrix (also known as Zbus matrix). These two matrices are related by

We shall discuss the formation of the Ybus matrix first. This will be followed by the discussion of the formation of the Zbus matrix.

Section I: Formation of Bus Admittance Matrix

Node Elimination by Matrix Partitioning

Node Elimination by Kron Reduction

Inclusion of Line Charging Capacitors

Formation of Bus Admittance Matrix

Consider the voltage source VS with a source (series) impedance of ZS as shown in Fig. 3.1 (a). Using Norton's theorem this circuit can be replaced by a current source IS with a parallel admittance of YS as shown in Fig. 3.1 (b). The relations between the original system and the Norton equivalent are

We shall use this Norton's theorem for the formulation of the Ybus matrix.

Fig. 3.1 (a) Voltage source with a source impedance and (b) its Norton equivalent.

For the time being we shall assume the short line approximation for the formulation of the bus admittance matrix. We shall thereafter relax this assumption and use the π -representation of the network for power flow studies. Consider the 4-bus power system shown in Fig. 3.2. This contains two generators G1 and G2 that are connected through transformers T1 and T2 to buses 1 and 2. Let us denote the synchronous reactances of G1 and G2 by XG1 and XG2 respectively and the leakage reactances of T1 and T2 by XT1 and XT2 respectively. Let Zij, i = 1, ..., 4 and j = 1, ... , 4 denote the line impedance between buses i and j .

(3.1)

Fig. 3.2 Single-line diagram of a simple power network.

Then the system impedance diagram is as shown in Fig. 3.3 where Z11 = j ( XG1 + XT1 ) and Z22 = j ( XG2 + XT2 ).

Fig. 3.3 Impedance diagram of the power network of Fig. 3.2.

In this figure the nodes with the node voltages of V1 to V4 indicate the buses 1 to 4 respectively. Bus 0 indicates the reference node that is usually the neutral of the Y-connected system. The impedance diagram is converted into an equivalent admittance diagram shown in Fig. 3.4. In this diagram Yij = 1/ Zij , i = 1,..., 4 and j = 1, ... , 4. The voltage sources EG1 and EG2 are converted into the equivalent current sources I1 and I2 respectively using the Norton's theorem discussed before.

Fig. 3.4 Equivalent admittance diagram of the impedance of Fig. 3.3.

Contd from previous slide

We would like to determine the voltage-current relationships of the network shown in Fig. 3.4. It is to be noted that this relation can be written in terms of the node (bus) voltages V1 to V4 and injected currents I1 and I2 as follows

or,

(3.3)

(3.4)

It can be easily seen that we get (3.1) from (3.3) and (3.4).

Consider node (bus) 1 that is connected to the nodes 2 and 3. Then applying KCL at this node we get

In a similar way application of KCL at nodes 2, 3 and 4 results in the following equations

Combining (3.5) to (3.8) we get

Comparing (3.9) with (3.3) we can write

In general the format of the Ybus matrix for an n -bus power system is as follows

where

It is to be noted that Ybus is a symmetric matrix in which the sum of all the elements of the k th column is Ykk .

Example 3.1

Consider the impedance diagram of Fig. 3.2 in which the system parameters are given in per unit by

Z11 = Z22 = j0.25, Z12 = j0.2, Z13 = j0.25, Z23 = Z34 = j0.4 and Z24 = j0.5

(3.5)

(3.6)

(3.7)

(3.8)

(3.9)

(3.10)

(3.11)

(3.12)

The system admittance can then be written in per unit as

Y11 = Y22 = j4, Y12 = j5, Y13 = j4, Y23 = Y34 = j2.5 and Y24 = j2

The Ybus is then given from (3.10) as

per unit

Consequently the bus impedance matrix is given by

per unit

It can be seen that like the Ybus matrix the Zbus matrix is also symmetric.

Let us now assume that the voltages EG1 and EG2 are given by

The current sources I1 and I2 are then given by

We then get the node voltages from (3.4) as

p.u.

Solving the above equation we get the node voltages as

per unit

Node Elimination by Matrix Partitioning

Sometimes it is desirable to reduce the network by eliminating the nodes in which the current do not enter or leave. Let (3.3) be written as

(3.13)

In the above equation IA is a vector containing the currents that are injected, Ix is a null vector and the Ybus matrix is portioned with the matrices K , L and M . Note that the Ybus matrix contains both L and LT due to its symmetric nature.

We get the following two sets of equations from (3.13)

Substituting (3.15) in (3.14) we get

Therefore we obtain the following reduced bus admittance matrix

Example 3.2

Let us consider the system of Example 3.1. Since there is no current injection in either bus 3 or bus 4, from the Ybus computed we can write

We then have

per unit

Substituting I1 = 460 per unit and I2 = 490 per unit we shall get the same values of V1 and V2 as given in Example 3.1.

Inspecting the reduced Ybus matrix we can state that the admittance between buses 1 and 2 is - j 6.8978. Therefore the self admittance (the admittance that is connected in shunt) of the buses 1 and 2 is - j4 per unit (= - j 10.8978 + j 6.8978). The reduced admittance diagram obtained by eliminating nodes 3 and 4 is shown in Fig. 3.5. It is to be noted that the impedance between buses 1 and 2 is the Thevenin impedance between these two buses. The value of this impedance is 1/( - j 6.8978) = j 0.145 per unit.

Fig. 3.5 Reduced admittance diagram after the elimination of buses 3 and 4.

Node Elimination by Kron Reduction

Consider an equation of the form

(3.14)

(3.15)

(3.16)

where A is an ( n X n ) real or complex valued matrix, x and b are vectors in either Rn or Cn . assume that the b vector has a zero element in the n th row such that (3.18) is given as

We can then eliminate the kth row and kth column to obtain a reduced ( n - 1) number of equations of the form

The elimination is performed using the following elementary operations

Example 3.3

Let us consider the same system of Example 3.1. We would like to eliminate the last two rows and columns. Let us first eliminate the last row and last column. Some of the values are given below

In a similar way we can calculate the other elements. Finally eliminating the last row and last column, as all these elements are zero, we get the new Ybus matrix as

Further reducing the last row and the last column of the above matrix using (3.21), we obtain the reduced Ybus matrix given in Example 3.2.

Inclusion of Line Charging Capacitors

So far we have assumed that the transmission lines are modeled with lumped series impedances without the shunt capacitances. However in practice, the Ybus matrix contains the shunt admittances for load flow analysis in which the transmission lines are represented by its π -equivalent. Note that

(3.18)

(3.19)

(3.20)

(3.21)

whether the line is assumed to be of medium length or long length is irrelevant as we have seen in Chapter 2 how both of them can be represented in a p -equivalent.

Consider now the power system of Fig. 3.2. Let us assume that all the lines are represented in an equivalent- π with the shunt admittance between the line i and j being denoted by Ychij . Then the equivalent admittance at the two end of this line will be Ychij/2. For example the shunt capacitance at the two ends of the line joining buses 1 and 3 will be Ych13/2. We can then modify the admittance diagram Fig. 3.4 as shown in Fig. 3.6. The Ybus matrix of (3.10) is then modified as

(3.22)

where

Fig. 3.6 Admittance diagram of the power system Fig. 3.2 with line charging capacitors.

Section II: Elements Of The Bus Impedance And Admittance Matrices

Equation (3.1) indicates that the bus impedance and admittance matrices are inverses of each other. Also since Ybus is a symmetric matrix, Zbus is also a symmetric matrix. Consider a 4-bus system for which the voltage-current relations are given in terms of the Ybus matrix as

We can then write

(3.23)

(3.24)

This implies that Y11 is the admittance measured at bus-1 when buses 2, 3 and 4 are short circuited. The admittance Y11 is defined as the self admittance at bus-1. In a similar way the self admittances

of buses 2, 3 and 4 can also be defined that are the diagonal elements of the Ybus matrix. The off diagonal elements are denoted as the mutual admittances . For example the mutual admittance between buses 1 and 2 is defined as

The mutual admittance Y12 is obtained as the ratio of the current injected in bus-1 to the voltage of bus-2 when buses 1, 3 and 4 are short circuited. This is obtained by applying a voltage at bus-2 while shorting the other three buses.

The voltage-current relation can be written in terms of the Zbus matrix as

The driving point impedance at bus-1 is then defined as

i.e., the driving point impedance is obtained by injecting a current at bus-1 while keeping buses 2, 3 and 4 open-circuited. Comparing (3.26) and (3.28) we can conclude that Z11 is not the reciprocal of Y11 . The transfer impedance between buses 1 and 2 can be obtained by injecting a current at bus-2 while open-circuiting buses 1, 3 and 4 as

It can also be seen that Z12 is not the reciprocal of Y12 .

Section II: Modification of Bus Impedance Matrix

Adding a New Bus to the Reference Bus

Adding a New Bus to an Existing Bus through an Impedance

Adding an Impedance to the Reference Bus from an Existing Bus

Adding an Impedance between two Existing Buses

Direct Determination of Zbus Matrix

Modification of Bus Impedance Matrix

Equation (3.1) gives the relation between the bus impedance and admittance matrices. However it may be possible that the topology of the power system changes by the inclusion of a new bus or line.

(3.25)

(3.26)

(3.27)

(3.28)

(3.29)

In that case it is not necessary to recompute the Ybus matrix again for the formation of Zbus matrix. We shall discuss four possible cases by which an existing bus impedance matrix can be modified.

Let us assume that an n -bus power system exists in which the voltage-current relations are given in terms of the bus impedance matrix as

The aim is to modify the matrix Zorig when a new bus or line is connected to the power system.

Adding a New Bus to the Reference Bus

It is assumed that a new bus p ( p > n ) is added to the reference bus through an impedance Zp . The schematic diagram for this case is shown in Fig. 3.7. Since this bus is only connected to the reference bus, the voltage-current relations the new system are

Fig. 3.7 A new bus is added to the reference bus.

Adding a New Bus to an Existing Bus through an Impedance

This is the case when a bus, which has not been a part of the original network, is added to an existing bus through a transmission line with an impedance of Zb . Let us assume that p ( p > n ) is the new bus that is connected to bus k ( k < n ) through Zb. Then the schematic diagram of the circuit is as shown in Fig. 3.8. Note from this figure that the current Ip flowing from bus p will alter the voltage of the bus k . We shall then have

In a similar way the current Ip will also alter the voltages of all the other buses as

Furthermore the voltage of the bus p is given by

(3.30)

(3.31)

(3.32)

(3.33)

(3.34)

Therefore the new voltage current relations are

It can be noticed that the new Zbus matrix is also symmetric.

Fig. 3.8 A new bus is added to an existing bus through an impedance.

Adding an Impedance to the Reference Bus from an Existing Bus

To accomplish this we first assume that an impedance Zb is added from a new bus p to an existing bus k . This can be accomplished using the method discussed in Section 3.3.2. Then to add this bus k to the reference bus through Zb, we set the voltage Vp of the new bus to zero. However now we have an ( n + 1) X ( n + 1) Z bus matrix instead of an n X n matrix. We can then remove the last row and last column of the new Zbus matrix using the Kron's reduction given in (3.21).

Adding an Impedance between two Existing Buses

Let us assume that we add an impedance Zb between two existing buses k and j as shown in Fig. 3.9. Therefore the current injected into the network from the bus k side will be Ik - Ib instead of Ik. Similarly the current injected into the network from the bus j side will be Ij + Ib instead of Ij. Consequently the voltage of the ith bus will be

Similarly we have

and

Fig. 3.9 An impedance is added between two existing buses.

We shall now have to eliminate Ib from the above equations. To do that we note from Fig. 3.9 that

(3.35)

(3.36)

(3.37)

(3.38)

Substituting (3.37) and (3.38) in (3.39) we get

We can then write the voltage current relations as

where

We can now eliminate the last row and last column using the Kron's reduction given in (3.21).

Direct Determination of Zbus Matrix

We shall now use the methods given in Sections 3.3.1 to 3.3.4 for the direct determination of the Zbus matrix without forming the Ybus matrix first. To accomplish this we shall consider the system of Fig. 3.2 and shall use the system data given in Example 3.1. Note that for the construction of the Z bus matrix we first eliminate all the voltage sources from the system.

Step-1 : Start with bus-1. Assume that no other buses or lines exist in the system. We add this bus to

the reference bus with the impedance of j 0.25 per unit. Then the Zbus matrix is

Step-2 : We now add bus-2 to the reference bus using (3.31). The system impedance diagram is shown in Fig. 3.10. We then can modify (3.43) as

Fig. 3.10 Network of step-2.

Step-3 : We now add an impedance of j 0.2 per unit between buses 1 and 2 as shown in Fig. 3.11. The interim Z bus matrix is then obtained by applying (3.41) on (3.44) as

Eliminating the last row and last column using the Kron's reduction of (3.31) we get

(3.39)

(3.40)

(3.41)

(3.42)

(3.43)

(3.44)

Step-4: We now

add bus-3 to bus-1 through an impedance of j 0.25 per unit as shown in Fig. 3.12. The application of (3.35) on (3.45) will then result in the following matrix

Fig. 3.11 Network of step-3.

Fig. 3.12 Network of step-4.

Step-5 : Connect buses 2 and 3 through an impedance of j 0.4 per unit as shown in Fig. 3.13. The

interim Zbus matrix is then formed from (3.41) and (3.46) as

Fig. 3.13 Network of step-5. Using the Kron's reduction we get the following matrix Step-6 : We now add a new bus-4 to bus-2 through an impedance of j 0.5 as shown in Fig. 3.14.

Then the application of (3.35) on (3.47) results in the following matrix

(3.45)

(3.46)

(3.47)

Fig. 3.14 Network of step-6. Step-7 : Finally we add buses 3 and 4 through an impedance of j 0.4 to obtain the network of Fig. 3.3

minus the voltage sources. The application of (3.41) on (3.48) results in the interim Zbus matrix of Eliminating the 5th row and column through Kron's reduction we get the final Zbus as The Zbus matrix given in (3.49) is the as that given in Example 3.1 which is obtained by inverting the Ybus matrix. Section IV: Thevenin Impedance And Zbus Matrix

To establish relationships between the elements of the Zbus matrix and Thevenin equivalent, let us consider the following example Example 3.4 Consider the two bus power system shown in Fig. 3.15. It can be seen that the open-circuit voltages of buses a and b are Va and Vb respectively. From (3.11) we can write the Ybus matrix of the system as

Fig. 3.15 Two-bus power system of Example 3.4.

The determinant of the above matrix is

(3.48)

(3.49)

Therefore the Zbus matrix is Solving the last two equations we get Now consider the system of Fig. 3.15. The Thevenin impedance of looking into the system at bus- a is the parallel combination of Zaa and Zab + Zbb, i.e., Similarly the Thevenin impedance obtained by looking into the system at bus- b is the parallel combination of Zbb and Zaa + Zab, i.e., Hence the driving point impedances of the two buses are their Thevenin impedances. Let us now consider the Thevenin impedance while looking at the system between the buses a and b . From Fig. 3.15 it is evident that this Thevenin impedance is the parallel combination of Zab and Zaa + Zbb, i.e., With the values given in (3.50) we can write Comparing the last two equations we can write As we have seen in the above example in the relation V = Zbus I , the node or bus voltages Vi, i = 1, ... , n are the open circuit voltages. Let us assume that the currents injected in buses 1, ... , k - 1 and k + 1, ... , n are zero when a short circuit occurs at bus k . Then Thevenin impedance at bus k is

(3.50)

(3.51)

(3.52)

(3.53

From (3.51), (3.52) and (3.54) we can surmise that the driving point impedance at each bus is the Thevenin impedance. Let us now find the Thevenin impedance between two buses j and k of a power system. Let the open circuit voltages be defined by the voltage vector V° and corresponding currents be defined by I° such that Now suppose the currents are changed by ΔI such that the voltages are changed by ΔV . Then Comparing (3.55) and (3.56) we can write Let us now assume that additional currents ΔIk and ΔIk are injected at the buses k and j respectively while the currents injected at the other buses remain the same. Then from (3.57) we can write We can therefore write the following two equations form (3.58) The above two equations can be rewritten as Since Zjk = Zkj the network can be drawn as shown in Fig. 3.16. By inspection we can see that the open circuit voltage between the buses k and j is and the short circuit current through these two buses is Also during the short circuit Vk- Vj = 0. Therefore combining (3.59) and (3.60) we get

(3.54)

(3.55)

(3.56)

(3.57)

(3.58)

(3.59)

(3.60)

(3.61)

(3.62)

Combining (3.61) to (3.63) we find the Thevenin impedance between the buses k and j as The above equation agrees with our earlier derivation of the two bus network given in (3.53).

Fig 3.16 Thevenin equivalent between buses k and j

Chapter 4: Load Flow Studies

Overview

Load flow studies are one of the most important aspects of power system planning and operation. The load flow gives us the sinusoidal steady state of the entire system - voltages, real and reactive power generated and absorbed and line losses. Since the load is a static quantity and it is the power that flows through transmission lines, the purists prefer to call this Power Flow studies

rather than load flow studies. We shall however stick to the original nomenclature of load flow.

Through the load flow studies we can obtain the voltage magnitudes and angles at each bus in the steady state. This is rather important as the magnitudes of the bus voltages are required to be held within a specified limit. Once the bus voltage magnitudes and their angles are computed using the load flow, the real and reactive power flow through each line can be computed. Also based on the difference between power flow in the sending and receiving ends, the losses in a particular line can also be computed. Furthermore, from the line flow we can also determine the over and under load conditions.

The steady state power and reactive power supplied by a bus in a power network are expressed in terms of nonlinear algebraic equations. We therefore would require iterative methods for solving these equations. In this chapter we shall discuss two of the load flow methods. We shall also delineate how to interpret the load flow results.

Section I: Real And Reactive Power Injected in a Bus

For the formulation of the real and reactive power entering a b us, we need to define the following quantities. Let the voltage at the i th bus be denoted by

Also let us define the self admittance at bus- i as

(3.63)

(3.64)

(4.1)

(4.2)

Similarly the mutual admittance between the buses i and j can be written as

Let the power system contains a total number of n buses. The current injected at bus- i is given as

It is to be noted we shall assume the current entering a bus to be positive and that leaving the bus to be negative. As a consequence the power and reactive power entering a bus will also be assumed to be positive. The complex power at bus- i is then given by

Note that

Therefore substituting in (4.5) we get the real and reactive power as

Section II: Classification Of Buses

For load flow studies it is assumed that the loads are constant and they are defined by their real and reactive power consumption. It is further assumed that the generator terminal voltages are tightly regulated and therefore are constant. The main objective of the load flow is to find the voltage magnitude of each bus and its angle when the powers generated and loads are pre-specified. To facilitate this we classify the different buses of the power system shown in the chart below.

(4.3)

(4.4)

(4.5)

(4.6)

Classification Of Buses

Load Buses : In these buses no generators are connected and hence the generated real power PGi and reactive power QGi are taken as zero. The load drawn by these buses are defined by real power -PLi and reactive power -QLi in which the negative sign accommodates for the power flowing out of the bus. This is why these buses are sometimes referred to as P-Q bus. The objective of the load flow is to find the bus voltage magnitude |Vi| and its angle δi.

Voltage Controlled Buses : These are the buses where generators are connected. Therefore the power generation in such buses is controlled through a prime mover while the terminal voltage is controlled through the generator excitation. Keeping the input power constant through turbine-governor control and keeping the bus voltage constant using automatic voltage regulator, we can specify constant PGi and | Vi | for these buses. This is why such buses are also referred to as P-V buses. It is to be noted that the reactive power supplied by the generator QGi depends on the system configuration and cannot be specified in advance. Furthermore we have to find the unknown angle δi of the bus voltage.

Slack or Swing Bus : Usually this bus is numbered 1 for the load flow studies. This bus sets the angular reference for all the other buses. Since it is the angle difference between two voltage sources that dictates the real and reactive power flow between them, the particular angle of the slack bus is not important. However it sets the reference against which angles of all the other bus voltages are measured. For this reason the angle of this bus is usually chosen as 0° . Furthermore it is assumed that the magnitude of the voltage of this bus is known.

Now consider a typical load flow problem in which all the load demands are known. Even if the generation matches the sum total of these demands exactly, the mismatch between generation and load will persist because of the line I 2R losses. Since the I 2R loss of a line depends on the line current which, in turn, depends on the magnitudes and angles of voltages of the two buses connected to the line, it is rather difficult to estimate the loss without calculating the voltages and angles. For this reason a generator bus is usually chosen as the slack bus without specifying its real power. It is assumed that the generator connected to this bus will supply the balance of the real power required and the line losses.

Section III: Preparation Of Data For Load Flow

Let real and reactive power generated at bus- i be denoted by PGi and QGi respectively. Also let us denote the real and reactive power consumed at the i th th bus by PLi and QLi respectively. Then the net real power injected in bus- i is

Let the injected power calculated by the load flow program be Pi, calc . Then the mismatch between the actual injected and calculated values is given by

In a similar way the mismatch between the reactive power injected and calculated values is given by

The purpose of the load flow is to minimize the above two mismatches. It is to be noted that (4.6) and (4.7) are used for the calculation of real and reactive power in (4.9) and (4.10). However since the magnitudes of all the voltages and their angles are not known a priori, an iterative procedure must be used to estimate the bus voltages and their angles in order to calculate the mismatches. It is expected that mismatches ΔPi and ΔQi reduce with each iteration and the load flow is said to have converged when the mismatches of all the buses become less than a very small number.

For the load flow studies we shall consider the system of Fig. 4.1, which has 2 generator and 3 load buses. We define bus-1 as the slack bus while taking bus-5 as the P-V bus. Buses 2, 3 and 4 are

(4.8)

(4.9)

(4.10)

P-Q buses. The line impedances and the line charging admittances are given in Table 4.1. Based on this data the Y bus matrix is given in Table 4.2. This matrix is formed using the same procedure as given in Section 3.1.3. It is to be noted here that the sources and their internal impedances are not considered while forming the Ybus matrix for load flow studies which deal only with the bus voltages.

Fig. 4.1 The simple power system used for load flow studies.

Table 4.1 Line impedance and line charging data of the system of Fig. 4.1.

Line(bus to bus) Impedance Line charging ( Y /2)

1-2 0.02 + j 0.10 j 0.030

1-5 0.05 + j 0.25 j 0.020

2-3 0.04 + j 0.20 j 0.025

2-5 0.05 + j 0.25 j 0.020

3-4 0.05 + j 0.25 j 0.020

3-5 0.08 + j 0.40 j 0.010

4-5 0.10 + j 0.50 j 0.075

Table 4.2 Ybus matrix of the system of Fig. 4.1.

1 2 3 4 5

1 2.6923 - j 13.4115 - 1.9231 + j 9.6154 0 0 - 0.7692 + j 3.8462

2 - 1.9231 + j 9.6154 3.6538 - j 18.1942 - 0.9615 + j 4.8077 0 - 0.7692 + j 3.8462

3 0 - 0.9615 + j 4.8077 2.2115 - j 11.0027 - 0.7692 + j 3.8462 - 0.4808 + j 2.4038

4 0 0 - 0.7692 + j 3.8462 1.1538 - j 5.6742 - 0.3846 + j 1.9231

5 - 0.7692 + j 3.8462 - 0.7692 + j 3.8462 - 0.4808 + j 2.4038 - 0.3846 + j 1.9231 2.4038 - j 11.8942

The bus voltage magnitudes, their angles, the power generated and consumed at each bus are given in Table 4.3. In this table some of the voltages and their angles are given in boldface letters. This indicates that these are initial data used for starting the load flow program. The power and reactive power generated at the slack bus and the reactive power generated at the P-V bus are unknown. Therefore each of these quantities are indicated by a dash ( - ). Since we do not need these quantities for our load flow calculations, their initial estimates are not required. Also note from Fig. 4.1 that the slack bus does not contain any load while the P-V bus 5 has a local load and this is indicated in the load column.

Table 4.3 Bus voltages, power generated and load - initial data.

Bus no. Bus voltage Power generated Load

Magnitude (pu) Angle (deg) P (MW) Q (MVAr) P (MW) P (MVAr)

1 1.05 0 - - 0 0

2 1 0 0 0 96 62

3 1 0 0 0 35 14

4 1 0 0 0 16 8

5 1.02 0 48 - 24 11

Section IV: Load Flow by Gauss-Seidel Method

Updating Load Bus Voltages

Updating P-V Bus Voltages

Convergence of the Algorithm

The basic power flow equations (4.6) and (4.7) are nonlinear. In an n -bus power system, let the number of P-Q buses be np and the number of P-V (generator) buses be ng such that n = np + ng + 1. Both voltage magnitudes and angles of the P-Q buses and voltage angles of the P-V buses are unknown making a total number of 2np + ng quantities to be determined. Amongst the known quantities are 2np numbers of real and reactive powers of the P-Q buses, 2ng numbers of real powers and voltage magnitudes of the P-V buses and voltage magnitude and angle of the slack bus. Therefore there are sufficient numbers of known quantities to obtain a solution of the load flow problem. However, it is rather difficult to obtain a set of closed form equations from (4.6) and (4.7). We therefore have to resort to obtain iterative solutions of the load flow problem.

At the beginning of an iterative method, a set of values for the unknown quantities are chosen. These are then updated at each iteration. The process continues till errors between all the known and actual quantities reduce below a pre-specified value. In the Gauss-Seidel load flow we denote the initial voltage of the i th bus by Vi

(0) , i = 2, ... , n . This should read as the voltage of the i th bus at the 0th iteration, or initial guess. Similarly this voltage after the first iteration will be denoted by Vi

(1) . In this Gauss-Seidel load flow the load buses and voltage controlled buses are treated differently. However in both these type of buses we use the complex power equation given in (4.5) for updating the voltages. Knowing the real and reactive power injected at any bus we can expand (4.5) as

We can rewrite (4.11) as

In this fashion the voltages of all the buses are updated. We shall outline this procedure with the help of the system of Fig. 4.1, with the system data given in Tables 4.1 to 4.3. It is to be noted that the real and reactive powers are given respectively in MW and MVAr. However they are converted into per unit quantities where a base of 100 MVA is chosen.

Updating Load Bus Voltages

Let us start the procedure with bus-2 of the 5 bus 7 line system given in fig: 4.1. Since this is load bus, both the real and reactive power into this bus is known. We can therefore write from (4.12)

(4.11)

(4.12)

(4.13)

From the data given in Table 4.3 we can write

It is to be noted that since the real and reactive power is drawn from this bus, both these quantities appear in the above equation with a negative sign. With the values of the Y bus elements given in Table 4.2 we get V2

1 = 0.9927 < − 2.5959° .

The first iteration voltage of bus-3 is given by

Note that in the above equation since the update for the bus-2 voltage is already available, we used the 1st iteration value of this rather than the initial value. Substituting the numerical data we get V3

(1) = 0.9883 < − 2. 8258° . Finally the bus-4 voltage is given by Solving we get V4

(1) = 0. 9968 < −3.4849° . Updating P-V Bus Voltages

It can be seen from Table 4.3 that even though the real power is specified for the P-V bus-5, its reactive power is unknown. Therefore to update the voltage of this bus, we must first estimate the reactive power of this bus. Note from Fig. 4.11 that And hence we can write the kth iteration values as For the system of Fig. 4.1 we have This is computed as 0.0899 per unit. Once the reactive power is estimated, the bus-5 voltage is updated as

It is to be noted that even though the power generation in bus-5 is 48 MW, there is a local load that is consuming half that amount. Therefore the net power injected by this bus is 24 MW and consequently the injected power P5, inj in this case is taken as 0.24 per unit. The voltage is calculated as V5

(1) = 1.0169 < − 0.8894° . Unfortunately however the magnitude of the voltage obtained above is not equal to the magnitude given in Table 4.3. We must therefore force this voltage magnitude to be equal to that specified. This is accomplished by

(4.14)

(4.15)

(4.16)

(4.17)

(4.18)

(4.19)

This will fix the voltage magnitude to be 1.02 per unit while retaining the phase of − 0.8894 ° . The corrected voltage is used in the next iteration.

Convergence of the Algorithm

As can be seen from Table 4.3 that a total number of 4 real and 3 reactive powers are known to us. We must then calculate each of these from (4.6) and (4.7) using the values of the voltage magnitudes and their angle obtained after each iteration. The power mismatches are then calculated from (4.9) and (4.10). The process is assumed to have converged when each of ΔP2 , ΔP3, ΔP4 , ΔP5 , ΔQ2 , ΔQ3 and ΔQ4 is below a small pre-specified value. At this point the process is terminated.

Sometimes to accelerate computation in the P-Q buses the voltages obtained from (4.12) is multiplied by a constant. The voltage update of bus- i is then given by

where λ is a constant that is known as the acceleration factor . The value of λ has to be below 2.0 for the convergence to occur. Table 4.4 lists the values of the bus voltages after the 1st iteration and number of iterations required for the algorithm to converge for different values of λ. It can be seen that the algorithm converges in the least number of iterations when λ is 1.4 and the maximum number of iterations are required when λ is 2. In fact the algorithm will start to diverge if larger values of acceleration factor are chosen. The system data after the convergence of the algorithm will be discussed later.

Table 4.4 Gauss-Seidel method: bus voltages after 1 st iteration and number of iterations required for convergence for different values of l .

k Bus voltages (per unit) after 1st

iteration No of iterations

for convergence V2 V3 V4 V5

1 0.9927 2.6 0.9883 2.83 0.9968 3.48 1.02 0.89 28

2 0.9874 5.22 0.9766 8.04 0.9918 14.02 1.02 4.39 860

1.8 0.9883 4.7 0.9785 6.8 0.9903 11.12 1.02 3.52 54

1.6 0.9893 4.17 0.9807 5.67 0.9909 8.65 1.02 2.74 24

1.4 0.9903 3.64 0.9831 4.62 0.9926 6.57 1.02 2.05 14

1.2 0.9915 3.11 0.9857 3.68 0.9947 4.87 1.02 1.43 19

Section V: Solution of a Set of Nonlinear Equations by Newton-Raphson Method

In this section we shall discuss the solution of a set of nonlinear equations through Newton-Raphson method. Let us consider that we have a set of n nonlinear equations of a total number of n variables x1 , x2 , ... , xn. Let these equations be given by where f1, ... , fn are functions of the variables x1 , x2 , ... , xn. We can then define another set of functions g1 , ... , gn as given below

(4.20)

(4.21)

(4.22)

Let us assume that the initial estimates of the n variables are x1

(0) , x2(0) , ... , xn

(0) . Let us add corrections Δx1(0) ,

Δx2(0) , ... , Δxn

(0) to these variables such that we get the correct solution of these variables defined by The functions in (4.23) then can be written in terms of the variables given in (4.24) as We can then expand the above equation in Taylor 's series around the nominal values of x1

(0) , x2(0) ,

... , xn(0) . Neglecting the second and higher order terms of the series, the expansion of gk , k = 1, ... ,

n is given as

where is the partial derivative of gk evaluated at x2(1) , ... , xn

(1) . Equation (4.26) can be written in vector-matrix form as

The square matrix of partial derivatives is called the Jacobian matrix J with J (1) indicating that the matrix is evaluated for the initial values of x2

(0) , ... , xn(0) . We can then write the solution of (4.27) as

Since the Taylor 's series is truncated by neglecting the 2nd and higher order terms, we cannot expect to find the correct solution at the end of first iteration. We shall then have

(4.23)

(4.24)

(4.25)

(4.26)

(4.27)

(4.28)

, k= 1, ....n

These are then used to find J (1) and Δgk

(1) , k = 1, ... , n . We can then find Δx2(1) , ... , Δxn

(1) from an equation like (4.28) and subsequently calculate x2

(1) , ... , xn(1). The process continues till Δgk , k = 1,

... , n becomes less than a small quantity.

Section VI: Load Flow By Newton-Raphson Method

Load Flow Algorithm

Formation of the Jacobian Matrix

Solution of Newton-Raphson Load Flow

Let us assume that an n -bus power system contains a total np number of P-Q buses while the number of P-V (generator) buses be ng such that n = np + ng + 1. Bus-1 is assumed to be the slack bus. We shall further use the mismatch equations of ΔPi and ΔQi given in (4.9) and (4.10) respectively. The approach to Newton-Raphson load flow is similar to that of solving a system of nonlinear equations using the Newton-Raphson method: At each iteration we have to form a

Jacobian matrix and solve for the corrections from an equation of the type given in (4.27). For the load flow problem, this equation is of the form where the Jacobian matrix is divided into submatrices as It can be seen that the size of the Jacobian matrix is ( n + np − 1) x ( n + np −1). For example for the 5-bus problem of Fig. 4.1 this matrix will be of the size (7 x 7). The dimensions of the submatrices are as follows:

J11: (n 1) (n 1), J12: (n 1) np, J21: np (n 1) and J22: np np The sub-matrices are

(4.29)

(4.30)

(4.31)

Load Flow Algorithm

The Newton-Raphson procedure is as follows: Step-1: Choose the initial values of the voltage magnitudes |V| (0) of all np load buses and n − 1 angles δ (0) of the voltages of all the buses except the slack bus. Step-2: Use the estimated |V|(0) and δ (0) to calculate a total n − 1 number of injected real power

Pcalc(0) and equal number of real power mismatch ΔP (0) .

Step-3: Use the estimated |V| (0) and δ (0) to calculate a total np number of injected reactive power

Qcalc(0) and equal number of reactive power mismatch ΔQ (0) .

Step-3: Use the estimated |V| (0) and δ (0) to formulate the Jacobian matrix J (0) . Step-4: Solve (4.30) for δ (0) and Δ |V| (0) ÷ |V| (0). Step-5 : Obtain the updates from

Step-6: Check if all the mismatches are below a small number. Terminate the process if yes.

Otherwise go back to step-1 to start the next iteration with the updates given by (4.36) and (4.37).

Formation of the Jacobian Matrix

We shall now discuss the formation of the submatrices of the Jacobian matrix. To do that we shall use the real and reactive power equations of (4.6) and (4.7). Let us rewrite them with the help of (4.2) as

(4.32)

(4.33)

(4.34)

(4.36)

(4.37)

(4.38)

A. Formation of J11

Let us define J11 as

It can be seen from (4.32) that Lik's are the partial derivatives of Pi with respect to δk. The derivative Pi (4.38) with respect to k for i ≠ k is given by

Similarly the derivative Pi with respect to k for i = k is given by

Comparing the above equation with (4.39) we can write

B. Formation of J21

Let us define J21 as From (4.34) it is evident that the elements of J21 are the partial derivative of Q with respect to δ . From (4.39) we can write Similarly for i = k we have

(4.39)

(4.40)

(4.41)

(4.42)

(4.43)

(4.44)

(4.45)

The last equality of (4.45) is evident from (4.38). C. Formation of J12 Let us define J12 as As evident from (4.33), the elements of J21 involve the derivatives of real power P with respect to magnitude of bus voltage |V| . For i ≠ k , we can write from (4.38) For i = k we have Formation of J22 For the formation of J22 let us define For i ≠ k we can write from (4.39) Finally for i = k we have We therefore see that once the submatrices J11 and J21 are computed, the formation of the submatrices J12 and J22 is fairly straightforward. For large system this will result in considerable saving in the computation time.

(4.46)

(4.47)

(4.48)

(4.49)

(4.50)

(4.51)

Solution of Newton-Raphson Load Flow

The Newton-Raphson load flow program is tested on the system of Fig. 4.1 with the system data and initial conditions given in Tables 4.1 to 4.3. From (4.41) we can write

Similarly from (4.39) we have

Hence from (4.42) we get

In a similar way the rest of the components of the matrix J11

(0) are calculated. This matrix is given by

For forming the off diagonal elements of J21 we note from (4.44) that

Also from (4.38) the real power injected at bus-2 is calculated as

Hence from (4.45) we have

Similarly the rest of the elements of the matrix J21 are calculated. This matrix is then given as

For calculating the off diagonal elements of the matrix J12 we note from (4.47) that they are negative of the off diagonal elements of J21 . However the size of J21 is (3 X 4) while the size of J12 is (4 X 3). Therefore to avoid this discrepancy we first compute a matrix M that is given by

The elements of the above matrix are computed in accordance with (4.44) and (4.45). We can then define

Furthermore the diagonal elements of J12 are overwritten in accordance with (4.48). This matrix is then given by

Finally it can be noticed from (4.50) that J22 = J11 (1:3, 1:3). However the diagonal elements of J22 are then overwritten in accordance with (4.51). This gives the following matrix

From the initial conditions the power and reactive power are computed as

Consequently the mismatches are found to be

Then the updates at the end of the first iteration are given as

The load flow converges in 7 iterations when all the power and reactive power mismatches are below 10−6 .

Section VII: Load Flow Results

In this section we shall discuss the results of the load flow. It is to be noted here that both Gauss-Seidel and Newton-Raphson methods yielded the same result. However the Newton-Raphson method converged faster than the Gauss-Seidel method. The bus voltage magnitudes, angles of each bus along with power generated and consumed at each bus are given in Table 4.4. It can be seen from this table that the total power generated is 174.6 MW whereas the total load is 171 MW. This indicates that there is a line loss of about 3.6 MW for all the lines put together. It is to be noted that the real and reactive power of the slack bus and the reactive power of the P-V bus are computed from (4.6) and (4.7) after the convergence of the load flow.

Table 4.4 Bus voltages, power generated and load after load flow convergence.

Bus no.

Bus voltage Power generated Load

Magnitude (pu) Angle (deg) P (MW) Q (MVAr) P (MW) P (MVAr)

1 1.05 0 126.60 57.11 0 0

2 0.9826 5.0124 0 0 96 62

3 0.9777 7.1322 0 0 35 14

4 0.9876 7.3705 0 0 16 8

5 1.02 3.2014 48 15.59 24 11

The current flowing between the buses i and k can be written as

Therefore the complex power leaving bus- i is given by

Similarly the complex power entering bus- k is

Therefore the I 2 R loss in the line segment i-k is

The real power flow over different lines is listed in Table 4.5. This table also gives the I2 R loss along various segments. It can be seen that all the losses add up to 3.6 MW, which is the net difference between power generation and load. Finally we can compute the line I2X drops in a similar fashion. This drop is given by

However we have to consider the effect of line charging separately.

Table 4.5 Real power flow over different lines.

Power dispatched Power received Line loss (MW)

from (bus) amount (MW) in (bus) amount (MW)

1 101.0395 2 98.6494 2.3901

1 25.5561 5 25.2297 0.3264

2 17.6170 3 17.4882 0.1288

3 0.7976 4 0.7888 0.0089

5 15.1520 2 14.9676 0.1844

5 18.6212 3 18.3095 0.3117

5 15.4566 4 15.2112 0.2454

Total = 3.5956

Consider the line segment 1-2. The voltage of bus-1 is V1 = 1.05 < 0° per unit while that of bus-2 is V2 = 0.9826 < − 5.0124° per unit. From (4.52) we then have

per unit Therefore the complex power dispatched from bus-1 is

where the negative signal indicates the power is leaving bus-1. The complex power received at bus-2 is

MW

(4.52)

(4.53)

(4.54)

(4.55)

(4.56)

Therefore out of a total amount of 101.0395 MW of real power is dispatched from bus-1 over the line segment 1-2, 98.6494 MW reaches bus-2. This indicates that the drop in the line segment is 2.3901 MW. Note that

MW

where R12 is resistance of the line segment 1-2. Therefore we can also use this method to calculate the line loss.

Now the reactive drop in the line segment 1-2 is

MV Ar

We also get this quantity by subtracting the reactive power absorbed by bus-2 from that supplied by bus-1. The above calculation however does not include the line charging. Note that since the line is modeled by an equivalent- p , the voltage across the shunt capacitor is the bus voltage to which the shunt capacitor is connected. Therefore the current I 12 flowing through line segment is not the current leaving bus-1 or entering bus-2 - it is the current flowing in between the two charging capacitors. Since the shunt branches are purely reactive, the real power flow does not get affected by the charging capacitors. Each charging capacitor is assumed to inject a reactive power that is the product of the half line charging admittance and square of the magnitude of the voltage of that at bus. The half line charging admittance of this line is 0.03. Therefore line charging capacitor will inject

MV Ar

at bus-1. Similarly the reactive injected at bus-2 will be

MV Ar

The power flow through the line segments 1-2 and 1-5 are shown in Fig. 4.2.

(a)

(b)

Fig. 4.2 Real and reactive power flow through (a) line segment 1-2 and (b) line segment 1-5. The thin lines indicate reactive power flow while the thick lines indicate real power flow.

Section VIII: Load Flow Programs In Matlab

The load flow programs are developed in MATLAB. Altogether there are 4 mfiles that are attached with this chapter. The program listings and descriptions of these mfiles are given below. It must however be emphasized that these are not general purpose programs and are written only for the examples of this chapter.

Forming Ybus Matrix

Gauss- Seidel Load Flow

Solving Nonlinear Equations using Newton- Raphson

Newton - Raphson Load Flow

Forming Ybus Matrix

This is a function that can be called by various programs. The function can be invoked by the statement

[yb,ych]=ybus;

where 'yb' and 'ych' are respectively the Ybus matrix and a matrix containing the line charging admittances. It is assumed that the system data of Table 4.1 are given in matrix form and the matrix that contains line impedances is 'zz', while 'ych' contains the line charging information. This program is stored in the file ybus.m. The program listing is given below.

Chapter 5: Economic Operation of Power Systems

Overview

A good business practice is the one in which the production cost is minimized without sacrificing the quality. This is not any different in the power sector as well. The main aim here is to reduce the production cost while maintaining the voltage magnitudes at each bus. In this chapter we shall discuss the economic operation strategy along with the turbine-governor control that are required to maintain the power dispatch economically.

A power plant has to cater to load conditions all throughout the day, come summer or winter. It is therefore illogical to assume that the same level of power must be generated at all time. The power generation must vary according to the load pattern, which may in turn vary with season. Therefore the economic operation must take into account the load condition at all times. Moreover once the economic generation condition has been calculated, the turbine-governor must be controlled in such a way that this generation condition is maintained. In this chapter we shall discuss these two aspects.

Section I: Economic Operation Of Power System

Economic Distribution of Loads between the Units of a Plant

Generating Limits

Economic Sharing of Loads between Different Plants

In an early attempt at economic operation it was decided to supply power from the most efficient plant at light load conditions. As the load increased, the power was supplied by this most efficient plant till the point of maximum efficiency of this plant was reached. With further increase in load, the next most efficient plant would supply power till its maximum efficiency is reached. In this way the power would be supplied by the most efficient to the least efficient plant to reach the peak demand. Unfortunately however, this method failed to minimize the total cost of electricity generation. We must therefore search for alternative method which takes into account the total cost generation of all the units of a plant that is supplying a load.

Economic Distribution of Loads between the Units of a Plant

To determine the economic distribution of a load amongst the different units of a plant, the variable operating costs of each unit must be expressed in terms of its power output. The fuel cost is the main cost in a thermal or nuclear unit. Then the fuel cost must be expressed in terms of the power output. Other costs, such as the operation and maintenance costs, can also be expressed in terms of the power output. Fixed costs, such as the capital cost, depreciation etc., are not included in the fuel cost.

The fuel requirement of each generator is given in terms of the Rupees/hour. Let us define the input cost of an unit- i , fi in Rs./h and the power output of the unit as Pi . Then the input cost can be expressed in terms of the power output as

Rs./h

(5.1)

The operating cost given by the above quadratic equation is obtained by approximating the power in MW versus the cost in Rupees curve. The incremental operating cost of each unit is then computed as Let us now assume that only two units having different incremental costs supply a load. There will be a reduction in cost if some amount of load is transferred from the unit with higher incremental cost to the unit with lower incremental cost. In this fashion, the load is transferred from the less efficient unit to the more efficient unit thereby reducing the total operation cost. The load transfer will continue till the incremental costs of both the units are same. This will be optimum point of operation for both the units. The above principle can be extended to plants with a total of N number of units. The total fuel cost will then be the summation of the individual fuel cost fi , i = 1, ... , N of each unit, i.e., Let us denote that the total power that the plant is required to supply by PT , such that where P1 , ... , PN are the power supplied by the N different units. The objective is minimize fT for a given PT . This can be achieved when the total difference dfT becomes zero, i.e., Now since the power supplied is assumed to be constant we have Multiplying (5.6) by λ and subtracting from (5.5) we get The equality in (5.7) is satisfied when each individual term given in brackets is zero. This gives us Also the partial derivative becomes a full derivative since only the term fi of fT varies with Pi, i = 1, ..., N . We then have

Rs./MWh (5.2)

(5.3)

(5.4)

(5.5)

(5.6)

(5.7)

(5.8)

Generating Limits

It is not always necessary that all the units of a plant are available to share a load. Some of the units may be taken off due to scheduled maintenance. Also it is not necessary that the less efficient units are switched off during off peak hours. There is a certain amount of shut down and start up costs associated with shutting down a unit during the off peak hours and servicing it back on-line during the peak hours. To complicate the problem further, it may take about eight hours or more to restore the boiler of a unit and synchronizing the unit with the bus. To meet the sudden change in the power demand, it may therefore be necessary to keep more units than it necessary to meet the load demand during that time. This safety margin in generation is called spinning reserve . The optimal load dispatch problem must then incorporate this startup and shut down cost for without endangering the system security.

The power generation limit of each unit is then given by the inequality constraints

The maximum limit Pmax is the upper limit of power generation capacity of each unit. On the other hand, the lower limit Pmin pertains to the thermal consideration of operating a boiler in a thermal or nuclear generating station. An operational unit must produce a minimum amount of power such that the boiler thermal components are stabilized at the minimum design operating temperature.

Example 5.2

let us consider a generating station that contains a total number of three generating units. The fuel costs of these units are given by

Rs./h

Rs./h

Rs./h

The generation limits of the units are

The total load that these units supply varies between 90 MW and 1250 MW. Assuming that all the three units are operational all the time, we have to compute the economic operating settings as the load changes.

The incremental costs of these units are

(5.10)

Rs./MWh

Rs./MWh

Rs./MWh

At the minimum load the incremental cost of the units are

Rs./MWh

Rs./MWh

Rs./MWh

Since units 1 and 3 have higher incremental cost, they must therefore operate at 30 MW each. The incremental cost during this time will be due to unit-2 and will be equal to 26 Rs./MWh. With the generation of units 1 and 3 remaining constant, the generation of unit-2 is increased till its incremental cost is equal to that of unit-1, i.e., 34 Rs./MWh. This is achieved when P2 is equal to 41.4286 MW, at a total power of 101.4286 MW.

An increase in the total load beyond 101.4286 MW is shared between units 1 and 2, till their incremental costs are equal to that of unit-3, i.e., 43.5 Rs./MWh. This point is reached when P1 = 41.875 MW and P2 = 55 MW. The total load that can be supplied at that point is equal to 126.875. From this point onwards the load is shared between the three units in such a way that the incremental costs of all the units are same. For example for a total load of 200 MW, from (5.4) and (5.9) we have

Solving the above three equations we get P1 = 66.37 MW, P2 = 80 MW and P3 = 50.63 MW and an incremental cost ( λ ) of 63.1 Rs./MWh. In a similar way the economic dispatch for various other load settings are computed. The load distribution and the incremental costs are listed in Table 5.1 for various total power conditions.

Table 5.1 Load distribution and incremental cost for the units of Example 5.1

PT (MW) P1 (MW) P2 (MW) P3 (MW) λ (Rs./MWh)

90 30 30 30 26

101.4286 30 41.4286 30 34

120 38.67 51.33 30 40.93

126.875 41.875 55 30 43.5

150 49.62 63.85 36.53 49.7

200 66.37 83 50.63 63.1

300 99.87 121.28 78.85 89.9

400 133.38 159.57 107.05 116.7

500 166.88 197.86 135.26 143.5

600 200.38 236.15 163.47 170.3

700 233.88 274.43 191.69 197.1

800 267.38 312.72 219.9 223.9

906.6964 303.125 353.5714 250 252.5

1000 346.67 403.33 250 287.33

1100 393.33 456.67 250 324.67

1181.25 431.25 500 250 355

1200 450 500 250 370

1250 500 500 250 410

At a total load of 906.6964, unit-3 reaches its maximum load of 250 MW. From this point onwards then, the generation of this unit is kept fixed and the economic dispatch problem involves the other two units. For example for a total load of 1000 MW, we get the following two equations from (5.4) and (5.9)

Solving which we get P1 = 346.67 MW and P2 = 403.33 MW and an incremental cost of 287.33 Rs./MWh. Furthermore, unit-2 reaches its peak output at a total load of 1181.25. Therefore any further increase in the total load must be supplied by unit-1 and the incremental cost will only be borne by this unit. The power distribution curve is shown in Fig. 5.1.

Fig.5.1 Power distribution between the units of Example 5.2.

Example 5.3

Consider two generating plant with same fuel cost and generation limits. These are given by

For a particular time of a year, the total load in a day varies as shown in Fig. 5.2. Also an additional cost of Rs. 5,000 is incurred by switching of a unit during the off peak hours and switching it back on

during the during the peak hours. We have to determine whether it is economical to have both units operational all the time.

Fig. 5.2 Hourly distribution of load for the units of Example 5.2.

Since both the units have identical fuel costs, we can switch of any one of the two units during the off peak hour. Therefore the cost of running one unit from midnight to 9 in the morning while delivering 200 MW is

Rs.

Adding the cost of Rs. 5,000 for decommissioning and commissioning the other unit after nine hours, the total cost becomes Rs. 167,225.

On the other hand, if both the units operate all through the off peak hours sharing power equally, then we get a total cost of

Rs.

which is significantly less that the cost of running one unit alone.

Economic Sharing of Loads between Different Plants

So far we have considered the economic operation of a single plant in which we have discussed how a particular amount of load is shared between the different units of a plant. In this problem we did not have to consider the transmission line losses and assumed that the losses were a part of the load supplied. However if now consider how a load is distributed between the different plants that are joined by transmission lines, then the line losses have to be explicitly included in the economic dispatch problem. In this section we shall discuss this problem.

When the transmission losses are included in the economic dispatch problem, we can modify (5.4) as

where PLOSS is the total line loss. Since PT is assumed to be constant, we have

In the above equation dPLOSS includes the power loss due to every generator, i.e.

Also minimum generation cost implies dfT = 0 as given in (5.5). Multiplying both (5.12) and (5.13) by λ and combining we get

(5.11)

(5.12)

(5.13)

Adding (5.14) with (5.5) we obtain

The above equation satisfies when

Again since

from (5.16) we get

where Li is called the penalty factor of load- i and is given by

Example 5.4

Consider an area with N number of units. The power generated are defined by the vector

Then the transmission losses are expressed in general as

where B is a symmetric matrix given by

The elements Bij of the matrix B are called the loss coefficients . These coefficients are not constant

but vary with plant loading. However for the simplified calculation of the penalty factor Li these coefficients are often assumed to be constant.

When the incremental cost equations are linear, we can use analytical equations to find out the economic settings. However in practice, the incremental costs are given by nonlinear equations that may even contain nonlinearities. In that case iterative solutions are required to find the optimal generator settings.

(5.14)

(5.15)

(5.16)

(5.17)

(5.18)

(5.19)

Section II: Automatic Generation Control

Load Frequency Control

Automatic Generation Control

Electric power is generated by converting mechanical energy into electrical energy. The rotor mass, which contains turbine and generator units, stores kinetic energy due to its rotation. This stored kinetic energy accounts for sudden increase in the load. Let us denote the mechanical torque input by Tm and the output electrical torque by Te . Neglecting the rotational losses, a generator unit is said to be operating in the steady state at a constant speed when the difference between these two elements of torque is zero. In this case we say that the accelerating torque

(5.20)

is zero.

When the electric power demand increases suddenly, the electric torque increases. However, without any feedback mechanism to alter the mechanical torque, Tm remains constant. Therefore the accelerating torque given by (5.20) becomes negative causing a deceleration of the rotor mass. As the rotor decelerates, kinetic energy is released to supply the increase in the load. Also note that during this time, the system frequency, which is proportional to the rotor speed, also decreases. We can thus infer that any deviation in the frequency for its nominal value of 50 or 60 Hz is indicative of the imbalance between Tm and Te. The frequency drops when Tm < Te and rises when Tm > Te .

The steady state power-frequency relation is shown in Fig. 5.3. In this figure the slope of the ΔPref line is negative and is given by

where R is called the regulating constant . From this figure we can write the steady state power frequency relation as

Fig. 5.3 A typical steady-state power-frequency curve.

Suppose an interconnected power system contains N turbine-generator units. Then the steady-state power-frequency relation is given by the summation of (5.22) for each of these units as

(5.21)

(5.22)

In the above equation, ΔPm is the total change in turbine-generator mechanical power and ΔPref is the total change in the reference power settings in the power system. Also note that since all the generators are supposed to work in synchronism, the change is frequency of each of the units is the same and is denoted by Δf. Then the frequency response characteristics is defined as

We can therefore modify (5.23) as

Example 5.5

Consider an interconnected 50-Hz power system that contains four turbine-generator units rated 750 MW, 500 MW, 220 MW and 110 MW. The regulating constant of each unit is 0.05 per unit based on its own rating. Each unit is operating on 75% of its own rating when the load is suddenly dropped by 250 MW. We shall choose a common base of 500 MW and calculate the rise in frequency and drop in the mechanical power output of each unit.

The first step in the process is to convert the regulating constant, which is given in per unit in the base of each generator, to a common base. This is given as

(5.26)

We can therefore write

Therefore

p.u.

We can therefore calculate the total change in the frequency from (5.25) while assuming ΔPref = 0, i.e., for no change in the reference setting. Since the per unit change in load - 250/500 = - 0.5 with the negative sign accounting for load reduction, the change in frequency is given by

(5.23)

(5.24)

Then the change in the mechanical power of each unit is calculated from (5.22) as

It is to be noted that once ΔPm2 is calculated to be - 79.11 MW, we can also calculate the changes in the mechanical power of the other turbine-generators units as

This implies that each turbine-generator unit shares the load change in accordance with its own rating.

Load Frequency Control

Modern day power systems are divided into various areas. For example in India , there are five regional grids, e.g., Eastern Region, Western Region etc. Each of these areas is generally interconnected to its neighboring areas. The transmission lines that connect an area to its neighboring area are called tie-lines . Power sharing between two areas occurs through these tie-lines. Load frequency control, as the name signifies, regulates the power flow between different areas while holding the frequency constant.

As we have in Example 5.5 that the system frequency rises when the load decreases if ΔPref is kept at zero. Similarly the frequency may drop if the load increases. However it is desirable to maintain the frequency constant such that Δf=0 . The power flow through different tie-lines are scheduled - for example, area- i may export a pre-specified amount of power to area- j while importing another pre-specified amount of power from area- k . However it is expected that to fulfill this obligation, area- i absorbs its own load change, i.e., increase generation to supply extra load in the area or decrease generation when the load demand in the area has reduced. While doing this area- i must however maintain its obligation to areas j and k as far as importing and exporting power is concerned. A conceptual diagram of the interconnected areas is shown in Fig. 5.4.

Fig. 5.4 Interconnected areas in a power system.

We can therefore state that the load frequency control (LFC) has the following two objectives:

Hold the frequency constant ( Δf = 0) against any load change. Each area must contribute to absorb any load change such that frequency does not deviate.

Each area must maintain the tie-line power flow to its pre-specified value.

The first step in the LFC is to form the area control error (ACE) that is defined as

where Ptie and Psch are tie-line power and scheduled power through tie-line respectively and the constant Bf is called the frequency bias constant .

The change in the reference of the power setting ΔPref, i , of the area- i is then obtained by the feedback of the ACE through an integral controller of the form

where Ki is the integral gain. The ACE is negative if the net power flow out of an area is low or if the frequency has dropped or both. In this case the generation must be increased. This can be achieved by increasing ΔPref, i . This negative sign accounts for this inverse relation between ΔPref, i and ACE. The tie-line power flow and frequency of each area are monitored in its control center. Once the ACE is computed and ΔPref, i is obtained from (5.28), commands are given to various turbine-generator controls to adjust their reference power settings.

Example 5.6

Consider a two-area power system in which area-1 generates a total of 2500 MW, while area-2 generates 2000 MW. Area-1 supplies 200 MW to area-2 through the inter-tie lines connected between the two areas. The bias constant of area-1 ( β1 ) is 875 MW/Hz and that of area-2 ( β2 ) is 700 MW/Hz. With the two areas operating in the steady state, the load of area-2 suddenly increases by 100 MW. It is desirable that area-2 absorbs its own load change while not allowing the frequency to drift. The area control errors of the two areas are given by

and

Since the net change in the power flow through tie-lines connecting these two areas must be zero, we have

Also as the transients die out, the drift in the frequency of both these areas is assumed to be constant, i.e.,

If the load frequency controller (5.28) is able to set the power reference of area-2 properly, the ACE of the two areas will be zero, i.e., ACE1 = ACE2 = 0. Then we have

This will imply that Δf will be equal to zero while maintaining ΔPtie1 =ΔPtie2 = 0. This signifies that area-2 picks up the additional load in the steady state.

Coordination Between LFC And Economic Dispatch

Both the load frequency control and the economic dispatch issue commands to change the power setting of each turbine-governor unit. At a first glance it may seem that these two commands can be conflicting. This however is not true. A typical automatic generation control strategy is shown in Fig.

(5.27)

(5.28)

5.5 in which both the objective are coordinated. First we compute the area control error. A share of this ACE, proportional to αi , is allocated to each of the turbine-generator unit of an area. Also the share of unit- i , γi X Σ( PDK - Pk ), for the deviation of total generation from actual generation is computed. Also the error between the economic power setting and actual power setting of unit- i is computed. All these signals are then combined and passed through a proportional gain Ki to obtain the turbine-governor control signal.

Fig. 5.5 Automatic generation control of unit-i.

Chapter 6: Short Circuit Studies - Symmetrical Faults

Introduction

Short circuits occur in power system due to various reasons like, equipment failure, lightning strikes, falling of branches or trees on the transmission lines, switching surges, insulation failures and other electrical or mechanical causes. All these are collectively called faults in power systems.

A fault usually results in high current flowing through the lines and if adequate protection is not taken, may result in damages in the power apparatus.

In this chapter we shall discuss the effects of symmetrical faults on the system. Here the term symmetrical fault refers to those conditions in which all three phases of a power system are grounded at the same point. For this reason the symmetrical faults sometimes are also called three-line-to-ground (3LG) faults.

Section I: Transients in R-L Circuits

DC Sourse

AC Sourse

Fault in an AC Circuit

Transients in R-L Circuits

In this section we shall consider transients in a circuit that contains a resistor and inductor (R - L circuit). Consider the circuit shown in Fig. 6.1 that contains an ideal source ( νs ), a resistor ( R ), an inductor ( L ) and a switch ( S ). It is assumed that the switch is open and is closed at an instant of time t = 0. This implies that the current i is zero before the closing of the switch. We shall first discuss the effect of closing the switch on the line current (i ) when the source is dc. Following this we shall study the effect when the source is ac and will show that the shape of the transient current changes with the changes in the phase of the source voltage waveform at the instant of closing the switch.

Fig 6.1 A Simple R - L Circuit

DC Source

Let us assume that the source voltage is dc and is given by νs = Vdc . Then the line current is given by the differential equation

The solution of the above equation is written in the form

Since the initial current i (0) = 0 and since νs ( τ ) = Vdc for 0 £ t < ¥ , we can rewrite the above equation as

where T = L / R is the time constant of the circuit.

Let us assume R = 1Ω , L = 10 mH and Vdc= 100 V. Then the time response of the current is as shown in Fig. 6.2. It can be seen that the current reaches at steady state value of 100 A. The time constant of the circuit is 0.01 s. This is defined by the time in which the current i ( t ) reaches 63.2% of its final value and is obtained by substituting t = T. Note that the slope of the curve is given by

Fig 6.2 Current in the R-L circuit when the source is dc

AC Source

The current response remains unchanged when the voltage source is dc. This however is not the case when the circuit is excited by an ac source. Let us assume that the source voltage is now given by

where α is the

(6.1)

(6.2)

(6.3)

(6.4)

(6.5)

phase angle of the applied voltage. We shall show that the system response changes with a change in α .

The solution of (6.2) for the source voltage given in (6.5) is

The system response for V m = 100 V and α = 45° is shown in Fig. 6.3. In this figure both iac and i dc are also shown. It can be seen that iac is the steady state waveform of the circuit, while idc dies out once the initial transient phase is over. Fig. 6.4 shows the response of the current for different values of a . Since the current is almost inductive, it can be seen that the transient is minimum when α = 90° , i.e., the circuit is switched on almost at the zero-crossing of the current. On the other hand, the transient is maximum when α = 0° , i.e., almost at the peak of the current.

Fig 6.3 Transient in current and its ac and dc components at the instant of switch closing

Fig 6.4 Transient in current for different values of α

(6.6)

A (6.7)

A (6.8)

,

Fault in an AC Circuit

Now consider the single-phase circuit of Fig. 6.5 where Vs = 240 V (rms), the system frequency is 50 Hz, R = 0.864 Ω, L = 11 mH ( ωL = 3.46 Ω ) and the load is R-L comprising of an 8.64 W resistor and a 49.5 mH inductor ( ωL = 15.55Ω ). With the system operating in the steady state, the switch S is suddenly closed creating a short circuit. The current (i ) waveform is shown in Fig. 6.6. The current phasor before the short circuit occurs is

A

This means that the pre-fault current has a peak value of 15.97 A.

Fig. 6.5 A single-phase circuit in which a source supplies a load through a source impedance.

Fig 6.6 The current waveform of the circuit of Fig 6.5 before and after the closing of the switch S

Once the fault occurs and the system is allowed to reach the steady state, the current phasor is given by

This current has a peak value of 95.28 A. However it can be seen that the current rises suddenly and the first peak following the fault is 124 A which is about 30% higher than the post-fault steady-state value. Also note that the peak value of the current will vary with the instant of the occurrence of the fault. However the peak value of the current is nearly 8 times the pre-fault current value in this case. In general, depending on the ratio of source and load impedances, the faulted current may shoot up anywhere between 10 and 20 times the pre-fault current.

Short Circuit in an Unloaded Synchronous Generator

Fig. 6.7 shows a typical response of the armature current when a three-phase symmetrical short circuit occurs at the terminals of an unloaded synchronous generator.

Fig. 6.7 Armature current of a synchronous generator as a short circuit occurs at its terminals.

It is assumed that there is no dc offset in the armature current. The magnitude of the current decreases exponentially from a high initial value. The instantaneous expression for the fault current is given by

where Vt is the magnitude of the terminal voltage, α is its phase angle and

is the direct axis subtransient reactance

is the direct axis transient reactance

is the direct axis synchronous reactance

with . The time constants are

is the direct axis subtransient time constant

is the direct axis transient time constant

In the expression of (6.9) we have neglected the effect of the armature resistance hence α = π/2. Let us assume that the fault occurs at time t = 0. From (6.9) we get the rms value of the current as

which is called the subtransient fault current. The duration of the subtransient current is dictated by

the time constant Td . As the time progresses and Td < t < Td , the first exponential term of (6.9) will start decaying and will eventually vanish. However since t is still nearly equal to zero, we have the following rms value of the current

This is called the transient fault current. Now as the time progress further and the second exponential term also decays, we get the following rms value of the current for the sinusoidal steady state

(6.9)

(6.10)

(6.11)

(6.12)

In addition to the ac, the fault currents will also contain the dc offset. Note that a symmetrical fault occurs when three different phases are in three different locations in the ac cycle. Therefore the dc offsets in the three phases are different. The maximum value of the dc offset is given by

where TA is the armature time constant.

Section III: Symmetrical Fault in a Power System

Calculation of Fault Current Using Impedance Diagram

Calculation of Fault Current Using Zbus Matrix

Calculation of Fault Current Using Impedance Diagram

Let us first illustrate the calculation of the fault current using the impedance diagram with the help of the following examples

Example 6.1

Consider the power system of Fig. 6.8 in which a synchronous generator supplies a synchronous motor. The motor is operating at rated voltage and rated MVA while drawing a load current at a power factor of 0.9 (lagging) when a three phase symmetrical short circuit occurs at its terminals. We shall calculate the fault current that flow from both the generator and the motor.

We shall choose a base of 50 MVA, 20 kV in the circuit of the generator. Then the motor synchronous reactance is given by

per unit

Also the base impedance in the circuit of the transmission line is

Ω

Fig. 6.8 A generator supplying a motor load though a transmission line.

Therefore the impedance of the transmission line is

per unit

The impedance diagram for the circuit is shown in Fig. 6.9 in which the switch S indicates the fault.

Fig. 6.9 Impedance diagram of the circuit of Fig. 6.8.

The motor draws a load current at rated voltage and rated MVA with 0.9 lagging power factor. Therefore

(6.13)

per unit

Then the subtransient voltages of the motor and the generator are

per unit

per unit

Hence the subtransient fault currents fed by the motor and the generator are

per unit

per unit

and the total current flowing to the fault is

per unit

Note that the base current in the circuit of the motor is

A

Therefore while the load current was 1603.8 A, the fault current is 7124.7 A.

Example 6.2

We shall now solve the above problem differently. The Thevenin impedance at the circuit between the terminals A and B of the circuit of Fig. 6.9 is the parallel combination of the impedances j 0.4 and j 0.5148. This is then given as

per unit

Since voltage at the motor terminals before the fault is 1.0 per unit, the fault current is

per unit

If we neglect the pre-fault current flowing through the circuit, then fault current fed by the motor and the generator can be determined using the current divider principle, i.e.,

per unit

per unit

If, on the other hand, the pre-fault current is not neglected, then the fault current supplied by the motor and the generator are

per unit

per unit

Calculation of Fault Current Using Zbus Matrix

Consider the circuit of Fig. 3.3 which is redrawn as shown in Fig. 6.10.

Fig. 6.10 Network depicting a symmetrical fault at bus-4.

We assume that a symmetrical fault has occurred in bus-4 such that it is now connected to the reference bus. Let us assume that the pre-fault voltage at this bus is Vf . To denote that bus-4 is short circuit, we add two voltage sources Vf and - Vf together in series between bus-4 and the reference bus. Also note that the subtransient fault current If flows from bus-4 to the reference bus. This implies that a current that is equal to - If is injected into bus-4. This current, which is due to the source - Vf will flow through the various branches of the network and will cause a change in the bus voltages. Assuming that the two sources and Vf are short circuited. Then - Vf is the only source left in the network that injects a current - If into bus-4. The voltages of the different nodes that are caused by the voltage - Vf and the current - If are then given by where the prefix Δ indicates the changes in the bus voltages due to the current - If . From the fourth row of (6.14) we can write Combining (6.14) and (6.15) we get We further assume that the system is unloaded before the fault occurs and that the magnitude and phase angles of all the generator internal emfs are the same. Then there will be no current circulating anywhere in the network and the bus voltages of all the nodes before the fault will be same and equal to Vf . Then the new altered bus voltages due to the fault will be given from (6.16) by

Example 6.3

(6.14)

(6.15)

(6.16)

Let us consider the same system as discussed in Example 3.1 except that we assume that the internal voltages of both the generators are equal to 1.0 < 0° . Then the current injected in both bus-1 and 2 will be given by 1.0/ j 0.25 = - j 4.0 per unit. We therefore get the pre-fault bus voltages using the Zbus matrix given in Example 3.1 as

Now the altered bus voltages for a symmetrical fault in bus-4 are given from (6.17) as

per unit

per unit

per unit

per unit

Also since the Thevenin impedance looking into the network at bus-4 is Z44 (see Section 3.4), the subtransient fault current flowing from bus-4 is

pu

Section IV: Circuit Breaker Selection

A typical circuit breaker operating time is given in Fig. 6.11. Once the fault occurs, the protective devices get activated. A certain amount of time elapses before the protective relays determine that there is overcurrent in the circuit and initiate trip command. This time is called the detection time.

The contacts of the circuit breakers are held together by spring mechanism and, with the trip command, the spring mechanism releases the contacts. When two current carrying contacts part, a voltage instantly appears at the contacts and a large voltage gradient appears in the medium between the two contacts. This voltage gradient ionizes the medium thereby maintaining the flow of current. This current generates extreme heat and light that is called electric arc. Different mechanisms are used for elongating the arc such that it can be cooled and extinguished. Therefore the circuit breaker has to withstand fault current from the instant of initiation of the fault to the time the arc is extinguished.

Fig. 6.11 Typical circuit breaker operating time.

Two factors are of utmost importance for the selection of circuit breakers. These are:

The maximum instantaneous current that a breaker must withstand and

The total current when the breaker contacts part.

In this chapter we have discussed the calculation of symmetrical subtransient fault current in a network. However the instantaneous current following a fault will also contain the dc component. In a high power circuit breaker selection, the subtransient current is multiplied by a factor of 1.6 to determine the rms value of the current the circuit breaker must withstand. This current is called the momentary current . The interrupting current of a circuit breaker is lower than the momentary

current and will depend upon the speed of the circuit breaker. The interrupting current may be asymmetrical since some dc component may still continue to decay.

Breakers are usually classified by their nominal voltage, continuous current rating, rated maximum voltage, K -factor which is the voltage range factor, rated short circuit current at maximum voltage and operating time. The K -factor is the ratio of rated maximum voltage to the lower limit of the range of the operating voltage. The maximum symmetrical interrupting current of a circuit breaker is given by K times the rated short circuit current.

Overview

An unbalanced three-phase system can be resolved into three balanced systems in the sinusoidal steady state. This method of resolving an unbalanced system into three balanced phasor system has been proposed by C. L. Fortescue. This method is called resolving symmetrical components of the original phasors or simply symmetrical components.

In this chapter we shall discuss symmetrical components transformation and then will present how unbalanced components like Y- or Δ -connected loads, transformers, generators and transmission lines can be resolved into symmetrical components. We can then combine all these components together to form what are called sequence networks .

Section I: Symmetrical Components

Symmetrical Component Transformation

Real and Reactive Power

Orthogonal Transformation

Symmetrical Components

A system of three unbalanced phasors can be resolved in the following three symmetrical components:

Positive Sequence: A balanced three-phase system with the same phase sequence as the original sequence.

Negative sequence: A balanced three-phase system with the opposite phase sequence as the original sequence.

Zero Sequence: Three phasors that are equal in magnitude and phase.

Fig. 7.1 depicts a set of three unbalanced phasors that are resolved into the three sequence components mentioned above. In this the original set of three phasors are denoted by Va , Vb and Vc , while their positive, negative and zero sequence components are denoted by the subscripts 1, 2 and 0 respectively. This implies that the positive, negative and zero sequence components of phase-a are denoted by Va1 , Va2 and Va0 respectively. Note that just like the voltage phasors given in Fig. 7.1 we can also resolve three unbalanced current phasors into three symmetrical components.

Fig. 7.1 Representation of (a) an unbalanced network, its (b) positive sequence, (c) negative sequence and (d) zero sequence.

Symmetrical Component Transformation

Before we discuss the symmetrical component transformation, let us first define the α -operator. This has been given in (1.34) and is reproduced below.

Note that for the above operator the following relations hold Also note that we have Using the a -operator we can write from Fig. 7.1 (b) Similarly from Fig. 7.1 (c) we get Finally from Fig. 7.1 (d) we get Therefore, The symmetrical component transformation matrix is then given by

(7.1)

(7.2)

(7.3)

(7.4)

(7.5)

(7.6)

(7.7)

(7.8)

(7.9)

(7.10)

(7.11)

Defining the vectors V a012 and V abc as we can write (7.4) as where C is the symmetrical component transformation matrix and is given by

The original phasor components can be obtained from the inverse symmetrical component transformation, i.e.,

Finally, if we define a set of unbalanced current phasors as Iabc and their symmetrical components as Ia012 , we can then define

Example 7.1

Let us consider a set of balanced voltages given in per unit by

These imply

Then from (7.7) we get

(7.12)

(7.13)

(7.14)

(7.15)

We then see that for a balanced system the zero and negative sequence voltages are zero. Also the positive sequence voltage is the same as the original system, i.e.,

Example 7.2

All the quantities given in this example are in per unit. Let us now consider the following set of three unbalanced voltages

If we resolve them using (7.4) we then have

Therefore we have

Furthermore note that

Real and Reactive Power The three-phase power in the original unbalanced system is given by where I* is the complex conjugate of the vector I . Now from (7.10) and (7.15) we get From (7.11) we get Therefore from (7.17) we get We then find that the complex power is three times the summation of the complex power of the three phase sequences. Example 7.3 Let us consider the voltages given in Example 7.2. Let us further assume that these voltages are line-to-neutral voltages and they supply a balanced Y-connected load whose per phase impedance is ZY = 0.2 + j 0.8 per unit. Then the per unit currents in the three phases are

(7.16)

(7.17)

(7.18)

pu

pu

pu

Then the real and reactive power consumed by the load is given by

Now using the transformation (7.15) we get

pu

From the results given in Example 7.2 and from the above values we can compute the zero sequence complex power as

pu

The positive sequence complex power is

pu

Finally the negative sequence complex power is

pu

Adding the three complex powers together we get the total complex power consumed by the load as

pu

Orthogonal Transformation

Instead of the transformation matrix given in (7.13), let us instead use the transformation matrix We then have Note from (7.19) and (7.20) that C -1 = ( CT )* . We can therefore state C( CT )* = I3 , where I3 is (3 × 3) identity matrix. Therefore the transformation matrices given in (7.19) and (7.20) are orthogonal. Now since

(7.19)

(7.20)

we can write from (7.17)

We shall now discuss how different elements of a power system are represented in terms of their sequence components. In fact we shall show that each element is represented by three equivalent circuits, one for each symmetrical component sequence.

Section II: Sequence Circuits for Loads

In this section we shall construct sequence circuits for both Y and Δ -connected loads separately.

Sequence Circuit for a Y-Connected Load

Sequence Circuit for a Δ-Connected Load

Sequence Circuit for a Y-Connected Load

Consider the balanced Y-connected load that is shown in Fig. 7.2. The neutral point (n) of the windings are grounded through an impedance Zn . The load in each phase is denoted by ZY . Let us consider phase-a of the load. The voltage between line and ground is denoted by Va , the line-to-neutral voltage is denoted by Van and voltage between the neutral and ground is denoted by Vn . The neutral current is then Therefore there will not be any positive or negative sequence current flowing out of the neutral point.

Fig. 7.2 Schematic diagram of a balanced Y-connected load. The voltage drop between the neutral and ground is Now We can write similar expression for the other two phases. We can therefore write

(7.21)

(7.22)

(7.23)

(7.24)

(7.25)

Pre-multiplying both sides of the above equation by the matrix C and using (7.8) we get Now since We get from (7.26) We then find that the zero, positive and negative sequence voltages only depend on their respective sequence component currents. The sequence component equivalent circuits are shown in Fig. 7.3. While the positive and negative sequence impedances are both equal to ZY , the zero sequence impedance is equal to If the neutral is grounded directly (i.e., Zn = 0), then Z0 = ZY . On the other hand, if the neutral is kept floating (i.e., Zn = ∞ ), then there will not be any zero sequence current flowing in the circuit at all.

Fig. 7.3 Sequence circuits of Y-connected load: (a) positive, (b) negative and (c) zero sequence.

Sequence Circuit for a Δ -Connected Load

Consider the balanced Δ -connected load shown in Fig. 7.4 in which the load in each phase is denoted by ZΔ . The line-to-line voltages are given by Adding these three voltages we get

Fig. 7.4 Schematic diagram of a balanced Δ -connected load.

(7.26)

(7.27)

(7.28)

(7.29)

Denoting the zero sequence component Vab , Vbc and Vca as Vab0 and that of Iab , Ibc and Ica as Iab0 we can rewrite (7.30) as Again since We find from (7.31) Vab0 = Iab0 = 0. Hence a Δ -connected load with no mutual coupling has not any zero sequence circulating current. Note that the positive and negative sequence impedance for this load will be equal to ZΔ . Example 7.4 Consider the circuit shown in Fig. 7.5 in which a Δ -connected load is connected in parallel with a Y-connected load. The neutral point of the Y-connected load is grounded through an impedance. Applying Kirchoff's current law at the point P in the circuit we get

The above expression can be written in terms of the vector Vabc as

Since the load is balanced we can write

Fig. 7.5 Parallel connection of balanced D and Y-connected loads.

Pre-multiplying both sides of the above expression by the transformation matrix C we get

Now since

(7.30)

(7.31)

we get

Separating the three components, we can write from the above equation

Suppose now if we convert the Δ -connected load into an equivalent Y, then the composite load will be a parallel combination of two Y-connected circuits - one with an impedance of ZY and the other with an impedance of ZΔ /3. Therefore the positive and the negative sequence impedances are given by the parallel combination of these two impedances. The positive and negative sequence impedance is then given by

Now refer to Fig. 7.5. The voltage Vn is given by

From Fig. 7.5 we can also write Ia = IaΔ + IaY . Therefore

This implies that Ia0 = Iay0 and hence Vn = 3Zn Ia0 . We can then rewrite the zero sequence current expression as

It can be seen that the ZΔ term is absent from the zero sequence impedance.

Section III: Sequence Circuits for Synchronous Generator

The three-phase equivalent circuit of a synchronous generator is shown in Fig. 1.16. This is redrawn in Fig. 7.6 with the neutral point grounded through a reactor with impedance Zn . The neutral current is then given by

(7.32)

Fig. 7.6 Equivalent circuit of a synchronous generator with grounded neutral.

The derivation of Section 1.3 assumes balanced operation which implies Ia + Ib + Ic = 0. As per (7.32) this assumption is not valid any more. Therefore with respect to this figure we can write for phase-a voltage as

Similar expressions can also be written for the other two phases. We therefore have

Pre-multiplying both sides of (7.34) by the transformation matrix C we get

Since the synchronous generator is operated to supply only balanced voltages we can assume that Ean0 = Ean2 = 0 and Ean1 = Ean . We can therefore modify (7.35) as

We can separate the terms of (7.36) as

Furthermore we have seen for a Y-connected load that Va1 = V an1 , Va2 = Van2 since the neutral current does not affect these voltages. However Va0 = Van0 + Vn . Also we know that Vn = - 3ZnIa0 . We can therefore rewrite (7.37) as

(7.33)

(7.34)

(7.35)

(7.36)

(7.37)

(7.38)

(7.39)

The sequence diagrams for a synchronous generator are shown in Fig. 7.7.

Fig. 7.7 Sequence circuits of synchronous generator: (a) positive, (b) negative and (c) zero sequence.

(7.40)

Section IV: Sequence Circuits for Symmetrical Transmission Line

The schematic diagram of a transmission line is shown in Fig. 7.8. In this diagram the self impedance of the three phases are denoted by Zaa , Zbb and Zcc while that of the neutral wire is denoted by Znn . Let us assume that the self impedances of the conductors to be the same, i.e., Since the transmission line is assumed to be symmetric, we further assume that the mutual inductances between the conductors are the same and so are the mutual inductances between the conductors and the neutral, i.e The directions of the currents flowing through the lines are indicated in Fig. 7.8 and the voltages between the different conductors are as indicated.

Fig. 7.8 Lumped parameter representation of a symmetrical transmission line.

Applying Kirchoff's voltage law we get Again Substituting (7.42) and (7.43) in (7.41) we get Since the neutral provides a return path for the currents Ia , Ib and Ic , we can write

(7.41)

(7.42)

(7.43)

(7.44)

Therefore substituting (7.45) in (7.44) we get the following equation for phase-a of the circuit

Denoting

(7.46) can be rewritten as

Since (7.47) does not explicitly include the neutral conductor we can define the voltage drop across the phase-a conductor as

Combining (7.47) and (7.48) we get

Similar expression can also be written for the other two phases. We therefore get

Pre-multiplying both sides of (7.50) by the transformation matrix C we get

Now

Hence

(7.45)

(7.46)

(7.47)

(7.48)

(7.49)

(7.50)

(7.51)

Therefore from (7.51) we get The positive, negative and zero sequence equivalent circuits of the transmission line are shown in Fig. 7.9 where the sequence impedances are

Fig. 7.9 Sequence circuits of symmetrical transmission line: (a) positive, (b) negative and (c) zero sequence.

Section V: Sequence Circuits for Transformers

Y-Y Connected Transformer

Δ - Δ Connected Transformer

Y- Δ Connected Transformer

In this section we shall discuss the sequence circuits of transformers. As we have seen earlier that the sequence circuits are different for Y- and Δ -connected loads, the sequence circuits are also different for Y and Δ connected transformers. We shall therefore treat different transformer connections separately.

Y-Y Connected Transformer

Fig. 7.10 shows the schematic diagram of a Y-Y connected transformer in which both the neutrals are grounded. The primary and secondary side quantities are denoted by subscripts in uppercase letters and lowercase letters respectively. The turns ratio of the transformer is given by α = N1 : N2 .

(7.52)

Fig. 7.10 Schematic diagram of a grounded neutral Y-Y connected transformer.

The voltage of phase-a of the primary side is

Expanding VA and VAN in terms of their positive, negative and zero sequence components, the above equation can be rewritten as

Noting that the direction of the neutral current In is opposite to that of IN , we can write an equation similar to that of (7.53) for the secondary side as

Now since the turns ratio of the transformer is α = N1 : N2 we can write

Substituting in (7.54) we get

Multiplying both sides of the above equation by a results in

Finally combining (7.53) with (7.55) we get

Separating out the positive, negative and zero sequence components we can write

(7.53)

(7.54)

(7.55)

(7.56)

Fig. 7.11 Zero sequence equivalent circuit of grounded neutral Y-Y connected transformer.

From (7.57) and (7.58) we see that the positive and negative sequence relations are the same as that we have used for representing transformer circuits given in Fig. 1.18. Hence the positive and negative sequence impedances are the same as the transformer leakage impedance Z . The zero sequence equivalent circuit is shown in Fig. 7.11.

The total zero sequence impedance is given by

The zero sequence diagram of the grounded neutral Y-Y connected transformer is shown in Fig. 7.12 (a) in which the impedance Z0 is as given in (7.60). If both the neutrals are solidly grounded, i.e., Zn = ZN = 0, then Z0 is equal to Z . The single line diagram is still the same as that shown in Fig. 7.12 (a). If however one of the two neutrals or both neutrals are ungrounded, then we have either Zn = ∞ or ZN = ∞ or both. The zero sequence diagram is then as shown in Fig. 7.12 (b) where the value of Z0 will depend on which neutral is kept ungrounded.

Fig. 7.12 Zero sequence diagram of (a) grounded neutral and (b) ungrounded neutral Y-Y connected transformer.

Δ - Δ Connected Transformer

The schematic diagram of a Δ - Δ connected transformer is shown in Fig. 7.13. Now we have

Again

(7.57)

(7.58)

(7.59)

(7.60)

(7.61)

Fig. 7.13 Schematic diagram of a Δ - Δ connected transformer.

Therefore from (7.61) we get

The sequence components of the line-to-line voltage VAB can be written in terms of the sequence com ponents of the line-to-neutral voltage as

Therefore combining (7.62)-(7.64) we get

Hence we get

Thus the positive and negative sequence equivalent circuits are represented by a series impedance that is equal to the leakage impedance of the transformer. Since the Δ -connected winding does not provide any path for the zero sequence current to flow we have

However the zero sequence current can sometimes circulate within the Δ windings. We can then draw the zero sequence equivalent circuit as shown in Fig. 7.14.

Fig. 7.14 Zero sequence diagram of Δ - Δ connected transformer

Y- Δ Connected Transformer

The schematic diagram of a Y- Δ connected transformer is shown in Fig. 7.15. It is assumed that the Y-connected side is grounded with the impedance ZN . Even though the zero sequence current in the primary Y-connected side has a path to the ground, the zero sequence current flowing in the Δ -connected secondary winding has no path to flow in the line. Hence we have Ia0 = 0. However the circulating zero sequence current in the Δ winding magnetically balances the zero sequence current of the primary winding.

(7.62)

(7.63)

(7.64)

(7.65)

(7.66)

Fig. 7.15 Schematic diagram of a Y- Δ connected transformer.

The voltage in phase-a of both sides of the transformer is related by

Also we know that

We therefore have

Separating zero, positive and negative sequence components we can write

The positive sequence equivalent circuit is shown in Fig. 7.16 (a). The negative sequence circuit is the same as that of the positive sequence circuit except for the phase shift in the induced emf. This is shown in Fig. 7.16 (b). The zero sequence equivalent circuit is shown in Fig. 7.16 (c) where Z0 = Z + 3ZN . Note that the primary and the secondary sides are not connected and hence there is an open circuit between them. However since the zero sequence current flows through primary windings, a return path is provided through the ground. If however, the neutral in the primary side is not grounded, i.e., ZN = ∞ , then the zero sequence current cannot flow in the primary side as well. The sequence diagram is then as shown in Fig. 7.16 (d) where Z0 = Z .

(7.67)

(7.68)

(7.69)

(7.70)

Fig. 7.16 Sequence diagram of a Y- Δ connected transformer: (a) positive sequence, (b) negative sequence, (c) zero sequence with grounded Y-connection and (d) zero sequence with ungrounded

Y-connection.

Section VI: Sequence Networks

The sequence circuits developed in the previous sections are combined to form the sequence networks. The sequence networks for the positive, negative and zero sequences are formed separately by combining the sequence circuits of all the individual elements. Certain assumptions are made while forming the sequence networks. These are listed below.

1. Apart from synchronous machines, the network is made of static elements.

2. The voltage drop caused by the current in a particular sequence depends only on the impedance of that part of the network.

3. The positive and negative sequence impedances are equal for all static circuit components, while the zero sequence component need not be the same as them. Furthermore subtransient positive and negative sequence impedances of a synchronous machine are equal.

4. Voltage sources are connected to the positive sequence circuits of the rotating machines.

5. No positive or negative sequence current flows between neutral and ground.

Example 7.5

Example 7.5

Let us consider the network shown in Fig 7.17 which is essentially the same as that discussed in Example 1.2. The values of the various reactances are not important here and hence are not given in this figure. However various points of the circuit are denoted by the letters A to G . This has been done to identify the impedances of various circuit elements. For example, the leakage reactance of the transformer T1 is placed between the points A and B and that of transformer T2 is placed between D and E .

The positive sequence network is shown in Fig. 7.18. This is essentially same as that shown in Fig. 1.24. The negative sequence diagram, shown in Fig. 7.19, is almost identical to the positive sequence diagram except that the voltage sources are absent in this circuit. The zero sequence network is shown in Fig. 7.20. The neutral point of generator G1 is grounded. Hence a path from point A to the ground is provided through the zero sequence reactance of the generator. The primary side of the transformer T1 is Δ -connected and hence there is discontinuity in the circuit after point A . Similar connections are also made for generator G2 and transformer T2 . The transmission line impedances are placed between the points B - C , C - D and C - F . The secondary side of transformer T3 is ungrounded and hence there is a break in the circuit after the point F . However the primary side of T3 is grounded and so is the neutral point of generator G3 . Hence the zero sequence components of these two apparatus are connected to the ground.

Fig. 7.17 Single-line diagram of a 3-machine power system.

Fig. 7.18 Positive sequence network of the power system of Fig. 7.17.

Fig. 7.19 Negative sequence network of the power system of Fig. 7.17.

Fig. 7.20 Zero sequence network of the power system of Fig. 7.17.

Unsymmetrical Faults

Introduction

The sequence circuits and the sequence networks developed in the previous chapter will now be used for finding out fault current during unsymmetrical faults.

Three Types of Faults

Calculation of fault currents

Let us make the following assumptions:

The power system is balanced before the fault occurs such that of the three sequence networks only the positive sequence network is active. Also as the fault occurs, the sequence networks are connected only through the fault location.

The fault current is negligible such that the pre-fault positive sequence voltages are same at all nodes and at the fault location.

All the network resistances and line charging capacitances are negligible.

All loads are passive except the rotating loads which are represented by synchronous machines.

Based on the assumptions stated above, the faulted network will be as shown in Fig. 8.1 where the voltage at the faulted point will be denoted by Vf and current in the three faulted phases are Ifa , I fb and I fc .

We shall now discuss how the three sequence networks are connected when the three types of faults discussed above occur.

Fig. 8.1 Representation of a faulted segment.

Single-Line-to-Ground Fault

Let a 1LG fault has occurred at node k of a network. The faulted segment is then as shown in Fig. 8.2 where it is assumed that phase-a has touched the ground through an impedance Zf . Since the system is unloaded before the occurrence of the fault we have

(8.1)

Fig. 8.2 Representation of 1LG fault.

Also the phase-a voltage at the fault point is given by From (8.1) we can write Solving (8.3) we get This implies that the three sequence currents are in series for the 1LG fault. Let us denote the zero, positive and negative sequence Thevenin impedance at the faulted point as Z kk0 , Z kk1 and Z kk2 respectively. Also since the Thevenin voltage at the faulted phase is Vf we get three sequence circuits that are similar to the ones shown in Fig. 7.7. We can then write Then from (8.4) and (8.5) we can write Again since We get from (8.6)

The Thevenin equivalent of the sequence network is shown in Fig. 8.3.

(8.2)

(8.3)

(8.4)

(8.5)

(8.6)

(8.7)

Fig. 8.3 Thevenin equivalent of a 1LG fault.

Example 8.1

A three-phase Y-connected synchronous generator is running unloaded with rated voltage when a 1LG fault occurs at its terminals. The generator is rated 20 kV, 220 MVA, with subsynchronous reactance of 0.2 per unit. Assume that the subtransient mutual reactance between the windings is 0.025 per unit. The neutral of the generator is grounded through a 0.05 per unit reactance. The equivalent circuit of the generator is shown in Fig. 8.4. We have to find out the negative and zero sequence reactances.

Fig. 8.4 Unloaded generator of Example 8.1.

Since the generator is unloaded the internal emfs are

Since no current flows in phases b and c, once the fault occurs, we have from Fig. 8.4

Then we also have

From Fig. 8.4 and (7.34) we get

Therefore

From (7.38) we can write Z1 = j ω ( Ls + Ms ) = j 0.225. Then from Fig. 7.7 we have

Also note from (8.4) that

Therefore from Fig. 7.7 we get

Comparing the above two values with (7.37) and (7.39) we find that Z 0 indeed is equal to j ω ( Ls - 2 Ms ) and Z2 is equal to j ω ( Ls + Ms ). Note that we can also calculate the fault current from (8.7) as

Line-to-Line Fault

The faulted segment for an L-L fault is shown in Fig. 8.5 where it is assumed that the fault has occurred at node k of the network. In this the phases b and c got shorted through the impedance Zf . Since the system is unloaded before the occurrence of the fault we have

Fig. 8.5 Representation of L-L fault.

Also since phases b and c are shorted we have

Therefore from (8.8) and (8.9) we have

We can then summarize from (8.10)

(8.8)

Therefore no zero sequence current is injected into the network at bus k and hence the zero sequence remains a dead network for an L-L fault. The positive and negative sequence currents are negative of each other.

Now from Fig. 8.5 we get the following expression for the voltage at the faulted point

Again

Moreover since I fa0 = I fb0 = 0 and I fa1 = - I fb2 , we can write

Therefore combining (8.12) - (8.14) we get

Equations (8.12) and (8.15) indicate that the positive and negative sequence networks are in parallel. The sequence network is then as shown in Fig. 8.6. From this network we get

Fig. 8.6 Thevenin equivalent of an LL fault.

Example 8.2

Let us consider the same generator as given in Example 8.1. Assume that the generator is unloaded when a bolted ( Zf = 0) short circuit occurs between phases b and c. Then we get from (8.9) I fb = - I fc . Also since the generator is unloaded, we have I fa = 0. Therefore from (7.34) we get

Also since V bn = V cn , we can combine the above two equations to get

Then

We can also obtain the above equation from (8.16) as

(8.13)

Also since the neutral current I n is zero, we can write V a = 1.0 and

Hence the sequence components of the line voltages are

Also note that

which are the same as obtained before.

Double- Line -to Ground Fault

The faulted segment for a 2LG fault is shown in Fig. 8.7 where it is assumed that the fault has occurred at node k of the network. In this the phases b and c got shorted through the impedance Zf to the ground. Since the system is unloaded before the occurrence of the fault we have the same

condition as (8.8) for the phase-a current. Therefore

Fig. 8.7 Representation of 2LG fault.

Also voltages of phases b and c are given by

Therefore

We thus get the following two equations from (8.19)

(8.19)

(8.21)

Substituting (8.18) and (8.20) in (8.21) and rearranging we get

Also since I fa = 0 we have

The Thevenin equivalent circuit for 2LG fault is shown in Fig. 8.8. From this figure we get

The zero and negative sequence currents can be obtained using the current divider principle as

Fig. 8.8 Thevenin equivalent of a 2LG fault.

Example 8.3

Let us consider the same generator as given in Examples 8.1 and 8.2. Let us assume that the generator is operating without any load when a bolted 2LG fault occurs in phases b and c. The equivalent circuit for this fault is shown in Fig. 8.9. From this figure we can write

Fig. 8.9 Equivalent circuit of the generator in Fig. 8.4 for a 2LG fault in phases b and c.

Combining the above three equations we can write the following vector-matrix form

Solving the above equation we get

Hence

We can also obtain the above values using (8.24)-(8.26). Note from Example 8.1 that

Then

Now the sequence components of the voltages are

Also note from Fig. 8.9 that

and Vb = Vc = 0. Therefore

which are the same as obtained before.

FAULT CURRENT COMPUTATION USING SEQUENCE NETWORKS

In this section we shall demonstrate the use of sequence networks in the calculation of fault currents using sequence network through some examples

Example 8.4

Consider the network shown in Fig. 8.10. The system parameters are given below

Generator G : 50 MVA, 20 kV, X" = X1 = X2 = 20%, X0 = 7.5%

Motor M : 40 MVA, 20 kV, X" = X1 = X2 = 20%, X0 = 10%, Xn = 5%

Transformer T1 : 50 MVA, 20 kV Δ /110 kVY, X = 10%

Transformer T2 : 50 MVA, 20 kV Δ /110 kVY, X = 10%

Transmission line: X1 = X2 = 24.2 Ω , X0 = 60.5 Ω

We shall find the fault current for when a (a) 1LG, (b) LL and (c) 2LG fault occurs at bus-2.

Fig. 8.10 Radial power system of Example 8.4.

Let us choose a base in the circuit of the generator. Then the per unit impedances of the generator are:

The per unit impedances of the two transformers are

The MVA base of the motor is 40, while the base MVA of the total circuit is 50. Therefore the per unit impedances of the motor are

For the transmission line

Therefore

Let us neglect the phase shift associated with the Y/ Δ transformers. Then the positive, negative and zero sequence networks are as shown in Figs. 8.11-8.13.

Fig. 8.11 Positive sequence network of the power system of Fig. 8.10.

Fig. 8.12 Negative sequence network of the power system of Fig. 8.10.

Fig. 8.13 Zero sequence network of the power system of Fig. 8.10.

From Figs. 8.11 and 8.12 we get the following Ybus matrix for both positive and negative sequences

Inverting the above matrix we get the following Zbus matrix

Again from Fig. 8.13 we get the following Ybus matrix for the zero sequence

Inverting the above matrix we get

Hence for a fault in bus-2, we have the following Thevenin impedances

Alternatively we find from Figs. 8.11 and 8.12 that

(a) Single-Line-to-Ground Fault : Let a bolted 1LG fault occurs at bus-2 when the system is unloaded with bus voltages being 1.0 per unit. Then from (8.7) we get

per unit

Also from (8.4) we get

per unit

Also I fb = I fc = 0. From (8.5) we get the sequence components of the voltages as

Therefore the voltages at the faulted bus are

(b) Line-to-Line Fault : For a bolted LL fault, we can write from (8.16)

per unit

Then the fault currents are

Finally the sequence components of bus-2 voltages are

Hence faulted bus voltages are

(c) Double-Line-to-Ground Fault : Let us assumes that a bolted 2LG fault occurs at bus-2. Then

Hence from (8.24) we get the positive sequence current as

per unit

The zero and negative sequence currents are then computed from (8.25) and (8.26) as

per unit

per unit

Therefore the fault currents flowing in the line are

Furthermore the sequence components of bus-2 voltages are

Therefore voltages at the faulted bus are

Example 8.5

Let us now assume that a 2LG fault has occurred in bus-4 instead of the one in bus-2. Therefore

Also we have

Hence

per unit

Also

per unit

per unit

Therefore the fault currents flowing in the line are

We shall now compute the currents contributed by the generator and the motor to the fault. Let us denote the current flowing to the fault from the generator side by Ig , while that flowing from the motor by Im . Then from Fig. 8.11 using the current divider principle, the positive sequence currents contributed by the two buses are

per unit

per unit

Similarly from Fig. 8.12, the negative sequence currents are given as

per unit

per unit

Finally notice from Fig. 8.13 that the zero sequence current flowing from the generator to the fault is 0. Then we have

per unit

Therefore the fault currents flowing from the generator side are

and those flowing from the motor are

It can be easily verified that adding Ig and Im we get If given above.

In the above two examples we have neglected the phase shifts of the Y/ Δ transformers. However according to the American standard, the positive sequence components of the high tension side lead those of the low tension side by 30° , while the negative sequence behavior is reverse of the positive sequence behavior. Usually the high tension side of a Y/ Δ transformer is Y-connected. Therefore as we have seen in Fig. 7.16, the positive sequence component of Y side leads the positive sequence component of the Δ side by 30° while the negative sequence component of Y side lags that of the Δ side by 30° . We shall now use this principle to compute the fault current for an unsymmetrical fault.

Let us do some more examples

Example 8.6

Let us consider the same system as given in Example 8.5. Since the phase shift does not alter the zero sequence, the circuit of Fig. 8.13 remains unchanged. The positive and the negative sequence circuits must however include the respective phase shifts. These circuits are redrawn as shown in Figs. 8.14 and 8.15.

Note from Figs. 8.14 and 8.15 that we have dropped the √3 α vis-à-vis that of Fig. 7.16. This is because the per unit impedances remain unchanged when referred to the either high tension or low tension side of an ideal transformer. Therefore the per unit impedances will also not be altered.

Fig. 8.14 Positive sequence network of the power system of Fig. 8.10 including transformer phase shift.

Fig. 8.15 Negative sequence network of the power system of Fig. 8.10 including transformer phase shift.

Since the zero sequence remains unaltered, these currents will not change from those computed in Example 8.6. Thus

and per unit

Now the positive sequence fault current from the generator Iga1 , being on the Y-side of the Y/ Δ transformer will lead I ma1 by 30° . Therefore

per unit

per unit

Finally the negative sequence current I ga2 will lag I ma2 by 30° . Hence we have

per unit

per unit

Therefore

Also the fault currents flowing from the motor remain unaltered. Also note that the currents flowing into the fault remain unchanged. This implies that the phase shift of the Y/ Δ transformers does not affect the fault currents.

Example 8.7

Let us consider the same power system as given in Example 1.2, the sequence diagrams of which are given in Figs. 7.18 to 7.20. With respect to Fig. 7.17, let us define the system parameters as:

Generator G1 : 200 MVA, 20 kV, X" = 20%, X0 = 10%

Generator G2 : 300 MVA, 18 kV, X" = 20%, X0 = 10%

Generator G3 : 300 MVA, 20 kV, X " = 25%, X0 = 15%

Transformer T1 : 300 MVA, 220Y/22 kV, X = 10%

Transformer T2 : Three single-phase units each rated 100 MVA, 130Y/25 kV, X = 10%

Transformer T3 : 300 MVA, 220/22 kV, X = 10%

Line B-C : X1 = X2 = 75 Ω , X0 = 100 Ω

Line C-D : X1 = X2 = 75 Ω , X0 = 100 Ω

Line C-F : X1 = X2 = 50 Ω , X0 = 75 Ω

Let us choose the circuit of Generator 3 as the base, the base MVA for the circuit is 300. The base voltages are then same as those shown in Fig. 1.23. Per unit reactances are then computed as shown below.

Generator G1 : , X0 = 0.15

Generator G2 : , X0 = 0.0656

Generator G3 : , X0 = 0.15

Transformer T1 :

Transformer T2 :

Transformer T3 :

Line B-C : ,

Line C-D : ,

Line C-F : ,

Neglecting the phase shifts of Y/ Δ connected transformers and assuming that the system is unloaded, we shall find the fault current for a 1LG fault at bus-1 (point C of Fig. 7.17).

From Figs. 7.18 and 7.19, we can obtain the positive and negative sequence Thevenin impedance at point C as (verify)

X1 = X2 = j 0.2723 per unit

Similarly from Fig. 7.20, the Thevenin equivalent of the zero sequence impedance is

X0 = j 0.4369 per unit

Therefore from (8.7) we get

per unit

Then the fault current is Ifa = 3 Ifa0 = 3.0565 per unit.

Chapter 9: Transient Stability Introduction

The first electric power system was a dc system built by Edison in 1882. The subsequent power systems that were constructed in the late 19th century were all dc systems. However despite the initial popularity of dc systems by the turn of the 20th century ac systems started to outnumber them. The ac systems were though to be superior as ac machines were cheaper than their dc counterparts and more importantly ac voltages are easily transformable from one level to other using transformers. The early stability problems of ac systems were experienced in 1920 when insufficient damping caused spontaneous oscillations or hunting. These problems were solved using generator damper winding and the use of turbine-type prime movers.

The stability of a system refers to the ability of a system to return back to its steady state when subjected to a disturbance. As mentioned before, power is generated by synchronous generators that operate in synchronism with the rest of the system. A generator is synchronized with a bus when both of them have same frequency, voltage and phase sequence. We can thus define the power system stability as the ability of the power system to return to steady state without losing synchronism. Usually power system stability is categorized into Steady State, Transient and Dynamic Stability.

Steady State Stability studies are restricted to small and gradual changes in the system operating

conditions. In this we basically concentrate on restricting the bus voltages close to their nominal values. We also ensure that phase angles between two buses are not too large and check for the overloading of the power equipment and transmission lines. These checks are usually done using power flow studies.

Transient Stability involves the study of the power system following a major disturbance. Following a large disturbance the synchronous alternator the machine power (load) angle changes due to sudden acceleration of the rotor shaft. The objective of the transient stability study is to ascertain whether the load angle returns to a steady value following the clearance of the disturbance.

The ability of a power system to maintain stability under continuous small disturbances is investigated under the name of Dynamic Stability (also known as small-signal stability). These small disturbances occur due random fluctuations in loads and generation levels. In an interconnected power system, these random variations can lead catastrophic failure as this may force the rotor angle to increase steadily. In this chapter we shall discuss the transient stability aspect of a power system.

Section I: Power-Angle Relationship

The power-angle relationship has been discussed in Section 2.4.3. In this section we shall consider this relation for a lumped parameter lossless transmission line. Consider the single-machine-infinite-bus (SMIB) system shown in Fig. 9.1. In this the reactance X includes the reactance of the

transmission line and the synchronous reactance or the transient reactance of the generator. The sending end voltage is then the internal emf of the generator. Let the sending and receiving end voltages be given by

Fig. 9.1 An SMIB system. We then have The sending end real power and reactive power are then given by This is simplified to Since the line is loss less, the real power dispatched from the sending end is equal to the real power received at the receiving end. We can therefore write where Pmax = V1 V2 / X is the maximum power that can be transmitted over the transmission line. The power-angle curve is shown in Fig. 9.2. From this figure we can see that for a given power P0 . There are two possible values of the angle δ - δ0 and δmax . The angles are given by

Example 9.1

A generator is connected to a constant voltage bus through an external reactance of 0.3 per unit. The synchronous reactance of the generator is 0.2 per unit and the voltage magnitude of the constant voltage bus is 1.0 per unit with its angle being 0° . The generator delivers 0.9 per unit power to the constant voltage bus when the angle of its terminal voltage is 15° . We have to determine the magnitude and angle of its internal emf.

Let us denote the angle of the terminal voltage by δt , while its magnitude by Vt. Since the generator delivers 0.9 per unit power to the constant voltage bus of magnitude 1.0 per unit through a 0.3 per unit reactance, we can write from (9.4)

(9.1)

(9.2)

(9.3)

(9.4)

(9.5)

Fig. 9.2 A typical power-angle curve. Solving the above equation we get Vt = 1.0432 per unit. Therefore the current flowing through the line is

per unit

The internal emf of the generator is then given by

Then the magnitude of the internal emf is 1.1082 per unit angle its angle δ is 23.96° . It can be readily verified that the power delivered is

per unit Section II: Swing Equation

Let us consider a three-phase synchronous alternator that is driven by a prime mover. The equation of motion of the machine rotor is given by

where

J is the total moment of inertia of the rotor mass in kgm2

Tm is the mechanical torque supplied by the prime mover in N-m

Te is the electrical torque output of the alternator in N-m

θ is the angular position of the rotor in rad

Neglecting the losses, the difference between the mechanical and electrical torque gives the net accelerating torque Ta . In the steady state, the electrical torque is equal to the mechanical torque, and hence the accelerating power will be zero. During this period the rotor will move at synchronous speed ωs in rad/s.

The angular position θ is measured with a stationary reference frame. To represent it with respect to the synchronously rotating frame, we define

(9.6)

(9.7)

where δ is the angular position in rad with respect to the synchronously rotating reference frame. Taking the time derivative of the above equation we get Defining the angular speed of the rotor as we can write (9.8) as We can therefore conclude that the rotor angular speed is equal to the synchronous speed only when dδ / dt is equal to zero. We can therefore term dδ / dt as the error in speed. Taking derivative of (9.8), we can then rewrite (9.6) as Multiplying both side of (9.11) by ωm we get where Pm , Pe and Pa respectively are the mechanical, electrical and accelerating power in MW. We now define a normalized inertia constant as Substituting (9.12) in (9.10) we get

In steady state, the machine angular speed is equal to the synchronous speed and hence we can replace ωr in the above equation by ωs. Note that in (9.13) Pm , Pe and Pa are given in MW. Therefore dividing them by the generator MVA rating Srated we can get these quantities in per unit. Hence dividing both sides of (9.13) by Srated we get

Equation (7.14) describes the behaviour of the rotor dynamics and hence is known as the swing equation. The angle δ is the angle of the internal emf of the generator and it dictates the amount of power that can be transferred. This angle is therefore called the load angle.

Example 9.2

A 50 Hz, 4-pole turbogenerator is rated 500 MVA, 22 kV and has an inertia constant (H) of 7.5. Assume that the generator is synchronized with a large power system and has a zero accelerating power while delivering a power of 450 MW. Suddenly its input power is changed to 475 MW. We have

(9.8)

(9.9)

(9.10)

(9.11)

(9.12)

(9.13)

per unit (9.14)

to find the speed of the generator in rpm at the end of a period of 10 cycles. The rotational losses are assumed to be zero.

We then have

Noting that the generator has four poles, we can rewrite the above equation as

The machines accelerates for 10 cycles, i.e., 20 × 10 = 200 ms = 0.2 s, starting with a synchronous speed of 1500 rpm. Therefore at the end of 10 cycles

Speed = 1500 + 43.6332 ´ 0.2 = 1508.7266 rpm.

Section III: Equal Area Criterion

The real power transmitted over a lossless line is given by (9.4). Now consider the situation in which the synchronous machine is operating in steady state delivering a power Pe equal to Pm when there is a fault occurs in the system. Opening up of the circuit breakers in the faulted section subsequently clears the fault. The circuit breakers take about 5/6 cycles to open and the subsequent post-fault transient last for another few cycles. The input power, on the other hand, is supplied by a prime mover that is usually driven by a steam turbine. The time constant of the turbine mass system is of the order of few seconds, while the electrical system time constant is in milliseconds. Therefore, for all practical purpose, the mechanical power is remains constant during this period when the electrical transients occur. The transient stability study therefore concentrates on the ability of the power system to recover from the fault and deliver the constant power Pm with a possible new load angle δ .

Consider the power angle curve shown in Fig. 9.3. Suppose the system of Fig. 9.1 is operating in the steady state delivering a power of Pm at an angle of δ0 when due to malfunction of the line, circuit breakers open reducing the real power transferred to zero. Since Pm remains constant, the accelerating power Pa becomes equal to Pm . The difference in the power gives rise to the rate of change of stored kinetic energy in the rotor masses. Thus the rotor will accelerate under the constant influence of non-zero accelerating power and hence the load angle will increase. Now suppose the circuit breaker re-closes at an angle δc. The power will then revert back to the normal operating curve. At that point, the electrical power will be more than the mechanical power and the accelerating power will be negative. This will cause the machine decelerate. However, due to the inertia of the rotor masses, the load angle will still keep on increasing. The increase in this angle may eventually stop and the rotor may start decelerating, otherwise the system will lose synchronism. Note that

Fig. 9.3 Power-angle curve for equal area criterion.

Hence multiplying both sides of (9.14) by and rearranging we get Multiplying both sides of the above equation by dt and then integrating between two arbitrary angles δ0 and δc we get Now suppose the generator is at rest at δ0. We then have dδ / dt = 0. Once a fault occurs, the machine starts accelerating. Once the fault is cleared, the machine keeps on accelerating before it reaches its peak at δc , at which point we again have dδ / dt = 0. Thus the area of accelerating is given from (9.15) as In a similar way, we can define the area of deceleration. In Fig. 9.3, the area of acceleration is given by A1 while the area of deceleration is given by A2 . This is given by

Now consider the case when the line is reclosed at δc such that the area of acceleration is larger than the area of deceleration, i.e., A1 > A2 . The generator load angle will then cross the point δm , beyond which the electrical power will be less than the mechanical power forcing the accelerating power to be positive. The generator will therefore start accelerating before is slows down completely and will eventually become unstable. If, on the other hand, A1 < A2 , i.e., the decelerating area is larger than the accelerating area, the machine will decelerate completely before accelerating again. The rotor inertia will force the subsequent acceleration and deceleration areas to be smaller than the first ones and the machine will eventually attain the steady state. If the two areas are equal, i.e., A1 = A2 , then the accelerating area is equal to decelerating area and this is defines the boundary of the stability limit. The clearing angle δc for this mode is called the Critical Clearing Angle and is denoted by δcr. We then get from Fig. 9.3 by substituting δc = δcr

We can calculate the critical clearing angle from the ab move equation. Since the critical clearing angle depends on the equality of the areas, this is called the equal area criterion.

Example 9.3:

Consider the system of Example 9.1. Let us assume that the system is operating with Pm = Pe = 0.9 per unit when a circuit breaker opens inadvertently isolating the generator from the infinite bus. During this period the real power transferred becomes zero. From Example 9.1 we have calculated δ0 = 23.96 ° = 0.4182 rad and the maximum power transferred as

(9.15)

(9.16)

(9.17)

(9.18)

per unit

We have to find the critical clearing angle. From (9.15) the accelerating area is computed as by note that Pe = 0 during this time. This is then given by

To calculate the decelerating area we note that δm = π - 0.4182 = 2.7234 rad. This area is computed by noting that Pe = 2.2164 sin(δ ) during this time. Therefore

Equating A1 = A2 and rearranging we get

Now a frequently asked question is what does the critical clearing angle mean?

Since we are interested in finding out the maximum time that the circuit breakers may take for opening, we should be more concerned about the critical clearing time rather than clearing angle. Furthermore, notice that the clearing angle is independent of the generalized inertia constant H . Hence we can comment that the critical clearing angle in this case is true for any generator that has a d-axis transient reactance of 0.20 per unit. The critical clearing time, however, is dependent on H and will vary as this parameter varies.

To obtain a description for the critical clearing time, let us consider the period during which the fault occurs. We then have Pe = 0. We can therefore write from

Integrating the above equation with the initial acceleration being zero we get

Further integration will lead to

Replacing δ by δcr and t by tcr in the above equation, we get the critical clearing time as

Example 9.4:

In Example 9.2, let us choose the system frequency as 50 Hz such that ωs is 100π. Also let us choose H as 4 MJ/MVA. Then with δcr being 1.5486 rad, δ0 being 0.4182 rad and Pm being 0.9 per unit, we get the following critical clearing time from (9.20)

s

To illustrate the response of the load angle δ , the swing equation is simulated in MATLAB. The swing equation of (9.14) is then expressed as

where Δωr is the deviation for the rotor speed from the synchronous speed ωs . It is to be noted that the swing equation of (9.21) does not contain any damping. Usually a damping term, that is proportional to the machine speed Δωr, is added with the accelerating power. Without the damping the load angle will exhibit a sustained oscillation even when the system remains stable when the fault cleared within the critical clearing time.

Fig. 9.4 Stable and unstable system response as a function of clearing time.

Fig. 9.4 depicts the response of the load angle δ for two different values of load angle. It is assumed that the fault occurs at 0.5 s when the system is operating in the steady state delivering 0.9 per unit power. The load angle during this time is constant at 23.96° . The load angle remains stable, albeit the sustained oscillation when the clearing time tcl is 0.253 s. The clearing angle during this time is 88.72° . The system however becomes unstable when the clearing time 0.2531s and the load angle increases asymptotically. The clearing time in this case is 88.77° . This is called the Loss of Synchronism. It is to be noted that such increase in the load angle is not permissible and the protection device will isolate the generator from the system.

The clearing time of (8.20) is derived based on the assumption that the electrical power Pe becomes zero during the fault as in (8.19). This need not be the case always. In that even we have to resort to finding the clearing time using the numerical integration of the swing equation. See example 9.5 to illustrates the point.

Example 9.5

Consider the system in which a generator is connected to an infinite bus through a double circuit transmission line as shown in Fig. 9.5. The per unit system reactances that are converted in a common base, are also shown in this figure. Let us assume that the infinite bus voltage is 1< 0° . The generator is delivering 1.0 per unit real power at a lagging power factor of 0.9839 to the infinite bus. While the generator is operating in steady state, a three-phase bolted short circuit occurs in the transmission line connecting buses 2 and 4 - very near to bus 4. The fault is cleared by opening the circuit breakers at the two ends of this line. We have to find the critical clearing time for various values of H .

(9.21)

Fig. 9.5 Schematic diagram of the power system of Example 9.4.

Let the current flowing through the line be denoted by I . Then the power delivered to the infinite bus is

From the above equation we get

The total impedance during the time when both the lines are operational, the impedance between the generator and the infinite bus is j(0.3+0.1+0.1) = j 0.5 per unit. Then the generator internal voltage is

Therefore the machine internal voltage is E' = 1.2 per unit its angle is 24.625° or 0.4298 rad.

The pre-fault equivalent circuit is shown in Fig. 9.6 (a). From this figure we can write the power transfer equation as

Once the fault is cleared by opening of the breakers connected near buses 2 and 4, only of one the two lines will be operational. Therefore during the post fault period, the impedance between the generator and the infinite bus is j (0.3+0.2+0.6) = j 0.5 per unit as shown in Fig. 9.6 (b). Then the post-fault power transfer equation is given by

Fig. 9.6 Equivalent circuit: (a) pre-fault and (b) post-fault.

It is to be noted that since one of the two lines is functional during the fault, the power transfer during the fault will not be zero. The equivalent circuit during the fault is shown in Fig. 9.7. Since the fault has occurred very near to bus-4, we can assume that this bus has been short circuited. We shall find the Thevenin equivalent of the portion of the circuit to the right of the points A and B .

Fig. 9.7 Faulted equivalent circuit.

The circuit between buses 2 and 3 has been converted into an equivalent Y using Δ-Y transformation. This is shown in Fig. 9.8 (a). From this figure we find the Thevenin impedance as

per unit

Also the Thevenin voltage is then given by

per unit

The reduced circuit is shown in Fig. 9.8 (b). From this circuit we can write the following power transfer equation during the fault

Fig. 9.8 Equivalent circuit during the fault: (a) D -Y transformed and (b) Thevenin equivalent.

Three power-angle curves are shown in Fig. 9.9. From this figure we find that

rad

Fig. 9.9 Power-angles curves for the three modes of operation of the system of Example 9.4.

The accelerating area is given by

and the decelerating area is

Equating the two areas we get

As mentioned earlier, the critical clearing angle depends on the system network configuration. The critical clearing time, however, is dependent on H and will vary as this parameter varies. Usually numerical methods are employed for finding out the clearing time. We shall however demonstrate the effect of H through a MATLAB program. The program uses the built-in ordinary differential equation solver though which the swing equations are solved. The results obtained are listed in Table 9.1. It can be seen that as the value of H increases, the clearing time also increases, even though the clearing angle remains the same. This is obvious as the value of H increases, the response of the system becomes more sluggish due to larger inertia. Hence, the rotor takes more time to accelerate.

Table 9.1 Effect of H on critical clearing time

H ( MJ/MVA) Approximate Critical Clearing Time (s)

2 0.2783

4 0.3936

6 0.4821

8 0.5566

10 0.6223

Section IV: Multimachine Stability Oscillations in a Two Area System

Consider Fig. 9.10, which depicts a number of weights that are suspended by elastic strings. The weights represent generators and the electric transmission lines being represented by the strings. Note that in a transmission system, each transmission line is loaded below its static stability limit. Similarly, when the mechanical system is in static steady state, each string is loaded below its break point. At this point one of the strings is suddenly cut. This will result in transient oscillations in the coupled strings and all the weights will wobble. In the best possible case, this may result in the coupled system settling down to a new steady state. On the other hand, in the worst possible scenario this may result in the breaking of one more additional string, resulting in a chain reaction in which more strings may break forcing a system collapse. In a similar way, in an interconnected electric power network, the tripping of a transmission line may cause a catastrophic failure in which a large number of generators are lost forcing a blackout in a large area.

Modern power systems are interconnected and operate close to their transient and steady state stability limits. In large interconnected systems, it is common to find a natural response of a group of closely coupled machines oscillating against other groups of machines. These oscillations have a frequency range of 0.1 Hz to 0.8 Hz. The lowest frequency mode involves all generators of the system. This oscillation groups the system into two parts - with generators in one part oscillating against those of the the other part. The higher frequency modes are usually localized with small groups oscillating against each other. Unfortunately, the inter-area oscillation can be initiated by a small disturbance in any part of the system. These small frequency oscillations fall under the category of dynamic stability and are analysed in linear domain through the liberalisation of the entire interconnected systems model.

Inter-area oscillations manifest wherever the power system is heavily interconnected. The oscillations, unless damped, can lead to grid failure and total system collapse. Low frequency oscillations in the range of 0.04 Hz to 0.06 Hz were observed in the Pacific North West region as early as 1950. Improper speed governor control of hydro units created these oscillations. The Northern and Southern regions of WSCC were interconnected by a 230 kV line in 1964. Immediately the system experienced a 0.1 Hz oscillation resulting in over 100 instances of opening of the tie line in the first nine months of operation. Some system damping was provided through the modification in the hydro turbine governors.

A 500 kV pacific intertie and another ± 400 kV HVDC system was commissioned in 1968. This raised the frequency of oscillation from 0.1 Hz to 0.33 Hz and these oscillations could no longer be controlled through governor action alone. In late 1980's a new intertie joined the WSCC system to Alberta and British Columbia in Canada . As a result of this interconnection, the two different oscillation frequencies manifested - one at 0.29 Hz and the other at 0.45 Hz.

Ontario Hydro is one of the largest utilities in North America . Due to the vast and sparsely populated topology of Canada , the operating span of Ontario hydro is over 1000 km from East to West and from North to South. The Ontario Hydro system is connected to the neighbouring Canadian provinces and the North Western region of the United States . In 1959 Ontario Hydro was connected to Michigan in the South and Quebec Hydro in the East. As a result of this connection, a 0.25 Hz oscillation was observed and a result of this it was decided to remove the tie with Quebec and retain the tie to Michigan . The Western portion of Ontario was connected to neighbouring Manitoba in 1956

and then Manitoba was connected to its neighbour Saskatchewan in 1960. This resulted in oscillation in the frequency range 0.35 Hz to 0.45 Hz often tripping the tie. As a result of this, Ontario Hydro decided to commission power system stabilizers for all their generating units since early 1960's. It has also sponsored extensive research in this area.

Through research it was established that the action of automatic voltage regulators caused these oscillations. An automatic voltage regulator (AVR) regulates the generator terminal voltage and also helps in the enhancement of transient stability by reducing the peak of the first swing following any disturbance. However, its high gain contributed to negative damping to the system. The knowledge of this relation resulted in the commissioning of power system stabilizers. It was observed that these oscillations were results of the periodic interchange of kinetic energy between the generator rotors. A power system stabilizer (PSS) provides a negative feedback of the changes in rotor kinetic energy when it is connected to the excitation system thereby providing damping to these small oscillations. The PSS has been a subject of extensive research. The team of Dr. P. Kundur, then with Ontario Hydro, and his co-workers has done extensive research in the area of PSS tuning and its characteristics. Through their vast experience and extensive research, they reported the enhancement of inter-area and local modes through PSS reported in. Since a power system is piece-wise linear, its system characteristics changes with operating point. Therefore an adaptive controller that can tune with the changes in the system has been developed and reported in. It was shown that the adaptive PSS is effective in damping large as well as small disturbances.

The power flow between generators, as evident from (9.4), is dependent on the angle between those generators. The stable operating point of the power system is where the generated power at each station is matched by the electrical power sent out from that station. When there is a mismatch between electrical power out and the generated mechanical shaft power, the generator will accelerate at a rate determined by the power mismatch and the machine inertia as given in (9.14).

Oscillations in s Two Area System

Consider the simple power system shown in Fig. 9.11 in which two machines are operating. Let us assume that starting with the initial angles δ1 and δ2 with respect to some reference at nominal frequency, machine 1 accelerates while machine 2 decelerates from this nominal frequency. We then have

where the subscripts 1 and 2 refer to machines 1 and 2 respectively. Let us assume that the transmission line is loss less. Then in the simple case where the power from machine 1 flows to machine 2, we get

where δ12 = δ1 - δ2 .

Fig. 9.11 Single-line diagram of a two-machine power system.

Now since the system is lossless, (9.26) will also imply that Pm1 = - Pm2 . This means that in the steady state, the power generated at machine 1 is absorbed through machine 2. Combining (9.25) and (9.26) we get

(9.25)

(9.26)

Let us now assume that H1 = H2 = H , V1 = V2 = 1.0 per unit and Pm1 = 0. We then get from (9.27)

where the oscillation frequency ω is given by

Thus the weighted difference of angles will approximate simple harmonic motion for small changes in δ12 and the frequency will decrease for an increase in inertia H or impedance X . Another aspect can

be seen by adding the system to give

Thus the overall acceleration of the machine group will depend on the overall balance between power generated and consumed. Usually there are governors on the generators to reduce generated power if the system frequency increases.

Introduction

Ideal Shunt Compensator

Improving Volatage Profile

Improving Power-Angle Characteristics

Improving Stability Margin

Improving Damping to Power Oscillations

Ideal Series Compensator

Impact of Series Compensator on Voltage Profile

Improving Power-Angle Characteristics

An Alternate Method of Voltage Injection

Improving Stability Margin

Comparisons of the Two Modes of Operation

Power Flow Control and Power Swing Damping

Introduction

The two major problems that the modern power systems are facing are voltage and angle stabilities. There are various approaches to overcome the problem of stability arising due to small signal oscillations in an interconnected power system. As mentioned in the previous chapter, installing power system stabilizers with generator excitation control system provides damping to these oscillations. However, with the advancement in the power electronic technology, various reactive power control equipment are increasingly used in power transmission systems.

A power network is mostly reactive. A synchronous generator usually generates active power that is specified by the mechanical power input. The reactive power supplied by the generator is dictated by the network and load requirements. A generator usually does not have any control over it. However the lack of reactive power can cause voltage collapse in a system. It is therefore important to supply/absorb excess reactive power to/from the network. Shunt compensation is one possible approach of providing reactive power support.

A device that is connected in parallel with a transmission line is called a shunt compensator , while a device that is connected in series with the transmission line is called a series compensator . These

(9.27)

(9.28)

are referred to as compensators since they compensate for the reactive power in the ac system. We shall assume that the shunt compensator is always connected at the midpoint of transmission system, while the series compensator can be connected at any point in the line. We shall demonstrate that such connections in an SMIB power system improves

voltage profile

power-angle characteristics

stability margin

damping to power oscillations

A static var compensator ( SVC ) is the first generation shunt compensator. It has been around since 1960s. In the beginning it was used for load compensation such as to provide var support for large industrial loads, for flicker mitigation etc. However with the advancement of semiconductor technology, the SVC started appearing in the transmission systems in 1970s. Today a large number of SVCs are connected to many transmission systems all over the world. An SVC is constructed using the thyristor technology and therefore does not have gate turn off capability.

With the advancement in the power electronic technology, the application of a gate turn off thyristor (GTO) to high power application became commercially feasible. With this the second generation shunt compensator device was conceptualized and constructed. These devices use synchronous voltage sources for generating or absorbing reactive power. A synchronous voltage source (SVS) is constructed using a voltage source converter (VSC). Such a shunt compensating device is called static compensator or STATCOM . A STATCOM usually contains an SVS that is driven from a dc storage capacitor and the SVS is connected to the ac system bus through an interface transformer. The transformer steps the ac system voltage down such that the voltage rating of the SVS switches are within specified limit. Furthermore, the leakage reactance of the transformer plays a very significant role in the operation of the STATCOM.

Like the SVC, a thyristor controlled series compensator ( TCSC ) is a thyristor based series compensator that connects a thyristor controlled reactor ( TCR ) in parallel with a fixed capacitor. By varying the firing angle of the anti-parallel thyristors that are connected in series with a reactor in the TCR, the fundamental frequency inductive reactance of the TCR can be changed. This effects a change in the reactance of the TCSC and it can be controlled to produce either inductive or capacitive reactance.

Alternatively a static synchronous series compensator or SSSC can be used for series compensation. An SSSC is an SVS based all GTO based device which contains a VSC. The VSC is driven by a dc capacitor. The output of the VSC is connected to a three-phase transformer. The other end of the transformer is connected in series with the transmission line. Unlike the TCSC, which changes the impedance of the line, an SSSC injects a voltage in the line in quadrature with the line current. By making the SSSC voltage to lead or lag the line current by 90 ° , the SSSC can emulate the behavior of an inductance or capacitance.

In this chapter, we shall discuss the ideal behavior of these compensating devices. For simplicity we shall consider the ideal models and broadly discuss the advantages of series and shunt compensation.

Section I: Ideal Shunt Compensator

The ideal shunt compensator is an ideal current source. We call this an ideal shunt compensator because we assume that it only supplies reactive power and no real power to the system. It is needless to say that this assumption is not valid for practical systems. However, for an introduction, the assumption is more than adequate. We shall investigate the behavior of the compensator when connected in the middle of a transmission line. This is shown in Fig. 10.1, where the shunt compensator, represented by an ideal current source, is placed in the middle of a lossless transmission line. We shall demonstrate that such a configuration improves the four points that are mentioned above.

Fig 10.1 Schematic diagram of an ideal, midpoint shunt compensation

Improving Voltage Profile

Let the sending and receiving voltages be given by and respectively. The ideal shunt compensator is expected to regulate the midpoint voltage to

against any variation in the compensator current. The voltage current characteristic of the compensator is shown in Fig. 10.2. This ideal behavior however is not feasible in practical systems where we get a slight droop in the voltage characteristic. This will be discussed later.

Fig. 10.2 Voltage-current characteristic of an ideal shunt compensator.

Under the assumption that the shunt compensator regulates the midpoint voltage tightly as given by (10.1), we can write the following expressions for the sending and receiving end currents

Again from Fig. 10.1 we write

Combing (10.2)-(10.4) and solving we get We thus have to generate a current that is in phase with the midpoint voltage and has a magnitude of (4V / XL ){1 - cos( δ /2)}. The apparent power injected by the shunt compensator to the ac bus is then Since the real part of the injected power is zero, we conclude that the ideal shunt compensator injects only reactive power to the ac system and no real power.

Improving Power-Angle Characteristics The apparent power supplied by the source is given by

(10.5)

(10.6)

Similarly the apparent power delivered at the receiving end is

Hence the real power transmitted over the line is given by

Combining (10.6)-(10.8), we find the reactive power consumed by the line as

The power-angle characteristics of the shunt compensated line are shown in Fig. 10.3. In this figure Pmax = V2/X is chosen as the power base.

Fig. 10.3 Power-angle characteristics of ideal shunt compensated line.

Fig. 10.3 depicts Pe - δ and QQ - δ characteristics. It can be seen from fig 10.4 that for a real power transfer of 1 per unit, a reactive power injection of roughly 0.5359 per unit will be required from the shunt compensator if the midpoint voltage is regulated as per (10.1). Similarly for increasing the real power transmitted to 2 per unit, the shunt compensator has to inject 4 per unit of reactive power. This will obviously increase the device rating and may not be practical. Therefore power transfer enhancement using midpoint shunt compensation may not be feasible from the device rating point of view.

Fig. 10.4 Variations in transmitted real power and reactive power injection by the shunt compensator

with load angle for perfect midpoint voltage regulation.

Let us now relax the condition that the midpoint voltage is regulated to 1.0 per unit. We then obtain some very interesting plots as shown in Fig. 10.5. In this figure, the x-axis shows the reactive power available from the shunt device, while the y-axis shows the maximum power that can be transferred over the line without violating the voltage constraint. There are three different P-Q relationships given for three midpoint voltage constraints. For a reactive power injection of 0.5 per unit, the power transfer can be increased from about 0.97 per unit to 1.17 per unit by lowering the midpoint voltage to 0.9 per unit. For a reactive power injection greater than 2.0 per unit, the best power transfer capability is obtained for VM = 1.0 per unit. Thus there will be no benefit in reducing the voltage constraint when the shunt device is capable of injecting a large amount of reactive power. In practice, the level to which the midpoint voltage can be regulated depends on the rating of the installed shunt device as well the power being transferred.

Fig. 10.5 Power transfer versus shunt reactive injection under midpoint voltage constraint.

Improving Stability Margin

This is a consequence of the improvement in the power angle characteristics and is one of the major benefits of using midpoint shunt compensation. As mentioned before, the stability margin of the system pertains to the regions of acceleration and deceleration in the power-angle curve. We shall use this concept to delineate the advantage of mid point shunt compensation. Consider the power angle curves shown in Fig. 10.6.

The curve of Fig. 10.6 (a) is for an uncompensated system, while that of Fig. 10.6 (b) for the compensated system. Both these curves are drawn assuming that the base power is V2/X . Let us assume that the uncompensated system is operating on steady state delivering an electrical power equal to Pm with a load angle of δ0 when a three-phase fault occurs that forces the real power to zero. To obtain the critical clearing angle for the uncompensated system is δcr , we equate the accelerating area A1 with the decelerating area A2 , where

with δmax = π - δ0 . Equating the areas we obtain the value of δcr as Let us now consider that the midpoint shunt compensated system is working with the same mechanical power input Pm. The operating angle in this case is δ1 and the maximum power that can be transferred in this case is 2 per unit. Let the fault be cleared at the same clearing angle δcr as before. Then equating areas A3 and A4 in Fig. 10.6 (b) we get δ2, where

Example 10.1 Let an uncompensated SMIB power system is operating in steady state with a mechanical power input Pm equal to 0.5 per unit. Then δ0 = 30° = 0.5236 rad and δmax = 150° = 2.6180 rad. Consequently, the critical clearing angle is calculated as (see Chapter 9) δcr = 79.56° = 1.3886 rad. Let us now put an ideal shunt compensator at the midpoint. The pre-fault steady state operating angle of the compensated system can be obtained by solving 2 sin(δ/2) = 0.5, which gives δ1 = 28.96° = 0.5054 rad. Let us assume that we use the same critical clearing angle as obtained above for clearing a fault in the compensated system as well. The accelerating area is then given by A3 = 0.4416. Equating with area A4 we get a nonlinear equation of the form

Fig 10.6 Power-angel curve showing clearing angles: (a) for uncompensated sytem and (b) for

compensated system Solving the above equation we get δ2 = 104.34° = 1.856 rad. It is needless to say that the stability margin has increased significantly in the compensated system. Improving Damping to Power Oscillations The swing equation of a synchronous machine is given by (9.14). For any variation in the electrical quantities, the mechanical power input remains constant. Assuming that the magnitude of the midpoint voltage of the system is controllable by the shunt compensating device, the accelerating power in (9.14) becomes a function of two independent variables, | VM | and δ . Again since the mechanical power is constant, its perturbation with the independent variables is zero. We then get the following small perturbation expression of the swing equation

where Δ indicates a perturbation around the nominal values. If the mid point voltage is regulated at a constant magnitude, Δ| VM | will be equal to zero. Hence the above equation will reduce to The 2nd order differential equation given in (10.13) can be written in the Laplace domain by neglecting the initial conditions as The roots of the above equation are located on the imaginary axis of the s-plane at locations ± j ωm where

This implies that the load angle will oscillate with a constant frequency of ωm . Obviously, this solution is not acceptable. Thus in order to provide damping, the mid point voltage must be varied according to in sympathy with the rate of change in Δδ . We can then write

where KM is a proportional gain. Substituting (10.15) in (10.12) we get Provided that KM is positive definite, the introduction of the control action (10.15) ensures that the roots of the second order equation will have negative real parts. Therefore through the feedback, damping to power swings can be provided by placing the poles of the above equation to provide the necessary damping ratio and undamped natural frequency of oscillations. Example 10.2

Consider the SMIB power system shown in Fig. 10.7. The generator is connected to the infinite bus through a double circuit transmission line. At the midpoint bus of the lines, a shunt compensator is connected. The shunt compensator is realized by the voltage source VF that is connected to the midpoint bus through a pure inductor XF , also known as an interface inductor . The voltage source

VF is driven such that it is always in phase with the midpoint voltage VM . The current IQ is then purely inductive, its direction being dependent on the relative magnitudes of the two voltages. If the magnitude of the midpoint voltage is higher than the voltage source VF , inductive current will flow from the ac system to the voltage source. This implies that the source is absorbing var in this configuration. On the other hand, the source will generate var if its magnitude is higher than that of the midpoint voltage. The system is simulated in MATLAB. The three-phase transmission line equations are simulated using their differential equations, while the generator is represented by a pure voltage source. The second order swing equation is simulated in which the mechanical power input is chosen such that the initial operating angle of the generator voltage is (0.6981 rad). The instantaneous electrical power is computed from the dot product of the three-phase source current vector and source voltage vector. The system parameters chosen for simulation are:

(10.12)

(10.13)

(10.14

(10.15)

(10.16)

Fig. 10.7 SMIB system used in the numerical example.

Sending end voltage, VS = 1 < 40° per unit, Receiving end voltage, VR = 1 < 0° per unit, System Frequency ωs = 100 π rad/s, Line reactance, X=0.5 per unit, Interface reactance, XF per unit, Generator inertia constant, H=4.0 MJ/MVA. Two different tests are performed. In the first one, the midpoint voltage is regulated to 1 per unit using a proportional-plus-integral (PI) controller. The magnitude of the midpoint voltage is first calculated using the d-q transformation of the three phase quantities. The magnitude is then compared with the set reference (1.0) and the error is passed through the PI controller to determine the magnitude of the source voltage, ie,

The source voltage is then generated by phase locking it with the midpoint voltage using

Fig. 10.8 depicts the system quantities when the system is perturbed for its nominal operating condition. The proportional gain ( KP ) is chosen as 2.0, while the integral gain ( KI ) is chosen as 10. In Fig. 10.8 (a) the a-phase of the midpoint voltage, source voltage and the injected current are shown once the system transients die out. It can be seen that the source and midpoint voltages are phase aligned, while the injected current is lagging these two voltages by 90° . Furthermore, the midpoint voltage magnitude is tightly regulated. Fig. 10.8 (b) depicts the perturbation in the load angle and the injected reactive power. It can be seen that the load angle undergoes sustained oscillation and this oscillation is in phase with the injected reactive power. This implies that, by tightly regulating the midpoint voltage though a high gain integral controller, the injected reactive power oscillates in sympathy with the rotor angle. Therefore to damp out the rotor oscillation, a controller must be designed such that the injected reactive power is in phase opposition with the load angle. It is to be noted that the source voltage also modulates in sympathy with the injected reactive power. This however is not evident from Fig. 10.8 (a) as the time axis has been shortened here.

(10.17)

(a) (b)

Fig. 10.8 Sustained oscillation in rotor angle due to strong regulation of midpoint voltage.

To improve damping, we now introduce a term that is proportional to the deviation of machine speed in the feedback loop such that the control law is given by

The values of proportional gain KP and integral gain KI chosen are same as before, while the value of CP chosen is 50. With the system operating on steady state, delivering power at a load angle of 40° for 50 ms, breaker B (see Fig. 10.7) opens inadvertently. The magnitude of the midpoint voltage is shown in Fig. 10.9 (a). It can be seen that the magnitude settles to the desired value of 1.0 per unit once the initial transients die down. Fig. 10.9 (b) depicts perturbations in load angle and reactive power injected from their prefault steady state values. It can be seen that these two quantities have a phase difference of about 90 ° and this is essential for damping of power oscillations.

(a) (b)

Fig 10.9 System response with the damping controller

Ideal Series Compensator

Let us assume that the series compensator is represented by an ideal voltage source. This is shown in Fig. 10.10. Let us further assume that the series compensator is ideal, i.e., it only supplies reactive power and no real power to the system. It is needless to say that this assumption is not valid for practical systems. However, for an introduction, the assumption is more than adequate. It is to be noted that, unlike the shunt compensator, the location of the series compensator is not crucial, and it can be placed anywhere along the transmission line.

Fig. 10.10 Schematic diagram of an ideal series compensated system.

Impact of Series Compensator on Voltage Profile

In the equivalent schematic diagram of a series compensated power system is shown in Fig. 10.10, the receiving end current is equal to the sending end current, i.e., IS = IR . The series voltage VQ is injected in such a way that the magnitude of the injected voltage is made proportional to that of the line current. Furthermore, the phase of the voltage is forced to be in quadrature with the line current. We then have

The ratio λ/X is called the compensation level and is often expressed in percentage. This compensation level is

usually measured with respect to the transmission line reactance. For example, we shall refer the compensation level as 50% when λ = X /2. In the analysis presented below, we assume that the injected voltage lags the line current. The implication of the voltage leading the current will be discussed later.

Applying KVL we get

Assuming VS = V < δ and VR = V <0° , we get the following expression for the line current

When we choose VQ = λ IS e- j90° , the line current equation becomes

Thus we see that λ is subtracted from X . This choice of the sign corresponds to the voltage source acting as a pure capacitor. Hence we call this as the capacitive mode of operation . In contrast, if we choose VQ = λIS e+j90° , λ is added to X , and this mode is referred to as the inductive mode of operation . Since this voltage injection using (10.20) add λ to or subtract λ from the line reactance, we shall refer it as voltage injection in constant reactance mode. We shall consider the implication of series voltage injection on the transmission line voltage through the following example.

(10.20)

Example 10.3

Consider a lossless transmission line that has a 0.5 per unit line reactance (X). The sending end and receiving end voltages are given by 1< δ and 1< 0° per unit respectively where δ is chosen as 30° . Let us choose λ = 0.5 and operation in the capacitive mode. For this line, this implies a 30% level of line impedance compensation. The line current is then given from (10.21) as IS = 1.479 7 < 15° per unit and the injected voltage calculated from (10.20) is VQ = 0.2218 < - 75° per unit. The phasor diagrams of the two end voltages, line current and injected voltage are shown in Fig. 10.11 (a). We shall now consider a few different cases. Let us assume that the series compensator is placed in the middle of the transmission line. We then define two voltages, one at either side of the series compensator. These are: Voltage on the left: VQL = VS - jXIS / 2 = 0.9723 < 8.45° per unit Voltage on the right: VQR = VR + jXIS / 2 = 0.9723 < 21.55° per unit The difference of these two voltages is the injected voltage. This is shown in Fig. 10.11 (b), where

the angle θ = 8.45° . The worst case voltage along the line will then be at the two points on either side of the series compensator where the voltage phasors are aligned with the line current phasor. These two points are equidistant from the series compensator. However, their particular locations will be dependent on the system parameters. As a second case, let us consider that the series compensator is placed at the end of the transmission line, just before the infinite bus. We then have the following voltage Voltage on the left of the compensator: VQL = VR + VQ = 1.0789 < - 11.46° per unit This is shown in Fig. 10.11 (c). The maximum voltage rise in the line is then to the immediate left of the compensator, i.e., at VQL . The maximum voltage drop however still occurs at the point where the voltage phasor is aligned with the line current phasor.

As a third case, let us increase the level of compensation from 30% to 70% (i.e., change λ from 0.15 to 0.35). We however, do not want to change the level of steady state power transfer. The relation between power transfer and compensation level will be discussed in the next subsection. It will however suffice to say that this is accomplished by lowering the value of the angle δ of the sending end voltage to 12.37° . Let us further assume that the series compensator is placed in the middle of the transmission line. We then have VQL = 1.0255 < - 8.01° per unit and VQR = 1.0255 < 20.38° per unit. This is shown in Fig. 10.11 (d). It is obvious that the voltage along the line rises to a maximum level at either side of the series compensator.

Improving Power-Angle Characteristics

Noting that the sending end apparent power is VS IS* , we can write

Similarly the receiving end apparent power is given by

Hence the real power transmitted over the line is given by

The power-angle characteristics of a series compensated power system are given in Fig. 10.12. In this figure the base power is chosen as V2 / X . Three curves are shown, of which the curve P0 is the power-angle curve when the line is not compensated. Curves which have maximum powers greater than the base power pertain to capacitive mode of operation. On the other hand, all curves the inductive mode of operation will have maximum values less than 1. For example, in Fig. 10.12, the curve P1 is for capacitive mode and the curve P2 is for inductive mode of operation.

Fig. 10.12 Power-angle characteristics in constant reactance mode.

Let us now have a look at the reactive power. For simplicity let us restrict our attention to capacitive mode of operation only as this represents the normal mode of operation in which the power transfer over the line is enhanced. From (10.20) and (10.21) we get the reactive power supplied by the compensator as Solving the above equation we get

(10.25)

In Fig. 10.13, the reactive power injected by the series compensator is plotted against the maximum power transfer as the compensation level changes from 10% to 60%. As the compensation level increases, the maximum power transfer also increases. However, at the same time, the reactive injection requirement from the series compensator also increases. It is interesting to note that at 50% compensation level, the reactive power injection requirement from a series compensator is same that from shunt compensator that is regulating the midpoint voltage to 1.0 per unit.

Fig. 10.13 Reactive power injection by a series compensator versus maximum power transfer as the

level of compensation changes in constant reactance mode. An Alternate Method of Voltage Injection So far we have assumed that the series compensator injects a voltage that is in quadrature with the line current and its magnitude is proportional to the magnitude of the line current. A set of very interesting equations can be obtained if the last assumption about the magnitude is relaxed. The injected voltage is then given by We can then write the above equation as

i.e., the voltage source in quadrature with the current is represented as a pure reactance that is either inductive or capacitive. Since in this form we injected a constant voltage in quadrature with the line current, we shall refer this as constant voltage injection mode. The total equivalent inductance of

the line is then Defining VS = V < δ and VR < 0° , we can then write the power transfer equation as

Since | VQ | / | IS | = XQ , we can modify the above equation as

(10.26)

Consider the phasor diagram of Fig. 10.14 (a), which is for capacitive operation of the series

compensator. From this diagram we get

Similarly from the inductive operation phasor diagram shown in Fig. 10.14 (b), we get

Substituting the above two equations in (10.28) and rearranging we get

where the positive sign is for capacitive operation.

Fig. 10.14 Phasor diagram of series compensated system: (a) capacitive operation and (b) inductive

operation.

The power-angle characteristics of this particular series connection are given in Fig. 10.15. In this figure the base power is chosen as V2/X . Three curves are shown, of which the curve P0 is the power-angle curve when the line is not compensated. Curves which have maximum powers greater than the base power pertain to capacitive mode of operation. On the other hand, all curves the inductive mode of operation will have maximum values less than 1. For example, in Fig. 10.15, the curve P1 is for capacitive mode and the curve P2 is for inductive mode of operation.

Fig. 10.15 Power-angle characteristics for constant voltage mode.

The reactive power supplied by the compensator in this case will be

(10.30)

Improving Stability Margin

From the power-angle curves of Figs. 10.13 and 10.15 it can be seen that the same amount of power can be transmitted over a capacitive compensated line at a lower load angle than an uncompensated system. Furthermore, an increase in the height in the power-angle curve means that a larger amount of decelerating area is available for a compensated system. Thus improvement in stability margin for a capacitive series compensated system over an uncompensated system is obvious.

Comparisons of the Two Modes of Operation

As a comparison between the two different modes of voltage injection, let us first consider the constant reactance mode of voltage injection with a compensation level of 50%. Choosing V2 / X as the base power, the power-angle characteristic reaches a maximum of 2.0 per unit at a load angle π / 2. Now | VQ | in constant voltage mode is chosen such that the real power is 2.0 per unit at a load angle of π / 2. This is accomplished using (10.29) where we get

per unit

The power-angle characteristics of the two different modes are now drawn in Fig. 10.16 (a). It can be seen that the two curves match at π / 2. However, the maximum power for constant voltage case is about 2.1 per unit and occurs at an angle of 67° .

Fig. 10.16 (b) depicts the line current for the two cases. It can be seen that the increase in line current in either case is monotonic. This is not surprising for the case of constant reactance mode since as the load angle increases, both real power and line currents increase. Now consider the case of constant voltage control. When the load angle moves backwards from π /2 to 67° , the power moves from 2.0 per unit to its peak value of 2.1 per unit. The line current during this stage decreases from about 2.83 to 2.50 per unit. Thus, even though the power through the line increases, the line current decreases.

Power Flow Control and Power Swing Damping

One of the major advantages of series compensation is that through its use real power flow over transmission corridors can be effectively controlled. Consider, for example, the SMIB system shown in Fig. 10.17 in which the generator and infinite bus are connected through a double circuit transmission line, labeled line-1 and line-2. Of the two transmission lines, line-2 is compensated by a series compensator. The compensator then can be utilized to regulate power flow over the entire system.

(a) (b)

Fig. 10.16 Power-angle and line current-angle characteristics of the two different methods of voltage injection: solid line showing constant reactance mode and dashed line showing constant voltage

mode.

For example, let us consider that the system is operating in the steady state delivering a power of Pm0 at a load angle of δ0 . Lines 1 and 2 are then sending power Pe1 and Pe2 respectively, such that Pm0 = Pe1 + Pe2 . The mechanical power input suddenly goes up to Pm1 . There are two ways of controlling the power in this situation:

Regulating Control: Channeling the increase in power through line-1. In this case the series compensator maintains the power flow over line-2 at Pe2 . The load angle in this case goes up in sympathy with the increase in Pe1 .

Tracking Control: Channeling the increase in power through line-2. In this case the series

compensator helps in maintaining the power flow over line-1 at Pe1 while holding the load angle to δ0 .

Let us illustrate these two aspects with the help of a numerical example.

Example 10.4

Let us consider the system of Fig. 7.8 where the system parameters are given by

System Frequency = 50 Hz, | V S | = | V R | = 1.0 per unit, X = 0.5 per unit and d 0 = 30 ° /

It is assumed that the series compensator operates in constant reactance mode with a compensation level of 30%. We then have

Pe1 = 1.0 per unit, Pe2 = 1.43 per unit, Pm = 2.43 per unit

The objective of the control scheme here is to maintain the power through line-2 to a pre-specified value, Pref . To accomplish this a proportional-plus-integral (PI) controller is placed in the feedback loop of Pe2 . In addition, to improve damping a term that is proportional to the deviation of machine speed is introduced in the feedback loop. The control law is then given by

where CL = λ / X is the compensation level. For the simulation studies performed, the following controller parameters are chosen

KP = 0.1, KI = 1.0 and CP = 75

Regulating Control: With the system operating in the nominal steady state, the mechanical power input is suddenly raised by 10%. It is expected that the series compensator will hold the power through line-2 constant at line-2 at Pe2 such that entire power increase is channeled through line-1. We then expect that the power Pe1 will increase to 1.243 per unit and the load angle to go up to 0.67 rad. The compensation level will then change to 13%.

The time responses for various quantities for this test are given in Fig. 10.18. In Fig. 10.18 (a), the power through the two line is plotted. It can be seen that while the power through line-2 comes back to its nominal value following the transient, the power through the other line is raised to expected level. Similarly, the load angle and the compensation level reach their expected values, as shown in Figs. 10.18 (b) and (c), respectively. Finally, Fig. 10.18 (d) depicts the last two cycles of phase-a of the line current and injected voltage. It can be clearly seen that these two quantities are in quadrature, with the line current leading the injected voltage.

(10.31)

Fig. 10.18 System response with regulating power flow controller.

Tracking Control: With the system operating in the nominal steady state, the mechanical power input is suddenly raised by 25%. It is expected that the series compensator will make the entire power increase to flow through line-2 such that both Pe1 and load angle are maintained constant at their nominal values. The power Pe2 through line-2 will then increase to about 2.04 per unit and the compensation level will change to 51%.

The time responses for various quantities for this test are given in Fig. 10.19. It can be seen that while the power through line-1 comes back to its nominal value following the transient, the power through the other line is raised to level expected. Similarly, the load angle comes back to its nominal value and the compensation level is raised 51%, as shown in Figs. 10.19 (b) and (c), respectively. Finally, Fig. 7.19 (d) depicts the last two cycles of phase-a of the line current and injected voltage.

Fig. 10.19 System response with regulating power flow controller.