Computer Aided Power System Analysis

Module 1 Introduction Electricity is the most preferred used form of energy used in industry, homes, businesses and transportation. It can be easily and efficiently transported from the production centers to the point of use. It is highly flexible in use as it can be converted to any desired form like mechanical, thermal, light, chemical etc. An electrical power system is made up of many components connected together to form a large, complex system that is capable of generating, transmitting and distributing electrical energy over large areas.

1.1

The structure of electrical power system

The basic structure of modern day power system is shown in Fig. 1.1. A power system is usually divided into three parts: generation, transmission and distribution system.

1.1.1

Generation

Electricity is produced by converting the mechanical energy into electrical energy. In majority of cases, the mechanical energy is either obtained from thermal energy or provided by the flowing water. The main sources of thermal energy sources are coal, natural gas, nuclear fuel and oil. The use of non-fossil fuels such as wind, solar, tidal, and geothermal and biogas in electricity generation is also increasing. Hydro-power is the main non-thermal source of mechanical energy used in electricity generation. The conversion of mechanical to electrical energy is done using synchronous generators in majority of power plants. Few wind generation systems use induction generators. The power is usually generated at low voltage, between 11 and 35 KV, and then fed into the transmission system using a step up transformer.

1.1.2

Transmission system

The electricity is generated in bulk in the generating stations and then transmitted over long distances to the load points. The transmission system interconnects all the generating stations and major load centers in the system. It forms the back bone of the power system. Since the power loss in a transmission line is proportional to the square of line current, the transmission lines operate at the 1

Figure 1.1: Structure of a Power System highest voltage levels, usually 220 KV and upwards. Usually the transmission network has a meshed structure in order to provide many alternate routes for the power to flow from the generators to the load points. This improves the reliability of the system. High voltage transmission lines are terminated at substations. Very large industrial customers may be provided power directly from these substations. At these substations, the voltage is stepped down to a lower level and fed into the sub-transmission system. This part of the transmission system connects the high voltage substation through step down transformers to distribution substation. Typically the sub-transmission voltage levels are from 66 KV to 132 KV. Some large industrial consumers may be served directly from the sub-transmission system. The transmission lines connect the neighboring power systems at transmission levels, thus forming 2

a grid. The grid is the network of multiple generating resources and several layers of transmission network. The interconnections of power systems offer the following advantages. (a) Quality: The voltage profile of the transmission network improves as more generators contribute to the system, resulting in an increased total system capability. This also improves the frequency behavior of the system following any load perturbation due to increased inertia of the system. (b) Economy: In interconnected systems, it is possible to reduce the total set of generating plants required to maintain the desired level of generation reserve. This results in reduction of operational and investment costs. Also, operational (including plant start-ups and shut down) and generation scheduling of units can be more economically coordinated. (c) Security: In case of emergency, power can be made available from the neighboring systems and each system can benefit even when individual spinning reserves may not be sufficient for isolated operation.

1.1.3

Distribution system

The distribution represents the final stage of power transfer to the individual consumer. The distribution network is generally connected in a radial structure. The primary distribution voltage is typically between 11 KV and 33 KV. Small industrial customers are supplied by primary feeders at this voltage level. The secondary distribution feeders supply residential and commercial at 415/240 V. Small generating plants located near the load centers are usually connected to sub-transmission or distribution system directly. A power system operates in a normal state, if the following conditions are satisfied: • The bus voltages are within the prescribed limits. • The system frequency is within the specified limits. • The active and reactive power balance exists in the system. However, the system load varies continuously and hence, in order to ensure satisfactory system operation, proper controls have to be provided in a power system.

1.1.4

Power system control structure

The various elements of power system operation and control are shown in Table 1.1 along with the time-scale of operation.

1.2

Power system control

A properly designed and operated power system must meet the following requirements: 3

Table 1.1: Various elements of power system operation and control

1 2 3 4 5 6 7

Operation and control action Relaying execution control, system voltage control System frequency control tie-line power control Economic dispatch System security analysis Unit commitment Maintenance scheduling System planning

Time period Multi seconds Few seconds to few minutes Few minutes to few hours Few minutes to few hours Few hours to few weeks One month to one year One year to 10 years

(a) The system must have adequate capability to meet the continuously varying active and reactive power demand of system load. This requires maintaining and approximately controlling adequate spinning reserve of active and reactive power at all time instants. (b) The system should be designed and operated so as to supply electrical energy at minimum cost and with minimum adverse ecological impact. (c) The electrical power supplied to the consumers must meet certain minimum quality standards with respect to the following: i) The network frequency should be maintained within a range of ±3 percent of its ‘nominal’ value. ii) The voltage magnitudes should be maintained within a range of ±10 percent of the corresponding ‘nominal’ value at each network bus bar. iii) The supply should meet a desired level of reliability to ensure supply continuity as far as possible. (d) It should maintain scheduled tie-line flow and contractual power exchange. To meet the requirements at points i), ii) and iii) above, several levels of controls incorporating a large number of devices are needed. These controls are as shown in Fig. 1.2. A brief explanation of various power system controllers is given next.

1.2.1

Generating unit controls

The controls provided in generating units consist of prime mover control and excitation controls as shown in Fig. 1.3. The controls are also called as local frequency control (LFC) and automatic voltage control (AVC). These controllers are set for a particular operating condition and maintain the frequency and voltage magnitude within the specified limits following small changes in load demand. If the input to the prime mover is constant, then an increase in the active power of load at the generator terminals results in a drop in the prime mover speed. This then, causes a reduction in the frequency. On 4

Figure 1.2: Controls in a Power System the other hand, an increase in reactive power demand at the generator results in the reduction of terminal voltage, if the excitation (generation field current) is kept constant. As the time constant of excitation system is much smaller than that of prime mover system, the coupling between LFC and AVC loop is negligible and hence they are considered independently.

1.2.2

Load frequency control

In LFC, two feedback loops namely, primary and secondary loops are provided. Both the loops help in maintaining the real power balance by adjusting the turbine input power. The primary LFC loop senses the generator speed and accordingly controls the turbine input. This is a faster loop 5

Figure 1.3: Generator controls and operates in the order of seconds. But this loop provides only a coarse frequency control. The secondary LFC loop which senses the system frequency and tie-time power, fine tunes the frequency back to the nominal value. This is a slower loop and may take minutes to eliminate frequency error.

1.2.3

Automatic voltage control

In AVC, the bus voltage is measured and compared to a reference. The resulting error voltage is then amplified and applied to the excitation control system. The output of the exciter controls the generator field current. An increase in the reactive power load of the generator causes the terminal voltage to decrease and this results in generation of voltage error signal. The amplified error signal then increases the exciter field current which in turn increases the exciter terminal voltage. This increases the generator field current, which results in an increase in the generated emf. The reactive power generation of the generator is thus increased and the terminal voltage is brought back to its nominal value. The generation control maintains the active power balance in the system. It also controls the division of load active power between the generators in the system to ensure economic operation.

1.2.4

Economic dispatch

Economic operation and planning of electric energy generating system has been accorded due importance by the power system operators. Power systems need to be operated economically to make 6

electrical energy cost-effective to the consumer and profitable for the operator. The operational economics that deals with power generation and delivery can be divided into two sub-problems. One dealing with minimum cost of power generation and other dealing with delivery of power with minimum power loss. The problem of minimum production cost is solved using economic dispatch. The main aim of economic dispatch problem is to minimize the total cost of generating real power at different plants in the system while maintaining the real power balance in the system. For system having hydro-plants, a coordinated dispatch of hydro-thermal units is carried out. The economic dispatch and minimum loss problems can be solved by means of optimal power flow (OPF) method. The OPF calculations involve a sequence of load flow solutions in which certain controllable parameters are automatically adjusted to satisfy the network constraints while minimizing a specified objective function. The power system control objectives are dependent on the operating state of the system. Under normal operating conditions, the controller tries to operate the system as economically as possible with voltages and frequency maintained close to nominal values. But abnormal conditions like outage of a larger generator, of a major transmission line or sudden increase or reduction of system load can cause havoc in the system, if not properly controlled. Different operating objectives have to be met in order to restore the system to normal operation after the occurrence of such contingencies.

1.2.5

Security analysis and contingency evaluation

For the analysis of power system security and development of approximate control systems, the system operating conditions are classified into five states: normal, alert, emergency, in extremis and restorative. The state and the transitions between them are shown in Fig. 1.4.

Figure 1.4: Power system state transition diagram Normal state: In this state, all the system variables are within the normal range with no equipment being overloaded. The system is in a secure state with both ‘equality’ (total system 7

generation eqauls total system load) and ‘inequality’(bus voltages and equipment currents within the limits) constraints being satisfied. In this state, a single contingency cannot disrupt the system security and cannot cause any variable to violate the limit. The system has adequate spinning reserve. Alert state: If the security level of the system falls below some specified threshold, the system then enters the alert state and is termed as ‘insecure’. The system variables are still within limits. This state may be brought about by a single contingency, large increase in system load or adverse weather conditions. Preventive control steps taken to restore generation or to eliminate disturbace can help in restoring the system to the normal state. If these restorative steps do not succeed, the system remains in the alert state. Occurence of a contingency with the system already in alert state, may cause overloading of equipments and the system may enter emergency state. If the disturbance is very severe, the system may enter into extremis state directly from alert state. Emergency state: If the preventive controls fail or if a severe disturbance occurs, the system enters emergency state. The transition to this state can occur either from normal state or alert state. In this state the balance between generation and load is still maintained (equality constraints still satisfied) and the system remains in synchronism. Some components are however overloaded(some inequality constrints violated). Failure of these components results in system disintegration. Emergency control actions like disconnection of faulted section, re-routing of power excitation control, fast valving, and load curtailment have to be taken. It is most urgent that the system be restored to normal or alert state by means of these actions. In-extremis state: If the emergency control actions fail when the system is in emergency state, then the system enters into in-extremis state. The system starts to disintegrate into sections or islands. Some of these islands may still have sufficient generation to meet the load. The components are overloaded and the active power balance is also disrupted. Overloaded generators start tripping leading to cascade outages and possible ‘blackout’. Control actions, such as load shedding and controlled system operation are taken to save as much of the system as possible from a widespread blackout. Restorative state: The restorative state represents a condition in which control action is being taken to restart the tripped generators and restore the interconnections. The system transition can be either to normal or alert state depending on system conditions. The sequence of events that result in system transition from normal to in-extremis state may take from few seconds to several minutes. Bringing the system back to normal through the restorative state is an extremely time consuming process and may last for hours or may be days. A large generator may require many hours from restart to synchronization. The switched off loads can be picked up gradually and resynchronization of operating islands to the grids is also a time consuming process. The control actions may be initiated from the central energy control centre either through operators or automatically.

1.2.6

Unit Commitment 8

The total load in the power system varies throughout a day and its value also changes with the day of the week and season. Hence, it is not economical to run all the units available all the time. Thus, the problem of unit commitment is to determine in advance, the start and the shut down sequence of the available generators such that the load demand is met and the cost of generation is minimum.

1.2.7

Maintenance scheduling

Preventive maintenance has to be carried out on power system components to ensure that they continue to operate efficiently and reliably. Generators are usually put on maintenance once every year. Their maintenance has to be so scheduled such that the available generation is sufficient to meet the system load demand. The problem of maintenance scheduling deals with the sequencing of generator maintenance such that sufficient generation is always available to meet the load demand and the cost of maintenances and cost of lost generation is minimum.

1.3

System planning

To meet ever increasing load demand, either new power systems have to be built or the existing power systems are expanded by adding new generators and transmission lines. Many analyses must be performed to design and study the performance of the system and plan expansion. To study the system feasibility and performance, the following analyses need to be carried out: (a) Load flow analysis (b) Fault analysis/short circuit studies (c) Stability studies (d) Contingency analysis

1.3.1

Load flow analysis

The load flow analysis involves the steady state solution of the power system network to determine power flows and bus voltages of a transmission network for specified generation and loading conditions. These calculations are required for the study of steady state and dynamic performance of the system. The system is assumed to be balanced and hence, single phase representation is used. These studies are important in planning and designing future expansion of power system and also in determining the best operation of the existing systems.

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1.3.2

Fault studies

In these studies the line currents and bus voltages of a system are calculated during various types of faults. Faults on power system are divided into balanced and unbalanced faults. Three phase symmetrical faults are balanced faults in which the system retains its balanced nature. The unbalanced faults are single line to ground fault, line to line fault and double line to ground fault. The fault currents values are useful in relay setting and co-ordination as well as for selecting the proper rating of the circuit breakers.

1.3.3

Stability studies

The stability studies ascertain the impact of disturbances on the electrochemical dynamic behavior of the power system. These studies are of two types; small signal stability study and transient stability study. The small signal stability studies deal with the behavior of a system following any small disturbances like small change in load, small change in AVR gain etc. As the disturbance is small, the equations that describe the dynamics of the power system are linearized for the purpose of analysis. The system is small signal stable for a particular operating point, if following a small disturbance it returns to essentially the same steady state operating condition. Transient stability study deals with the response of a power system subjected to a large disturbance such a short circuit, line tripping or loss of large genration. In this study the equations describing system dynamics are solved using numerical techniques. The power system transient stability problem is then defined as that of assessing whether or not the system will reach an acceptable steady state operating point following a large disturbance. We will now start with the study of load flow analysis technique from the next lecture.

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Module 2 Load Flow Analysis AC power flow analysis is basically a steady-state analysis of the AC transmission and distribution grid. Essentially, AC power flow method computes the steady state values of bus voltages and line power flows from the knowledge of electric loads and generations at different buses of the system under study. In this module, we will look into the power flow solution of the AC transmission grid only (the solution methodology of AC distribution grid will not be covered). Further, we will also study the power flow solution technique when an HVDC link is embedded into an AC transmission grid. Also, we will be considering only a balanced system in which the transmission lines and loads are balanced (the impedances are equal in all the three phases) and the generator produces balanced three phase voltages (magnitudes are equal in all the phases while the angular difference between any two phases is 120 degree).

2.1

Modeling of power system components

Basically, an AC transmission grid consists of, i) synchronous generator, ii) loads, iii) transformer and iv) transmission lines. For the purpose of power flow solution, synchronous generators are not represented explicitly, rather their presence in implicitly modeled. We will look into the implicit representation of synchronous generators a little later. However, the other three components are modeled explicitly and their representations are discussed below.

2.1.1

Loads

As we all know, loads can be classified into three categories; i) constant power, ii) constant impedance and iii) constant current. However, within the normal operating range of the voltage almost all the loads behave as constant power loads. As the objective of the AC power flow analysis is to compute the normal steady-state values of the bus voltages, the loads are always represented as constant power loads. Hence, at any bus ‘k’ (say), the real and reactive power loads are specified as 100 MW and 50 MVAR (say) respectively. An important point needs to be mentioned here. As the loads are always varying with time (the customers are always switching ‘ON’ and ‘OFF’ the loads), any specific value of load (MW and/or MVAR) is valid only at a particular time instant. Hence, AC 11

power flow analysis is always carried out for the load and generator values at a particular instant.

2.1.2

Transmission line

In a transmission grid, the transmission lines are generally of medium length or of long length. A line of medium length is always represented by the nominal-π model as shown in Fig. 2.1, where z¯ is the total series impedance of the line and Bc is the total shunt charging susceptance of the line. On the other hand, a long transmission line is most accurately represented by its distributed parameter model. However, for steady-state analysis, a long line can be accurately represented by the equivalent-π model, which predicts accurate behavior of the line with respect to its terminal measurements taken at its two ends. The equivalent-π model is shown in Fig. 2.2.

Figure 2.1: Normal π model of a line connected between buses ‘i’ and ‘j’

Figure 2.2: Equivalent π model of a long transmission line connected between buses ‘i’ and ‘j’ In Fig. 2.2, 12

z¯ is the characteristic impedance of the line y ¯ √ γ = z¯y¯ is the propagation constant z¯ = series impedance of the line per unit length y¯ = shunt admittance of the line per unit length L = length of the line z¯c =

Hence, for power system analysis, a transmission line (medium or long) is always represented by a π circuit.

2.1.3

Transformer

For power system steady-state and fault studies, generally the exciting current of the transformer is neglected as it is quite low compared to the normal load current flowing through the transformer. Therefore, a two winding transformer connected between buses ‘i’ and ‘j’ is represented by its per unit leakage impedance as shown in Fig. 2.3.

Figure 2.3: Equivalent Equivalent circuit of a two winding transformer It is to be noted that in Fig. 2.3, the transformer tap ratio is 1:1. For a regulating transformer with transformation ratio 1:t, the equivalent circuit of the transformer is shown in Fig. 2.4. Sometimes the transformer ratio is also represented as a:1. In that case, the equivalent circuit is as shown in Fig. 2.5. Please note that in Figs. 2.4 and 2.5, the quantities ‘t’ and ‘a’ are real (i.e. the transformer is changing only the voltage magnitude, not its angle). Further, in these two figures, the quantity y¯ is the per unit admittance of the transformer. Also, Fig. 2.5 can be derived from Fig. 2.4 by noting t = 1/a and by interchanging the buses ‘i’ and ‘j’. With the models of above components in place, we are now in a position to start systematic study of an ‘n’ bus power system. Towards that goal, we first must understand the concept of injected power and injected current, which is our next topic.

2.2

Concept of injected power and current

As the name suggests, the injected power (current) indicates the power (current) which is fed ‘in’ to a bus. To understand this concept, let us consider Fig. 2.6. In part (a) of this figure, a generator is connected at bus ‘k’ supplying both real and reactive power to the bus and thus, the injected real and reactive power are taken to be equal to the real (reactive) power supplied by the generator. The 13

Figure 2.4: Equivalent circuit of a regulating transformer with transformation ratio 1:t

Figure 2.5: Equivalent circuit of a regulating transformer with transformation ratio a:1 corresponding injected current is also taken to be equal to the current supplied by the generator. On the other hand, for a load connected to bus ‘k’ (as shown in Fig. 2.6(b)), physically the real (reactive) power consumed by the load flows away from the bus and thus, the injected real (reactive) power is taken to be the negative of the real (reactive) power consumed by the load. Similarly, the corresponding injected current I¯k is also taken as the negative of the load current. If both a generator 14

and a load are connected at a particular bus (as depicted in Fig. 2.6(c)), then the net injected real (reactive) power supplied to the bus is equal to the generator real (reactive) power minus the real (reactive) power consumed by the load. Similarly, the net injected current in this case is taken to be the difference of the generator current and the load current.

Figure 2.6: Illustration of injected power To summarize, if Pk , Qk , and I¯k denote the injected real power, reactive power and complex current at bus ‘k’ respectively, • Pk = PG ; Qk = QG and I¯k = I¯G if only a generator is connected to the bus ‘k’. • Pk = −PL ; Qk = −QL and I¯k = −I¯L if only a load is connected to the bus ‘k’. • Pk = PG − PL ; Qk = QG − QL and I¯k = I¯G − I¯L if both generator and load are connected to the bus ‘k’. • Pk = 0 ; Qk = 0 ; I¯k = 0 If neither generator nor load is connected to the bus ‘k’. With this concept of injected power and current, we are now in a position to start analysis of any general ‘n’ bus power system. The first step towards this goal is to derive the bus admittance matrix, which we will take up next.

2.3

¯ BUS) Formation of bus admittance matrix (Y

Let us consider a 5 bus network as shown in Fig. 2.7. In this network, all the transmissions are represented by π models. Therefore, the equivalent circuit of the above network is shown in Fig. 2.8. In Fig. 2.8, I¯k ; k = 1, 2, 3, 4, 5 are the injected currents at bus ‘k’. Further, the quantity y¯ij denotes the series admittance of the line ‘i-j’ whereas the quantity y¯ijs denotes the half line charging susceptance of the line ‘i-j’. Now applying ‘KCL’ at each bus ‘k’ one obtains,

I¯1 = y¯T 1 (V¯1 − V¯2 ) = y¯T 1 V¯1 − y¯T 1 V¯2 15

(2.1)

Figure 2.7: A sample 5 bus network

Figure 2.8: Equivalent circuit of Fig. 2.7

I¯2 = y¯T 1 (V¯2 − V¯1 ) + V¯2 y¯23s + (V¯2 − V¯3 )¯ y23 + V¯2 y¯24s + (V¯2 − V¯4 )¯ y24 = −¯ yT 1 V¯1 + (¯ yT 1 + y¯23s + y¯23 + y¯24s + y¯24 )V¯2 − y¯23 V¯3 − y¯24 V¯4 16

(2.2)

I¯3 = (V¯3 − V¯2 )¯ y23 + V¯3 y¯23s + (V¯3 − V¯5 )t¯ yT 2 + t(t − 1)¯ yT 2 V¯3 + (V¯3 − V¯4 )¯ y34 + V¯3 y¯34s = −V¯2 y¯23 + {¯ y23 + y¯23s + t¯ yT 2 + t(t − 1)¯ yT 2 + y¯34 + y¯34s } V¯3 − y¯34 V¯4 − t¯ yT 2 V¯5 (2.3)

I¯4 = (V¯4 − V¯2 )¯ y24 + y¯24s V¯4 + (V¯4 − V¯3 )¯ y34 + y¯34s V¯4 = −V¯2 y¯24 − V¯3 y¯34 + (¯ y24 + y¯24s + y¯34 + y¯34s )V¯4

(2.4)

I¯5 = (V¯5 − V¯3 )t¯ yT 2 + (1 − t)¯ yT 2 V¯5 = −V¯3 t¯ yT 2 + {t¯ yT 2 + (1 − t)¯ yT 2 } V¯5

(2.5)

Equations (2.1) - (2.5) can be represented in a matrix form as,

⎡I¯1 ⎤ ⎡Y¯11 ⎢ ⎥ ⎢ ⎢¯ ⎥ ⎢¯ ⎢I2 ⎥ ⎢Y21 ⎢ ⎥ ⎢ ⎢¯ ⎥ ⎢¯ ⎢I3 ⎥ = ⎢Y31 ⎢ ⎥ ⎢ ⎢I¯ ⎥ ⎢Y¯ ⎢ 4 ⎥ ⎢ 41 ⎢¯ ⎥ ⎢¯ ⎢I5 ⎥ ⎢Y51 ⎣ ⎦ ⎣

Y¯12 Y¯22 Y¯32 Y¯42 Y¯52

Y¯13 Y¯23 Y¯33 Y¯43 Y¯53

Y¯14 Y¯24 Y¯34 Y¯44 Y¯54

Y¯15 ⎤⎥ ⎡⎢V¯1 ⎤⎥ ⎥⎢ ⎥ Y¯25 ⎥⎥ ⎢⎢V¯2 ⎥⎥ ⎥⎢ ⎥ Y¯35 ⎥⎥ ⎢⎢V¯3 ⎥⎥ Y¯45 ⎥⎥ ⎢⎢V¯4 ⎥⎥ ⎥⎢ ⎥ Y¯55 ⎥⎦ ⎢⎣V¯5 ⎥⎦

(2.6)

Where, Y¯11 = y¯T 1 ;

Y¯12 = −¯ yT 1 ; Y¯13 = Y¯14 = Y¯15 = 0; Y¯21 = −¯ yT 1 ; ¯ ¯ ¯ Y22 = (¯ yT 1 + y¯23s + y¯23 + y¯24s + y¯24 ); Y23 = −¯ y23 ; Y24 = −¯ y24 ; Y¯25 = 0; Y¯31 = 0; Y¯32 = −¯ y23 ; Y¯33 = {¯ y23 + y¯23s + t¯ yT 2 + t(t − 1)¯ yT 2 + y¯34 + y¯34s } ; ¯ ¯ ¯ ¯ ¯ Y34 = −¯ y34 ; Y35 = −t¯ yT 2 ; Y41 = 0; Y42 = −¯ y24 ; Y43 = −¯ y34 ; Y¯44 = (¯ y24 + y¯24s + y¯34 + y¯34s ); Y¯45 = 0; Y¯51 = Y¯52 = 0; Y¯53 = −t¯ yT 2 ; Y¯54 = 0; Y¯55 = {t¯ yT 2 + (1 − t)¯ yT 2 } Equation (2.6) can be written as,

¯IBUS = Y ¯ BUS V ¯ BUS

(2.7)

Where,

¯IBUS = [I¯1 , I¯2 ⋯ I¯5 ]T → (5 × 1) is the vector of bus injection currents ¯ BUS = [V¯1 , V¯2 ⋯ V¯5 ]T → (5 × 1) is the vector of bus voltages measured with respect to the V ground

¯ BUS → (5 × 5) is the bus admittance matrix Y ¯ BUS it can be observed that for i = 1, 2, ⋯⋯ 5; Furthermore, from the elements of the Y Y¯ii = sum total of all the admittances connected at bus ‘i’ Y¯ij = negative of the admittance connected between bus ‘i’ and ‘j’ (if these two buses are physically connected with each other) Y¯ij = 0; if there is no physical connection between buses ‘i’ and ‘j’

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Similarly, for a ‘n’ bus power system, the relation given in equation (2.7) holds good, where, ¯IBUS = [I¯1 , I¯2 ⋯ I¯n ]T → (n × 1) is the vector of bus injection currents

¯ BUS = [V¯1 , V¯2 ⋯ V¯n ]T → (n × 1) is the vector of bus voltages V ¯ BUS → (n × n) is the bus admittance matrix Y ¯ BUS matrix are calculated in the same way as described above. Furthermore, the elements of the Y ¯ BUS matrix when there is no mutual coupling So far, we have considered the formation of the Y among the elements of the network. In the next lecture, we will look into the procedure for forming ¯ BUS matrix in the presence of mutual coupling between the elements. the Y

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2.4

¯ BUS matrix in the presence of mutually Formation of Y coupled elements

¯c connected between nodes ‘u’ and ‘v’ is Let us consider Fig. 2.9. In this figure, the impedance Z ¯d connected between nodes ‘x’ and ‘y’ through a mutual mutually coupled with the impedance Z ¯m . The currents through the impedances, the voltages across the impedances and the impedance Z injected currents at all the four nodes are also shown in Fig. 2.9.

Figure 2.9: Two mutually coupled impedances From Fig. 2.9, the relationship between the voltages and currents associated with the two impedances can be written as,

Z¯c Z¯m I¯c V¯c ][ ] [¯ ] = [ ¯ Zm Z¯d I¯d Vd Or,

−1 V¯c Z¯c Z¯m I¯c V¯c Z¯d −Z¯m 1 ] = ] [ [¯ ] = [ ¯ ] ] [ [ 2 V¯d Zm Z¯d Id −Z¯m Z¯c V¯d Z¯c Z¯d − Z¯m

Or,

I¯c Y¯c Y¯m V¯c ][ ] [¯ ] = [¯ Ym Y¯d V¯d Id

(2.8)

Where,

Y¯c =

Z¯d ; 2 Z¯c Z¯d − Z¯m

Y¯d =

Z¯c ; 2 Z¯c Z¯d − Z¯m

and Y¯m = −

Z¯m 2 Z¯c Z¯d − Z¯m

⎡¯ ⎤ ⎢Vu ⎥ ⎢ ⎥ ¯ ¯ ¯ Vc Vu − Vv 1 −1 0 0 ⎢⎢V¯v ⎥⎥ Now from Fig. 2.9, [ ] = [ ]=[ ]⎢ ⎥ V¯d V¯x − V¯y 0 0 1 −1 ⎢⎢V¯x ⎥⎥ ⎢¯ ⎥ ⎢ Vy ⎥ ⎣ ⎦ Or, ⎡¯ ⎤ ⎢Vu ⎥ ⎢ ⎥ ⎢V¯ ⎥ V¯c 1 −1 0 0 ⎢ v⎥ [ ¯ ] = [C] ⎢ ¯ ⎥ where, C = [ ] ⎢ ⎥ Vd 0 0 1 −1 ⎢ Vx ⎥ ⎢¯ ⎥ ⎢ Vy ⎥ ⎣ ⎦ 19

(2.9)

(2.10)

Again, from Fig. 2.9,

⎡¯ ⎤ ⎡ ¯ ⎤ ⎡ ⎢Iu ⎥ ⎢ Ic ⎥ ⎢ ⎥ ⎢ ⎢ ⎥ ⎢ ⎢I¯ ⎥ ⎢ −I¯ ⎥ ⎢ ⎢ v⎥ ⎢ c ⎥ ⎢ ⎢¯ ⎥ = ⎢ ¯ ⎥ = ⎢ ⎢Ix ⎥ ⎢ Id ⎥ ⎢ ⎥ ⎢ ⎢ ⎥ ⎢ ⎢¯ ⎥ ⎢ ¯ ⎥ ⎢ ⎢Iy ⎥ ⎢ −Id ⎥ ⎢ ⎦ ⎣ ⎣ ⎦ ⎣

⎤ 1 0⎥ ⎥ ¯c T I −1 0 ⎥⎥ I¯c ⎥ [ ¯ ] = [C] [ ¯ ] Id 0 1 ⎥⎥ Id ⎥ 0 −1 ⎥⎦

(2.11)

From equations (2.8) and (2.10),

I¯c Y¯c Y¯m V¯c Y¯c [¯ ] = [¯ ] [ ] = [ Id Ym Y¯d V¯d Y¯m

Or,

⎡¯ ⎤ ⎢Vu ⎥ ⎢ ⎥ ⎢V¯ ⎥ Y¯m ⎢ v⎥ [C] ⎢¯ ⎥ ] ⎢ Vx ⎥ Y¯d ⎢ ⎥ ⎢¯ ⎥ ⎢ Vy ⎥ ⎣ ⎦

⎡¯ ⎤ ⎡¯ ⎤ ⎢Vu ⎥ ⎢Iu ⎥ ⎢ ⎥ ⎢ ⎥ ⎢V¯ ⎥ ⎢I¯ ⎥ ¯c T I Y¯c Y¯m T ⎢ v⎥ ⎢ v⎥ [C] [C] [ ¯ ] = ⎢ ¯ ⎥ = [[C] ] [ ¯ ⎢¯ ⎥ ] ⎢ Vx ⎥ ⎢Ix ⎥ Ym Y¯d Id ⎢ ⎥ ⎢ ⎥ ⎢¯ ⎥ ⎢¯ ⎥ ⎢ Vy ⎥ ⎢Iy ⎥ ⎣ ⎦ ⎣ ⎦

(2.12)

Now,

¯c T Y [C] [ ¯ Ym

⎡ ⎢ ⎢ ⎢ ¯ Ym ⎢ [C] ⎢ = ] ¯ ⎢ Yd ⎢ ⎢ ⎢ ⎣

⎤ 1 0⎥ ⎥ −1 0 ⎥⎥ Y¯c Y¯m 1 −1 0 0 ⎥[ ][ ] 0 1 ⎥⎥ Y¯m Y¯d 0 0 1 −1 ⎥ 0 −1 ⎥⎦

(2.13)

Or,

¯c T Y [C] [ ¯ Ym

⎡ ⎢ ⎢ ⎢ ¯ Ym ⎢ [C] ⎢ ] = ¯ ⎢ Yd ⎢ ⎢ ⎢ ⎣

⎤ 1 0⎥ ⎥ −1 0 ⎥⎥ ⎥[ 0 1 ⎥⎥ ⎥ 0 −1 ⎥⎦

Y¯c −Y¯c Y¯m −Y¯m

Y¯m −Y¯m ] Y¯d −Y¯d (2.14)

Or,

¯c T Y [C] [ ¯ Ym

⎡ ¯ ⎤ ⎢ Yc −Y¯c Y¯m −Y¯m ⎥ ⎢ ⎥ ⎢ −Y¯ Y¯m Y¯c −Y¯m Y¯m ⎥⎥ ⎢ c ⎥ ] [C] = ⎢ ¯ ⎢ Ym −Y¯m Y¯d Y¯d −Y¯d ⎥⎥ ⎢ ⎢ ¯ ⎥ ⎢ −Ym Y¯m −Y¯d Y¯d ⎥⎦ ⎣ 20

(2.15)

Hence, from equations (2.12) and (2.15)

⎤⎡ ⎤ ⎡¯ ⎤ ⎡ ¯ ⎢Iu ⎥ ⎢ Yc −Y¯c Y¯m −Y¯m ⎥ ⎢V¯u ⎥ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎢I¯ ⎥ ⎢ −Y¯ Y¯c −Y¯m Y¯m ⎥⎥ ⎢⎢V¯v ⎥⎥ ⎢ v⎥ ⎢ c ⎥⎢ ⎥ ⎢¯ ⎥ = ⎢ ¯ ⎢Ix ⎥ ⎢ Ym −Y¯m Y¯d −Y¯d ⎥⎥ ⎢⎢V¯x ⎥⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢¯ ⎥ ⎢ ¯ ⎢Iy ⎥ ⎢ −Ym Y¯m −Y¯d Y¯d ⎥⎦ ⎢⎣V¯y ⎥⎦ ⎣ ⎦ ⎣

(2.16)

From equation (2.16),

I¯u = Y¯c V¯u − Y¯c V¯v + Y¯m V¯x − Y¯m V¯y = Y¯c V¯u − Y¯c V¯v + Y¯m V¯x − Y¯m V¯y + Y¯m V¯u − Y¯m V¯u = Y¯c (V¯u − V¯v ) + (−Y¯m )(V¯u − V¯x ) + Y¯m (V¯u − V¯y )

(2.17)

Or,

I¯u = I¯uv + I¯ux + I¯uy

(2.18)

Similarly,

I¯v = −Y¯c V¯u + Y¯c V¯v − Y¯m V¯x + Y¯m V¯y = −Y¯c V¯u + Y¯c V¯v − Y¯m V¯x + Y¯m V¯y + Y¯m V¯v − Y¯m V¯v = Y¯c (V¯u − V¯v ) + (−Y¯m )(V¯v − V¯y ) + Y¯m (V¯v − V¯x )

(2.19)

Or,

I¯v = I¯vu + I¯vy + I¯vx

(2.20)

I¯x = Y¯m V¯u − Y¯m V¯v + Y¯d V¯x − Y¯d V¯y = Y¯m V¯u − Y¯m V¯v + Y¯d V¯x − Y¯d V¯y + Y¯m V¯x − Y¯m V¯x = Y¯d (V¯x − V¯y ) + (−Y¯m )(V¯x − V¯u ) + Y¯m (V¯x − V¯u )

(2.21)

Or,

I¯x = I¯xy + I¯xu + I¯xv

(2.22)

Equations (2.18), (2.20) and (2.22) can be represented by the partial networks shown in Figs. 2.10, 2.11 and 2.12 respectively. Combining Figs. 2.10, 2.11 and 2.12, Fig. 2.13 is obtained. Again from the last row of equation (2.16),

I¯y = −Y¯m V¯u + Y¯m V¯v − Y¯d V¯x + Y¯d V¯y = −Y¯m V¯u + Y¯m V¯v − Y¯d V¯x + Y¯d V¯y + Y¯m V¯y − Y¯m V¯y = Y¯d (V¯y − V¯x ) + (−Y¯m )(Y¯y − V¯v ) + Y¯m (Y¯y − V¯u )

(2.23)

It can be observed that equation (2.23) is also represented by Fig. 2.13. Therefore, the voltage21

Figure 2.10: Partial network corresponding to equation (2.18) current relationship of equation (2.16) is adequately represented by Fig. 2.13. Thus, Fig. 2.13 can be considered as an equivalent circuit of Fig. 2.9. As Fig. 2.13 does not contain any mutual ¯ BUS formulation can be adopted for this circuit also. admittance, usual method for Y Fig. 2.13 shows the most general case in which all the four nodes are distinct from each other. However, in many cases mutual coupling exists between two elements which have one common node between them. The equivalent circuit for this case can also be derived from Fig. 2.13. For example, in Fig. 2.13, if nodes ‘v’ and ‘y’ are common (say ‘w’), then the equivalent circuit becomes as shown in Fig. 2.14. Moreover, if the nodes ‘u’ and ‘x’ are also common (say ‘s’), then the equivalent circuit ¯ BUS formulation can be adopted for these two is shown in Fig. 2.15. Again, the usual method for Y circuits also. We are now in a position to write down the basic power flow equation, which we will take up in the next lecture.

22

Figure 2.11: Partial network corresponding to equation (2.20)

Figure 2.12: Partial network corresponding to equation (2.22)

23

Figure 2.13: Combined network of Figs. 2.10, 2.11 and 2.12

Figure 2.14: Equivalent circuit with one common node

24

Figure 2.15: Equivalent circuit with two common nodes

25

2.5

Basic power flow equation

From equation (2.7), for a ‘n’ bus system,

⎡¯ ⎤ ⎡¯ ⎢ I1 ⎥ ⎢ Y11 ⎢ ⎥ ⎢ ⎢ I¯ ⎥ ⎢ Y¯ ⎢ 2 ⎥ ⎢ 21 ⎢ ⎥=⎢ ⎢⋮⎥ ⎢ ⋮ ⎢ ⎥ ⎢ ⎢¯ ⎥ ⎢¯ ⎢In ⎥ ⎢Yn1 ⎣ ⎦ ⎣ Or,

⎤⎡ ⎤ Y¯12 ⋯ Y¯1n ⎥ ⎢ V¯1 ⎥ ⎥⎢ ⎥ Y¯22 ⋯ Y¯2n ⎥⎥ ⎢⎢ V¯2 ⎥⎥ ⎥⎢ ⎥ ⋮ ⋮ ⋮ ⎥⎥ ⎢⎢ ⋮ ⎥⎥ ⎥⎢ ⎥ Y¯n2 ⋯ Y¯nn ⎥⎦ ⎢⎣V¯n ⎥⎦

(2.24)

n

I¯i = ∑ Y¯ij V¯j

(2.25)

j=1

Complex power injected at bus ‘i’ is given by,

S¯i = Pi + jQi = V¯i I¯i∗ Now, V¯i = Vi ejθi ;

V¯j = Vj ejθj ;

(2.26)

Y¯ij = Yij ejαij ;

n

Hence, S¯i = Pi + jQi = Vi ejθi [∑ Yij Vj ej(θj + αij ) ]

∗

j=1

Or, n

Pi = ∑ Vi Vj Yij cos(θi − θj − αij )

(2.27)

j=1 n

Qi = ∑ Vi Vj Yij sin(θi − θj − αij )

(2.28)

j=1

Equations (2.27) and (2.28) are known as the basic load flow equations. It can be seen that for any ith bus, there are two equations. Therefore, for a ‘n’-bus power system, there are altogether ‘2n’ load-flow equations. Now, from equations (2.27) and (2.28) it can be seen that there are four variables (Vi , θi , Pi and Qi ) associated with the ith bus. Thus for the ‘n’-bus system, there are a total of ‘4n’ variables. As there are only ‘2n’ equations available, out of these ‘4n’ variables, ‘2n’ quantities need to be specified and remaining ‘2n’ quantities are solved from the ‘2n’ load-flow equations. As ‘2n’ variables are to be specified in a ‘n’ bus system, for each bus, two quantities need to be specified. For this purpose, the buses in a system are classified into three categories and in each category, two different quantities are specified as described below. 1. PQ Bus: At these buses loads are connected and therefore, these buses are also termed as load buses. Generally the values of loads (real and reactive) connected at these buses are known and hence, at these buses Pi and Qi are specified (or known). Consequently, Vi and θi need to be calculated for these buses. 26

2. PV Bus: Physically, these buses are the generator buses. Generally, the real power supplied by the generator is known (as we say that the generation is supplying 100 MW) and also, the magnitude of the terminal voltage of the generator is maintained constant at a pre-specified value by the exciter (provided that the reactive power supplied or absorbed by the generator is within the limits). Thus, at a PV bus, Pi and Vi are specified and consequently, Qi and θi need to be calculatd. 3. Slack Bus: To calculate the angles θi (as discussed above), a reference angle (θi = 0) needs to be specified so that all the other bus voltage angles are calculated with respect to this reference angle. Moreover, physically, total power supplied by all the generation must be equal to the sum of total load in the system and system power loss. However, as the system loss cannot be computed before the load flow problem is solved, the real power output of all the generators in the system cannot be pre-specified. There should be at least one generator in the system which would supply the loss (plus its share of the loads) and thus for this generator, the real power output can’t be pre-specified. However, because of the exciter action, Vi for this generator can still be specified. Hence for this generator, Vi and θi (= 0) are specified and the quantities Pi and Qi are calculated. This generator bus is designated as the slack bus. Usually, the largest generator in the system is designated as the slack bus. To summarise, the details of different types of buses in a ‘n’ bus, ‘m’ generator power system are shown in Table 2.1. Now, please note that in a load flow problem, the quantities Pi and Qi (Qi at Table 2.1: Classification of buses Type

Total no. of buses

Specified quantity

Solution quantity

PQ

n-m

P i , Qi

Vi , θi

PV

m-1

Pi , Vi

Qi , θi

Slack

1

Vi , θi

P i , Qi

PV while Pi and Qi at slack buses) are not directly solved. Only the quantities Vi and θi are directly solved (Vi for all PQ buses while θi for all PV and PQ buses). This is because of the fact that once Vi and θi at all PV and PQ buses are solved, then the voltage magnitudes and angles at all the buses are known (Vi , θi at the slack bus are already specified) and subsequently, using equations (2.27) and (2.28), Pi and Qi at any bus can be calculated. Therefore, in a ‘n’ bus, ‘m’ generator system, the unknown quantities are: Vi (total ‘n-m’ of them) and θi (total ‘n-1’ of them). Therefore, total number of unknown quantities is ‘2n-m-1’. On the other hand, the specified quantities are: Pi (total ‘n-1’ of them) and Qi (total ‘n-m’ of them). Hence total number of specified quantities is also ‘2n-m-1’. As the number of unknown quantities is equal to the number of specified quantities, the load-flow problem is well-posed. 27

Equations (2.27) and (2.28) represent a set of simultaneous, non-linear, algebraic equations. As the set of equations is non-linear, no closed form, analytical solution for these equations exist. Hence, these equations can only be solved by using suitable numerical iterative techniques. For solving the load flow problem, various iterative methods exist. These are: 1. Gauss-seidel method 2. Newton Raphron (polar) technique 3. Newton Raphron (rectangular) technique 4. Fast-decoupled load flow We will discuss these methods one by one and we start with the Gauss-Seidel method.

2.6

Basic Gauss Seidel solution method

Before discussing the Gauss-Seidel load flow (GSLF) technique, let we first review the basic GaussSeidel procedure for solving a set of non-linear algebraic equations. Let the following ‘n’ equations are given for the ‘n’ unknown quantities x1 , x2 , ⋯⋯ xn ;

⎫ f1 (x1 , x2 ⋯⋯ xn ) = 0 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ f2 (x1 , x2 ⋯⋯ xn ) = 0 ⎪ ⎪ ⎪ ⎪ ⋮ ⋮ ⎬ ⎪ ⎪ ⎪ ⋮ ⋮ ⎪ ⎪ ⎪ ⎪ ⎪ fn (x1 , x2 ⋯⋯ xn ) = 0 ⎪ ⎪ ⎭

(2.29)

It is to be noted that in equation (2.29), the function f1 , f2 , ⋯fn are all non-linear in nature and no particular form of these equations is assumed. Now, with some algebraic manipulation, from the first equation of equation set (2.29), the variable x1 can be represented in terms of the other variables. Similarly from the second equation, the variable x2 can be represented in terms of the other variables. Proceeding in the same way, from the nth equation, the variable xn can be expressed in terms of the other variables. Therefore, let,

x1 x2 ⋮ xk ⋮ xn

= g1 (x2 , x3 ⋯⋯ xn ) = g2 (x1 , x3 ⋯⋯ xn ) ⋮ = gk (x1 , x2 ⋯⋯ xk−1 , xk+1 ⋯⋯ xn ) ⋮ = gn (x1 , x2 ⋯⋯ xn−1 )

⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

(2.30)

To compute the variables x1 , x2 , ⋯⋯ xn from these equations g1 , g2 , ⋯⋯ gn , the first step is to (0) (0) (0) assume the initial values of these solution variables (x1 , x2 ⋯⋯ xn ). With these initially assumed 28

values, various steps of the basic Gauss-Seidel algorithm are as follows.

Basic Gauss-Seidel procedure Step 1: Set iteration count k = 1 Step 2: Update the variables ;

x(k) = g1 (x(k−1) , x(k−1) , ⋯⋯ x(k−1) ); n 1 2 3 (k−1) x(k) = g2 (x(k) , ⋯⋯ x(k−1) ); n 2 1 , x3

⋮

⋮

⋮

⋮

(k) (k) (k−1) (k−1) xp(k) = gp (x(k) ); 1 , x2 , ⋯⋯ xp−1 , xp+1 , ⋯⋯ xn (k) (k) x(k) = gn (x(k) n 1 , x2 , ⋯⋯ xn−1 ); (k)

(k)

(k−1)

∣ for all i = 1, 2, ⋯⋯ n; Step 3: Compute ei = ∣xi − xi (k) (k) (k) Step 4: Compute er = max(e1 , e2 , ⋯⋯ en ) ; Step 5: If er ≤  (tolerance limit), stop and print the solution. Else set k = k + 1 and go to step 2. It is to be noted that in step 2, for updating the variable xp , the most updated values of x1 , x2 , ⋯⋯ xp−1 (which are before xp in the sequence of the solution variables) are used while for the variables xp+1 , xp+2 , ⋯⋯ xn (which are after xp in the sequence of the solution variables), the values pertaining to previous iteration are used (as these variables have not been updated yet). Subsequently, in steps 3 and 4, the maximum absolute error between the solutions of the current iteration and previous iteration is calculated. If this maximum absolute error is less then a pre-specified tolerance value, then the algorithm is considered to be converged. Otherwise, the solution variables are again updated. With this background of basic Gauss-Seidel method, we are now in a position of discussing GSLF, which we will do next.

2.7

Gauss Seidel Load Flow technique

Let us now proceed for discussing GSLF. From equation (2.25),

⎡ ⎤ ⎢ ⎥ n 1 ⎢ ⎥ ⎢I¯i − ∑ Y¯ik V¯k ⎥. Now, from the relation I¯i = ∑ Y¯ik V¯k = Y¯ii V¯i + ∑ Y¯ik V¯k . Hence, V¯i = ⎥ Y¯ii ⎢⎢ k=1 k=1 k=1 ⎥ ≠i ≠i ⎣ ⎦ n

n

29

Pi − jQi . Thus, Pi + jQi = V¯i I¯i∗ we get, I¯i = V¯i∗ ⎡ ⎤ ⎢ P − jQ ⎥ n 1 ⎢ i i ¯ik V¯k ⎥⎥ ⎢ V¯i = − Y ∑ ⎥ Y¯ii ⎢⎢ V¯i∗ k=1 ⎥ ≠i ⎣ ⎦

(2.31)

Equation (2.31) is the basic equation for performing GSLF. It is to be noted that without loss of generality, it is assumed that the ‘m’ generators are connected to the first ‘m’ buses (bus ‘1’ being the slack bus) and remaining ‘(n-m)’ buses are load buses. Now, initially to understand the basic GSLF procedure, let us assume that m = 1, i.e., there is only one generator (which is also the slack bus) and the rest ‘(n-1)’ buses are all load buses. To perform load-flow computation, initial guesses of the bus voltages are necessary. As any power system is generally expected to operate at the normal steady-state operating condition (with the bus voltage magnitudes maintained between (0) 0.95 - 1.05 p.u.), all the unknown bus voltage are initialized to 1.0∠0o p.u (i.e. V¯j = 1.0∠0o for j = 2, 3, ⋯⋯ n). This process of initializing all bus voltage to 1.0∠0o is called flat start. With these initial bus voltages, the complete procedure for GSLF (having no PV bus) is as follows.

GSLF without PV bus Step 1: Set iteration count k = 1. Step 2: Update the bus voltages as;

⎡ ⎤ ⎢ P − jQ ⎥ n 1 ⎢ ⎥ 2 2 (k−1) ¯ ¯ ¯ ⎢ (k−1) ∗ − ∑ Y2j Vj ⎥ V = ⎢ ⎥ ¯ Y22 ⎢ {V¯2 } j =1 ⎥ ≠ 2 ⎣ ⎦ ⋮ ⋮ ⎡ ⎤ n 1 ⎢⎢ Pp − jQp p−1 ¯ ¯ (k) (k−1) ⎥ (k) ⎥ ¯ ¯ ¯ Vp = ⎢ ¯ (k−1) ∗ − ∑ Ypj Vj − ∑ Ypj Vj ⎥ ¯ ⎥ Ypp ⎢⎣ {Vp j=p+1 } j=1 ⎦ ⋮ ⋮ ⎡ ⎤ ⎢ Pn − jQn n−1 ⎥ 1 (k) (k) ⎢ ¯nj V¯j ⎥ V¯n = − Y ∑ ∗ ⎢ ⎥ ⎥ Y¯nn ⎢⎣ {V¯n(k−1) } j=1 ⎦ (k) 2

(k)

(k)

(k−1)

∣ for all i = 2, ⋯⋯ n; Step 3: Compute ei = ∣V¯i − V¯i (k) (k) (k) (k) Step 4: Compute e = max(e2 , e3 , ⋯⋯ en ) ; Step 5: If e(k) ≤  (tolerance limit), stop and print the solution. Else set k = k+1 and go to step 2. With the above understanding of the basic GSLF, we are now in a position to discuss the GSLF procedure for a system having multiple generators. Before we discuss the GSLF procedure, let us look into the procedure of initialisation of bus voltages (which is little different than assuming a flat start for all the bus volatges). For a system having multiple generators, the bus voltage 30

initialisation is carried out in a two step procedure; i) the load buses are initialised with flat start (0) (i.e. V¯j = 1.0∠0o for j = (m + 1), (m + 2), ⋯⋯ n) and ii) the magnitudes of the voltages of the PV buses are initialised with the corresponding specified voltage magnitudes while initialising all (0) these voltage angles to 0o (i.e. V¯j = Vjsp ∠0o for j = 2, 3, ⋯⋯ m, where Vjsp is the specified bus voltage magnitude of the j th generator). Now, as discussed earlier, the reactive power supplied or absorbed by a generator (QG ) is calculated by the load flow procedure. However any generator has a maximum and minimum limit on QG . If the QG from the generator is within these limits, then the generator excitation system is able to maintain the terminal voltage at the specified value. On the other hand, if the generator reaches its limit on QG (either maximum or minimum), then because of the insufficient amount of reactive power (either supplied or absorbed), the generator excitation system would not be able to maintain the terminal voltage magnitude at the specified value. In that case the generator bus would behave as a PQ bus (P being already specified for the generator and Q is set at either maximum or minimum limiting value of QG ). In power system terminology, this phenomenon (where the generator is behaving like a PQ bus) is termed as ‘PV to PQ switching’ which should also be accounted for in any load-flow solution methodology. This is incorporated in GSLF by the following procedure. At the beginning of each iteration, QG injection by each generator is calculated. If this calculated QG is found to be within the corresponding limits then this generator continues to behave as a PV bus. Hence ∣V¯i ∣ of this bus (at which the generator is connected ) is still maintained at the corresponding specified value and only the angle of this bus voltage is calculated in the present iteration. On the other hand if QG is found to exceed any limit (either maximum or minimum), then it is fixed at that limit and the bus is considered to act like a PQ bus. Thus, both the magnitude and angle of the bus voltage are calculated in the present iteration. With this background, the complete algorithm of GSLF involving multiple generator buses is as follows.

Complete GSLF algorithm (0)

(0)

Step 1: Initialise V¯j = Vjsp ∠0o for j = 2, 3, ⋯⋯ m and V¯j = 1.0∠0o for j = (m + 1), (m + 2), ⋯⋯ n. Set iteration count k = 1. Step 2: For i = 2, 3, ⋯⋯ m, carry out the following operations. a) Calculate, n

Q(k) = ∑ Vi(k−1) Vj(k−1) Yij sin (θi(k−1) − θj(k−1) − αij ) i j=1

(k)

b) If, Qmin ≤ Qi i is given by,

(k) (k) (k) (k) ≤ Qmax ; then assign ∣V¯i ∣ = Visp and θi = ∠ (Ai ). The quantity Ai i

(k) i

A

⎡ ⎤ n 1 ⎢⎢ Pi − jQi(k) i−1 ¯ ¯ (k) (k−1) ⎥ ⎥ ¯ ¯ = − ∑ Yij Vj ∗ − ∑ Yij Vj ⎢ ⎥ ⎥ Y¯ii ⎢⎣ {V¯i(k−1) } j=1 j=i+1 ⎦ 31

(k)

c) If Qi

≥ Qmax , then calculate i (k) i

V¯ (k)

d) If Qi

⎡ ⎤ i−1 n ⎥ 1 ⎢⎢ Pi − jQmax (k) (k−1) i ⎥ ¯ij V¯j − ∑ Y¯ij V¯j = − Y ∑ ∗ ⎢ ⎥ ⎥ Y¯ii ⎢⎣ {V¯i(k−1) } j=1 j=i+1 ⎦

≤ Qmin , then calculate i (k) i

V¯

⎡ ⎤ i−1 n ⎥ 1 ⎢⎢ Pi − jQmin (k) (k−1) i ⎥ ¯ij V¯j − ∑ Y¯ij V¯j = − Y ∑ ∗ ⎢ ⎥ ⎥ Y¯ii ⎢⎣ {V¯i(k−1) } j=1 j=i+1 ⎦

Step 3: For i = (m + 1), ⋯⋯ n, calculate (k) i

V¯

(k)

⎡ ⎤ i−1 n ⎥ 1 ⎢⎢ Pi − jQ(k) (k) (k−1) i ⎥ ¯ij V¯j − ∑ Y¯ij V¯j = − Y ∑ ∗ ⎢ ⎥ ⎥ Y¯ii ⎢⎣ {V¯i(k−1) } j=1 j=i+1 ⎦ (k)

(k−1)

∣ for all i = 2, ⋯⋯ n; Step 4: Compute ei = ∣V¯i − V¯i (k) (k) (k) (k) Step 5: Compute e = max(e2 , e3 , ⋯⋯ en ) ; Step 6: If e(k) ≤ , stop and print the solution. Else set k = k + 1 and go to step 2. We will illustrate the GSLF algorithm with an example in the next lecture.

32

2.7.1

Example of Gauss Seidel load flow technique

To illustrate the basic procedure of GSLF, let as consider a small 5-bus system as shown in Fig. 2.16. In this system, buses 1-3 are generator buses and buses 4-5 are load buses. Therefore, in this system, n = 5 and m = 3. Moreover, bus 1 is taken to be the slack bus and thus, buses 2-3 are considered to be PV buses. The bus data and line data of this system are given in Tables A.1 and A.2 respectively. From Table A.1, the injected real and reactive powers at different buses can be obtained as follows: P2 = 0.5 p.u, P3 = 1.0 p.u, P4 = −1.15 p.u, P5 = −0.85 p.u, Q4 = −0.6 p.u, and ¯ BUS of this system (computed from the line data given in Table Q5 = −0.4 p.u. Moreover, the Y A.2) is shown in equation (2.32). Note that in this equation, the real part (G) and the imaginary ¯ BUS matrix (Y ¯ BUS = G + jB) are shown separately. part (B) of the Y

Figure 2.16: The example 5 bus system

⎡ 3.2417 −1.4006 0 0 −1.8412⎤⎥ ⎢ ⎢ ⎥ ⎢−1.4006 3.2417 −1.8412 0 0 ⎥⎥ ⎢ ⎢ ⎥ G=⎢ 0 −1.8412 4.2294 −1.2584 −1.1298⎥⎥ ⎢ ⎢ 0 0 −1.2584 2.1921 −0.9337⎥⎥ ⎢ ⎢ ⎥ ⎢−1.8412 0 −1.1298 −0.9337 3.9047 ⎥⎦ ⎣ ⎡−13.0138 5.6022 0 0 7.4835 ⎤⎥ ⎢ ⎢ ⎥ ⎢ 5.6022 −13.0138 7.4835 ⎥ 0 0 ⎢ ⎥ ⎢ ⎥ B=⎢ 0 7.4835 −18.9271 7.1309 4.4768 ⎥⎥ ⎢ ⎢ 0 0 7.1309 −10.7227 3.7348 ⎥⎥ ⎢ ⎢ ⎥ ⎢ 7.4835 ⎥ 0 4.4768 3.7348 −15.5521 ⎣ ⎦ 33

(2.32)

For applying GSLF, initially the flat start profile is assumed. Please note that the flat voltage profile is followed for PQ buses. For PV buses, the initial voltage magnitude is taken to be equal to their corresponding specified voltage magnitude. However, the initial voltage angles are always assumed to be zero. Therefore, from the data given in Table A.1, all the 5 bus voltages are initialised to 1.0∠0o p.u. Now as the system contains both PV and PQ buses, we follow the ‘complete GSLF algorithm’. In step 2(a) of this algorithm, we first calculate the reactive power absorbed or generated by generators 2 and 3 (corresponding to i = 2 and i = 3). The calculated values of Q2 and Q3 at iteration 1 (in p.u.) are shown in Table 2.2 (denoted as Qcal in this table). Now, the data in Table A.1 show that the minimum and maximum reactive power limits for both these generators are -5 p.u. and 5 p.u respectively. Hence, the calculated values of Q2 and Q3 are well within the corresponding reactive power limits. Therefore, both bus 2 and bus 3 are continued to operate as PV buses and as a result, their voltage magnitudes are maintained at the corresponding specified values and only the voltage angles are calculated in step 2(b) (utilising the calculated values of Q2 and Q3 ). Subsequently in step 3, both the magnitude and angles of buses 4 and 5 are calculated. The results of iteration 1 are shown in Table 2.2. Finally in steps 4-5, the error is calculated, which is also shown in Table 2.2. The error is found to be more than the threshold value (taken to be equal to 1.0e−12 ) and therefore the algorithm goes back to step 2 again. The iteration wise result for first 6 iterations are shown in Tables 2.2 and 2.3. Please observed from these two tables that because of high Qmax G min and QG limits, the reactive powers supplied or observed by these two generators are always within these limits and thus bus 2 and 3 continue to act as PV buses from iteration to iteration. Also note from Tables 2.2 and 2.3 that the error reduces with iteration. Finally, the algorithm converges after 69 iterations and the final solution is shown in Table 2.4. Please note from this table that the final values of Q2 and Q3 are -18.51 and 68.87 MVAR respectively. These values are well within their corresponding reactive power limits and thus, the voltage magnitudes of bus 2 and 3 maintained at 1.0 p.u (pre-specified values). The finally computed values of real and reactive power injection at all the buses are also shown in Table 2.4, which are also found to be exactly equal to the specified injected values. Table 2.2: GSLF results in 5 bus system without any generator Q limit for iterations 1-3 Iteration = 1 Bus no. 1 2 3 4 5

Qcal

∣V ∣

Iteration = 2

θ

∣V ∣

Qcal

Iteration = 3

θ

Qcal

∣V ∣

θ

(p.u) (p.u) (deg) (p.u) (p.u) (deg) (p.u) (p.u) (deg) 1.0 0 1.0 0 1.0 0 -0.0720 1.0 2.0533 -0.0663 1.0 4.1038 -0.2022 1.0 3.7091 -0.0932 1.0 3.5968 0.2937 1.0 2.6494 0.5142 1.0 1.6234 0.9394 -3.2379 0.9167 -5.0548 0.9101 -6.0776 0.957 -2.5257 0.9482 -3.2562 0.946 -3.797 error = 0.081708 error = 0.037148 error = 0.017906

Let us now study the behaviour of GSLF when generator reactive power limit is violated. Towards this goal, let us assume that the maximum reactive power which can be supplied by generator 3 in 34

Table 2.3: GSLF results in 5 bus system without any generator Q limit for iterations 4-6 Iteration = 4 Bus no. 1 2 3 4 5

Qcal

∣V ∣

Iteration = 5

θ

∣V ∣

Qcal

Iteration = 6

θ

Qcal

∣V ∣

θ

(p.u) (p.u) (deg) (p.u) (p.u) (deg) (p.u) (p.u) (deg) 1.0 0 1.0 0 1.0 0 -0.2129 1.0 3.1318 -0.2071 1.0 2.6927 -0.2013 1.0 2.3782 0.5926 1.0 0.8641 0.6263 1.0 0.3228 0.6461 1.0 -0.0546 0.9082 -6.7844 0.9074 -7.266 0.9069 -7.5985 0.9452 -4.1735 0.9448 -4.435 0.9446 -4.6163 error = 0.013252 error = 0.0094475 error = 0.0065856

Table 2.4: Final Results of the 5 bus system with GSLF

Bus no. 1 2 3 4 5

∣V ∣

Without generator Q limit

θ

Pinj

Qinj

(p.u) (deg) (p.u) (p.u) 1.0 0 0.56743 0.26505 1.0 1.65757 0.5 -0.18519 1.0 -0.91206 1.0 0.68875 0.90594 -8.35088 -1.15 -0.6 0.94397 -5.02735 -0.85 -0.4 Total iteration = 69

50 MVAR (it supplies 68.87 MVAR when no limit is imposed on the generators). The iteration wise solutions for first 6 iterations of the load flow computation with this maximum limit are shown in Tables 2.5 and 2.6. Now, let us compare Tables 2.2 and 2.5. From these two tables it can be observed that the load flow solution with generator Q limit proceeds in identical fashion for first two iterations as in the case with no reactive power limit on the generators. However, from iteration 3 onwards the solution changes. In iteration 3, Q3 calculated is found to be equal to 51.42 MVAR. As a result, Q3 is limited to 50 MVAR and Bus 3 is converted to a PQ bus, and therefore its voltage magnitude is calculated using the expression shown in step 2(c). Please observe that this voltage magnitude is not maintained at 1.0 p.u (in fact, it becomes less than 1.0 p.u. because of insufficient reactive power). In the subsequent iteration also, calculated Q3 is always found to be more than the maximum limit and as a result, Q3 is always maintained at 50 MVAR thereby making ∣V3 ∣ < 1.0 p.u. The algorithm finally converges with a tolerance of 1.0e−12 p.u. after 66 iterations and the final solution are shown in Table 2.7. In this table, the GSLF results without any reactive power limit (as shown in Table 2.4) are also reproduced for comparison. Please note from Table 2.7 that because of cap on Q3 , the overall voltage profile of the system is little lower than that obtained with no limit on generator reactive power. As a second example, let us now consider the IEEE-14 bus system. The data of the IEEE 14 bus system are shown in Tables A.3 and A.4. The power flow solution without any limit on the 35

Table 2.5: GSLF results in 5 bus system with generator Q limit on bus 3 for iterations 1-3 Iteration = 1 Bus no.

Qcal

∣V ∣

Iteration = 2

θ

∣V ∣

Qcal

Iteration = 3

θ

Qcal

∣V ∣

θ

(p.u) (p.u) (deg) (p.u) (p.u) (deg) (p.u) (p.u) (deg) 1.0 0 1.0 0 1.0 0 -0.0720 1.0 2.0533 -0.0663 1.0 4.1038 -0.2022 1.0 3.7091 -0.0932 1.0 3.5968 0.2937 1.0 2.6494 0.5142 0.9955 1.6318 0.9394 -3.2379 0.9167 -5.0548 0.9071 -6.0918 0.957 -2.5257 0.9482 -3.2562 0.944 -3.8036 error = 0.081708 error = 0.037148 error = 0.019063

1 2 3 4 5

Table 2.6: GSLF results in 5 bus system with generator Q limit on bus 3 for iterations 4-6 Iteration = 1 Bus no.

Qcal

∣V ∣

Iteration = 2

θ

∣V ∣

Qcal

Iteration = 3

θ

Qcal

∣V ∣

θ

(p.u) (p.u) (deg) (p.u) (p.u) (deg) (p.u) (p.u) (deg) 1.0 0 1.0 0 1.0 0 -0.1784 1.0 3.1067 -0.1365 1.0 2.6692 -0.1082 1.0 2.3656 0.5337 0.991 0.9288 0.5178 0.9882 0.4399 0.5109 0.9864 0.1025 0.9012 -6.8046 0.8973 -7.2918 0.8947 -7.6271 0.9409 -4.1739 0.9388 -4.4292 0.9374 -4.6044 error = 0.012967 error = 0.0088975 error = 0.0061049

1 2 3 4 5

Table 2.7: Final Results of the 5 bus system with GSLF with generator Q limit

Bus no. 1 2 3 4 5

∣V ∣

Without generator Q limit

θ

Pinj

Qinj

∣V ∣

With generator Q limit

θ

Pinj

Qinj

(p.u) (deg) (p.u) (p.u) (p.u) (deg) (p.u) (p.u) 1.0 0 0.56743 0.26505 1.0 0 0.56979 0.33935 1.0 1.65757 0.5 -0.18519 1.0 1.69679 0.5 -0.04769 1.0 -0.91206 1.0 0.68875 0.9825 -0.63991 1.0 0.5 0.90594 -8.35088 -1.15 -0.6 0.88918 -8.35906 -1.15 -0.6 0.94397 -5.02735 -0.85 -0.4 0.93445 -4.98675 -0.85 -0.4 Total iteration = 69 Total iteration = 66

generator reactive power is shown in Table 2.8. Note that all the terminal voltage of the generators are maintained at their corresponding specified values. Also observe that the generator connected at bus 6 supplies a reactive power of 37.27 MVAR. Now assume that generator 6 is constrained to supply only 30 MVAR. With this Q limit, the power flow solution is also shown in Table 2.8. From these result followings salient point can be noted: a. Reactive power supplied by generator 6 is limited at 30 MVAR. b. As a result, ∣V6 ∣ goes down to 1.05497 p.u. (from the specified value of 1.07 p.u.). 36

Table 2.8: Final Results of the 14 bus system with GSLF Without generator Q limit

Bus no.

∣V ∣

1 2 3 4 5 6 7 8 9 10 11 12 13 14

(p.u) 1.06 1.045 1.04932 1.03299 1.04015 1.07 1.02076 1.0224 1.0201 1.0211 1.04144 1.0526 1.04494 1.01249

θ

Pinj

(deg) (p.u) 0 2.37259 -5.17113 0.183 -14.54246 -1.19 -10.39269 -0.4779 -8.76418 -0.07599 -12.52265 0.112 -13.44781 0 -13.47154 0 -13.60908 -0.29499 -13.69541 -0.09 -13.22158 -0.03501 -13.42868 -0.06099 -13.50388 -0.135 -14.60128 -0.14901 Total iteration = 676

Qinj

∣V ∣

(p.u) -0.3308 -0.166 -0.08762 -0.039 -0.01599 0.37278 0 -0.129 -0.16599 -0.05799 -0.018 -0.01599 -0.05799 -0.05001

(p.u) 1.06 1.045 1.04697 1.02902 1.03615 1.05497 1.01266 1.01391 1.0118 1.01154 1.02915 1.03787 1.03063 1.00136

With generator Q limit

θ

Pinj

(deg) (p.u) 0 2.37188 -5.17845 0.183 -14.55556 -1.19 -10.35987 -0.4779 -8.71027 -0.07599 -12.45871 0.112 -13.49478 0 -13.5185 0 -13.66101 -0.29499 -13.73679 -0.09 -13.21814 -0.03501 -13.39145 -0.06099 -13.48166 -0.135 -14.64504 -0.14901 Total iteration = 718

Qinj (p.u) -0.31249 -0.1066 -0.08762 -0.039 -0.01599 0.3 0 -0.129 -0.16599 -0.05799 -0.018 -0.01599 -0.05799 -0.05001

c. Because of the limit on generator reactive power, the overall voltage profile is on the lower side as compared to that obtained without any Q limit. As the last example the 30 bus system is considered. The data of this system are given in Tables A.5 and A.6. Again initially the load flow solution has been computed without any limit on the generator reactive power and the result are shown in Table 2.9. Subsequently Q limits have been imposed on both the generator connected at bus 11 (20 MVAR) and the generator connected at bus 13 (30 MVAR). However, with a tolerance of 10e−12 p.u. GSLF algorithm fails to converge even after 10,000 iteration. When the tolerance is reduced to 10e−6 p.u., the algorithm converges in 348 iteration and the result are again shown in Table 2.9. As can be seen from these results, for both the generation, the reactive power supplied have been fixed at their corresponding limits and as a result, the overall voltage profile of the system has gone down. From these results it is observed that the convergence characteristics of the GSLF technique is quite poor. Usually the number of iteration taken by GSLF is quite large and moreover in many cases, GSLF even fails to converge. To overcome these difficulties of GSLF, Newton- Raphson(NR) techniques have been developed, which are our next topics of discussion. These are two versions of NR techniques, namely, i) NR in polar co-ordinate and ii) NR in rectangular co-ordinate. We will study both these versions one by one and will start with the NR in polar co-ordinate from the next lecture.

37

Table 2.9: Final Results of the 30 bus system with GSLF Without generator Q limit

Bus no.

∣V ∣

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

(p.u) 1.05 1.0338 1.03128 1.02578 1.0058 1.02178 1.00111 1.023 1.04608 1.03606 1.0913 1.04859 1.0883 1.03346 1.02825 1.0359 1.0306 1.01873 1.01626 1.02041 1.02305 1.02343 1.0165 1.00939 1.00048 0.9825 1.00379 1.02049 0.98353 0.97181

θ

Pinj

(deg) (p.u) 0 2.38673 -4.97945 0.3586 -7.96653 -0.024 -9.58235 -0.076 -13.60103 -0.6964 -11.50296 0 -13.9994 -0.628 -12.56853 -0.45 -13.04088 0 -14.88589 -0.058 -11.16876 0.1793 -13.74947 -0.112 -12.56078 0.1691 -14.71704 -0.062 -14.86737 -0.082 -14.50539 -0.035 -14.98291 -0.09 -15.58107 -0.032 -15.81066 -0.095 -15.63819 -0.022 -15.35955 -0.175 -15.35222 0 -15.41998 -0.032 -15.81043 -0.087 -15.84004 0 -16.27422 -0.035 -15.59587 0 -12.1474 0 -16.87497 -0.024 -17.79427 -0.106 Total iteration = 851

Qinj (p.u) -0.29842 -0.05698 -0.012 -0.016 0.05042 0 -0.109 0.12343 0 -0.02 0.24018 -0.075 0.31043 -0.016 -0.025 -0.018 -0.058 -0.009 -0.034 -0.007 -0.112 0 -0.016 -0.067 0 -0.023 0 0 -0.009 -0.019

38

∣V ∣

With generator Q limit

θ

Pinj

Qinj

(p.u) (deg) (p.u) (p.u) 1.05 0 2.3865 -0.29386 1.0338 -4.98084 0.35861 -0.04562 1.03045 -7.95523 -0.02399 -0.012 1.02477 -9.56929 -0.07598 -0.016 1.0058 -13.60836 -0.6964 0.05532 1.02084 -11.49304 0.00003 0 1.00055 -13.99792 -0.628 -0.109 1.023 -12.57567 -0.45 0.15111 1.04006 -13.02865 0.00001 0 1.03117 -14.8866 -0.05797 -0.02001 1.07807 -11.12256 0.1793 0.2 1.04456 -13.75755 -0.11198 -0.075 1.08311 -12.55854 0.1691 0.3 1.02931 -14.73168 -0.062 -0.016 1.02403 -14.88097 -0.08199 -0.025 1.03148 -14.51198 -0.035 -0.018 1.02583 -14.98712 -0.09 -0.058 1.01422 -15.59618 -0.03199 -0.009 1.01159 -15.8251 -0.09499 -0.034 1.01568 -15.64961 -0.022 -0.007 1.01825 -15.36524 -0.17496 -0.11201 1.01867 -15.3581 0 0 1.01226 -15.43637 -0.032 -0.016 1.00517 -15.8277 -0.087 -0.067 0.99748 -15.86849 0 0 0.97944 -16.30534 -0.035 -0.023 1.00158 -15.62827 0 0 1.01959 -12.14235 0 0 0.98126 -16.91314 -0.024 -0.009 0.96951 -17.83675 -0.106 -0.019 Total iteration = 348 (totlerance 10e−6 )

2.8

Basic Newton - Raphson (NR) Techniques

Before discussing the application of NR technique in load flow solution, let us first review the basic procedure of solving a set of non-linear algebraic equation by means of NR algorithm. Let there be ‘n’ equations in ‘n’ unknown variables x1 , x2 , ⋯⋯ xn as given below,

f1 (x1 , x2 , ⋯⋯ xn ) f2 (x1 , x2 , ⋯⋯ xn ) ⋮ ⋮ fn (x1 , x2 , ⋯⋯ xn )

⎫ = b1 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ = b2 ⎪ ⎪ ⎪ ⎪ ⋮ ⋮ ⎬ ⎪ ⎪ ⎪ ⋮ ⋮ ⎪ ⎪ ⎪ ⎪ ⎪ = bn ⎪ ⎪ ⎭

(2.33)

In equation (2.33), the quantities b1 , b2 , ⋯⋯ bn as well as the functions f1 , f2 , ⋯⋯ fn are known. To solve equation (2.33), first we take an initial guess of the solution and let these initial guesses be (0) (0) (0) denoted as, x1 , x2 , ⋯⋯ xn . Subsequently, first order Taylor’s series expansion (neglecting the higher order terms) is carried out for these equation around the initial guess of solution. Also let (0) (0) (0) T the vector of initial guess be denoted as x(0) = [x1 , x2 , ⋯⋯ xn ] . Now, application of Taylor’s expansion on the equations of set (2.33) yields,

⎫ ∂f1 ∂f1 ∂f1 ⎪ ∆x1 + ∆x2 + ⋯⋯ + ∆xn = b1 ⎪ ⎪ ⎪ ⎪ ∂x1 ∂x2 ∂xn ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ∂f2 ∂f2 ∂f2 (0) (0) (0) ⎪ f2 (x1 , x2 , ⋯⋯ xn ) + ∆x1 + ∆x2 + ⋯⋯ + ∆xn = b2 ⎪ ⎪ ⎬ ∂x1 ∂x2 ∂xn ⎪ ⋮ ⋮ ⋮ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⋮ ⋮ ⋮ ⎪ ⎪ ⎪ ⎪ ⎪ ∂fn ∂fn ∂fn ⎪ (0) (0) (0) ⎪ fn (x1 , x2 , ⋯⋯ xn ) + ∆x1 + ∆x2 + ⋯⋯ + ∆xn = bn ⎪ ⎪ ⎪ ∂x1 ∂x2 ∂xn ⎭ (0) (0) f1 (x(0) 1 , x2 , ⋯⋯ xn ) +

(2.34)

Equation (2.34) can be written as,

⎡ ∂f1 ⎢ ⎡ f1 (x(0) ) ⎤ ⎢⎢ ∂x1 ⎢ ⎥ ⎢ ⎢ ⎥ ⎢ ∂f ⎢ ⎥ (0) ⎢ f2 (x ) ⎥ ⎢⎢ 2 ⎢ ⎥ + ⎢ ∂x1 ⎢ ⎥ ⎢ ⋮ ⎢ ⎥ ⎢ ⋮ ⎢ ⎥ ⎢fn (x(0) )⎥ ⎢⎢ ∂f ⎣ ⎦ ⎢ n ⎢ ⎣ ∂x1

∂f1 ∂x2 ∂f2 ∂x2 ⋮ ∂fn ∂x2

∂f1 ⎤ ⎥ ∂xn ⎥⎥ ⎡⎢ ∆x1 ⎤⎥ ⎡⎢ b1 ⎤⎥ ⎥⎢ ⎥ ⎢ ⎥ ∂f2 ⎥⎥ ⎢ ⎥ ⎢ ⎥ ⋯ ⎥ ⎢⎢ ∆x2 ⎥⎥ = ⎢⎢ b2 ⎥⎥ ∂xn ⎥⎥ ⎢ ⋮ ⎥⎥ ⎢⎢ ⋮ ⎥⎥ ⋮ ⎥⎥ ⎢⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢bn ⎥ ∆x ⎥ n ∂fn ⎥ ⎣ ⎦ ⎣ ⎦ ⋯ ⎥ ∂xn ⎦ ⋯

(2.35)

In equation (2.35), the matrix containing the partial derivative terms is known as the Jacobin 39

matrix (J). As can be seen, it is a square matrix. Hence, from equation (2.35),

⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎢ ∆m1 ⎥ ⎢ b1 − f1 (x(0) ) ⎥ ⎢ ∆x1 ⎥ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎢ ⎢ ∆x ⎥ (0) ) ⎥ −1 ⎢ ∆m2 ⎥ −1 ⎢ b2 − f2 (x ⎥ ⎢ 2⎥ ⎥ ⎥ = [J] ⎢ ⎥ = [J] ⎢ ⎢ ⎢ ⋮ ⎥ ⎥ ⎢ ⎢ ⋮ ⎥ ⋮ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎢∆mn ⎥ ⎢bn − fn (x(0) )⎥ ⎢∆xn ⎥ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣

(2.36)

Equation (2.36) is the basic equation for solving the ‘n’ algebraic equations given in equation (2.33). The steps of solution are as follow: Step 1: Assume a vector of initial guess x(0) and set iteration counter k = 0. Step 2: Compute f1 (x(k) ), f2 (x(k) ), ⋯⋯ fn (x(k) ). Step 3: Compute ∆m1 , ∆m2 , ⋯⋯ ∆mn . Step 4: Compute error =max [∣∆m1 ∣ , ∣∆m2 ∣ , ⋯⋯ ∣∆mn ∣] Step 5: If error ≤  (pre - specified tolerance), then the final solution vector is x(k) and print the results. Otherwise go to step 6. Step 6: Form the Jacobin matrix analytically and evaluate it at x = x(k) . T Step 7: Calculate the correction vector ∆x = [∆x1 , ∆x2 , ⋯⋯ ∆xn ] by using equation (2.36). Step 8: Update the solution vector x(k+1) = x(k) +∆x and update k = k+1 and go back to step 2. With this basic understanding of NR technique, we will now discuss the application of NR technique for load flow solution. We will first discuss the Newton Raphson load- flow (NRLF) in polar co-ordinates.

2.9

Newton Raphson load flow (NRLF) in polar co-ordinates

For NRLF techniques, the starting equations are same as those in equations (2.27) and (2.28), which are reproduced below: n

Pi = ∑ Vi Vj Yij cos(θi − θj − αij )

(2.37)

j=1 n

Qi = ∑ Vi Vj Yij sin(θi − θj − αij )

(2.38)

j=1

Now, as before let us again assume that in a ‘n’ bus, ‘m’ machine system, the first ‘m’ buses are the generator buses with bus 1 being the slack bus. Therefore, the unknown quantities are; θ2 , θ3 , ⋯⋯ θn (total ‘n-1’ quantities) and Vm+1 , Vm+2 , ⋯⋯ Vn (total ‘n-m’ quantities). Thus the total number of unknown quantities is n − 1 + n − m = 2n − m − 1. Against these unknown quantities, the sp sp specified quantities are; P2sp , P3sp , ⋯⋯ Pnsp (total ‘n-1’ quantities) and Qsp m+1 , Qm+2 , ⋯⋯ Qn (total ‘n-m’ quantities). Hence, the total number of specified quantities is also (2n − m − 1). Let the T T vectors of unknown quantities be denoted as θ = [θ2 , θ3 , ⋯⋯ θn ] and V = [Vm+1 , Vm+2 , ⋯⋯ Vn ] . T Similarly let the vector of the specified quantities be denoted as Psp = [P2sp , P3sp , ⋯⋯ Pnsp ] and sp sp Qsp = [Qsp m+1 , Qm+2 , ⋯⋯ Qn ]. Also note from equations (2.37) and (2.38) that the real and reactive power injections at any bus are functions of θ and V. Thus, these injection quantities can be 40

written as Pi = Pi (θ, V) for i = 2, 3, ⋯⋯ n and Qi = Qi (θ, V) for i = (m + 1), (m + 2), ⋯⋯ n. (0) For proceeding with NRLF, we assume initial guesses of the bus voltage angles (θ ) and the bus voltage magnitudes (V(0) ). Subsequently, Taylor’s series expansion of equations (2.37) and (2.38) yields (following the same procedure as in the basic N-R technique),

⎡ ∂P2 ⎢ ⎢ ∂θ ⎢ 2 ⎢ ⎢ ⋮ ⎢ ⎢ ∂Pn ⎢ ⎢ ∂θ ⎢ 2 ⎢ ⎢ ∂Qm+1 ⎢ ⎢ ∂θ ⎢ 2 ⎢ ⎢ ⋮ ⎢ ⎢ ∂Qn ⎢ ⎢ ∂θ ⎣ 2

∂P2 ∂P2 ∂θn ∂Vm+1 ⋮ ⋮ ∂Pn ∂Pn ⋯ ∂θn ∂Vm+1 ∂Qm+1 ∂Qm+1 ⋯ ∂θn ∂Vm+1 ⋮ ⋮ ∂Qn ∂Qn ⋯ ∂θn ∂Vm+1 ⋯

∂P2 ⎤⎥ ⎡ (0) sp ⎤ ⎡ ∆θ2 ⎤⎥ ⎢⎢ P2 − P2 (θ , V(0) ) ⎥⎥ ⎢ ⎥ ∂Vn ⎥ ⎢ ⎥ ⎢ ⎥ ⋮ ⎥ ⎥⎢ ⎥ ⋮ ⎥⎥ ⎢⎢ ⋮ ⎥⎥ ⎢⎢ ⎥ ⎥ ⋮ ∂Pn ⎥⎥ ⎢⎢ ⋮ ⎥⎥ ⎢⎢ ⎥ (0) sp ⎥ ⎢ ⋯ (0) ⎥ ⎢ ⎥ (θ ) P − P , V ⎥ n ∂Vn ⎥ ⎢ ∆θn ⎥ ⎢⎢ n ⎥ ⎥ = ⎢ sp ⎥⎢ ⎥ (0) (0) ∂Qm+1 ⎥⎥ ⎢⎢∆Vm+1 ⎥⎥ ⎢Q , V )⎥⎥ m+1 − Qm+1 (θ ⎢ ⋯ ⎥ ∂Vn ⎥⎥ ⎢⎢ ⋮ ⎥⎥ ⎢⎢ ⋮ ⎥ ⎥ ⎢ ⎥⎢ ⎥ ⋮ ⎥⎥ ⎢⎢ ⋮ ⎥⎥ ⎢ ⎥ ⋮ ⎥ ⎥ ⎢⎢ ∂Qn ⎥⎥ ⎢⎢ ⎥ ⎢ Qsp − Q (θ (0) , V(0) ) ⎥⎥ ∆V ⋯ m+1 ⎦ ⎣ n n ⎦ ∂Vn ⎥⎦ ⎣ ⋯

(2.39)

(0)

In equation (2.39), the quantity Pi (θ , V(0) ) is nothing but the calculated value of Pi with (0) (0) vectors θ , V(0) . As a result, commonly, the quantity Pi (θ , V(0) ) is denoted as Pical . With these notations, equation (2.39) can be written as,

Psp − Pcal J1 J2 ∆θ ∆P ][ ] = [ sp [ ]=[ ] cal J3 J4 ∆V Q −Q ∆Q

(2.40)

In equation (2.40) the vectors Pcal and Qcal are defined as; Pcal = [P2cal , P3cal , ⋯⋯ Pncal ] and cal cal sp are of dimension (n − 1) × 1 Qcal = [Qcal m+1 , Qm+2 , ⋯⋯ Qn ]. Also note that the vectors θ and P each and the vectors V and Qsp are of dimension (n − m) × 1 each. Therefore, from equations (2.39) and (2.40), T

⎡ ∂P2 ∂P2 ∂P2 ⎤ ⎢ ⎥ ⋯ ⎢ ∂θ2 ∂θ3 ∂θn ⎥⎥ ⎢ ⎢ ⎥ ⎢ ∂P3 ∂P3 ⎥ ∂P 3 ⎢ ⎥ ∂P ⎢ ⋯ ⎥ = ⎢ ∂θ2 ∂θ3 J1 = ∂θn ⎥⎥ ∂θ ⎢ ⎢ ⋮ ⋮ ⋮ ⎥⎥ ⎢ ⎢ ∂Pn ∂Pn ∂Pn ⎥⎥ ⎢ ⋯ ⎢ ⎥ ⎣ ∂θ2 ∂θ3 ∂θn ⎦

(2.41)

⎡ ∂P2 ∂P2 ∂P2 ⎤ ⎢ ⎥ ⋯ ⎢ ∂Vm+1 ∂Vm+2 ∂Vn ⎥⎥ ⎢ ⎢ ⎥ ⎢ ∂P3 ⎥ ∂P ∂P 3 3 ⎢ ⎥ ∂P ⎢ ⋯ ⎥ = ⎢ ∂Vm+1 ∂Vm+2 J2 = ∂Vn ⎥⎥ ∂V ⎢ ⎢ ⋮ ⋮ ⋮ ⎥⎥ ⎢ ⎢ ∂Pn ∂Pn ∂Pn ⎥⎥ ⎢ ⋯ ⎢ ⎥ ⎣ ∂Vm+1 ∂Vm+2 ∂Vn ⎦

(2.42)

41

⎡ ∂Qm+1 ∂Qm+1 ∂Qm+1 ⎤ ⎥ ⎢ ⋯ ⎢ ∂θ2 ∂θ3 ∂θn ⎥⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ∂Qm+2 ∂Qm+2 ∂Q m+2 ⎥ ⎢ ∂Q ⎢ ⋯ ⎥ J3 = = ⎢ ∂θ2 ∂θ3 ∂θn ⎥⎥ ∂θ ⎢ ⎢ ⋮ ⋮ ⋮ ⎥⎥ ⎢ ⎢ ∂Qn ∂Qn ⎥⎥ ∂Qn ⎢ ⋯ ⎥ ⎢ ⎣ ∂θ2 ∂θ3 ∂θn ⎦

(2.43)

⎡ ∂Qm+1 ∂Qm+1 ∂Qm+1 ⎤ ⎥ ⎢ ⋯ ⎢ ∂Vm+1 ∂Vm+2 ∂Vn ⎥⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ∂Qm+2 ∂Qm+2 ∂Q m+2 ⎥ ⎢ ∂Q ⎢ ⋯ ⎥ J4 = = ⎢ ∂Vm+1 ∂Vm+2 ∂Vn ⎥⎥ ∂V ⎢ ⎢ ⋮ ⋮ ⋮ ⎥⎥ ⎢ ⎢ ∂Qn ∂Qn ∂Qn ⎥⎥ ⎢ ⋯ ⎥ ⎢ ⎣ ∂Vm+1 ∂Vm+2 ∂Vn ⎦

(2.44)

In equations (2.41) to (2.44) the sizes of the various matrices are as follows: J1 → (n−1)×(n−1), J2 → (n − 1) × (n − m), J3 → (n − m) × (n − 1) and J4 → (n − m) × (n − m). Now, equation (2.40) can be written in compact form as,

[J] [∆X] = [∆M ]

(2.45)

J1 J2 ] is known as the Jacobian matrix, the vector ∆X = J3 J4 ∆θ Psp − Pcal [ ] is known as the correction vector and the vector ∆M = [ sp ] is known as the ∆V Q − Qcal mismatch vector. Further, the size of the matrix J is (2n − m − 1) × (2n − m − 1) while the sizes of both the vectors ∆X and ∆M is (2n − m − 1) × 1. In equation (2.45), the matrix J = [

Equation (2.45) forms the basis of the NRLF (polar) algorithm, which is described below. Please note that in the algorithm described below it is assumed that there is no generator which violates its reactive power generation or absorption limit. The case of violation of reactive power generation or absorption limit would be dealt with a little later. Basic NRLF (polar) algorithm (0)

(0)

Step 1: Initialise V¯j = Vjsp ∠0o for j = 2, 3, ⋯⋯ m and V¯j = 1.0∠0o for j = (m + 1), (m + 2), ⋯⋯ n. Let the vectors of the initial voltage magnitudes and angles be denoted as V(0) and θ (0) respectively. Step 2: Set iteration counter k = 1. (k−1) Step 3: Compute the vectors Pcal and Qcal with the vectors θ and V(k−1) thereby forming T the vector ∆M . Let this vector be represented as ∆M = [∆M1 , ∆M2 , ⋯⋯ ∆M2n−m−1 ] . Step 4: Compute error = max (∣∆M1 ∣ , ∣∆M2 ∣ , ⋯⋯ ∣∆M2n−m−1 ∣). (k−1) Step 5: If error ≤  (pre - specified tolerance), then the final solution vectors are θ and 42

V(k−1) and print the results. Otherwise go to step 6.

(k−1)

Step 6: Evaluate the Jacobian matrix with the vectors θ and V(k−1) . Step 7: Compute the correction vector ∆X by solving equation (2.45). (k) Step 8: Update the solution vectors θ = θ (k−1) + ∆θ and V(k) = V(k−1) + ∆V . Update k = k + 1 and go back to step 3. In the above algorithm, the Jacobian matrix needs to be evaluated at each iteration. Therefore, the element of the Jacobian matrix needs to be found out analytically. This is discussed next.

Formation of Jacobian matrix elements for NRLF (polar) technique To derive the elements of the Jacobian matrix, let us revisit equations (2.37) and (2.38). n

n

Pi = ∑ Vi Vj Yij cos(θi − θj − αij ) = V Gii + ∑ Vi Vj Yij cos(θi − θj − αij ) 2 i

j=1

j =1 ≠i

n

n

j=1

j =1 ≠i

Qi = ∑ Vi Vj Yij sin(θi − θj − αij ) = −Vi2 Bii + ∑ Vi Vj Yij sin(θi − θj − αij )

(2.46)

(2.47)

In the above two equations, the relations Gii = Yii cos(αii ) and Bii = Yii sin(αii ) have been used. From the expressions of the Pi and Qi in equations (2.46) and (2.47) respectively, the elements of the Jacobian matrix can be calculated as follows. Matrix J1 (=

∂P ) (in this case, i = 2, 3, ⋯⋯ n, ∂θ

j = 2, 3, ⋯⋯ n)

n ∂Pi = − ∑ Vi Vk Yik sin(θi − θk − αik ); ∂θj k=1

j=i

(2.48)

≠i

∂Pi = Vi Vj Yij sin(θi − θj − αij ); ∂θj Matrix J2 (=

∂P ) (in this case, i = 2, 3, ⋯⋯ n, ∂V

j≠i

(2.49)

j = (m + 1), (m + 2), ⋯⋯ n)

n ∂Pi = 2Vi Gii + ∑ Vk Yik cos(θi − θk − αik ); ∂Vj k=1

j=i

(2.50)

≠i

∂Pi = Vi Yij cos(θi − θj − αij ); ∂Vj 43

j≠i

(2.51)

Matrix J3 (=

∂Q ) (in this case, i = (m + 1), (m + 2), ⋯⋯ n, ∂θ

j = 2, 3, ⋯⋯ n)

n ∂Qi = ∑ Vi Vk Yik cos(θi − θk − αik ); ∂θj k = 1

j=i

(2.52)

≠i

∂Qi = −Vi Vj Yij cos(θi − θj − αij ); ∂θj Matrix J4 (=

∂Q ) (in this case, i = (m + 1), (m + 2), ⋯⋯ n, ∂V

j≠i

(2.53)

j = (m + 1), (m + 2), ⋯⋯ n).

n ∂Qi = −2Vi Bii + ∑ Vk Yik sin(θi − θk − αik ); ∂Vj k=1

j=i

(2.54)

≠i

∂Qi = Vi Yij sin(θi − θj − αij ); ∂Vj

j≠i

(2.55)

With these expressions of Jacobian elements given in equations (2.48)-(2.55), the Jacobian matrix can be evaluated at each iteration as discussed earlier. Now, in the basic NRLF (polar) algorithm described earlier, the generator Q-limits have not been considered. To accommodate the generator Q-limits, at the beginning of each iteration, reactive power absorbed or produced by each generator is calculated. If the calculated reactive power is within the specified limits, the generator is retained as PV bus, otherwise the generator bus is converted to a PQ bus, with the voltage at this bus no longer held at the specified value. The detailed algorithm is as follows. Complete NRLF (polar) algorithm (0)

(0)

Step 1: Initialise V¯j = Vjsp ∠0o for j = 2, 3, ⋯⋯ m and V¯j = 1.0∠0o for j = (m + 1), (m + 2), ⋯⋯ n. Let the vectors of the initial voltage magnitudes and angles be denoted as V(0) and θ (0) respectively. Step 2: Set iteration counter k = 1. Step 3: For i = 2, 3, ⋯⋯ m, carry out the following operations. a) Calculate, n

= ∑ Vi(k−1) Vj(k−1) Yij sin (θi(k−1) − θj(k−1) − αij ) Q(k) i j=1

(k)

(k)

b) If, Qmin ≤ Qi ≤ Qmax ; then assign ∣V¯i ∣ = Vispec and the ith bus is retained as PV bus i i for k th iteration. (k) (k) sp max min c) If Qi > Qmax , then assign Qsp or, if Qi < Qmin . In i i = Qi i , then assign Qi = Qi 44

both the cases, this bus is converted to PQ bus. Hence, its voltage magnitude becomes an unknown for the present iteration (thereby introducing an extra unknown quantity) and to solve for this extra unknown quantity, an extra equation is required, which is obtained by the new value of Qsp i (as th shown above). Therefore, when the i bus is converted to a PQ bus, the dimensions of both ∆V and ∆Q vectors increases by one. In general, if l generator buses (l ≤ (m − 1)) violate their corresponding reactive power limits at step 3, then the dimensions of both ∆V and ∆Q vectors increases from (n − m) to (n − m + l). However, the dimensions of both ∆P and ∆θ vectors remain the same. Therefore, the size of matrix J2 becomes (n − 1) × (n − m + l), that of matrix J3 becomes (n − m + l) × (n − 1) and the matrix J4 becomes of size (n − m + l) × (n − m + l). The size of matrix J1 , however, does not change. Hence, the size of the matrix J becomes (2n − m − 1 − l) × (2n − m − 1 − l) while the sizes of both the vectors ∆X and ∆M (in equation (2.45)) becomes (2n − m − 1 − l) × 1. Of course, if there is no generator reactive power limit violation, then l = 0. (k−1) Step 4: Compute the vectors Pcal and Qcal with the vectors θ and V(k−1) thereby forming T the vector ∆M . Let this vector be represented as ∆M = [∆M1 , ∆M2 , ⋯⋯∆M2n−m−1−l ] . Step 5: Compute error = max (∣∆M1 ∣ , ∣∆M2 ∣ , ⋯⋯ ∣∆M2n−m−1−l ∣). (k−1) Step 6: If error ≤  (pre - specified tolerance), then the final rotation vectors are θ and (k−1) V and print the results. Otherwise go to step 7. (k−1) Step 7: Evaluate the Jacobian matrix with the vectors θ and V(k−1) . Step 8: Compute the correction vector ∆X by solving equation (2.45). (k) Step 9: Update the solution vectors θ = θ (k−1) + ∆θ and V(k) = V(k−1) + ∆V . Update k = k + 1 and go back to step 3. In the next lecture, we will look at an example of NRLF (polar) technique.

45

2.9.1

Example for NRLF (polar) technique

As an example, let us consider again the 5-bus system shown in Fig. 2.16. In this System, n = 5 and m = 3 (as the number of generator in 3). Initially, let us assume that there is no violation of reactive power limit at any generator. Therefore, the sizes of The Jacobian sub-matinees are as follows; J1 → (4 × 4), J2 → (4 × 2), J3 → (2 × 4) and J4 → (2 × 2). Therefore, the size of the combined Jacobian matrix J is (6×6). As before, the NRLF algorithm starts with flat voltage profile and with this assumed voltage profile, the different quantities in the first iteration are calculated as shown in Table 2.10: Table 2.10: Initial calculation with NRLF (polar) in the 5 bus system

Pcal = [−0.0444 −0.1776 −0.0333 −0.0444] × 10−14 ; T

T

Qcal = [−0.143 −0.143] ; T

∆M = [0.5 1.0 −1.15 −0.85 −0.457 −0.257] ;

error = 1.15;

As the error is more than the tolerance value ( = 10−12 p.u), the algorithm proceeds and in the next few steps, the Jacobian matrix, correction vector and the updated values of the error are calculated as shown in Table 2.11. As the error is still more than the tolerance value, the algorithm continues and enters 2nd iteration. The values of the relevant qualities in second iteration are shown in Table 2.12 below. Since the error is still more than the tolerance value, the algorithm continues and finally converges after 4 iterations. The final converged solution is shown in Table 2.13. Comparsion of Tables 2.13 and 2.4 (GSLF results) shows that the final results obtained by these two methods are identically same. However, NRLF achieves this solution in 4 iterations as opposed at 69 iterations required by GSLF. The convergence behavior of both GSLF and NRLF are shown in Fig. 2.17. From this figure it can be observed that, NRLF has certainly much better convergence characteristics as compared to GSLF. Now let us consider the case where reactive power generation of generator 3 is limited to 50 MVAR. The algorithm again starts with a flat start and the initial calculation are same as shown in Table 2.10 earlier. As Q3 calculated has not crossed the limit, the program proceeds in the same way as shown in Table 2.11 till the end of first iteration. With the voltage magnitudes and angles obtained at the end of first iteration, the reactive power generated by all the machines are again calculated and the value of Q3 is found to be 53.42 MVAR (0.5342 p.u). Hence, bus 3 (generator 3) is converted to a PQ bus and hence, the PQ buses in the system now are (4, 5, 3). The vectors of calculated injected real and reactive powers, mismatch vector and final mismatch at the end of 1st iteration are shown in Table 2.14. Note that without any generator reactive power violation, the size of ∆M vector was (6 × 1) and with generator limit violation in Q3 , the size has now increased to (7 × 1). At the end of 1st iteration, n = 5 and m = 2 (as the number of generator buses is 2). Hence, the sizes of the various Jacobian sub-matrices are: J1 → (4 × 4), J2 → (4 × 3), J3 → (3 × 4) and 46

Table 2.11: Calculation at 1st iteration with NRLF (polar) in the 5 bus system without any generator Q limit violation

⎡ 13.0858 ⎢ ⎢ ⎢−7.4835 ⎢ J1 = ⎢ ⎢ 0 ⎢ ⎢ ⎢ 0 ⎣ J3 = [

−7.4835 0 0 ⎤⎥ ⎥ 19.0911 −7.1309 −4.4768⎥⎥ ⎥; −7.1309 10.8657 −3.7348⎥⎥ ⎥ −4.4768 −3.7348 15.6951 ⎥⎦

0 1.2584 −2.1921 0.9337 ]; 0 1.1298 0.9337 −3.9047

⎡ 13.0858 ⎢ ⎢ ⎢−7.4835 ⎢ ⎢ ⎢ 0 ⎢ J=⎢ ⎢ 0 ⎢ ⎢ ⎢ 0 ⎢ ⎢ ⎢ 0 ⎣

⎤ ⎡ 0 0 ⎥ ⎢ ⎥ ⎢ ⎢−1.2584 −1.12982⎥ ⎥ ⎢ ⎥ J2 = ⎢ ⎢ 2.1921 −0.9337 ⎥ ⎥ ⎢ ⎥ ⎢ ⎢−0.9337 3.9047 ⎥ ⎦ ⎣

10.5797 −3.7348 J4 = [ ]; −3.7348 15.4091

−7.4835 0 0 0 0 ⎤⎥ ⎥ 19.0911 −7.1309 −4.4768 −1.2584 −1.1298⎥⎥ ⎥ −7.1309 10.8657 −3.7348 2.1921 −0.9337⎥⎥ ⎥; −4.4768 −3.7348 15.6951 −0.9337 3.9047 ⎥⎥ ⎥ 1.2584 −2.1921 0.9337 10.5797 −3.7348⎥⎥ ⎥ 1.1298 0.9337 −3.9047 −3.7348 15.4091 ⎥⎦ T

∆X = [0.0331 −0.0090 −0.1275 −0.0799 −0.0783 −0.0475] ; T

∆θ = [0.0331 −0.0090 −0.1275 −0.0799] ;

∆V = [−0.0783 −0.0475]

T

T

θ = [0 0.0331 −0.0090 −0.1275 −0.0799] ; T

V = [1.0000 1.0000 1.0000 0.9217 0.9525] ; T

Pcal = [0.5023 0.9293 −1.0413 −0.8128] ;

T

Qcal = [−0.5158 0.3472] ; T

∆M = [−0.0023 0.0707 −0.1087 −0.0372 −0.0842 −0.0528] ;

error = 0.1087;

J4 → (3 × 3). Thus, the size of the Jacobian matrix increases to (7 × 7). The Jacobian matrix, correction vector and the updated value of the mismatch as computed in the 2nd iteration are shown in Table 2.15. As the mismatch is still more than the tolerance value, the algorithm proceeds further and finally, the algorithm converges in 5 iterations. The final converged values are shown in Table 2.16. In this table also, the NRLF (polar) results without any reactive power limit (as shown in Table 2.13) are also reproduced for comparison. Moreover, it is observed that the final converged values are identically same as those calculated by the GSLF method (Table 2.7). The results with IEEE - 14 bus system are shown in Table 2.17 with and without limit on generator reactive power. For this case also, the limit on the generator at bus 6 has been maintained at 30 MVAR. Comparison of Tables 2.8 and 2.17 shows that the results obtained by GSLF and NRLF are identical, but due to quadratic convergence characteristics, the number of iteration required by 47

Table 2.12: Calculation at 2nd iteration with NRLF (polar) in the 5 bus system without any generator Q limit violation

⎡ 13.1998 ⎢ ⎢ ⎢−7.3995 ⎢ J1 = ⎢ ⎢ 0 ⎢ ⎢ ⎢ 0 ⎣ J3 = [

−7.5543 0 0 ⎤⎥ ⎥ 18.3929 −6.6638 −4.3296⎥⎥ ⎥; −6.3896 9.6258 −3.2363⎥⎥ ⎥ −4.1771 −3.3143 14.3457 ⎥⎦

0 1.9289 −2.9037 0.9748 ]; 0 1.3756 0.6628 −4.3554

⎡ 13.1998 ⎢ ⎢ ⎢−7.3995 ⎢ ⎢ ⎢ 0 ⎢ J=⎢ ⎢ 0 ⎢ ⎢ ⎢ 0 ⎢ ⎢ ⎢ 0 ⎣

⎡ 0 0 ⎤⎥ ⎢ ⎥ ⎢ ⎢−0.4065 −0.8098⎥ ⎥ ⎢ ⎥; J2 = ⎢ ⎢ 0.8909 −1.0234⎥ ⎥ ⎢ ⎥ ⎢ ⎢−0.7191 2.8658 ⎥ ⎦ ⎣

9.3239 −3.3977 J4 = [ ]; −3.5957 14.4487

−7.5543 0 0 0 0 ⎤⎥ ⎥ 18.3929 −6.6638 −4.3296 −0.4065 −0.8098⎥⎥ ⎥ −6.3896 9.6258 −3.2363 0.8909 −1.0234⎥⎥ ⎥; −4.1771 −3.3143 14.4567 −0.7191 2.8658 ⎥⎥ ⎥ 1.9289 −2.9037 0.9748 9.3239 −3.3977⎥⎥ ⎥ 1.3756 0.6628 −4.3554 −3.5957 14.4487 ⎥⎦ T

∆X = [−0.0041 −0.0068 −0.0179 −0.0154 −0.0783 −0.0084] ; T

∆θ = [0.0331 −0.0090 −0.1275 −0.0799] ;

T

∆V = [−0.0783 −0.0475] ; T

θ = [0 0.0290 −0.0158 −0.1453 −0.0876] ; T

V = [1.0000 1.0000 1.0000 0.9063 0.9441] ; T

Pcal = [0.5000 0.9988 −1.1474 −0.8501] ;

T

Qcal = [−0.5980 −0.3990] ; T

∆M = [−0.0000 0.0012 −0.0026 0.0001 −0.0020 −0.0010] ;

error = 0.0026 ;

NRLF to reach the same solution is much less compared to that taken by GSLF for a tolerance of 10e−12 p.u. The results for 30-bus system are shown in Table 2.18. Without any generator Q-limit, the results obtained by GSLF and NRLF (polar) are identical, although NRLF (polar) takes only 4 iterations against 851 iterations taken by GSLF. Also,as mentioned earlier, with a tolerance of 10e−12 (p.u), GSLF does not converge for Q limit on gen 11 and 13 (20 and 30 MVAR respectively). However, NRLF (polar) does not face any such difficulty in convergence in this case and the algorithm converges with 7 iterations in this case. The corresponding results are also shown in Table 2.18. In the next lecture, we will discuss another version of NRLF, namely, the rectangular version, in which, all the complex quantities are represented in the rectangular co-ordinates instead of polar co-ordinates as is done in the case of NRLF (polar) technique.

48

Table 2.13: Final Results of the 5 bus system with NRLF (polar) without any generator Q limit violation Bus no. 1 2 3 4 5

∣V ∣

Without generator Q limit

θ

Pinj

Qinj

(p.u) (deg) (p.u) (p.u) 1.0 0 0.56743 0.26505 1.0 1.65757 0.5 -0.18519 1.0 -0.91206 1.0 0.68875 0.90594 -8.35088 -1.15 -0.6 0.94397 -5.02735 -0.85 -0.4 Total iteration = 4

Figure 2.17: Convergence characteristics of GSLF and NRLF Table 2.14: Calculations at the end of 1st iteration with NRLF (polar) in the 5 bus system for limit on Q3 T

Pcal = [0.5023 0.9293 −1.0413 −0.8128] ;

T

Qcal = [−0.5158 −0.3472 0.5342] ; T

∆M = [−0.0023 0.0707 −0.1087 0.0372 −0.0842 −0.0528 −0.0342] ; error = 0.1087 ;

49

Table 2.15: Calculations at 2nd iteration with NRLF (polar) in the 5 bus system for limit on Q3

⎡ 0 0 −1.5250⎤⎥ −7.5543 0 0 ⎤⎥ ⎢ ⎥ ⎢ ⎥ ⎢−0.4065 −0.8098 5.1587 ⎥ 18.3929 −6.6638 −4.3296⎥⎥ ⎥ ⎢ ⎥; ⎥ ; J2 = ⎢ ⎥ ⎢ ⎥ −6.3896 9.6258 −3.2363⎥ ⎢ 0.8909 −1.0234 −1.9289⎥ ⎥ ⎢ ⎥ ⎢−0.7191 2.8658 −1.3756⎥ −4.1771 −3.3143 14.3457 ⎥⎦ ⎦ ⎣ ⎡ 9.3239 −3.3977 −6.3896⎤ ⎡ 0 1.9289 −2.9037 0.9748 ⎤⎥ ⎥ ⎢ ⎢ ⎥ ⎢ ⎥ ⎢ ⎢ ⎥ ⎢ 1.3756 0.6628 −4.3554⎥ ; J4 = ⎢−3.5957 14.4487 −4.1771⎥⎥; J3 = ⎢ 0 ⎥ ⎢ ⎥ ⎢ ⎢−7.2296 −4.5456 19.4614 ⎥ ⎢2.1541 −3.3002 0.3747 0.7713 ⎥ ⎦ ⎣ ⎦ ⎣ ⎡ 13.1998 −7.5543 0 0 0 0 −1.5250⎤⎥ ⎢ ⎢ ⎥ ⎢−7.3995 18.3929 −6.6638 −4.3296 −0.4065 −0.8098 5.1587 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ 0 −6.3896 9.6258 −3.2363 0.8909 −1.0234 −1.9289⎥⎥ ⎢ ⎢ −4.1771 −3.3143 14.4567 −0.7191 2.8658 −1.3756⎥⎥; J=⎢ 0 ⎢ ⎥ ⎢ 0 ⎥ 1.9289 −2.9037 0.9748 9.3239 −3.3977 −6.3896 ⎢ ⎥ ⎢ ⎥ ⎢ 0 ⎥ 1.3756 0.6628 −4.3554 −3.5957 14.4487 −4.1771 ⎢ ⎥ ⎢ ⎥ ⎢ 2.1542 −3.3002 0.3747 0.7713 −7.2296 −4.5456 19.4614 ⎥ ⎣ ⎦

⎡ 13.1998 ⎢ ⎢ ⎢−7.3995 ⎢ J1 = ⎢ ⎢ 0 ⎢ ⎢ ⎢ 0 ⎣

T

∆X = [−0.0033 −0.0022 −0.0174 −0.0069 −0.0310 −0.0173 −0.0167] ; T

∆θ = [−0.0033 −0.0022 −0.0174 −0.0069] ;

T

∆V = [−0.0310 −0.0173 −0.0167] ; T

θ = [0 0.0297 −0.0112 −0.1449 −0.0868] ; T

V = [1.0000 1.0000 0.9833 0.8907 0.9352] ; T

Pcal = [0.5002 0.9949 −1.1421 −0.8498] ;

T

Qcal = [−0.5964 −0.3984 0.5015] ; T

∆M = [−0.0002 0.0051 −0.0073 0.0002 −0.0036 −0.0016 −0.0015] ; error = 0.0073;

50

Table 2.16: Final Results of the 5 bus system with NRLF (polar) with generator Q limit

Bus no. 1 2 3 4 5

∣V ∣

Without generator Q limit

θ

Pinj

Qinj

∣V ∣

With generator Q limit

θ

Pinj

Qinj

(p.u) (deg) (p.u) (p.u) (p.u) (deg) (p.u) (p.u) 1.0 0 0.56743 0.26505 1.0 0 0.56979 0.33935 1.0 1.65757 0.5 -0.18519 1.0 1.69679 0.5 -0.04769 1.0 -0.91206 1.0 0.68875 0.9825 -0.63991 1.0 0.5 0.90594 -8.35088 -1.15 -0.6 0.88918 -8.35906 -1.15 -0.6 0.94397 -5.02735 -0.85 -0.4 0.93445 -4.98675 -0.85 -0.4 Total iteration = 4 Total iteration = 5

Table 2.17: Final Results of the 14 bus system with NRLF (Polar) Without generator Q limit

Bus no.

∣V ∣

1 2 3 4 5 6 7 8 9 10 11 12 13 14

(p.u) 1.06 1.045 1.04932 1.03299 1.04015 1.07 1.02076 1.0224 1.0201 1.0211 1.04144 1.0526 1.04494 1.01249

θ

Pinj

(deg) (p.u) 0 2.37259 -5.17113 0.183 -14.54246 -1.19 -10.39269 -0.4779 -8.76418 -0.07599 -12.52265 0.112 -13.44781 0 -13.47154 0 -13.60908 -0.29499 -13.69541 -0.09 -13.22158 -0.03501 -13.42868 -0.06099 -13.50388 -0.135 -14.60128 -0.14901 Total iteration = 4

Qinj

∣V ∣

(p.u) -0.3308 -0.166 -0.08762 -0.039 -0.01599 0.37278 0 -0.129 -0.16599 -0.05799 -0.018 -0.01599 -0.05799 -0.05001

(p.u) 1.06 1.045 1.04697 1.02902 1.03615 1.05497 1.01266 1.01391 1.0118 1.01154 1.02915 1.03787 1.03063 1.00136

51

With generator Q limit

θ

Pinj

(deg) (p.u) 0 2.37188 -5.17845 0.183 -14.55556 -1.19 -10.35987 -0.4779 -8.71027 -0.07599 -12.45871 0.112 -13.49478 0 -13.5185 0 -13.66101 -0.29499 -13.73679 -0.09 -13.21814 -0.03501 -13.39145 -0.06099 -13.48166 -0.135 -14.64504 -0.14901 Total iteration = 6

Qinj (p.u) -0.31249 -0.1066 -0.08762 -0.039 -0.01599 0.3 0 -0.129 -0.16599 -0.05799 -0.018 -0.01599 -0.05799 -0.05001

Table 2.18: Final Results of the 30 bus system with NRLF (Polar) Without generator Q limit

Bus no.

∣V ∣

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

(p.u) 1.05 1.0338 1.03128 1.02578 1.0058 1.02178 1.00111 1.023 1.04608 1.03606 1.0913 1.04859 1.0883 1.03346 1.02825 1.0359 1.0306 1.01873 1.01626 1.02041 1.02305 1.02343 1.0165 1.00939 1.00048 0.9825 1.00379 1.02049 0.98353 0.97181

θ

Pinj

Qinj

∣V ∣

With generator Q limit

θ

Pinj

Qinj

(deg) (p.u) (p.u) (p.u) (deg) (p.u) (p.u) 0 2.38673 -0.29842 1.05 0 2.3865 -0.29386 -4.97945 0.3586 -0.05698 1.0338 -4.98084 0.35861 -0.04562 -7.96653 -0.024 -0.012 1.03045 -7.95523 -0.02399 -0.012 -9.58235 -0.076 -0.016 1.02477 -9.56929 -0.07598 -0.016 -13.60103 -0.6964 0.05042 1.0058 -13.60836 -0.6964 0.05532 -11.50296 0 0 1.02084 -11.49304 0.00003 0 -13.9994 -0.628 -0.109 1.00055 -13.99792 -0.628 -0.109 -12.56853 -0.45 0.12343 1.023 -12.57567 -0.45 0.15111 -13.04088 0 0 1.04006 -13.02865 0.00001 0 -14.88589 -0.058 -0.02 1.03117 -14.8866 -0.05797 -0.02001 -11.16876 0.1793 0.24018 1.07807 -11.12256 0.1793 0.2 -13.74947 -0.112 -0.075 1.04456 -13.75755 -0.11198 -0.075 -12.56078 0.1691 0.31043 1.08311 -12.55854 0.1691 0.3 -14.71704 -0.062 -0.016 1.02931 -14.73168 -0.062 -0.016 -14.86737 -0.082 -0.025 1.02403 -14.88097 -0.08199 -0.025 -14.50539 -0.035 -0.018 1.03148 -14.51198 -0.035 -0.018 -14.98291 -0.09 -0.058 1.02583 -14.98712 -0.09 -0.058 -15.58107 -0.032 -0.009 1.01422 -15.59618 -0.03199 -0.009 -15.81066 -0.095 -0.034 1.01159 -15.8251 -0.09499 -0.034 -15.63819 -0.022 -0.007 1.01568 -15.64961 -0.022 -0.007 -15.35955 -0.175 -0.112 1.01825 -15.36524 -0.17496 -0.11201 -15.35222 0 0 1.01867 -15.3581 0 0 -15.41998 -0.032 -0.016 1.01226 -15.43637 -0.032 -0.016 -15.81043 -0.087 -0.067 1.00517 -15.8277 -0.087 -0.067 -15.84004 0 0 0.99748 -15.86849 0 0 -16.27422 -0.035 -0.023 0.97944 -16.30534 -0.035 -0.023 -15.59587 0 0 1.00158 -15.62827 0 0 -12.1474 0 0 1.01959 -12.14235 0 0 -16.87497 -0.024 -0.009 0.98126 -16.91314 -0.024 -0.009 -17.79427 -0.106 -0.019 0.96951 -17.83675 -0.106 -0.019 Total iteration = 4 Total iteration = 7

52

2.10

NRLF in Rectangular co-ordinates [NRLF (Rect.)]

In rectangular co-ordinates, every complex quantity is expressed in terms of its real and imaginary parts. Hence, let V¯k = Vk ejθk = ek + jfk ; I¯i = ai + jci and Y¯ik = gik + jbik . n

n

k=1

k=1

Thus, from equation (2.25), I¯i = ∑ Y¯ik V¯k . Or, ai + jci = ∑(gik + jbik )(ek + jfk ). Or,

⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ k=1 ⎬ n ⎪ ⎪ = ∑ (bik ek + gik fk ) ⎪ ⎪ ⎪ ⎪ k=1 ⎭ n

ai = ∑ (gik ek − bik fk ) ci

(2.56)

Complex power injected at bus ‘i’ is, S¯i = Pi + jQi = V¯i I¯i∗ . Or, Pi + jQi = (ei + jfi )(ai − jci ). Or,

n

Pi = ai ei + ci fi = ∑[ei (gik ek − bik fk ) + fi (bik ek + gik fk )]

(2.57)

k=1

and

n

Qi = fi ai − ei ci = ∑[fi (gik ek − bik fk ) − ei (bik ek + gik fk )]

(2.58)

k=1

Equations (2.57) and (2.58) can be re-written as; n

Pi = gii (e2i + fi2 ) + ∑ [ei (gik ek − bik fk ) + fi (bik ek + gik fk )] k=1 ≠i

(2.59)

n

Qi = −bii (e2i + fi2 ) + ∑ [fi (gik ek − bik fk ) − ei (bik ek + gik fk )] k=1 ≠i

(2.60)

Now, again let us consider a ‘n’ bus system having ‘m’ generators (bus 1 being the slack bus). For each of the ‘(n-m)’ PQ buses, both voltage magnitude and angle are unknown. In other words, for these buses both real and imaginary parts of the voltages are unknown. For each of the ‘(m-1)’ PV buses, even though the voltage magnitude is known, real and imaginary parts of the voltage are not known as there can be many combination of ei and fi to give a specified value of Vi . Hence, for each of the PV buses also, the real and imaginary parts of the voltage are unknown. Therefore, total number of unknown quantities is = 2(n − m) + 2(m − 1) = (2n − 2). To solve for these (2n − 2) unknown quantities, (2n − 2) independent equations are also needed. Now, let us look at the specified quantities. As discussed earlier, for each of the ‘(n-1)’ buses, the quantity Pi is specified [given in equation (2.59)]. Similarly, for each of the ‘(n-m)’ buses, the quantity Q1 is also known [given in equation (2.60)]. Thus, total number of specified quantities is (n − 1 + n − m) = (2n − m − 1). Hence, still 2n − 2 − (2n − m − 1) = (m − 1) specified quantities are needed to make the NRLF (rectangular) a well posed problem. These additional specified quantities would be available from the specified voltage magnitudes at each of the ‘(m-1) buses. At each of these ‘(m-1)’ PV buses, following relation holds good between the specified quantity (Vi ) and the 53

unknown quantities (ei , fi ).

Vi2 = e2i + fi2

(2.61)

Thus, in NRLF rectangular co-ordinates method, the unknown quantities are; (ei , fi ) for i = 2, 3, ⋯⋯ n (total (2n − 2) in number). The specified quantities are; a) Pi ; for i = 2, 3, ⋯⋯ n; b) Qi ; for i = (m + 1)⋯⋯ n and c) Vi ; for i = 2, 3, ⋯⋯ m. Hence, the total number of specified quantities is also (2n − 2). The relations connecting the unknown quantities to the specified quantities are given by equations (2.59) - (2.61), which are solved by the standard Newton-Rabhosn technique to determine the unknown quantities. As in the case of NRLF (polar) method, in NRLF (rectangular) technique also, flat voltage profile is assumed at the starting. The standard, linearized form of Newton-Raphron solution of the equations (2.59) - (2.61) are given below (following the basic Newton-Raphron solution method discussed previously);

⎡ ∆P ⎤ ⎡J J ⎤ ⎢ ⎥ ⎢ 1 2⎥ ⎥ ⎢ ⎥ ⎢ ⎢ ∆Q ⎥ = ⎢J3 J4 ⎥ [∆e] ⎥ ∆f ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎢∆V2 ⎥ ⎢J5 J6 ⎥ ⎦ ⎣ ⎦ ⎣

(2.62)

In equation (2.62), the sizes of different vector are as follows: ∆e → (n − 1) × 1; ∆f → (n − 1) × 1; ∆P → (n − 1) × 1; ∆Q → (n − m) × 1; ∆V2 → (m − 1) × 1. From these sizes

∂P → ∂e ∂Q J4 → → ∂f

of the vectors, the sizes of different sub-matrices J1 ⋯⋯ J6 can be deduced as; J1 →

(n − 1) × (n − 1); (n − m) × (n − 1);

∂Q ∂P → (n − 1) × (n − 1); J3 → → (n − m) × (n − 1); ∂f ∂e ∂V 2 ∂V 2 J5 → → (m − 1) × (n − 1); J6 → → (m − 1) × (n − 1). ∂e ∂f

J2 →

Equation (2.62) can be succinctly written as,

[J] [∆X] = [∆M ]

(2.63)

In equation (2.63), as before, the quantities ∆M , ∆X and J are known as the mismatch vector, correction vector and the Jacobian matrix respectively. Note that the vector ∆M and ∆X each has a size of (2n − 2) × 1 whereas the Jacobian matrix has a size of (2n − 2) × (2n − 2). Equation (2.63) forms the basis of NRLF (rectangular) algorithm as discussed below. Note that the algorithm discussed below assumes that there is no violation of generator reactive power limits.

Basic NRLF (Rectangular) algorithm Step 1: Assume flat start profile and denote the initial real and imaginary parts of the bus voltages as e(0) and f (0) respectively. Step 2: Set iteration counter k = 1. Step 3: Compute the vectors Pcal and Qcal with the vectors e(k−1) and f (k−1) thereby forming T the vector ∆M . Let this vector be represented as ∆M = [∆M1 , ∆M2 , ⋯⋯ ∆M2n−2 ] . 54

Step 4: Compute error = max (∣∆M1 ∣ , ∣∆M2 ∣ , ⋯⋯ ∣∆M2n−2 ∣). Step 5: If error ≤  (pre - specified tolerance), then the final solution vectors are e(k−1) and f (k−1) and print the results. Otherwise go to step 6. Step 6: Evaluate the Jacobian matrix with the vectors e(k−1) and f (k−1) . Step 7: Compute the correction vector ∆X by solving equation (2.63). Step 8: Update the solution vectors e(k) = e(k−1) +∆e and f (k) = f (k−1) +∆f . Update k = k +1 and go back to step 3. As can be seen, in step 6, the Jacobian matrix needs to be evaluated at each iteration. For this purpose, the analytical expressions of the elements of the Jacobian matrix are needed, which are derived next. Derivation of Jacobian matrix elements for NRLF (rectangular) technique We derive the elements of the Jacobian matrix by utilising equations (2.59), (2.60) and (2.61). From these three equations, the Jacobian matrix elements can be obtained as follows: Matrix J1 (=

∂P ) (in this case, i = 2, 3, ⋯⋯ n, ∂e

j = 2, 3, ⋯⋯ n)

n ∂Pi = 2ei gii + ∑ (gik ek − bik fk ); ∂ej k=1

j=i

(2.64)

≠i

∂Pi = (ei gij + fi bij ); ∂ej Matrix J2 (=

∂P ) (in this case, i = 2, 3, ⋯⋯ n, ∂f

j≠i

(2.65)

j = 2, 3, ⋯⋯ n)

n ∂Pi = 2fi gii + ∑ (bik ek + gik fk ); ∂fj k=1

j=i

(2.66)

≠i

∂Pi = (fi gij − ei bij ); ∂fj Matrix J3 (=

j≠i

∂Q ) (in this case, i = (m + 1), (m + 2), ⋯⋯ n, ∂e n ∂Qi = −2ei bii + ∑ (−bik ek − gik fk ); ∂ej k=1

(2.67)

j = 2, 3, ⋯⋯ n) j=i

(2.68)

≠i

∂Qi = (fi gij − ei bij ); ∂ej 55

j≠i

(2.69)

Matrix J4 (=

∂Q ) (in this case, i = (m + 1), (m + 2), ⋯⋯ n, ∂f n ∂Qi = −2fi bii + ∑ (gik ek − bik fk ); ∂fj k=1

j = 2, 3, ⋯⋯ n) j=i

(2.70)

≠i

∂Qi = −(fi bij + ei gij ); ∂fj Matrix J5 (=

∂V 2 ) (in this case, i = 2, 3, ⋯⋯ m, ∂e ∂Vi2 = 2ei for j = i; ∂ej

Matrix J6 (=

∂Vi2 = 2fi for j = i; ∂fj

(2.71)

j = 2, 3, ⋯⋯ n)

and

∂V 2 ) (in this case, i = 2, 3, ⋯⋯ m, ∂f

j≠i

∂Vi2 = 0 for j ≠ i; ∂ej

(2.72)

j = 2, 3, ⋯⋯ n)

∂Vi2 and = 0 for j ≠ i; ∂fj

(2.73)

With these expressions of Jacobian elements, the matrix J can be evaluated at each iteration as discussed earlier. Now, in the above algorithm, the violations of generation reactive power limits have not been taken into account. As in the case of NRLF (polar), in this case also, if a generator violates its Qlimits at any particular iteration, it is treated as a PQ bus for that iteration. Otherwise, it continues to be treated as a PV bus. The detailed step-by-step procedure for taking the generation Q limit violation into account in given below. Complete NRLF (rectangular) algorithm (0)

(0)

Step 1: Initialise ej = Vjsp for j = 2, 3, ⋯⋯ m and ej = 1.0 for j = (m + 1), (m + 2), ⋯⋯ n. (0) Also initialise fj = 0.0 for j = 2, 3, ⋯⋯ n. Let the vectors of the initial real and imaginary parts of the voltages be denoted as e(0) and f (0) respectively. Step 2: Set iteration counter k = 1. (k) Step 3: For i = 2, 3, ⋯⋯ m, calculate Qi from equation (2.60) by using e(k−1) and f (k−1) . If (k) the calculated Qi is within its respective limit, then this bus would be retained as a PV bus in the current iteration. If either the lower limit or the upper limit is violated, then this bus is converted to a PQ bus. In general, if l generator buses (l ≤ (m − 1)) violate their corresponding reactive power limits at step 3, then the size of ∆Q vector increases from (n − m) to (n − m + l) and the size of ∆V 2 vector decreases from (m − 1) to (m − l − 1). As a result, the new sizes of the matrices J3 , J4 , J5 and J6 become (n − m + l) × (n − 1), (n − m + l) × (n − 1), (m − l − 1) × (n − 1) and 56

(m − l − 1) × (n − 1) respectively. The sizes of the matrices J1 and J2 remain same. Also, the total sizes of the mismatch vector, correction vector and the Jacobian matrix remain same. Step 4: Compute the calculated vectors P, Q and V2 from equations (2.59) - (2.61) with the vectors e(k−1) and f (k−1) thereby forming the vector ∆M . Let this vector be represented as T ∆M = [∆M1 , ∆M2 , ⋯⋯∆M2n−2 ] . Step 5: Compute error = max (∣∆M1 ∣ , ∣∆M2 ∣ , ⋯⋯ ∣∆M2n−2 ∣). Step 6: If error ≤  (pre - specified tolerance), then the final rotation vectors are e(k−1) and f (k−1) and print the results. Otherwise go to step 7. Step 7: Evaluate the Jacobian matrix with the vectors e(k−1) and f (k−1) . Step 8: Compute the correction vector ∆X by solving equation (2.63). Step 9: Update the solution vectors e(k) = e(k−1) +∆e and f (k) = f (k−1) +∆f . Update k = k +1 and go back to step 3. In the next lecture, we will look at an example of NRLF (rectangular) technique.

57

2.10.1

Example for NRLF (rectangular) technique

Taking the 5 bus system as an example again, the NRLF algorithm starts with the flat voltage profile and subsequently different quantitative are calculated as shown in Table 2.19. Table 2.19: Initial calculation with NRLF (rectangular) in the 5 bus system

Pcal = [0 −0.2220 0 0] × 10−15 ; T

T

Qcal = [−0.143 −0.143] ; T

∆M = [−0.5 1.0 −1.15 −0.85 −0.457 −0.257 0 0] ;

T

Vcal = [1.0 1.0] ; error = 1.15;

As the mismatch (= 1.15) is greater than the tolerance (= 1.0e−12 ), the algorithm proceeds further and calculates the Jacobian matrix. Following the discussion presented earlier, the sizes of various Jacobian sub matrices should be as follows: J1 → (4 × 4); J2 → (4 × 4); J3 → (2 × 4); J4 → (2×4); J5 → (2×4); J6 → (2×4) and J → (8×8). The computed Jacobian matrices are shown in Table 2.20. From this set observe that the sizes of the calculated Jacobian matrices are indeed the same as they are indicated above. After calculating the Jacobian matrix, the algorithm calculates the correction vector (∆X) and extracts the vectors ∆e and ∆f from ∆X . With these obtained vectors ∆e and ∆f , the updated vectors e and f are computed and lastly the mismatch vector (∆M ) and the final mismatch are calculated. All these calculations are also shown in Table 2.20. As the mismatch in still more than the tolerance (although it has reduced from the last value), the algorithm repeats all these calculation and finally converges with 5 iterations. The final load flow result is shown in Table 2.21. Observe that the result obtained with NRLF (rectangular) method are identically same as obtained by NRLF (polar) and GSLF techniques. The load flow solutions of IEEE-14 bus and 30-bus systems have also been computed with NRLF (rectangular) method. The results are shown in Tables 2.22 and 2.23 and respectively. Again confirm yourself that the results shown in these two tables are identically same as the corresponding results shown earlier with NRLF (polar) and GSLF methods. Now, let us study the behavior of the algorithm when generator Q-limit is considered. Towards this end, again let us first consider the 5-bus system and as before, let us consider the limit on Q3 to be 50 MVAR. TO begin with, the initial calculations for this case are identically same as that shown in Table 2.19. As no violation of Q3 is detected in this initial calculation, the algorithm proceeds as usual (without any consideration of generation Q-limit violation) and repeats the same calculations as shown in Table 2.20. As there is still no violation in Q3 , the algorithm advances to 2nd iteration and the calculations at the end of 2nd iteration are shown in Table 2.24. At the end of 2nd iteration, it is found that the calculated value of Q3 is 0.6824 p.u. and as a result, this bus is now converted from a PV bus to a PQ bus. Hence, the PQ buses are now (4, 5, 3) and only bus 2 is retained as a PV bus. The corresponding calculated vectors Qcal , Vcal and ∆M are shown in Table 2.24 along with the final value of the mismatch quantity. As this final value is still more than the tolerance, the algorithm enters the third iteration. 58

Table 2.20: Calculations at 1st iteration with NRLF (rectangular) in the 5 bus system

⎡ 3.2417 −1.8412 0 0 ⎤⎥ ⎢ ⎥ ⎢ ⎢−1.8412 4.2294 −1.2584 −1.1298⎥ ⎥ ⎢ ⎥; J1 = ⎢ ⎥ ⎢ 0 −1.2584 2.1921 −0.9337 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ 0 −1.1298 −0.9337 3.9047 ⎦ ⎣ 0 −7.1309 10.5797 −3.7348 J3 = [ ]; 0 −4.4768 −3.7348 15.4091 2 0 0 0 J5 = [ ]; 0 2 0 0 ⎡ 3.2417 ⎢ ⎢ ⎢−1.8412 ⎢ ⎢ ⎢ 0 ⎢ ⎢ ⎢ 0 J = ⎢⎢ ⎢ 0 ⎢ ⎢ 0 ⎢ ⎢ ⎢ 2.000 ⎢ ⎢ ⎢ 0 ⎣

⎡ 13.0858 ⎢ ⎢ ⎢−7.4835 ⎢ J2 = ⎢ ⎢ 0 ⎢ ⎢ ⎢ 0 ⎣

−7.4835 0 0 ⎤⎥ ⎥ 19.0911 −7.1309 −4.4768⎥⎥ ⎥; −7.1309 10.8657 −3.7348⎥⎥ ⎥ −4.4768 −3.7348 15.6951 ⎥⎦

0 1.2584 −2.1921 0.9337 J4 = [ ]; 0 1.1298 0.9337 −3.9047 J6 = [

0 0 0 0 ]; 0 0 0 0

−1.8412 0 0 13.0858 −7.4835 0 0 ⎤⎥ ⎥ 4.2294 −1.2548 −1.1298 −7.4835 19.0911 −7.1309 −4.4768⎥⎥ ⎥ −1.2548 2.1921 −0.9337 0 −7.1309 10.8657 −3.7348⎥⎥ ⎥ −1.1298 −0.9337 3.9047 0 −4.4768 −3.7348 15.6951 ⎥ ⎥; −7.1309 10.5797 −3.7348 0 1.2584 −2.1921 0.9337 ⎥⎥ ⎥ −4.4768 −3.7348 15.4091 0 1.1298 0.9337 −3.9047⎥⎥ ⎥ 0 0 0 0 0 0 0 ⎥⎥ ⎥ 2.0000 0 0 0 0 0 0 ⎥⎦ T

∆X = [0.0 −0.0 −0.0783 −0.0475 −0.0331 −0.009 −0.1275 −0.0799] ; T

∆e = [0.0 −0.0 −0.0783 −0.0475] ; T

e = [1.0 1.0 1.0 0.9217 0.9525] ;

T

∆f = [−0.0331 −0.009 −0.1275 −0.0799] ; T

f = [0 0.0331 −0.009 −0.1275 −0.0799] ; T

Pcal = [0.5041 1.0056 −1.1211 −0.8312] ; T

Qcal = [−0.4191 −0.3205] ;

T

Vcal = [1.0011 1.0001] ; T

∆M = [−0.0041 −0.0056 −0.0289 −0.0188 −0.1809 −0.0795 −0.0011 −0.0001] ; error = 0.1809;

In the third iteration the numbers of PQ buses is 3 and the number of PV buses is 1 and hence, the sizes of matrices J3 and J4 change to (3 × 4) and those of matrices J5 and J6 become (1 × 4). The sizes of the matrices J1 , J2 and J remain same as earlier. With this Jacobian matrix, the correction vector ∆X is calculated and from this vector ∆X , the vectors ∆e and ∆f are extracted. Subsequently, the vectors e and f are updated. It is found that Q3 still violates the limit and the mismatch is also found to be still more than the tolerance at the end of the third iteration. 59

Table 2.21: Final Results of the 5 bus system with NRLF (rectangular) with no generator Q limit Bus no. 1 2 3 4 5

e f V θ Pinj Qinj (p.u) (p.u) (p.u) (deg) (p.u) (p.u) 1 0 1 0 0.56743 0.26505 0.99958 0.02893 1 1.65757 0.5 -0.18519 0.99987 -0.01592 1 -0.91206 1 0.68875 0.89634 -0.13157 0.90594 -8.35088 -1.15 -0.6 0.94034 -0.08272 0.94397 -5.02735 -0.85 -0.4 Total iteration = 5

Table 2.22: Final Results of the 14 bus system with NRLF (rectangular) with no generator Q limit Bus no. 1 2 3 4 5 6 7 8 9 10 11 12 13 14

e (p.u) 1.06 1.04075 1.0157 1.01604 1.02801 1.04455 0.99277 0.99427 0.99146 0.99207 1.01383 1.02382 1.01605 0.97979

f V θ (p.u) (p.u) (deg) 0 1.06 0 -0.09419 1.045 -5.17113 -0.26348 1.04932 -14.54246 -0.18634 1.03299 -10.39269 -0.15849 1.04015 -8.76418 -0.232 1.07 -12.52265 -0.23739 1.02076 -13.44781 -0.23818 1.0224 -13.47154 -0.24002 1.0201 -13.60908 -0.24176 1.0211 -13.69541 -0.23819 1.04144 -13.22158 -0.24445 1.0526 -13.42868 -0.24401 1.04494 -13.50388 -0.25524 1.01249 -14.60128 Total iteration = 5

Pinj

Qinj

(p.u) 2.37259 0.183 -1.19 -0.4779 -0.07599 0.112 0 0 -0.29499 -0.09 -0.03501 -0.06099 -0.135 -0.14901

(p.u) -0.3308 -0.166 -0.08762 -0.039 -0.01599 0.37278 0 -0.129 -0.16599 -0.05799 -0.018 -0.01599 -0.05799 -0.05001

All the relevant calculations pertaining to 3rd iteration are shown in Table 2.25. As the mismatch is still more than the tolerance limit, the algorithm proceeds and finally converges with 5 iterations. The final results are shown in Table 2.26. Again the load flow solutions of the IEEE-14 bus and IEEE-30 bus system have been computed for the same generator reactive power limits as taken for GSLF and NRLF (polar) techniques. The final solutions are shown in Tables 2.27 and 2.28 respectively. Again cross-check for yourself that the final solution computed by this method are same as those computed by GSLF and NRLF (polar) techniques. We will now discuss the fast-decoupled load flow (FDLF) technique in the next lecture.

60

Table 2.23: Final Results of the 30 bus system with NRLF (rectangular) with no generator Q limit Bus no. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

e (p.u) 1.05 1.0299 1.02132 1.01147 0.97759 1.00126 0.97138 0.99849 1.0191 1.00129 1.07063 1.01854 1.06225 0.99956 0.99383 1.00288 0.99556 0.98129 0.97781 0.98263 0.98651 0.98691 0.97991 0.9712 0.96249 0.94313 0.96684 0.99764 0.94118 0.92532

f V θ Pinj Qinj (p.u) (p.u) (deg) (p.u) (p.u) 0 1.05 0 2.38673 -0.29842 -0.08973 1.0338 -4.97945 0.3586 -0.05698 -0.14293 1.03128 -7.96653 -0.024 -0.012 -0.17076 1.02578 -9.58235 -0.076 -0.016 -0.23652 1.0058 -13.60103 -0.6964 0.05042 -0.20376 1.02178 -11.50296 0 0 -0.24218 1.00111 -13.9994 -0.628 -0.109 -0.22261 1.023 -12.56853 -0.45 0.12343 -0.23604 1.04608 -13.04088 0 0 -0.26616 1.03606 -14.88589 -0.058 -0.02 -0.21138 1.0913 -11.16876 0.1793 0.24018 -0.24923 1.04859 -13.74947 -0.112 -0.075 -0.23668 1.0883 -12.56078 0.1691 0.31043 -0.26255 1.03346 -14.71704 -0.062 -0.016 -0.26383 1.02825 -14.86737 -0.082 -0.025 -0.25946 1.0359 -14.50539 -0.035 -0.018 -0.26644 1.0306 -14.98291 -0.09 -0.058 -0.27363 1.01873 -15.58107 -0.032 -0.009 -0.27689 1.01626 -15.81066 -0.095 -0.034 -0.27506 1.02041 -15.63819 -0.022 -0.007 -0.27098 1.02305 -15.35955 -0.175 -0.112 -0.27096 1.02343 -15.35222 0 0 -0.27028 1.0165 -15.41998 -0.032 -0.016 -0.27501 1.00939 -15.81043 -0.087 -0.067 -0.27308 1.00048 -15.84004 0 0 -0.27533 0.9825 -16.27422 -0.035 -0.023 -0.26987 1.00379 -15.59587 0 0 -0.21474 1.02049 -12.1474 0 0 -0.2855 0.98353 -16.87497 -0.024 -0.009 -0.29698 0.97181 -17.79427 -0.106 -0.019 Total iteration = 5

Table 2.24: Calculation at the end of 2nd iteration with NRLF (rectangular) in the 5 bus system for limit on Q3

T

Pcal = [0.5000 1.0016 −1.1502 −0.8500] ; T

Qcal = [−0.5944 −0.3988 0.6824] ;

Vcal = [1.0]; T

∆M = [0.0 −0.0016 0.0002 0.0 −0.0056 −0.0012 −0.1824 −0.0] ;

61

error = 0.1824;

Table 2.25: Calculations at 3rd iteration with NRLF (rectangular) in the 5 bus system with limit on

Q3

⎡ 13.3023 ⎡ 3.3566 −1.6230 0 0 ⎤⎥ ⎢ ⎢ ⎢ ⎥ ⎢ ⎢−7.4539 ⎢−1.9584 5.5167 −1.3701 −1.2000⎥ ⎢ ⎥ ⎢ ⎥ ; J2 = ⎢ J1 = ⎢ ⎢ 0 ⎥ ⎢ 0 −2.0669 2.2170 −1.3289⎥ ⎢ ⎢ ⎢ ⎥ ⎢ ⎢ 0 ⎥ ⎢ 0 −1.4328 −1.1870 4.0987 ⎦ ⎣ ⎣ ⎡ 0 ⎡ 0 −6.2313 8.4978 −3.2275⎤⎥ ⎢ ⎢ ⎢ ⎥ ⎢ ⎥ ⎢ −4.1175 −3.4358 13.8060 ⎥ ; J4 = ⎢⎢ 0 J3 = ⎢ 0 ⎢ ⎥ ⎢ ⎢1.9584 ⎢−7.4539 19.5569 −7.1104 −4.4586⎥ ⎣ ⎦ ⎣ J5 = [1.9992 0 0 0] ; ⎡ 3.3566 ⎢ ⎢ ⎢−1.9584 ⎢ ⎢ ⎢ 0 ⎢ ⎢ ⎢ 0 J = ⎢⎢ ⎢ 0 ⎢ ⎢ 0 ⎢ ⎢ ⎢−7.4539 ⎢ ⎢ ⎢ 1.992 ⎣

−7.5339 0 0 ⎤⎥ ⎥ 18.1609 −7.1104 −4.4586⎥⎥ ⎥; −6.2313 10.1633 −3.2275⎥⎥ ⎥ −4.1175 −3.4358 14.8052 ⎥⎦ 2.0669 −4.5371 1.3289 ⎤⎥ ⎥ 1.4328 1.1870 −5.8182⎥⎥; ⎥ −3.5351 1.3701 1.2000 ⎥⎦

J6 = [0.0581 0 0 0];

−1.6230 0 0 13.3023 −7.5339 0 0 ⎤⎥ ⎥ 5.5167 −1.3701 −1.2000 −7.4539 18.1609 −7.1104 −4.4586⎥⎥ ⎥ −2.0669 2.2170 −1.3289 0 −6.2313 10.1633 −3.2275⎥⎥ ⎥ −1.4328 −1.1870 4.0987 0 −4.1175 −3.4358 14.8052 ⎥ ⎥; −6.2313 8.4978 −3.2275 0 2.0669 −4.5371 1.3289 ⎥⎥ ⎥ −4.1175 −3.4358 13.8060 0 1.4328 1.1870 −5.8182⎥⎥ ⎥ 19.5569 −7.1104 −4.4586 1.9584 −3.5351 1.3701 1.2000 ⎥⎥ ⎥ 0 0 0 0.0581 0 0 0 ⎥⎦ T

∆X = [0.0000 −0.0171 −0.0169 −0.0095 0.0006 0.0047 0.0024 0.0015] ; T

∆e = [0.0000 −0.0171 −0.0169 −0.0095] ;

T

∆f = [0.0006 0.0047 0.0024 0.0015] ; T

e = [1.0 0.9996 0.9828 0.8801 0.9311] ; T

f = [0 0.0297 −0.0109 −0.1292 −0.0812] ; T

Pcal = [0.4999 1.0005 −1.1496 −0.8499] ; T

Qcal = [−0.5997 −0.4000 0.5031] ;

T

Vcal = [1.0] ; T

∆M = [0.0001 −0.0005 0.0004 0.0001 −0.0003 −0.0000 −0.0031 −0.0000] ; error = 0.0031;

62

Table 2.26: Final Results of the 5 bus system with NRLF (rectangular) with limit on Q3 Bus no. 1 2 3 4 5

e f V θ Pinj Qinj (p.u) (p.u) (p.u) (deg) (p.u) (p.u) 1 0 1 0 0.56979 0.33935 0.99956 0.02961 1 1.69679 0.5 -0.04769 0.98244 -0.01097 0.9825 -0.63991 1 0.5 0.87973 -0.12927 0.88918 -8.35906 -1.15 -0.6 0.93092 -0.08123 0.93445 -4.98675 -0.85 -0.4 Total iteration = 5

Table 2.27: Final Results of the 14 bus system with NRLF (rectangular) with generator Q limits Bus no. 1 2 3 4 5 6 7 8 9 10 11 12 13 14

e (p.u) 1.06 1.04073 1.01336 1.01224 1.0242 1.03013 0.9847 0.98582 0.98318 0.98261 1.00189 1.00965 1.00223 0.96883

f V θ (p.u) (p.u) (deg) 0 1.06 0 -0.09432 1.045 -5.17845 -0.26312 1.04697 -14.55556 -0.18505 1.02902 -10.35987 -0.15691 1.03615 -8.71027 -0.22759 1.05497 -12.45871 -0.23631 1.01266 -13.49478 -0.23701 1.01391 -13.5185 -0.23896 1.0118 -13.66101 -0.2402 1.01154 -13.73679 -0.23533 1.02915 -13.21814 -0.24037 1.03787 -13.39145 -0.24028 1.03063 -13.48166 -0.25317 1.00136 -14.64504 Total iteration = 8

63

Pinj

Qinj

(p.u) 2.37188 0.183 -1.19 -0.4779 -0.07599 0.112 0 0 -0.29499 -0.09 -0.03501 -0.06099 -0.135 -0.14901

(p.u) -0.31249 -0.1066 -0.08762 -0.039 -0.01599 0.3 0 -0.129 -0.16599 -0.05799 -0.018 -0.01599 -0.05799 -0.05001

Table 2.28: Final Results of the 30 bus system with NRLF (rectangular) with generator Q limits Bus no. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

e (p.u) 1.05 1.0299 1.02053 1.01051 0.97756 1.00036 0.97083 0.99845 1.01327 0.99655 1.05781 1.01458 1.05718 0.99547 0.98968 0.99856 0.99092 0.97686 0.97323 0.97802 0.98184 0.98228 0.97574 0.96705 0.95946 0.94003 0.96454 0.99677 0.93881 0.9229

f V θ Pinj Qinj (p.u) (p.u) (deg) (p.u) (p.u) 0 1.05 0 2.38676 -0.29389 -0.08977 1.0338 -4.98141 0.3586 -0.04555 -0.14263 1.03044 -7.95615 -0.024 -0.012 -0.17038 1.02477 -9.57038 -0.076 -0.016 -0.23666 1.0058 -13.6093 -0.6964 0.05533 -0.20342 1.02084 -11.49427 0 0 -0.24204 1.00055 -13.99903 -0.628 -0.109 -0.22276 1.023 -12.57695 -0.45 0.15116 -0.2345 1.04005 -13.0305 0 0 -0.26495 1.03117 -14.8887 -0.058 -0.02 -0.208 1.07807 -11.1244 0.1793 0.2 -0.24844 1.04456 -13.75943 -0.112 -0.075 -0.23554 1.08311 -12.56043 0.1691 0.3 -0.26178 1.02931 -14.73367 -0.062 -0.016 -0.26302 1.02403 -14.88301 -0.082 -0.025 -0.25851 1.03148 -14.51398 -0.035 -0.018 -0.26532 1.02583 -14.98919 -0.09 -0.058 -0.27271 1.01421 -15.59837 -0.032 -0.009 -0.2759 1.01159 -15.82733 -0.095 -0.034 -0.27402 1.01568 -15.65181 -0.022 -0.007 -0.26984 1.01825 -15.36745 -0.175 -0.112 -0.26983 1.01867 -15.36029 0 0 -0.26947 1.01226 -15.43846 -0.032 -0.016 -0.27419 1.00517 -15.82979 -0.087 -0.067 -0.27277 0.99748 -15.87039 0 0 -0.27501 0.97944 -16.30724 -0.035 -0.023 -0.26985 1.00158 -15.62998 0 0 -0.21448 1.01959 -12.14364 0 0 -0.2855 0.98126 -16.91488 -0.024 -0.009 -0.297 0.96951 -17.83848 -0.106 -0.019 Total iteration = 5

64

2.11

Fast-decoupled load-flow (FDLF) technique

An important and useful property of power system is that the change in real power is primarily governed by the charges in the voltage angles, but not in voltage magnitudes. On the other hand, the charges in the reactive power are primarily influenced by the charges in voltage magnitudes, but not in the voltage angles. To see this, let us note the following facts: (a) Under normal steady state operation, the voltage magnitudes are all nearly equal to 1.0. (b) As the transmission lines are mostly reactive, the conductances are quite small as compared to the susceptance (Gij << Bij ). (c) Under normal steady state operation the angular differences among the bus voltages are quite small (θi − θj ≈ 0 (within 5o − 10o )). (d) The injected reactive power at any bus is always much less than the reactive power consumed by the elements connected to this bus when these elements are shorted to the ground (Qi << Bii Vi2 ). With these facts at hand, let us re-visit the equations for Jacobian elements in Newton-Raphson (polar) method (equation (2.48) to (2.55)). From equations (2.50) and (2.51) we have, n ∂Pi = 2Vi Gii + ∑ Vk Yik cos(θi − θk − αik ) ∂Vj k=1 ≠i n

= 2Vi Gii + ∑ Vk Yik [cos(θi − θk ) cos αik + sin(θi − θk ) sin αik ] k=1 ≠i n

= 2Vi Gii + ∑ Vk [Gik cos(θi − θk ) + Bik sin(θi − θk )] ; k=1 ≠i

j=i

(2.74)

∂Pi = Vi Yij cos(θi − θj − αij ) ∂Vj = Vi Yij [cos(θi − θj ) cos αij + sin(θi − θj ) sin αij ] = Vi [Gij cos(θi − θj ) + Bij sin(θi − θj )] ;

j≠i

(2.75)

Now, Gii and Gij are quite small and negligible and also cos(θi − θj ) ≈ 1 and sin(θi − θj ) ≈ 0, as [(θi − θj ) ≈ 0]. Hence,

∂Pi ∂Pi ≈ 0 and ≈0 ∂Vi ∂Vj

Ô⇒ J2 ≈ 0

(2.76)

Similarly, from equations (2.52) and (2.53) we get, n ∂Qi = ∑ Vi Vk [Gik cos(θi − θk ) + Bik sin(θi − θk )] ; ∂θj k = 1

j=i

(2.77)

≠i

∂Qi = −Vi Vj [Gij cos(θi − θj ) + Bij sin(θi − θj )] ; ∂θj 65

j≠i

(2.78)

Again in light of the natures of the quantities Gii , Gij and (θi − θj ) as discussed above,

∂Qi ∂Qi ≈ 0 and ≈0 ∂θi ∂θj

Ô⇒ J3 ≈ 0

(2.79)

Substituting equations (2.76) and (2.79) into equation (2.40) one can get,

∆P J1 0 ∆θ [ ]=[ ][ ] ∆Q 0 J4 ∆V

(2.80)

In other words, ∆P depends only on ∆θ and ∆Q depends only on ∆V . Thus, there is a decoupling between ‘∆P - ∆θ ’ and ‘∆Q - ∆V ’ relations. Now, from equations (2.48) and (2.49) we get, n ∂Pi = − ∑ Vi Vk Yik sin(θi − θk − αik ); ∂θj k=1

j=i

≠i

n

= Vi Vi Yii sin(θi − θi − αii ) − ∑ Vi Vk Yik sin(θi − θk − αik );

j=i

k=1

= −Bii Vi2 − Qi ≈ −Bii Vi2 ;

j=i

[as Qi << Bii Vi2 ]

∂Pi = Vi Vj Yij sin(θi − θj − αij ); j ≠ i ∂θj = Vi Vj Yij [sin(θi − θj ) cos αij − cos(θi − θj ) sin αij ] ; = Vi Vj [Gij sin(θi − θj ) − Bij cos(θi − θj )] ; = −Vi Vj Bij ;

(2.81)

j≠i

j≠i

j≠i

(2.82)

Similarly, from equations (2.54) and (2.55) we get, n ∂Qi = −2Vi Bii + ∑ Vk Yik sin(θi − θk − αik ); ∂Vj k=1 ≠i n

or,

j=i

∂Qi Vi = −2Vi2 Bii + ∑ Vi Vk Yik sin(θi − θk − αik ); ∂Vj k=1

j=i

≠i

n

∂Qi Vi = −Vi2 Bii + ∑ Vi Vk Yik sin(θi − θk − αik ) = Qi − Vi2 Bii ; ∂Vj k=1 ∂Qi Vi = −Vi2 Bii ; j = i [as Qi << Bii Vi2 ] or, ∂Vj ∂Qi or, = −Vi Bii ; j = i ∂Vj or,

66

j=i

(2.83)

∂Qi = Vi Yij sin(θi − θj − αij ); j ≠ i ∂Vj = Vi Yij [sin(θi − θj ) cos αij − cos(θi − θj ) sin αij ] ; = Vi [Gij sin(θi − θj ) − Bij cos(θi − θj )] ; ≈ −Vi Bij ;

j≠i

j≠i

j≠i

(2.84) n

Combining equations (2.80)-(2.82) we get, ∆Pi = −Vi ∑ Vk Bik ∆θk . Or, k=1

n ∆Pi = − ∑ Vk Bik ∆θk Vi k=1

(2.85)

Now, as Vi ≈ 1.0 under normal steady state operating condition, equation (2.85) reduces to, n ∆Pi = − ∑ Bik ∆θk . Vi k=1

Or,

∆P = [−B] ∆θ . V

Or,

∆P ′ = [B ] ∆θ V

(2.86)

Matrix B is a constant matrix having a dimension of (n − 1) × (n − 1). Its elements are the negative of the imaginary part of the element (i, k) of the YBUS matrix where i = 2, 3, ⋯⋯ n and k = 2, 3, ⋯⋯ n. Again combining equations (2.80), (2.83) and (2.84) we get, ′

n

∆Qi = −Vi ∑ Bik ∆Vk . k=1

Or,

n ∆Qi = − ∑ Bik ∆Vk . Vi k=1

Or,

∆Q ′′ = [B ] ∆V V

(2.87)

Again, [B ] is also a constant matrix having a dimension of (n − m) × (n − m). Its elements are the negative of the imaginary part of the element (i, k) of the YBUS matrix where i = (m + 1), (m + ′ ′′ 2), ⋯⋯ n and k = (m + 1), (m + 2), ⋯⋯ n. As the matrixes [B ] and [B ] are constant, it is not necessary to invert these matrices in each iteration. Rather, the inverse of these matrices can be stored and used in every iteration, thereby making the algorithm faster. Further simplification in the FDLF algorithm can be made by, ′′

a. Ignoring the series resistances is calculating the elements of [B ]. Also, by omitting the ′ elements of [B ] that predominantly affect reactive power flows, i.e., shunt reactances and transformer off nominal in phase taps. ′

b. Omitting from [B ] the angle shifting effect of phase shifter, which predominantly affects real power flow. ′′

In the next lecture, we will look at an example of FDLF method. 67

2.11.1

Example for Fast-decoupled load-flow technique

As an example, the 5 bus system described earlier is considered again. Starting from the flat start, the initial calculations are shown in Table 2.29. As the mismatch is more than the tolerance, the ¯ BUS matrix of this system is shown in Table 2.30. In this table, the algorithm proceeds. The Y ¯ BUS (∶, m ∶ n) represents the elements (of the Y ¯ BUS matrix) corresponding to all rows notation Y (denoted by the notation ‘:’) and columns spanning from mth column to nth column (denoted by the ¯ BUS matrix, the matrices [B ′ ] and [B ′′ ] are constructed as shown in notation ‘m:n’). From this Y ′ Table 2.30. Note that the size of the matrix [B ] is (4 × 4) (corresponding to the non slack buses, ′′ i.e. 2, 3, 4 and 5) and the size of the [B ] matrix is (2 × 2) (corresponding to the PV buses 4 ′ ′′ and 5). Also note that the matrices [B ] and [B ] have been formed by taking the negative of the ¯ BUS matrix. With these constant matrices, imaginary parts of the corresponding elements of the Y the vectors ∆θ and ∆V are calculated and subsequently, the vectors θ and V have been updated. With these updated values of θ and V, the final mismatch (error) is again calculated as shown in Table 2.30. As the error is still more than the tolerance, the algorithm proceeds further and finally converges in 19 iterations for a tolerance value of 10−12 p.u. The final solution is shown in Table 2.31, which happens to be the same as the results obtained by the other methods described earlier.

Table 2.29: Initial calculation with FDLF in the 5 bus system

Pcal = [−0.0444 −0.1776 −0.0333 −0.0444] × 10−14 ; T

T

Qcal = [−0.1430 −0.1430] ; T

∆M = [0.5000 1.0000 −1.1500 −0.8500 −0.4570 −0.2570] ; error = 1.15 ;

Now, let us impose the reactive power limit on the generation at bus 3. Starting from the flat start, the calculation up to 1st iteration are same and hence are not shown here. The calculation pertaining to 2nd iteration are shown in Table 2.32. From this set, observe that the calculated value of Q3 has exceeded the limit of 50 MVAR and hence it should be now treated as PQ-bus. Thus, ′′ the dimension of the matrix [B ] increases to (3 × 3) (corresponding to the buses 4, 5 and 3). ′ ′′ ′ This new, augmented matrix is shown in Table 2.33. With these [B ] and [B ] matrices ([B ] matrix remains the same), the calculations are further carried out and the algorithm converges in 20 iterations. The final solution is shown in Table 2.31. Comparison of this result with the earlier results (obtained with other methods) shows that because of the approximations made in FDLF, the results obtained with FDLF are not identically the same with those obtained by the other methods but are very close. The results corresponding to 14-bus and 30-bus systems are shown in Tables 2.34 and 2.35 respectively. From these tables note that the number of iterations taken by FDLF is much higher than those required by NRLF. This is because of approximations adopted by FDLF (to achieve decoupling) due to which, the convergence is slower. However, because of the constant Jacobrian matrices, the 68

Table 2.30: Calculations at 1st iteration with FDLF in the 5 bus system

⎤ ⎡ 3.2417 − 13.0138i −1.4006 + 5.6022i 0 ⎥ ⎢ ⎥ ⎢ ⎢−1.4006 + 5.6022i 3.2417 − 13.0138i −1.8412 + 7.4835i⎥ ⎥ ⎢ ⎥ ⎢ ⎢ ¯ 0 −1.8412 + 7.4835i 4.2294 − 18.9271i ⎥⎥; YBUS (∶, 1 ∶ 3) = ⎢ ⎥ ⎢ ⎥ ⎢ 0 0 −1.2584 + 7.1309i ⎥ ⎢ ⎥ ⎢ ⎥ ⎢−1.8412 + 7.4835i 0 −1.1298 + 4.4768i ⎦ ⎣ ⎡ 0 −1.8412 + 7.4835i⎤⎥ ⎢ ⎥ ⎢ ⎥ ⎢ 0 0 ⎥ ⎢ ⎥ ⎢ ¯ BUS (∶, 4 ∶ 5) = ⎢⎢−1.2584 + 7.1309i −1.1298 + 4.4768i⎥⎥; Y ⎥ ⎢ ⎢ 2.1921 − 10.7227i −0.9337 + 3.7348i⎥ ⎥ ⎢ ⎥ ⎢ ⎢−0.9337 + 3.7348i 3.9047 − 15.5521i ⎥ ⎦ ⎣ ⎡ 13.0138 −7.4835 0 0 ⎤⎥ ⎢ ⎢ ⎥ ⎢−7.4835 18.9271 −7.1309 −4.4768⎥ 10.7227 −3.7348 ′ ′′ ⎢ ⎥ ⎥ ; [B ] = [ [B ] = ⎢ ]; ⎢ 0 −7.1309 10.7227 −3.7348⎥⎥ −3.7348 15.5521 ⎢ ⎢ ⎥ ⎢ 0 −4.4768 −3.7348 15.5521 ⎥⎦ ⎣ T

∆θ = [0.0306 −0.0136 −0.1492 −0.0944] ;

T

∆V = [−0.0528 −0.0292] ; T

θ = [0 0.0306 −0.0136 −0.1492 −0.0944] ; T

V = [1.0 1.0 1.0 0.9472 0.9708] ; T

Pcal = [0.5045 1.0488 −1.1725 −0.8975] ;

T

Qcal = [−0.2931 −0.1267] ; T

∆M = [−0.0045 −0.0488 0.0225 0.0475 −0.3069 −0.2733] ; error = 0.3069;

execution of each iteration is much faster (as the Jacobrian matrix need not be recomputed and reversed at each iteration) and hence, the total time taken by FDLF is quiet comparable to that needed by NRLF. From the two tables it is further noted that, in the absence of any generation reactive power limit, the results obtained by FDLF are almost identical to those obtained by other methods. However, in the presence of generation reactive power limits, FDLF results are quiet close to the results obtained by other methods, though not identical. We are now at the end of our discussion of AC load flow techniques. In the next lecture, we will study the method of load flow analysis of an AC system in which a HVDC link is also embedded (namely, the AC-DC load flow technique).

69

Table 2.31: Final Results of the 5 bus system with FDLF

Bus no. 1 2 3 4 5

∣V ∣

Without generator Q limit

θ

Pinj

∣V ∣

Qinj

With generator Q limit

θ

Pinj

Qinj

(p.u) (deg) (p.u) (p.u) (p.u) (deg) (p.u) (p.u) 1.0 0 0.56743 0.26505 1.0 0 0.56985 0.34069 1.0 1.65757 0.5 -0.18519 1.0 1.69742 0.5 -0.04522 1.0 -0.91206 1.0 0.68875 0.98219 -0.63507 1.0 0.49668 0.90594 -8.35088 -1.15 -0.6 0.88888 -8.35938 -1.15 -0.6 0.94397 -5.02735 -0.85 -0.4 0.93428 -4.9861 -0.85 -0.4 Total iteration = 19 Total iteration = 20

Table 2.32: Calculations at 2nd iteration with FDLF in the 5 bus system with limit on Q3 T

∆θ = [−0.0010 −0.0012 0.0026 0.0034] ;

T

∆V = [−0.0399 −0.0277] ; T

θ = [0 0.0296 −0.0148 −0.1465 −0.0910] ; T

V = [1.0 1.0 1.0 0.9074 0.9431] ; T

Pcal = [0.4999 1.0353 −1.1526 −0.9012] ;

T

Qcal = [−0.5830 −0.4020 0.6775] ; T

∆M = [0.0001 −0.0353 0.0026 0.0512 −0.0170 0.0020 −0.1775] ; error = 0.1775;

70

Table 2.33: Calculations at 3rd iteration with FDLF in the 5 bus system with limit on Q3

⎡ 13.0138 ⎢ ⎢ ⎢−7.4835 ′ ⎢ [B ] = ⎢ ⎢ 0 ⎢ ⎢ ⎢ 0 ⎣

−7.4835 0 0 ⎤⎥ ⎥ 18.9271 −7.1309 −4.4768⎥⎥ ⎥; −7.1309 10.7227 −3.7348⎥⎥ ⎥ −4.4768 −3.7348 15.5521 ⎥⎦ T

∆θ = [−0.0006 −0.0010 0.0008 0.0034] ;

⎡ 10.7227 −3.7348 −7.1309⎤ ⎥ ⎢ ⎥ ⎢ ′′ ⎢ [B ] = ⎢−3.7348 15.5521 −4.4768⎥⎥; ⎥ ⎢ ⎢−7.1309 −4.4768 18.9271 ⎥ ⎦ ⎣ T

∆V = [−0.0167 −0.0090 −0.0178] ; T

θ = [0 0.0290 −0.0157 −0.1457 −0.0875] ; T

V = [1.0 1.0 0.9822 0.8906 0.9341] ; T

T

Qcal = [−0.5939 −0.4125] ;

Pcal = [0.5268 0.9223 −1.1174 −0.8408] ;

T

∆M = [−0.0268 0.0777 −0.0326 −0.0092 −0.0061 0.0125] ; error = 0.0777;

Table 2.34: Final Results of the 14 bus system with FDLF Without generator Q limit

Bus no.

∣V ∣

1 2 3 4 5 6 7 8 9 10 11 12 13 14

(p.u) 1.06 1.045 1.04932 1.03299 1.04015 1.07 1.02076 1.0224 1.0201 1.0211 1.04144 1.0526 1.04494 1.01249

θ

Pinj

(deg) (p.u) 0 2.37259 -5.17113 0.183 -14.54246 -1.19 -10.39269 -0.4779 -8.76418 -0.07599 -12.52265 0.112 -13.44781 0 -13.47154 0 -13.60908 -0.29499 -13.69541 -0.09 -13.22158 -0.03501 -13.42868 -0.06099 -13.50388 -0.135 -14.60128 -0.14901 Total iteration = 123

Qinj

∣V ∣

(p.u) -0.3308 -0.166 -0.08762 -0.039 -0.01599 0.37278 0 -0.129 -0.16599 -0.05799 -0.018 -0.01599 -0.05799 -0.05001

(p.u) 1.06 1.045 1.04697 1.02902 1.03615 1.05497 1.01266 1.01391 1.0118 1.01154 1.02915 1.03787 1.03063 1.00136

71

With generator Q limit

θ

Pinj

(deg) (p.u) 0 2.37188 -5.17845 0.183 -14.55556 -1.19 -10.35987 -0.4779 -8.71026 -0.07599 -12.4587 0.112 -13.49478 0 -13.5185 0 -13.66102 -0.29499 -13.73679 -0.09 -13.21814 -0.03501 -13.39144 -0.06099 -13.48166 -0.135 -14.64504 -0.14901 Total iteration = 119

Qinj (p.u) -0.31249 -0.1066 -0.08762 -0.039 -0.01599 0.29999 0 -0.129 -0.16599 -0.05799 -0.018 -0.01599 -0.05799 -0.05001

Table 2.35: Final Results of the 30 bus system with FDLF Without generator Q limit

Bus no.

∣V ∣

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

(p.u) 1.05 1.0338 1.03128 1.02578 1.005 8 1.02178 1.00111 1.023 1.04608 1.03606 1.0913 1.04859 1.0883 1.03346 1.02825 1.0359 1.0306 1.01873 1.01626 1.02041 1.02305 1.02343 1.0165 1.00939 1.00048 0.9825 1.00379 1.02049 0.98353 0.97181

θ

Pinj

(deg) (p.u) 0 2.38673 -4.97945 0.3586 -7.96653 -0.024 -9.58235 -0.076 -13.60103 -0.6964 -11.50296 0 -13.9994 -0.628 -12.56853 -0.45 -13.04088 0 -14.88589 -0.058 -11.16876 0.1793 -13.74947 -0.112 -12.56078 0.1691 -14.71704 -0.062 -14.86737 -0.082 -14.50539 -0.035 -14.98291 -0.09 -15.58107 -0.032 -15.81066 -0.095 -15.63819 -0.022 -15.35955 -0.175 -15.35222 0 -15.41998 -0.032 -15.81043 -0.087 -15.84004 0 -16.27422 -0.035 -15.59587 0 -12.1474 0 -16.87497 -0.024 -17.79427 -0.106 Total iteration = 115

Qinj

∣V ∣

(p.u) -0.29842 -0.05698 -0.012 -0.016 0.05042 0 -0.109 0.12343 0 -0.02 0.24018 -0.075 0.31043 -0.016 -0.025 -0.018 -0.058 -0.009 -0.034 -0.007 -0.112 0 -0.016 -0.067 0 -0.023 0 0 -0.009 -0.019

(p.u) 1.05 1.0338 1.03026 1.02455 1.0058 1.0206 1.00041 1.023 1.03842 1.02995 1.07426 1.04384 1.0824 1.02853 1.0232 1.03054 1.02469 1.01323 1.01052 1.01457 1.01706 1.0175 1.01136 1.00418 0.99677 0.97872 1.00105 1.01937 0.98073 0.96897

72

With generator Q limit

θ

Pinj

(deg) (p.u) 0 2.38678 -4.98182 0.3586 -7.95403 -0.024 -9.56795 -0.076 -13.6111 -0.6964 -11.49172 0 -13.9986 -0.628 -12.5786 -0.45 -13.02612 0 -14.88705 -0.058 -11.11026 0.1793 -13.76458 -0.112 -12.56395 0.1691 -14.73984 -0.062 -14.88791 -0.082 -14.51633 -0.035 -14.98908 -0.09 -15.60231 -0.032 -15.83038 -0.095 -15.65374 -0.022 -15.36723 -0.175 -15.36018 0 -15.44311 -0.032 -15.83356 -0.087 -15.87684 0 -16.31432 -0.035 -15.63737 0 -12.14228 0 -16.92364 -0.024 -17.84827 -0.106 Total iteration = 112

Qinj (p.u) -0.29289 -0.0429 -0.012 -0.016 0.05654 0 -0.109 0.15797 0 -0.02 0.18809 -0.075 0.29991 -0.016 -0.025 -0.018 -0.058 -0.009 -0.034 -0.007 -0.112 0 -0.016 -0.067 0 -0.023 0 0 -0.009 -0.019

2.12

A.C.-D.C. LOAD FLOW

For solving the load flow problem of an A.C. system in which one or more HVDC links are present, either of the following two approaches are followed; a. Simultaneous solution technique b. Sequential solution technique In simultaneous solution technique, the equations pertaining to the A.C. system and the equations pertaining to the DC system are solved together. In the sequential method, the AC and DC systems are solved separately and the coupling between the AC and DC system in accomplished by injecting an equivalent amount of real and reactive power at the terminal AC buses. In other words, for an HVDC link existing between buses ‘i’ and ‘j’ of an AC system (rectifier at bus ‘i’ and inverter at bus ’j’), the effect of the DC link in incorporated into the AC system by injections (R) (R) (I) (I) PDCi and QDCi at the rectifier bus ’i’ and PDCj and QDCj at bus ’j’ (the super scripts ’R’ and ’I’ denote the rectifier and inverter respectively). Therefore the net injected power at bus ’i’ and (R) (R) (I) (I) ’j’ are: Pitotal = PACi + PDCi ; QTi otal = QACi + QDCi ; PjT otal = PACj + PDCj ; QTj otal = QACj + QDCj . With these net injected powers the AC system is again solved and subsequently, the equivalent in(R) (R) (I) (I) jected powers (PDCi , QDCi , PDCj , QDCi ) and the total injected powers (PiT otal , QTi otal , PjT otal , QTj otal ) are updated. This process of alternately solving AC and DC system quantities is continued till the changes in AC system and DC system quantities between two consecutive iterations become less then a threshold value. Although simultaneous technique gives the solution of the system without any to and fro switching between the AC and DC systems, the sequential solution technique is actually quite easy to implement as we will see later. Now let as look at the equations of the DC system.

2.12.1

DC system model

For deriving a suitable model of a HVDC system for steady state operation, few basic assumptions are adopted as described below; a. The three A.C. voltages at the terminal bus bar are balanced and sinusoidal. b. The converter operation is perfectly balanced. c. The direct current and voltages are smooth. d. The converter transformer is lossless and the magnetizing admittance is ignored.

With the above assumptions, the equivalent circuit of the converter (either rectifier or inverter) is shown in Fig. 2.18. In this figure, the notations are as follows; 73

Figure 2.18: Equivalent circuit of the converter under the steady state operation

Vt ∠θt a Es ∠θs I¯p , I¯s Vd , Id

⇒ ⇒ ⇒ ⇒ ⇒

Magnitude and angle of the terminal bus bar of the converter Converter transformer tap ratio Magnitude and angle of the secondary side of the converter transformer Primary and secondary current of the converter transformer respectively DC voltage and DC current respectively

It is to be noted that in Fig. 2.18, the angles are referred to the common reference of the entire AC-DC system. With the above notations, the basic equation governing the HVDC systems are as follow: For rectifier

√ 3 2 Vdr = Nr ar Etr cos φr = Vdor cos φr π 3 Vdr = Vdor cos α − Xcr Nr Id π

For inverter

√ 3 2 Vdi = Ni ai Eti cos φi = Vdoi cos φi π 3 Vdi = Vdoi cos γ − Xci Ni Id π

(2.88)

(2.89)

(2.90)

(2.91)

In the above equations, the subscripts ‘r’ and ‘i’ denote the rectifier and inverter side respectively. The quantity ‘N’ denotes the number of six-pulse bridges at any partienlar side and the angle ‘Φ’ denotes the angular difference between the terminal voltages and primary current of the transformer, i.e. the power factor of the converter as seen by the AC bus. ‘Xc ’ denotes the commutating reactance of the converter transformer and the angles ‘α’ and ‘γ ’ denote the firing angle of the rectifier and the extinction angle of the inverter respectively. 74

The rectifie and the inverter are interconnected though the following equation:

Vdr − Vdi = Id Rd

(2.92)

In equation (2.92), the quantity Rd denotes the DC link resistance. Equations (2.88)-(2.92) describe the operation of a two-terminal HVDC link. Now, as the basic objective of a HVDC link is to provide complete controllability of power over a transmission corridor, both the rectifies and the inverter stations are suitably controlled and thus, suitable control equations also need to be incorporated in the above model. We will discuss these control equations shortly. However, to solve the above equations, appropriate solution variables must be chosen. Now, for the reason of simplicity, following set of solution variables is chosen for each converter;

x¯ = [Vd Id a cos α φ]

T

(2.93)

Therefore, for a two terminal HVDC link, the complete set of solution vector is;

x¯c = [Vdr Vdi Id ar ai cos α cos γ φr φi ]

T

(2.94)

In equation (2.94), Id has been taken only once as the DC current is same at both the ends. From equation (2.94) it is observed that there are total 9 unknown variables which need to be solved to completely determine the HVDC link. However, we have only 5 independent equations as shown in equations (2.88)-(2.92). Therefore, out of 9 unknown variables, any 4 variables need to be specified and thereafter, remaining 5 variables can be solved using equations (2.88)-(2.92). These 4 variables can be specified using the control specification. There can be several combinations of control specification and some of their combination are; i) α, Pdr , γ , Vdi ; iii) ar , Pdr , ai , Vdi ; v) ar , Pdr , γ , ai ; vii) α, Id , γ , Vdi ;

ii) α, Pdr , ai , Vdi ; iv) ar , Pdr , γ , Vdi ; vi) ar , Pdr , α, γ ; viiii) α, Vdr , γ , Pdi ;

With any of these four specified control values, the remaining 5 variables can be solved from equations (2.88)-(2.92) by using standard Newton-Raphoson technique. However, for the sequential (R) (R) (I) (I) solution techniques, the quantities PDci , QDci , PDcj and QDcj can be competed in a much easier way by algebraic manipulation of equations (2.88)-(2.92). we will show this procedure by two of the eight combinations listed above.

75

Combination 1

In this case, α, Pdr , γ and Vdi are specified. With these known quantities, the calculation procedure is as follows: Step 1: We know, Pdr = Vdr Id .

Or, Pdr =

Or, Vdr2 − Vdr Vdi − Rd Pdr = 0.

Vdr =

Vdr (Vdr − Vdi ) (from equation (2.92)). Rd

Or,

Vdi ±

√ Vdi2 + 4Rd Pdr 2

(2.95)

From equation (2.95), two values of Vdr are obtained. Out of these two values, the value of Vdr which is greater than Vdi is chosen, i.e.

√ 1 Vdr = (Vdi + Vdi2 + 4Rd Pdr ) 2 Step 2: Id is calculated as,

Id =

Pdr Vdr

(2.96)

(2.97)

Step 3: Using equation (2.89), Vdor is calculated as,

3 Vdr + Xcr Nr Id π Vdor = cos α

(2.98)

Step 4: Using equation (2.88), ar and cos Φr are calculated as,

cos Φr =

Vdr Vdor

Vdor π ar = √ 3 2Nr Etr

(2.99)

(2.100)

In equation (2.100) Etr is known as in the sequential solution method, the terminal voltages are known from the immediate past solution of the AC system equations. (R) (R) Step 5: The quantities PDCi and QDCi are calculated as; (R) PDCi = Pdr

Q(R) DCi = Pdr tan φr

and

(2.101)

Step 6: From equation (2.91), Vdoi is calculated as,

3 Vdi + Xci Ni Id π Vdoi = cos γ 76

(2.102)

Step 7: Using equation (2.90), ai and cos Φi are calculated as,

cos φi =

Vdi Vdoi

(2.103)

πVdoi ai = √ 3 2Ni Eti (I)

(2.104)

(I)

Step 8: The quantities PDCj and QDCj are calculated as, (I) PDCj = Vdi Id (R)

(R)

(I) Q(I) DCj = PDCj tan φi

and (I)

(2.105)

(I)

With these values of PDCi , QDCi , PDCj and QDCj , the AC system equations are again solved to obtain the updated values of Etr and Eti and subsequently, steps (1)-(8) are repeated again to (R) (R) (I) (I) update the values of PDCi , QDCi , PDCj and QDCj . This alternate process of solving AC and DC system equations are repeated till convergence in obtained. Combination 8 In this case, α, γ , Pdi and Vdr are known. With these known quantities, the calculation procedure is as follows:

Vdr − Vdi Vdi Vdr − Vdi2 = . Step 1: We know Pdi = Vdi Id = Vdi Rd Rd Or, Vdi2 + Rd Pdi − Vdi Vdr = 0. Or, Vdi =

Vdr ±

√ Vdr2 − 4Rd Pdi 2

(2.106)

From the two values of Vdi in equation (2.106), the final value of Vdi is calculated as,

√ 1 Vdi = (Vdr + Vdr2 − 4Rd Pdi ) 2 Step 2: Id is calculated as,

Id =

(2.107)

Pdi Vdi

(2.108)

With these calculated values of Vdi and Id , steps (3)-(8) of combination-1 are followed to calculate (R) the Equivalent power injection values, where PDCi = Vdr Id . With these injected power values, the AC and DC systems are continued to be solved alternately till convergence in achieved. It is to be R noted that at the rectifier end, P(DCi) = −Pdr and Q(R) DCi = −Qdr as the rectifier draws both real and (I)

I reactive power from the grid. On the other hand, at the inverter end, P(DCj) = Pdi and QDCj = −Qdi as the inverter supplies real power to the AC grid and draws reactive power from the AC grid. In the next lecture, we will look at an example of AC-DC load flow method.

77

2.12.2

Example for A.C-D.C load flow

To illustrate the application of the above procedure, let us first consider the 5-bus system. In this system, it is now assumed that one bipolar HVDC link is connected between bus 4 and 5 (rectifier at bus 4 and inverter at bus 5). Other relevant data for this link are as follows; Rd = 10.0 Ω;

Nr = Ni = 2;

3 3 Xcr = Xci = 6.0 Ω. Further, let us also assume that the specified values have been π π

taken according to combination 1 and the values are as follows: Combination-1

α = 5o , Pdr = 100 MW; γ = 18o , Vdi = 250 kV With the above specification, the calculation procedure for the DC system is as follows. Initially, the flat start is assumed for all the buses in the system. Therefore, ∣V4 ∣ = ∣V5 ∣ = 1.0 p.u. Let us also assume that the base voltage of the AC system is 132 kV. Now, before commencing the AC load flow, the equivalent power injections (both real and reactive) at buses 4 and 5 need to be calculated. For this purpose, equations (2.95) - (2.105) are used to calculate the values of different DC variables as follows: Vdr = 253.938 kV; id = 393.8 Amp.; Qdr = 16.276 MVAR; Pdi = 98.45 MW; Qdi = 35.024 MVAR. Now, let us look at equations (2.95) - (2.105) more closely. Equations (2.96) - (2.99) show that the quantities Vdr , id , Vdor and cos Φr depend only on the DC system data. As the DC system data are constant, the calculatd values of these four quantities would also be constant (i.e. their values would not change from iteration to iteration). Similarly, from equations (2.102) and (2.103) it can be seen that the quantities Vdoi and cos Φi are also constant. As cos Φr and cos Φi are constant, (R) (I) from equations (2.101) and (2.105), QDCi and QDCj are also constant. Thus, the equivalent real and reactive power injections at buses 4 and 5 are constant (they need not be updated at every iteration) and hence these values can be pre-calculated and suitably adjusted into the injected real and reactive powers at buses 4 and 5 before solving the AC system equations. Thus, for the example at hand, the net injected real and reactive powers at bus 4 and bus 5 can be calculated as follows; P4 = −1.15 − 1.0 = −2.15 p.u., Q4 = −0.6 − 0.16276 = −0.76276 p.u., P5 = −0.85 + 0.9845 = 0.1345 p.u. and Q5 = −0.4 − 0.35024 = −0.75024 p.u. With these net injected real and reactive powers, the load flow solution of the AC system is computed and the final solution is shown in Table 2.36. It is to be noted that in Table 2.36, it has been assumed that no violation of reactive power limit has taken place for any of the generators. After the final solution of voltage magnitudes is obtained, the quantities ar and ai can be calculated from equations (2.100) and (2.104) as ar = 0.8714 and ai = 0.8149. Please note that in these two equations, the quantities Etr and Eti should be taken in actual values (i.e. in kV), not in per unit. Let us now turn our attention to combination 8. Following the same reasonong as described above, from equations (2.107) - (2.108) it can be observed that the quantities Vdi and Id are constant. Moreover, as steps (3)-(8) of combination 8 are same as in combination 1, it immediately follows that for combination 8 also, the equivalent real and reactive power injections (representing 78

Table 2.36: Final Results of AC-DC load flow of 5 bus system without any generator Q limit violation

Bus no. 1 2 3 4 5

∣V ∣

Without generator Q limit

θ

Pinj

Qinj

(p.u) (deg) (p.u) (p.u) 1.0 0 0.68984 0.46301 1.0 -0.63995 0.5 -0.17235 1.0 -4.91128 1.0 1.54134 0.82813 -17.48682 -2.15 -0.76277 0.91332 -3.89028 0.13449 -0.75025 Total iteration = 5

the DC system) are constant. Therefore, by pre-calculating these equivalent power injections and subsequently incorporating these calculated values into net bus power injections, standard AC load flow solution can be computed to obtain the solution of the composite AC-DC system. From the above discussion regarding combination 1 and 8, it may appear that the equivalent real and reactive power injections (representing the DC system) are always constant for any combination of the specified control variables. However, this is not true. Depending on the specified control variables, the equivalent real and reactive power injections may vary from iteration to iteration and therefore, they need to be calculated in every iteration. As an example, let us consider combination 3. For this combination, the various steps are as follows: Step 1: Initialise all the bus voltages with flat start. Hence, Etr and Eti are known. Step 2: From the specified values of Pdr and Vdi , calculate Vdr and Id using equations (2.96) and (2.97) respectively. Step 3: Calculate Vdor from equation (2.100). Step 4: Calculate cos α and cos Φr using equations (2.98) and (2.99) respectively. (R) Step 5: From the knowledge of Pdr and cos Φr , calculate QDCi using equation (2.101). Step 6: Calculate Vdoi from equation (2.104). Step 7: Calculate cos γ and cos Φi using equations (2.102) and (2.103) respectively. (I) (I) Step 8: Calculate PDCj and QDCj from equation (2.105). (R)

Please note that in steps 3-5, the quantities Vdor , cos Φr and QDCi are all dependent on the (I) rectifier side AC bus voltage, Etr . Similarly, in steps 6-8, the quantities Vdoi , cos Φi and QDCj are all (I) dependent on the inverter side AC bus voltage, Eti . The quantity PDCj however, depends only on the DC system quantities and hence remain constant. Thus, the equivalent reactive power injections at both rectifier and inverter side depend on the AC bus voltage magnitudes (although the equivalent real power injections at both the sides are independent of AC bus voltage magnitudes). Hence, the equivalent reactive power injections need to be updated at each iteration and with these updated power injection values, another iteration of AC load flow is carried out. This process is continued till convergence is achieved. To illustrate this procedure further, let us assume that the specified values corresponding to combination 3 are as follows: 79

ar = 1.0; Pdr = 100 MW; ai = 1.0; Vdi = 250 kV From the information of Pdr and Vdi , the quantities Pdi and Id are calculated as; Pdi = 98.45 MW and Id = 393.8 Amp. The calculated values for different other DC quantities corresponding to first 3 iterations are shown in Table 2.37. In this table, the symbols ‘In’ and ‘MM’ denote ‘iteration number’ and ‘mismatch’ respectively. Proceeding in this fashion, the algorithm finally converges in 70 iterations with a convergence threshold value of 1.0e−12 . The final converged values of different (R) DC quantities are as follows: Vdr = 253.938 kV; α = 19.82o ; γ = 34.84o ; QDCi = 41.52 MVAR; Q(I) DCj = 72.44 MVAR. The final converged values of the AC system quantities are shown in Table 2.38. Note that, as in the case of Table 2.36, in this case also, no generator reactive power violation has been assumed. Table 2.37: Calculated DC quantities for first three iterations in 5 bus system

In 0 1 2

V4

V5

Vdor

Q(R) DCi

φr

α

(p.u.) (p.u.) (kV) (rad.) (MVAR) 1.0 1.0 356.52 0.778 98.54 0.7843 0.8731 279.63 0.432 46.11 0.7682 0.8703 273.91 0.3842 40.43

Vdoi

φi

Q(I) DCj

(deg.) (kV) (rad.) (MVAR) 43.48 356.52 0.7937 100.09 22.33 311.29 0.638 73.04 19.20 310.29 0.634 72.37

γ

MM (deg.) 44.4 2.15 35.08 0.5578 34.82 0.038

Table 2.38: Final Results of AC-DC load flow of 5 bus system for combination 3 without any generator Q limit violation

Bus no. 1 2 3 4 5

∣V ∣

Without generator Q limit

θ

Pinj

(p.u) (deg) (p.u) 1.0 0 0.75382 1.0 -0.99374 0.5 1.0 -5.52481 1.0 0.82813 -18.69153 -2.15 0.91332 -3.65137 0.13449 Total iteration = 70

Qinj (p.u) 0.78420 -0.16893 2.17338 -1.01522 -1.12440

For further illustration, let us now consider the 30-bus system. In this system, it is now assumed that one bipolar HVDC link is connected between bus 9 and 28 (rectifier at bus 9 and inverter at bus 28). Other relevant data for this link are as follows; Rd = 10.0 Ω; Nr = Ni = 2;

3 3 Xcr = Xci = 6.0 Ω. π π

The load flow has been solved for combination-1, combination-3 and combination-8 (of specified quantities). The specified values which have been considered are as follows; Combination-1 80

α = 5o ; Pdr = 100 MW; γ = 18o ; Vdi = 250 kV Combination-3

ar = 0.75; Pdr = 100 MW; ai = 0.75; Vdi = 250 kV Combination-8

α = 5o ; Pdi = 100 MW; γ = 18o ; Vdr = 250 kV The results of the 30-bus system for combination 1 and 8 are shown in Table 2.39 for a tolerance of 10−12 p.u. Furthermore, the results corresponding to combination 3 are shown in Table 2.40. It is to be noted that for these results, no reactive power limit on the generators have been considered. The final solutions of corresponding DC system quantities are also shown in these tables for these three cases. Comparison of Tables 2.18, 2.39 and 2.40 shows that because of the reactive power absorption at both bus 9 and 28, the overall voltage profile of the system is lower in the presence of HVDC link. Moreover, when the equivalent injected real and reactive powers are constant (i.e. do not vary from iteration to iteration), the number of iterations taken by the algorithm is quiet comparable with that taken by the normal NRLF (polar) method (without any HVDC link). However, when these equivalent injected powers vary from iteration to iteration, the number of iterations taken by the sequential algorithm is appreciably more as compared that taken by the normal NRLF (polar) method (without any HVDC link). In the above, the detail calculation procedures for three combinations (1, 3 and 8) have been shown. For the remaining combinations, the DC quantities can be calculated following the procedure of either combination 1 or combination 3 and thus, these are not detailed here. Let us now turn our attention to simultaneous techniques. As discussed earlier, in the simultaneous technique, the AC and DC system equations are solved together. Now, in a N-bus, M-generator power system having a HVDC link between lens ‘k’ and ‘l’ (bus ‘k’ being the rectifier and bus ‘l’ being the inverter), the total number of unknown are (N − 1) + (N − M ) + 5 = 2N − M + 4. To solve these unknowns, we also have (N − 1) + (N − M ) + 5 = 2N − M + 4 equations. Therefore the size of Jacobrian matrix would be (2N −M +4)×(2N −M +4), as compared to the (2N −M −1)×(2N −M −1) Jacobrian matrix of the AC system. The additional 5 rows and 5 columns pertain to the DC equations which need to be evaluated in each iteration. Also, the Jacobian matrix also needs to be inverted in each iteration, thereby increasing the computation burden appreciably. Apart from that, depending upon the combination of specified quantities, the DC equations [(2.88)-(2.92)] need to be recasted appropriately before starting the solution procedure. Therefore, the simultaneous solution technique does not give any computational advantage vis-à-vis the sequential method and thus, this method is not further discussed here. 81

Table 2.39: Results of the 30 bus system with a bipolar HVDC link between bus 9 and 28

Bus no.

∣V ∣

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

(p.u) 1.05 1.0338 1.02345 1.01637 1.0058 1.01055 0.99437 1.023 1.01796 1.00713 1.0913 1.04132 1.0883 1.02408 1.01562 1.01841 1.00471 0.99979 0.9938 0.99623 0.99497 0.99578 1.00064 0.98942 0.98869 0.97048 0.99711 1.00723 0.97669 0.96489

With combination 1

θ

Pinj

Qinj

∣V ∣

With combination 8

θ

Pinj

Qinj

(deg) (p.u) (p.u) (p.u) (deg) (p.u) (p.u) 0 2.42039 -0.26099 1.05 0 2.42229 -0.26023 -5.0387 0.3586 0.06523 1.0338 -5.04273 0.3586 0.06857 -8.08233 -0.024 -0.012 1.02325 -8.08809 -0.024 -0.012 -9.72937 -0.076 -0.016 1.01613 -9.73654 -0.076 -0.016 -13.69708 -0.6964 0.10883 1.0058 -13.70427 -0.6964 0.11036 -11.35395 0 0 1.01026 -11.35703 0 0 -13.96765 -0.628 -0.109 0.99419 -13.97288 -0.628 -0.109 -12.18127 -0.45 0.47454 1.023 -12.18265 -0.45 0.48393 -20.00269 -1 -0.16277 1.01711 -20.13217 -1.01653 -0.16832 -19.52942 -0.058 -0.02 1.00633 -19.6175 -0.058 -0.02 -18.07884 0.1793 0.38778 1.0913 -18.20671 0.1793 0.39224 -16.16702 -0.112 -0.075 1.04109 -16.21744 -0.112 -0.075 -14.97002 0.1691 0.36696 1.0883 -15.02017 0.1691 0.36879 -17.34016 -0.062 -0.016 1.02379 -17.39439 -0.062 -0.016 -17.64022 -0.082 -0.025 1.01524 -17.69657 -0.082 -0.025 -17.83191 -0.035 -0.018 1.01792 -17.89732 -0.035 -0.018 -19.23196 -0.09 -0.058 1.00399 -19.31323 -0.09 -0.058 -19.02914 -0.032 -0.009 0.99925 -19.09708 -0.032 -0.009 -19.66267 -0.095 -0.034 0.99317 -19.73758 -0.095 -0.034 -19.69028 -0.022 -0.007 0.99555 -19.76851 -0.022 -0.007 -19.72658 -0.175 -0.112 0.99419 -19.81012 -0.175 -0.112 -19.62149 0 0 0.99501 -19.7033 0 0 -18.30125 -0.032 -0.016 1.00016 -18.3595 -0.032 -0.016 -18.8339 -0.087 -0.067 0.98881 -18.89458 -0.087 -0.067 -16.70952 0 0 0.98823 -16.73336 0 0 -17.15432 -0.035 -0.023 0.97001 -17.17859 -0.035 -0.023 -15.12179 0 0 0.99676 -15.12284 0 0 -9.39718 0.98449 -0.35025 1.00683 -9.35718 1 -0.35723 -16.41847 -0.024 -0.009 0.97633 -16.42047 -0.024 -0.009 -17.35084 -0.106 -0.019 0.96452 -17.35353 -0.106 -0.019 Total iteration = 4 Total iteration = 4

DC system solutions Vdr = 253.938 kV; idr = 393.8 Amp.; ar = 0.7088; Qdr = 16.276 MVAR; ai = 0.7389; Pdi = 98.45 MW; Qdi = 35.024 MVAR;

DC system solutions Vdi = 245.93 kV; idr = 406.613 Amp.; ar = 0.6988; Pdr = 101.65 MW; ai = 0.7275; Qdr = 16.831 MVAR; Qdi = 35.723 MVAR;

We are now at the end of our theoretical study of various load flow techniques. However, in production grade implementatin of these techniques, the sparsity of the linear equations (connecting the mismatch and solution vectors) is exploited to reduce the computation time as well as memory requirement. From the next lecture, we will study some methods for solution of sparse linear 82

Table 2.40: Further results of the 30 bus system with a bipolar HVDC link between bus 9 and 28

Bus no.

∣V ∣

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

(p.u) 1.05000 1.03380 1.02180 1.01437 1.00580 1.00804 0.99286 1.02300 1.00248 0.99624 1.09130 1.03787 1.08830 1.01972 1.01042 1.01181 0.99507 0.99250 0.98530 0.98713 0.98445 0.98541 0.99404 0.98099 0.98192 0.96358 0.99144 1.00345 0.97089 0.95901

With combination 3

θ

Pinj

(deg) (p.u) 0.00000 2.42241 -5.04581 0.35860 -8.07176 -0.02400 -9.71785 -0.07600 -13.71987 -0.69640 -11.32909 0.00000 -13.96633 -0.62800 -12.20102 -0.45000 -20.07763 -1.00000 -19.56171 -0.05800 -18.12405 0.17930 -16.26936 -0.11200 -15.06838 0.16910 -17.44774 -0.06200 -17.72802 -0.08200 -17.89806 -0.03500 -19.27534 -0.09000 -19.10546 -0.03200 -19.73134 -0.09500 -19.74984 -0.02200 -19.76760 -0.17500 -19.66188 0.00000 -18.37832 -0.03200 -18.89414 -0.08700 -16.75594 0.00000 -17.20702 -0.03500 -15.15766 -0.00000 -9.34449 0.98449 -16.46954 -0.02400 -17.41321 -0.10600 Total iteration = 23

Qinj (p.u) -0.25233 0.09175 -0.01200 -0.01600 0.12191 0.00000 -0.10900 0.55554 -0.33808 -0.02000 0.46906 -0.07500 0.39377 -0.01600 -0.02500 -0.01800 -0.05800 -0.00900 -0.03400 -0.00700 -0.11200 -0.00000 -0.01600 -0.06700 0.00000 -0.02300 -0.00000 -0.38369 -0.00900 -0.01900

DC system solutions Vdr = 253.938 kV; idr = 393.8 Amp.; α = 15.21o ; Qdr = 33.808 MVAR; γ = 18.31o ; Pdi = 98.45 MW; Qdi = 38.37 MVAR;

equations.

83

Module 3 Sparsity Technique 3.1

Sparse matrices

Sparse matrix is a matrix in which most (or, at least, significant number) of the elements are zero. In the context of power system analysis, the matrices associated with power flow solution are sparse. For example, let us consider the YBU S matrix. As we have already seen, the off-diagonal elements of YBU S matrix signifies the connectivity between the nodes. To be more precise, the element (i,j) of YBU S matrix is non-zero if there is a direct connection between node ‘i’ and node ‘j’, while it is zero if there is no direct connectivity between these two nodes. Now, in many power systems, generally any bus is connected to mostly 3-4 buses directly. Therefore, in a 100 buses system (say), there would be at best 4-5 non-zero terms (including the diagonal) in any row of the YBU S matrix, rest of the elements being zero. Therefore, out of (100 × 100) = 10, 000 elements, only about 500 terms would be non-zero and the other terms (elements) would be zero. Thus, in this case, the YBU S matrix is almost 95 percent sparse. For any larger system, the percentage of sparsity of the associated YBU S matrix would be even more. Because of the sparsity of the YBU S matrix, the Jacobian matrix for load flow solution is also sparse. To see that, please consider equations (2.48) - (2.55). From these equations it can be seen that all the elements of the Jacobian matrix depend on the element Yij . Therefore, if this element Yij is zero, the corresponding elements of the Jacobian matrix would also be zero. As most of the elements (Yij ) of the YBU S matrix are zero, it immediately follows that most of the elements of the Jacobian matrix would also be zero, thereby making the Jacobian matrix also quite sparse. Now, in each iteration of the NRLF technique, (we are considering the polar form here), the correction vector (∆X) is computed by inverting the Jacobian matrix and thereafter multiplying the inverse of the Jacobian matrix with the mismatch vector (∆M ) (please see equation (2.45)). However, even though the Jacobian matrix is sparse, its inverse is a full matrix. Hence, computation of the direct inverse of the sparse matrix involves a lot of computational burden. Therefore, it would be much less intensive if equation (2.45) can be solved exploiting the sparse nature of the Jacobian matrix. Apart from this, storing all the elements of a highly sparse matrix also consumes the memory unnecessarily. Therefore, if only the non-zero elements are stored in appropriate fashion, a lot of memory can be freed. Of course, with the storage of only the non-zero elements, the complexity of 84

programming will increase. However, for any general purpose load flow program, which is expected to handle any large size power system, enhancement in the complexity of programming is often a small cost as compared to the advantage of optimized memory utilization. Below we will discuss some schemes for solving a set of linear equations (note that equation (2.48) is a set of linear equations) utilizing the sparse nature of the Jacobian matrix and also some schemes for storing a sparse matrix. We will start with the Gaussian Elimination method for solving a set of linear equations.

3.2

Gaussian elimination technique

Let us consider a linear system of equations:

Ax = b

(3.1)

Where x and b are both (n×1) vectors and A is a (n×n) co-efficient matrix. The most obvious method for solving equation (3.1) is to invert matrix A, that is x = A−1 b. However, equation (3.1) can also be solved indirectly by converting the matrix A into an upper triangular form with appropriate changes reflected in the vector b and then by back substitution. To illustrate the basic procedure, let us consider a 4th order system as shown in equations (3.2)-(3.5).

a11 x1 + a12 x2 + a13 x3 + a14 x4 = b1

(3.2)

a21 x1 + a22 x2 + a23 x3 + a24 x4 = b2

(3.3)

a31 x1 + a32 x2 + a33 x3 + a34 x4 = b3

(3.4)

a41 x1 + a42 x2 + a43 x3 + a44 x4 = b4

(3.5)

The Gaussian elimination proceeds in certain sequential steps as described below: Step 1: a) Equation (3.2) is divided throughout by a11 .

x1 +

a13 a14 b1 a12 x2 + x3 + x4 = a11 a11 a11 a11

(3.6)

b) Multiply equation (3.6) by a21 , a31 , a41 (one by one) and subtract the resulting expression from equations (3.3), (3.4) and (3.5) respectively to yield:

(a22 −

a12 a21 a13 a21 a14 a21 b1 a21 ) x2 + (a23 − ) x3 + (a24 − ) x4 = b2 − a11 a11 a11 a11

(3.7)

(a32 −

a12 a31 a13 a31 a14 a31 b1 a31 ) x2 + (a33 − ) x3 + (a34 − ) x4 = b3 − a11 a11 a11 a11

(3.8)

85

(a42 −

a13 a41 a14 a41 b1 a41 a12 a41 ) x2 + (a43 − ) x3 + (a44 − ) x4 = b4 − a11 a11 a11 a11

(3.9)

Equations (3.6) to (3.9) can be written more compactly as,

x1 +

a12 a13 a14 b1 x2 + x3 + x4 = a11 a11 a11 a11

(3.10)

(1) (1) (1) a(1) 22 x2 + a23 x3 + a24 x4 = b2

(3.11)

(1) (1) (1) a(1) 32 x2 + a33 x3 + a34 x4 = b3

(3.12)

(1) (1) (1) a(1) 42 x2 + a43 x3 + a44 x4 = b4

(3.13)

Where, in equations (3.10) - (3.13)

a(1) jk = ajk −

aj1 a1k a11

for j, k = 2, 3, 4

(3.14)

Step 2: In this step we will work with equations (3.11) - (3.13). (1) a) Equation (3.11) is divided throughout by a22 .

x2 +

(1) a23

(1) a22

x3 +

a(1) 24

a(1) 22

(1)

x4 =

b2(1)

(3.15)

a(1) 22

(1)

b) Multiplying equation (3.15) by a32 and a42 (one by one) and subtracting the resulting expressions from equations (3.12) and (3.13) respectively one can obtain; (1) 33

[a

(1) 43

[a

− −

(1) a(1) 23 a32 (1) 22

a

(1) a(1) 23 a42 (1) 22

a

(1) 34

] x3 + [a

(1) 44

] x3 + [a

(1) (1) a24 a32

−

(1) 22

a

(1) (1) a24 a42

−

(1) 22

a

] x4 = [b

(1) 3

] x4 = [b

(1) 4

− −

b2(1)

(1) 22

a

b2(1)

(1) 22

a

a(1) 32 ]

(3.16)

a(1) 42 ]

(3.17)

Similar to step 1, equations (3.15) - (3.17) are re-written as,

x2 +

(1) a23 (1) a22

x3 +

a(1) 24

a(1) 22

x4 =

b2(1)

a(1) 22

(3.18)

(2) (2) a33 x3 + a34 x4 = b3(2)

(3.19)

(2) (2) a43 x3 + a44 x4 = b4(2)

(3.20)

Where, in equations (3.19) - (3.20), (2) jk

a

(1) jk

=a

−

(1) (1) aj2 a2k (1) a22

for j, k = 3, 4

Step 3: In this step we will work with equations (3.19) and (3.20). 86

(3.21)

(2)

a) Equation (3.19) is divided throughout by a33 .

x3 +

(2) a34

(2) a33

x4 =

b3(2)

(3.22)

(2) a33

(2)

b) Multiplying equation (3.22) by a43 and subtracting it from equation (3.20) one can obtain, (2) 44

[a

−

(2) a(2) 34 a43 (2) 33

a

] x4 = [b

(2) 4

−

b3(2)

(2) 33

a

(2) ] a43

(3.23)

Equation (3.23) contains only one unknown, x4 . Therefore, the value of x4 can be calculated from this equation. With the value of x4 thus calculated, x3 can be calculated from equation (3.22). Going back in this manner, x2 can be calculated from equation (3.18) (with the known values of x3 and x4 ) and lastly, the value of x1 can be calculated from equation (3.10) (with the known values of x2 , x3 and x4 ). The steps described in equations (3.6)-(3.23) can easily be expressed in terms of standard matrix operations. To see this, let us represent equations (3.2)-(3.5) in matrix notation as shown in equation (3.24). In this equation, it is assumed that a11 ≠ 0.

⎡ ⎢a11 ⎢ ⎢a ⎢ 21 ⎢ ⎢a31 ⎢ ⎢ ⎢a41 ⎣

a12 a22 a32 a42

a13 a23 a33 a43

⎤⎡ ⎤ ⎡ ⎤ a14 ⎥ ⎢x1 ⎥ ⎢b1 ⎥ ⎥⎢ ⎥ ⎢ ⎥ a24 ⎥⎥ ⎢⎢x2 ⎥⎥ ⎢⎢b2 ⎥⎥ ⎥⎢ ⎥ = ⎢ ⎥ a34 ⎥⎥ ⎢⎢x3 ⎥⎥ ⎢⎢b3 ⎥⎥ ⎥⎢ ⎥ ⎢ ⎥ a44 ⎥⎦ ⎢⎣x4 ⎥⎦ ⎢⎣b4 ⎥⎦

(3.24)

Starting with this matrix, the various steps for Gaussian elimination are as follows. Step M1 On equation (3.24), the operation R1/a11 (where ‘R1’ is the first row of the co-efficient matrix of equation (3.24)) is carried out to obtain equation (3.6) and the resulting matrix equation is shown in equation (3.25).

⎡ ⎤⎡ ⎤ ⎡ ⎤ ⎢ 1 a12 /a11 a13 /a11 a14 /a11 ⎥ ⎢x1 ⎥ ⎢b1 /a11 ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢a ⎥ ⎢x ⎥ ⎢ b ⎥ a a a ⎢ 21 ⎢ 2 ⎥ 22 23 24 ⎥ ⎢ 2 ⎥ ⎢ ⎥⎢ ⎥ = ⎢ ⎥ ⎢a31 ⎥ ⎢ ⎥ ⎢ b3 ⎥ a a a x 32 33 34 3 ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢a41 ⎥ ⎢ ⎥ ⎢ a a a x b 42 43 44 4 4 ⎥ ⎣ ⎦⎣ ⎦ ⎣ ⎦

(3.25)

Step M2 On equation (3.25), the operations (R2 − R1 ∗ a21 ), (R3 − R1 ∗ a31 ) and (R4 − R1 ∗ a41 ) are carried out (where ‘Ri’ denotes the ith (i = 1, 2, 3, 4) row of the co-efficient matrix of equation (3.25)) to obtain equations (3.10)-(3.13) and the resulting matrix equation is shown in equation (3.26). In 87

(1)

this equation, it is assumed that a22 ≠ 0.

⎡1 a /a a /a a /a ⎤ ⎡x ⎤ ⎡b /a ⎤ ⎥ ⎢ ⎢ 12 11 13 11 14 11 ⎥ ⎢ 1 ⎥ ⎥ ⎢ ⎥ ⎢ 1 11 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ (1) (1) ⎥ ⎥ ⎢x2 ⎥ ⎢ b(1) ⎢0 a(1) a a 2 22 23 24 ⎥ ⎥⎢ ⎥ ⎢ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ = (1) ⎥ (1) (1) ⎥ ⎢ ⎥ ⎢ ⎢0 a(1) b x a a ⎥ ⎢ 3⎥ ⎢ 3 ⎥ ⎢ 32 33 34 ⎥ ⎥⎢ ⎥ ⎢ ⎢ ⎢ (1) ⎥ ⎢ (1) (1) (1) ⎥ ⎢ ⎥ ⎢0 a42 a43 a44 ⎥⎦ ⎢⎣x4 ⎥⎦ ⎢⎣ b4 ⎥⎦ ⎣

(3.26)

Step M3 (1)

On equation (3.26), the operation ‘R2/a22 ’ is carried out (corresponding to equation (3.15)) to obtain the resulting matrix equation shown in equation (3.27).

⎡1 a /a a13 /a11 a14 /a11 ⎤⎥ ⎡⎢x1 ⎤⎥ ⎡⎢ b1 /a11 ⎤⎥ ⎢ 12 11 ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎢ (1) (1) ⎥ ⎢ (1) (1) (1) (1) ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎢0 ⎥ 1 a23 /a22 a24 /a22 ⎥ ⎢x2 ⎥ ⎢b2 /a22 ⎥⎥ ⎢ ⎢ ⎥ ⎢ ⎥ = ⎢ (1) ⎥ (1) (1) ⎥ ⎢0 a(1) ⎥ ⎢x ⎥ ⎢ b a a ⎥ ⎢ ⎥ ⎢ 3⎥ ⎢ 3 32 33 34 ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎢ ⎥ ⎢ ⎥ ⎢ (1) ⎥ (1) (1) (1) ⎥ ⎢0 a42 a43 a44 ⎥⎦ ⎢⎣x4 ⎥⎦ ⎢⎣ b4 ⎦ ⎣

(3.27)

Step M4 (1)

(1)

On equation (3.27), the operations (R3 − R2 ∗ a32 ) and (R4 − R2 ∗ a42 ) are carried out corresponding to the equations (3.18)-(3.21) and the resulting matrix equation is shown in equation (2) (3.28). In this equation, it is assumed that a33 ≠ 0.

⎡1 a /a a13 /a11 a14 /a11 ⎤⎥ ⎡⎢x1 ⎤⎥ ⎡⎢ b1 /a11 ⎤⎥ ⎢ 12 11 ⎢ ⎥ ⎥⎢ ⎥ ⎢ ⎢ ⎢ (1) (1) ⎥ (1) (1) (1) (1) ⎥ ⎢ ⎥ ⎢0 ⎥ ⎢ ⎥ ⎢ 1 a23 /a22 a24 /a22 ⎥ ⎢x2 ⎥ ⎢b2 /a22 ⎥⎥ ⎢ ⎢ ⎥ ⎢ ⎥ = ⎢ (2) ⎥ (2) (2) ⎢0 ⎥ ⎥ ⎢x ⎥ ⎢ b 0 a33 a34 ⎢ ⎥ ⎢ 3⎥ ⎢ 3 ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎥ ⎢ (2) (2) (2) ⎢0 ⎥ ⎢ ⎥ ⎥ ⎢ 0 a a x b 4 43 4 44 ⎣ ⎦⎣ ⎦ ⎣ ⎦

(3.28)

Step M5 (2)

On equation (3.28), the operation ‘R3/a33 ’ is carried out to obtain the matrix equation shown in equation (3.29).

⎡1 a /a a13 /a11 a14 /a11 ⎤⎥ ⎡⎢x1 ⎤⎥ ⎡⎢ b1 /a11 ⎤⎥ ⎢ 12 11 ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎢ (1) (1) ⎥ (1) (1) (1) (1) ⎥ ⎢ ⎥ ⎢0 1 a23 /a22 a24 /a22 ⎥⎥ ⎢⎢x2 ⎥⎥ ⎢⎢b2 /a22 ⎥⎥ ⎢ ⎢ ⎥ ⎢ ⎥ = ⎢ (2) (2) ⎥ (2) (2) ⎥ ⎢ ⎥ ⎢0 ⎢ ⎥ 0 1 a34 /a33 ⎢ ⎥ ⎢x3 ⎥ ⎢b3 /a33 ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ (2) ⎥ (2) (2) ⎢0 ⎥ ⎢ ⎥ ⎢ ⎥ 0 a43 a44 ⎦ ⎣x4 ⎦ ⎣ b4 ⎣ ⎦

(3.29)

Step M6 (2)

Lastly, on equation (3.29), the operation (R4 − R3 ∗ a43 ) is carried out to obtain the matrix 88

equation shown in equation (3.30).

⎡1 a /a a13 /a11 a14 /a11 ⎤⎥ ⎡⎢x1 ⎤⎥ ⎡⎢ b1 /a11 ⎤⎥ ⎢ 12 11 ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎢ (1) (1) ⎥ ⎢ (1) (1) (1) (1) ⎥ ⎢ ⎥ ⎥ ⎢ ⎥ ⎢ ⎢0 ⎥ b /a x 1 a /a a /a 22 ⎥ 23 22 24 22 ⎥ ⎢ 2 ⎥ ⎢ 2 ⎢ ⎥ ⎢ ⎥ ⎢ ⎢ ⎥ = (2) (2) ⎥ ⎢ ⎥ ⎢b(2) /a(2) ⎥ ⎢0 x 0 1 a /a ⎢ ⎢ 33 ⎥ 34 33 ⎥ ⎢ 3 ⎥ ⎥ ⎢ ⎥⎢ ⎥ ⎢ 3 ⎢ ⎥ ⎢ ⎥ ⎢ (3) ⎥ (3) ⎥ ⎢0 0 0 a44 ⎥⎦ ⎢⎣x4 ⎥⎦ ⎢⎣ b4 ⎦ ⎣ (3)

(2)

In equation (3.30), a44 = a44 −

(2) a(2) 34 a43 (2) 33

a

(3)

and b4

= b4(2) −

b(2) 3

(2) 33

a

(3.30)

(2) a43 . From this equation, the

unknowns can be easily solved by back-substitution starting from the last row of the final co-efficient matrix in equation (3.30). Thus, Gaussian elimination enables us to solve the unknown quantities in a systematic manner without inverting the co-efficient matrix. Therefore, by adopting the same procedure, the correction vector (∆M ) can be computed from equation (2.48) without having to invert the Jacobian matrix. When a large power system in analyzed, adopting Gaussian elimination reduces computational burden to a large extent (as compared to inversion of the Jacobian matrix). (1) (2) In the above procedure, the variables a11 , a22 and a33 have been assumed to be non-zero. These variables, by which the rows of the co-efficient matrix are divided, are called the ‘pivot variables’. However, during the elimination process, it is not necessary that the ‘pivot variables’ would be always non-zero. If any pivot variable turns out to be zero at any intermediate step, then the corresponding row is interchanged with the next row so that the new pivot variable is non-zero and the elimination process can continue. We will look into an example of Gaussian elimination procedure in the next lecture.

89

3.2.1

Example of Gaussian elimination

We wish to solve the following matrix equation by Gaussian elimination:

⎡ ⎢11 ⎢ ⎢23 ⎢ ⎢ ⎢22 ⎢ ⎢ ⎢12 ⎣

17 27 32 15

18 25 34 41

⎤⎡ ⎤ ⎡ ⎤ 16⎥ ⎢x1 ⎥ ⎢10⎥ ⎥⎢ ⎥ ⎢ ⎥ 28⎥⎥ ⎢⎢x2 ⎥⎥ ⎢⎢20⎥⎥ ⎥⎢ ⎥ = ⎢ ⎥ 36⎥⎥ ⎢⎢x3 ⎥⎥ ⎢⎢30⎥⎥ ⎥⎢ ⎥ ⎢ ⎥ 36⎥⎦ ⎢⎣x4 ⎥⎦ ⎢⎣40⎥⎦

(3.31)

Towards this goal, we proceed with the different steps as follows: Step M1 On equation (3.31), the operation R1/A(1, 1) (where, A(1, 1) = 11) is performed to yield,

⎤ ⎤⎡ ⎤ ⎡ ⎡ ⎢ 1 1.5455 1.6364 1.4545⎥ ⎢x1 ⎥ ⎢0.9091⎥ ⎥ ⎥⎢ ⎥ ⎢ ⎢ ⎢23 27 25 28 ⎥⎥ ⎢⎢x2 ⎥⎥ ⎢⎢ 20 ⎥⎥ ⎢ ⎥ ⎥⎢ ⎥ = ⎢ ⎢ ⎢22 32 34 36 ⎥⎥ ⎢⎢x3 ⎥⎥ ⎢⎢ 30 ⎥⎥ ⎢ ⎥ ⎥⎢ ⎥ ⎢ ⎢ ⎢12 15 41 36 ⎥⎦ ⎢⎣x4 ⎥⎦ ⎢⎣ 40 ⎥⎦ ⎣

(3.32)

Step M2 On equation (3.32), the operations (R2 − R1 ∗ A(2, 1)), (R3 − R1 ∗ A(3, 1)) and (R4 − R1 ∗ A(4, 1)) are carried out (where A(2, 1) = 23, A(3, 1) = 22 and A(4, 1) = 12) and the resulting matrix equation is given by;

⎡ ⎤⎡ ⎤ ⎡ ⎤ ⎢1 1.5455 1.6364 1.4545 ⎥ ⎢x1 ⎥ ⎢ 0.9091 ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢0 −8.5455 −12.6364 −5.4545⎥ ⎢x ⎥ ⎢−0.9091⎥ ⎢ ⎥ ⎢ 2⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ = ⎢ ⎥ ⎥ ⎢0 ⎢ ⎥ ⎢ ⎥ x −2.0 −2.0 4.0 10 3 ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢0 −3.5455 21.3636 18.5455 ⎥ ⎢x4 ⎥ ⎢ 20.0909 ⎥ ⎣ ⎦⎣ ⎦ ⎣ ⎦

(3.33)

Step M3 On equation (3.33), the operation R2/A(2, 2) (where, A(2, 2) = −8.5455) is carried out to get;

⎡ ⎤⎡ ⎤ ⎡ ⎤ ⎢1 1.5455 1.6364 1.4545 ⎥ ⎢x1 ⎥ ⎢ 0.9091 ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢0 1 1.4787 0.6383 ⎥⎥ ⎢⎢x2 ⎥⎥ ⎢⎢ 0.1064 ⎥⎥ ⎢ ⎢ ⎥⎢ ⎥ = ⎢ ⎥ ⎢0 −2.0 −2.0 4.0 ⎥⎥ ⎢⎢x3 ⎥⎥ ⎢⎢ 10 ⎥⎥ ⎢ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢0 −3.5455 21.3636 18.5455⎥ ⎢x4 ⎥ ⎢20.0909⎥ ⎣ ⎦⎣ ⎦ ⎣ ⎦

(3.34)

Step M4 On equation (3.34), the operations (R3 − R2 ∗ A(3, 2)) and (R4 − R2 ∗ A(4, 2)) are carried out to obtain (where A(3, 2) = −2.0 and A(4, 2) = −3.5455);

⎡ ⎤⎡ ⎤ ⎡ ⎤ ⎢1 1.5455 1.6364 1.4545 ⎥ ⎢x1 ⎥ ⎢ 0.9091 ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢0 1 1.4787 0.6383 ⎥⎥ ⎢⎢x2 ⎥⎥ ⎢⎢ 0.1064 ⎥⎥ ⎢ ⎢ ⎥⎢ ⎥ = ⎢ ⎥ ⎢0 0 0.9574 5.2766 ⎥⎥ ⎢⎢x3 ⎥⎥ ⎢⎢10.2128⎥⎥ ⎢ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢0 0 26.6065 20.8085⎥⎦ ⎢⎣x4 ⎥⎦ ⎢⎣29.4681⎥⎦ ⎣ 90

(3.35)

Step M5 On equation (3.35), the operation R3/A(3, 3) (where A(3, 3) = 0.9574) is carried out to obtain the matrix equation shown below:

⎤ ⎤⎡ ⎤ ⎡ ⎡ ⎢1 1.5455 1.6364 1.4545 ⎥ ⎢x1 ⎥ ⎢ 0.9091 ⎥ ⎥ ⎥⎢ ⎥ ⎢ ⎢ ⎢0 1 1.4787 0.6383 ⎥⎥ ⎢⎢x2 ⎥⎥ ⎢⎢ 0.1064 ⎥⎥ ⎢ ⎥ ⎥⎢ ⎥ = ⎢ ⎢ ⎢0 0 1 5.5114 ⎥⎥ ⎢⎢x3 ⎥⎥ ⎢⎢10.6672⎥⎥ ⎢ ⎥ ⎥⎢ ⎥ ⎢ ⎢ ⎢0 0 26.6065 20.8085⎥⎦ ⎢⎣x4 ⎥⎦ ⎢⎣29.4681⎥⎦ ⎣

(3.36)

Step M6 Lastly, on equation (3.36), the operation (R4 − R3 ∗ A(4, 3)) (where A(4, 3) = 26.6065) is carried out to get;

⎤ ⎤⎡ ⎤ ⎡ ⎡ ⎢1 1.5455 1.6364 1.4545 ⎥ ⎢x1 ⎥ ⎢ 0.9091 ⎥ ⎥ ⎥⎢ ⎥ ⎢ ⎢ ⎥ ⎢x ⎥ ⎢ 0.1064 ⎥ ⎢0 1 1.4787 0.6383 ⎥ ⎥ ⎢ 2⎥ ⎢ ⎢ ⎥ ⎥⎢ ⎥ = ⎢ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢0 10.6672 x 0 1 5.5114 3 ⎥ ⎥⎢ ⎥ ⎢ ⎢ ⎥ ⎥⎢ ⎥ ⎢ ⎢ ⎥ ⎢x4 ⎥ ⎢−254.3484⎥ ⎢0 0 0 −125.83 ⎦ ⎦⎣ ⎦ ⎣ ⎣

(3.37)

In equation (3.37), the co-efficient matrix has been converted to an upper-triangular matrix. From the last row of this equation, x4 can be calculated as x4 = 254.3484/125.83 = 2.0214. Back substituting this value of x4 in the third row of equation (3.37) one can obtain x3 = 10.6672 − 5.5114 × 2.0214 = −0.4735. Similarly, substitution of the values of x3 and x4 in the second row of equation (3.37) yields the of x2 as x2 = 0.1064 + 1.4787 × 0.4735 − 0.6383 × 2.0214 = −0.4837. Lastly, substituion of x2 , x3 and x4 in the first row of equation (3.37) gives x1 = 0.9091 + 1.5455 × 0.4837 + 1.6364 × 0.4735 − 1.4545 × 2.0214 = −0.5086.

3.3

Optimal order of elimination

We have seen that the Gaussian Elimination method is quite effective for solving a large set of spare linear equations without having to invert the co-efficient matrix. Moreover, at every stage, if the calculations pertaining to Gaussian elimination is carried out only using the non-zero terms, great saving in the computational burden can be achieved. However, if the elimination process is carried out in the normal sequence, at any stage of elimination, the original zero-elements may the concerted into a non-zero element. This is normally termed as ‘fill-in’ phenomenon. On the other hand, instead of following the normal sequence, if the elimination process is carried out in an appropriate order, then the occurrence of ‘fill-in’ can be avoided to a great extent. A simple example given below illustrates this point. In equation (3.38), an initial co-efficient matrix is shown at the left hand side (part ‘a’) and the structure of the co-efficient matrix after step-1 is shown at the right hand side (part ‘b’). It is to be noted that in this equation, only the positions of non-zero terms (denoted by ‘×’) and zero terms (denoted by ‘o’) are shown. As can be seen in equation (3.38), after step-1, all the original zero elements have been converted to non-zero terms (denoted by ‘⊗’), or, in other words, significant 91

level of ‘fill-in’ has occurred.

1 2 3 4

⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

1

2

3

4

× × × ×

× × × × o o o × o o o ×

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

1 2 3 4

a) Initial ‘A’ matrix

⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

1

2

3

4

1 × o × o ⊗ o ⊗

× ⊗ × ⊗

× ⊗ ⊗ ×

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

(3.38)

b) ‘A’ matrix after step 1

Now, if the original co-efficient matrix shown in part (a) of equation (3.38) is re-arranged as shown in part (a) of equation (3.39), then after step 1, there would be no ‘fill-in’ as can be observed in part (b) of equation (3.39).

4 2 3 1

⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

4

3

2

1

× o o o o × o × o × × ×

× × × ×

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

a) Rearranged ‘A’ matrix

4 2 3 1

⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

4

3

2

1

1 o o o o × o × o o × ×

× × × ×

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

(3.39)

b) Rearranged ‘A’ matrix after step 1

From the above example, it is apparent that if the rows are eliminated in an ‘optimal order’, then the number of ‘fill-in’ would be minimum. However, an ideal ‘optimal order’ is very difficult to develop and perhaps is impossible. As an alternative, various ‘near optimal ordering’ schemes have been developed. Some of them are discussed below: Scheme 1 In this scheme, before elimination, number the rows of the co-efficient matrix ‘A’ according to the number of non-zero, off-diagonal terms. Thus, the rows with only one off-diagonal, non-zero term are numbered first, those with two non-zero, off- diagonal terms are numbered second and so on. However, this scheme does not take into account the changes occurring in the co-efficient matrix during the elimination process. Therefore, this scheme in quite easy and straight forward to implement. Scheme 2 In this scheme the rows of the co-efficient matrix ‘A’ are numbered such that at each step of the elimination procedure, the row with the fewest number of non-zero off-diagonal terms would be operated next. If more than one row meets this criterion, then any one row is chosen. Therefore, this scheme requires the simulation of the elimination procedure to estimate the changes occurring in the co-efficient matrix in advance. Thus, this method takes longer time as compared to scheme 1 to compute the solution, but is definitely better than scheme 1. 92

Scheme 3 In this scheme, the rows are numbered in such a way so that the row which will introduce fewest nonzero off-diagonal terms would be operated upon next. If more than one row satisfies this criterion, choose any one row. Again, this scheme also requires the simulation of the elimination process to study its effects on the co-efficient matrix in advance. Hence, this method also takes longer time than scheme 1. Let us now look at another technique for solving a set of linear equations without the need of inverting the co-efficient matrix, namely, the triangular factorization or LU decomposition.

3.4

Triangular factorization:

In triangular factorization or decomposition method, a square matrix A is expressed as a product of two triangular matrices as A = LU, where L is a lower triangular matrix and U is an upper triangular matrix. As an example, let the matrix A be a 4 × 4 (N = 4) matrix. Upon triangular factorization (or ‘LU’ decomposition), the matrix A is represented as A = LU. Or,

⎡ ⎢a11 ⎢ ⎢a ⎢ 21 ⎢ ⎢a31 ⎢ ⎢ ⎢a41 ⎣

a12 a22 a32 a42

a13 a23 a33 a43

⎤ ⎡ ⎤ ⎡ 0 0 ⎥ ⎢β11 β12 β13 a14 ⎥ ⎢α11 0 ⎥ ⎢ ⎥ ⎢ 0 ⎥⎥ ⎢⎢ 0 β22 β23 a24 ⎥⎥ ⎢⎢α21 α22 0 ⎥=⎢ ⎥×⎢ a34 ⎥⎥ ⎢⎢α31 α32 α33 0 ⎥⎥ ⎢⎢ 0 0 β33 ⎥ ⎢ ⎥ ⎢ a44 ⎥⎦ ⎢⎣α41 α42 α43 α44 ⎥⎦ ⎢⎣ 0 0 0

⎤ β14 ⎥ ⎥ β24 ⎥⎥ ⎥ β34 ⎥⎥ ⎥ β34 ⎥⎦

(3.40)

With this decomposition, the equation Ax = b can be written as,

Ax = b or, (LU)x = b or, L(Ux) = b

(3.41)

Ly = b

(3.42)

Or, In equation (3.42), y = Ux. Expanding equation (3.42) we get,

⎡ ⎤⎡ ⎤ ⎡ ⎤ ⎢α11 0 0 0 ⎥ ⎢y1 ⎥ ⎢b1 ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢α α 0 ⎥⎥ ⎢⎢y2 ⎥⎥ ⎢⎢b2 ⎥⎥ ⎢ 21 22 0 ⎢ ⎥⎢ ⎥ = ⎢ ⎥ ⎢α31 α32 α33 0 ⎥ ⎢y3 ⎥ ⎢b3 ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢α41 α42 α43 α44 ⎥ ⎢y4 ⎥ ⎢b4 ⎥ ⎣ ⎦⎣ ⎦ ⎣ ⎦

(3.43)

From equation (3.43), the intermediate vector y can be calculated as,

b1 α11 i−1 1 = [bi − ∑ αij yj ] ; αii j=1

y1 = yi

93

i = 2, 3, ⋯ N

(3.44)

Again, expanding the expression y = Ux we get,

⎤⎡ ⎤ ⎡ ⎤ β14 ⎥ ⎢x1 ⎥ ⎢y1 ⎥ ⎥⎢ ⎥ ⎢ ⎥ β24 ⎥⎥ ⎢⎢x2 ⎥⎥ ⎢⎢y2 ⎥⎥ ⎥⎢ ⎥ = ⎢ ⎥ β34 ⎥⎥ ⎢⎢x3 ⎥⎥ ⎢⎢y3 ⎥⎥ ⎥⎢ ⎥ ⎢ ⎥ β34 ⎥⎦ ⎢⎣x4 ⎥⎦ ⎢⎣y4 ⎥⎦

⎡ ⎢β11 β12 β13 ⎢ ⎢0 β β ⎢ 22 23 ⎢ ⎢0 0 β33 ⎢ ⎢ ⎢0 0 0 ⎣

(3.45)

Now, With the knowledge of the intermediate vector y, from equation (3.45), the solution vector x can be calculated as,

yn βN N N 1 − [yi ∑ βij xj ] ; = βii j=1+1

xN = xi

i = (N − 1), (N − 2), ⋯ 1

(3.46)

We will now look into the basic procedure of obtaining the ‘LU’ decomposition in the next lecturel.

94

3.4.1

Method of LU decomposition

To solve for the matrices L and U for given matrix A, let us write the i, j th element of equation (3.40). In general, the element aij can be represented as (from equation (3.40)),

αi1 β1j + αi2 β2j + ⋯up to appropriate term = aij

(3.47)

The number of terms in the sum of equation (3.47) depends on whether ‘i’ or ‘j’ is the smaller number. We have in fact three distinct cases:

i < j;

αi1 β1j + αi2 β2j + ⋯ + αii βij = aij

(3.48)

i = j;

αi1 β1j + αi2 β2j + ⋯ + αii βjj = aij

(3.49)

i > j;

αi1 β1j + αi2 β2j + ⋯ + αij βjj = aij

(3.50)

In equations (3.48)-(3.50), total N 2 equations are available for a total of (N 2 + N ) unknown ‘α’ and ‘β ’ (the diagonals terms being repeated twice). As the number of unknown quantities is greater than the number of equations, we need to specify N of the unknown quantities arbitrarily and then the remaining unknown quantities can be solved. In fact, it is always possible to take

αii = 1;

i = 1, 2, ⋯ N

(3.51)

With the condition of equation (3.51), an elegant procedure called ‘Crout’s algorithm’ quite easily solves for the N 2 unknown ‘α′ and ‘β ′ by just rearranging the equations in a certain order. The algorithm is as follows: step 1: Set αii = 1 for i = 1, 2, 3 ⋯ N . step 2: For each j = 1, 2, 3 ⋯ N , perform the following operation: a) For i = 1, 2, 3 ⋯ N solve for βij as i−1

βij = aij − ∑ αik βkj

(3.52)

k=1

Note: when i = 1, the summation term in equation (3.52) is taken to be zero. b) For i = j + 1, j + 2, ⋯N solve for αij as; j−1 1 [aij − ∑ αik βkj ] αij = βjj k=1

(3.53)

These first and second operations both need to be carried out before going to next value of ‘j’. 95

3.4.2

Example of LU decomposition

⎡3 2 7⎤ ⎥ ⎢ ⎥ ⎢ ⎥ Decompose the matrix A = ⎢ ⎢2 3 1⎥ in ‘LU’ form through ‘Crout’s algorithm’. ⎥ ⎢ ⎢3 4 1⎥ ⎦ ⎣ ⎡3 2 7⎤ ⎡α 0 ⎤⎥ ⎡⎢β11 β12 β13 ⎤⎥ ⎥ ⎢ 11 0 ⎢ ⎥ ⎥⎢ ⎥ ⎢ ⎢ ⎥ = ⎢α21 α22 0 ⎥ ⎢ 0 β22 β23 ⎥ Solution: Let A = ⎢ 2 3 1 ⎥ ⎥⎢ ⎥ ⎢ ⎢ ⎥ ⎥⎢ ⎥ ⎢ ⎢ ⎥ ⎢3 4 1⎥ ⎢α31 α32 α33 ⎥ ⎢ 0 0 β 33 ⎦ ⎦⎣ ⎦ ⎣ ⎣ Applying ‘Crout’s algorithm; α11 = 1; α22 = 1; α33 = 1; For j = 1; i = 1 → β11 = a11 = 3; 1 (a21 − 0) = i = 2 → α21 = β11 1 i = 3 → α31 = (a31 − 0) = β11

2 ; 3 3 = 1; 3

For j = 2;

i = 1 → β12 = a12 = 2;

5 i = 2 → β22 = a22 − α21 β12 = ; 3 6 1 (a32 − α31 β12 ) = ; i = 3 → α32 = β22 5 For j = 3;

i = 1 → β13 = a13 = 7; i = 2 → β23 = a23 − α21 β13 = −

11 ; 13

8 i = 3 → β33 = a33 − α31 β13 − α32 β23 = − ; 5 ⎡3 2 7⎤ ⎡ 1 0 0⎤⎥ ⎡⎢3 2 7 ⎤⎥ ⎢ ⎥ ⎢ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎥ = ⎢2/3 1 0⎥ ⎢0 5/3 −11/3⎥ Hence, ⎢ 2 3 1 ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢3 4 1⎥ ⎢ 1 6/5 1⎥ ⎢0 0 −8/5 ⎥ ⎣ ⎦ ⎣ ⎦⎣ ⎦

3.5

Storage schemes for sparse matrices

There are several methods available in the literature for storing a sparse matrix. Some of these schemes are described here. 96

Random packing: In this method, every non-zero element of the matrix is stored in a primary array while its row and column indices are stored in two secondary arrays. As each element in individually identified, the elements can be stored in a random manner. For example, the sparse matrixes shown in equation (3.54) are stored in one primary array and two secondary arrays as shown below.

⎡ 0 0 0 0 0 ⎤⎥ ⎢ ⎥ ⎢ A = ⎢⎢ 0 0 2.67 0 3.12⎥⎥ ⎥ ⎢ ⎢−1.25 0.29 0 0 2.31⎥ ⎦ ⎣ Primary array (elements) X = Secondary array (row indices) i = Secondary array (column indices) j =

{0.29 {3 {2

3.12 2 5

(3.54)

-1.25 2.67 3 2 1 3

2.31 0} 3 0} 5 0}

In the above, the zero at the end of each array denotes the termination of the array.

Systematic packing: In case the elements of a sparse matrix are read or constructed or sorted in a systematic order, then there is no need to adopt both row and column indices for each element. Instead some alternative, more efficient schemes can be adopted as described below. a) The use of row address: In this scheme, the index of first non-zero element in each row is specified in a separate integer array. As an example, for the matrix A in equation (3.54) the elements can be represented as, Real array (elements) = {2.67 Integer array JA = {3 Integer array IST = {1

3.12 5 1

-1.25 1 3

0.29 2 6

2.31} 5} }

In this case, the array of row address IST has been constructed such that the number of non-zero elements in row ‘i’ is IST(I+1)-IST(I). Thus, for a matrix with ‘m’ rows, the array IST will have ‘(m+1)’ entries. For example, from the array ‘IST’, the number of non-zero elements in 1st row of the matrix A is IST(2) - IST(1) = 1 - 1 = 0, which is indeed true as observed from equation (3.54). Similarly, the array ‘IST’ indicates that the number of non-zero elements in 2nd row of the matrix A is IST(3) - IST(2) = 3 - 1 = 2 and from the first two elements of the array ‘JA’, these two elements are located at columns 3 and 5 of the matrix whereas the elements themselves are given by the first two elements of the real array (2.67 and 3.12) (which is again true from equation (3.54)). In a similar way, the elements and the locations of the elements in the third row of the matrix A can easily be identified from the above three arrays. Moreover, please note that, as the number of rows in the matrix A is 3, the total number of entries in array ‘IST’ is 4.

97

b) The use of dummy variables: In this scheme, the integer array IST is not used. Instead, dummy variables are introduced in the array JA itself to indicate the beginning of each row and the end of the matrix. For example, a zero entry (except the last one) in the array JA could indicate the presence of dummy variables and the dummy variables itself could specify the row number. Moreover, the non-zero elements would indicate the column number of the elements. Also, as before, the zero at the end of the array indicates the termination of the list. Hence the matrix A of equation (3.54) in this scheme looks like Real array (elements) Integer array JA

= {2 2.67 = {0 3

3.12 3 5 0

-1.25 1

0.29 2

2.31 0} 5 0}

c) Compound identifiers: In the random packing scheme it in possible to reduce the storage requirement by combining the two indices for each element so that these can be held in one integer storage. A suitable compound identifier could be (n × i + j) where n is an integer higher than the number of columns in the matrix and (i, j) denote the position of the non-zero element in the matrix. For example, the matrix A in equation (3.54) could look like Real array (elements) Integer array JA

= =

{2.67 {2003

3.12 -1.25 2005 3001

0.29 3002

2.31 3005

0} 9999}

In the above, ‘n’ has been chosen to be equal to 1000. Also, in the real array, the entry zero indicates the end of the array and the corresponding entry in the integer array is 9999 (to signifies the end of the integer array). However, unless compound identifiers yield necessary or highly desirable storage saving, it should not be used because of the following reasons: i) Extra programming would be required to interpret the identifiers correctly. ii) It should not be used for matrices whose order is so large that integer register overflow results. d) The use of mixed arrays: It is possible to use a single array to store both the non-zero elements of the matrix and the identifiers. For example, the matrix A in equation (3.54) could be stored as, Real array (B)

=

{-2

3

2.67

5

3.12

-3

1

-1.25

2

0.29

5

2.31

0}

In the above scheme, each non-zero element is preceded by its column number and each non-zero row is preceded by the negative of the corresponding row number. With this discussion, we are now at the and of our study of sparse linear systems. From the next lecture, we will start the discussion of short circuit analysis methods.

98

Module 4 Short Circuit analysis 4.1

¯ BUS formation without mutual coupling between eleZ ments

For a network with ‘m’ buses and a reference bus, one can write a relation between bus currents and bus voltages as ¯ BUS ] [V ¯ BUS ] [¯IBUS ] = [Y (4.1) Where,

¯IBUS is (m × 1) bus current injection vector ¯ BUS is (m × 1) bus voltage vector V ¯ BUS is (m × m) bus admittence matrix Y equation (4.1) can also be written as

¯ BUS ] [¯IBUS ] ¯ BUS ] = [Z [V

(4.2)

Where,

¯ BUS is m × m bus impedance matrix and is given by, Z

¯ BUS ] = [Y ¯ BUS ] [Z

−1

From equation (4.2) for the ith bus one can write

V¯i = Z¯i1 I¯1 + Z¯i2 I¯2 + ⋯Z¯ii I¯i + ⋯ + Z¯im I¯m

(4.3)

¯ij can be written as From equation (4.3), Z V¯i Z¯ij = ∣ I¯j I¯k = 0; ∀ k = 1, 2, ⋯ m, 99

(4.4) ≠j

V¯i Z¯ii = ∣ I¯i ¯

(4.5)

Ik = 0; ∀ k = 1, 2, ⋯ m, ≠ i

¯ BUS matrix Following points should be noted for the Z ¯ij is the off-diagonal element of Z ¯ BUS matrix and is called the • Z impedance’ between ith and j th bus. ¯ii is the diagonal element of Z ¯ BUS matrix and is called the • Z impedance’ of ith bus.

‘open-circuit transfer

‘open-circuit driving point

¯ BUS matrix is symmetrical, then the matrix Z ¯ BUS is also symmetrical i.e. Z¯ik = Z¯ki . • If the Y ¯ BUS matrix of • Since in a power network each bus is connected to very few other buses,the Y the network has large number of zero elements and is therefore, sparse in nature. The ZBUS , matrix on the other hand, is invariably a full matrix. ¯ BUS matrix of a network can be found out by inverting the Y ¯ BUS matrix of the network. The Z ¯ BUS matrix This is not an efficient method as every time there is a modification in the network, the Y ¯ BUS matrix. is modified and inversion has to be done again to obtain the modified the Z ¯ BUS building algorithm overcomes these problems. It avoids the inversion A step-by-step Z ¯ BUS . process and network modifications are easily incorporated in the existing Z Few terms need to be defined before the step by step process can be explained. These are : • Graph : The graph of a network describes the geometrical structure of the network showing the interconnections of network elements. • Tree : A tree of a graph is a connected sub graph that connects all the nodes without forming a closed path or a loop. A graph can have a number of distinct trees. • Branches : The elements of a tree are called branches. The number of branches ‘b’ of a tree with ‘n’ nodes, including reference, is given by

b=n−1

(4.6)

• Links : The elements of a graph not included in the tree of the graph are called links. Each link is associated with a loop. If ‘e’ is the number of elements in a graph, then the number of links ‘`’ is given by ` = e−b = e−n+1 (4.7) The above definitions are explained with the help of illustrations as shown below : Fig. 4.1 is a single line diagram of a power system. It has 4 buses, bus(1) to bus(4) and six elements element e1 to element e6 . In this figure, bus(0) is taken as the reference bus. Fig. 4.2 shows the graph of the network depicting the interconnection of the elements and the reference node. 100

Figure 4.1: Single Line Diagram of a Power System

Figure 4.2: A graph of the Power system of Fig. 4.1 A tree of the graph of Fig. 4.2 is shown in Fig. 4.3. The branches and the links have been shown with solid lines and dotted lines respectively. Following points should be noted from Fig. 4.3 : • The total number of nodes (including reference node) is 5 (i.e. n = 5) • The number of branches is b = n − 1 = 5 − 1 = 4 . As can be as in Fig. 4.3 where 101

Figure 4.3: A tree of the graph of Fig. 4.2

e1 , e2 , e5 , e6 , are such a set of branches that form a tree of the graph. • The total number of elements in the graph is e = 6 . • The number of links is ` = e − n + 1 = 6 − 5 + 1 = 2. The two links in the graph are e3 and e4 shown with dotted lines in Fig. 4.3.

Figure 4.4: Partial network with ‘m’ buses The bus impedance matrix is built up starting with a branch connected to the reference and subsequently the elements are added one by one till all the nodes and elements are considered. Let 102

¯ BUS matrix for a partial network with ‘m’ buses and a reference bus ‘0’, as us assume that the Z shown in Fig. 4.4, exists. The bus voltages and bus currents for the partial network satisfy the relation m ¯ BUS ¯m ¯m [V ] = [Z BUS ] [IBUS ]

(4.8)

Where, m ¯ BUS V is m × 1 bus voltage vector

¯Im BUS is m × 1 bus current injection vector ¯m Z BUS is m × m bus impedence matrix of the partial network

¯ BUS , one element at a time is added to the partial network, till all the elements are To build Z added to the network. The added element may be a branch or a link and hence the four possible element additions to a partial network are: a. Addition of a branch between a new node and the reference b. Addition of a branch between a new node and an existing node c. Addition of a link between an existing node and the reference d. Addition of a link between two existing nodes Let us now discuss these four cases one-by-one in detail.

4.1.1

Addition of a branch between a new node and the reference node (case 1):

Fig. 4.5 shows the addition of a branch between a new node ‘q’ and the reference ‘0’.The addition of ¯ BUS to (m + 1) × (m + 1) with the addition a new node to the partial network increases the size of Z of a new row and a new column corresponding to the new node ‘q’, Let the impedance of this branch be z¯q0 . The new network equation can be written as:

⎡ V¯1 ⎤ ⎡ Z¯11 ⎢ ⎥ ⎢ ⎢¯ ⎥ ⎢¯ ⎢ V2 ⎥ ⎢ Z21 ⎢ ⎥ ⎢ ⎢ ⎥ ⎢ ⎢ ⋮ ⎥ ⎢ ⋮ ⎢ ⎥ ⎢ ⎢ V¯ ⎥ ⎢ Z¯ ⎢ p ⎥ ⎢ p1 ⎢ ⎥=⎢ ⎢ ⋮ ⎥ ⎢ ⋮ ⎢ ⎥ ⎢ ⎢¯ ⎥ ⎢ ¯ ⎢Vm ⎥ ⎢Zm1 ⎢ ⎥ ⎢ ⎢⋯⎥ ⎢ ⋯ ⎢ ⎥ ⎢ ⎢ ⎥ ⎢ ⎢ V¯q ⎥ ⎢ Z¯q1 ⎣ ⎦ ⎣

Z¯12 ⋯ Z¯1p ⋯ Z¯1m Z¯22 ⋯ Z¯2p ⋯ Z¯2m Z¯p2 ⋯ Z¯pp ⋯ Z¯pm Z¯m2 ⋯ Z¯mp ⋯ Z¯mm ⋯ ⋯ ⋯ ⋯ ⋯ Z¯q2 ⋯ Z¯qp ⋯ Z¯qm 103

Z¯1q ⎤⎥ ⎡⎢ I¯1 ⎤⎥ ⎥⎢ ⎥ Z¯2q ⎥⎥ ⎢⎢ I¯2 ⎥⎥ ⎥⎢ ⎥ ⎥⎢ ⋮ ⎥ ⎥⎢ ⎥ Z¯pq ⎥⎥ ⎢⎢ I¯p ⎥⎥ ⎥⎢ ⎥ ⎥⎢ ⋮ ⎥ ⎥⎢ ⎥ ⎥⎢ ⎥ ¯ ⋮ Zmq ⎥⎥ ⎢⎢I¯m ⎥⎥ ⋯ ⋯ ⎥⎥ ⎢⎢ ⋯ ⎥⎥ ⎥⎢ ⎥ ⋮ Z¯qq ⎥⎦ ⎢⎣ I¯q ⎥⎦ ⋮ ⋮ ⋮ ⋮

(4.9)

Figure 4.5: Addition of a branch between a new node and the reference

¯ BUS . Only the elements The addition of branch does not change the elements of the original matrix Z of the added new row and column corresponding to q th bus need to be calculated. Further,since the ¯qi = Z¯iq , ∀ i = 1, 2, ⋯ m. power system elements are linear and bilateral, Z Now since,

V¯q ¯ Zqq = ∣ I¯q I¯k = 0; ∀ k = 1, 2, ⋯ m a current source of I¯q = 1 p.u is connected to the q th bus, with all the others buses open, and the voltage of q th bus (V¯q ) is computed, as shown in Fig. 4.6. From Fig. 4.6 one gets V¯q = z¯q0 I¯q , and thus with I¯q = 1 p.u.

V¯q Z¯qq = ∣ = z¯qo I¯q I¯k = 0; ∀ k = 1, 2, ⋯ m, ¯qi , a current source I¯i = 1 p.u. is connected between ith bus and the reference For finding out Z bus with all other buses open circuited as shown in Fig. 4.7. From Fig. 4.7 , V¯q = 0, and hence with I¯i = 1 p.u.

V¯q Z¯qi = ∣ I¯i I¯k = 0; ∀ k = 1, 2, ⋯ m,

≠i

=0

¯q1 , Z¯q2 , ⋯ Z¯mq and Z¯1q , Z¯2q , ⋯ Z¯qm are equal This implies that all the off-diagonal elements Z to zero. ¯ Bus matrix after addition of an element between the new bus ‘q ’ and the Hence, the modified Z 104

¯qq for Case 1 Figure 4.6: Calculation of Z

reference bus ‘0′ is given as,

⎡¯ ⎢ Z11 ⎢ ⎢ Z¯21 ⎢ ⎢ ⎢ ⋮ ⎢ ¯ BUS = ⎢⎢ Z¯p1 Z ⎢ ⎢ ⋮ ⎢ ⎢¯ ⎢Zm1 ⎢ ⎢ ⎢ 0 ⎣

4.1.2

Z¯12 ⋯ Z¯1p ⋯ Z¯1m Z¯22 ⋯ Z¯2p ⋯ Z¯2m Z¯p2 ⋯ Z¯pp ⋯ Z¯pm Z¯m2 ⋯ Z¯mp ⋯ Z¯mm 0 ⋯ 0 ⋯ 0

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ 0 ⎥⎥ ⎥ ⎥ ⎥ 0 ⎥⎥ ⎥ z¯qo ⎥⎦ 0 0

(4.10)

Addition of a branch between a new node and an existing node (Case 2):

Let a branch with impedance z¯pq be connected between an existing node ‘p’ and a new node ‘q ’ as ¯ Bus matrix increases by one to (m+1)×(m+1) due shown in Fig. 4.8. In this case also, the size of Z to the addition of a new node ’q ’ to the network. The modified network equations can be written 105

¯qi for case 1 Figure 4.7: Calculation of Z

as:

⎡ V¯1 ⎤ ⎢ ⎥ ⎢¯ ⎥ ⎢ V2 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⋮ ⎥ ⎢ ⎥ ⎢ V¯ ⎥ ⎢ p⎥ ⎢ ⎥= ⎢ ⋮ ⎥ ⎢ ⎥ ⎢¯ ⎥ ⎢Vm ⎥ ⎢ ⎥ ⎢⋯⎥ ⎢ ⎥ ⎢ ⎥ ⎢ V¯q ⎥ ⎣ ⎦

⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

Z¯11 Z¯21 ⋮ Z¯p1 ⋮ Z¯m1 ⋯ Z¯q1

Z¯12 ⋯ Z¯1p ⋯ Z¯1m Z¯22 ⋯ Z¯2p ⋯ Z¯2m Z¯p2 ⋯ Z¯pp ⋯ Z¯pm Z¯m2 ⋯ Z¯mp ⋯ Z¯mm ⋯ ⋯ ⋯ ⋯ ⋯ Z¯q2 ⋯ Z¯qp ⋯ Z¯qm

⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋯ ⋮

⎤ ⎡ I¯1 ⎤ ⎥⎢ ⎥ ⎥⎢ ¯ ⎥ ⎥ ⎢ I2 ⎥ ⎥⎢ ⎥ ⎥⎢ ⎥ ⎥⎢ ⋮ ⎥ ⎥⎢ ⎥ Z¯pq ⎥⎥ ⎢⎢ I¯p ⎥⎥ ⎥⎢ ⎥ ⎥⎢ ⋮ ⎥ ⎥⎢ ⎥ ⎥⎢ ⎥ ¯ Zmq ⎥⎥ ⎢⎢I¯m ⎥⎥ ⋯ ⎥⎥ ⎢⎢ ⋯ ⎥⎥ ⎥⎢ ⎥ Z¯qq ⎥⎦ ⎢⎣ I¯q ⎥⎦ Z¯1q Z¯2q

(4.11)

¯m Even after the addition of branch p-q, the original matrix Z Bus remains unchanged. Only the additional elements corresponding to the q th row and column need to be calculated. ¯qq one can write For calculating Z V¯q Z¯qq = ∣ I¯q I¯k = 0; ∀ k = 1, 2, ⋯ m ¯qq current source of I¯q = 1 p.u is connected to the q th bus, with all the others buses To evaluate Z open circuited, and the voltage of q th bus V¯q is computed, as shown in Fig. 4.9. From Fig. 4.9 with 106

Figure 4.8: Addition of a branch between an existing node ‘p’ and a new node ‘q’

I¯q = 1 p.u. and I¯k = 0, ∀ k = 1, 2, ⋯, m one can write, V¯1 = Z¯1q I¯q = Z¯1q ⎫ ⎪ ⎪ ⎪ ⎪ ¯ ¯ ¯ ¯ ⎪ V2 = Z2q Iq = Z2q ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⋮ ⎪ ⎪ ⎪ ¯ ¯ ¯ ¯ Vp = Zpq Iq = Zpq ⎬ ⎪ ⎪ ⎪ ⎪ ⋮ ⎪ ⎪ ⎪ ¯ ¯ ¯ ⎪ Vm = Zmq Iq = Zmq ⎪ ⎪ ⎪ ⎪ ⎪ ¯ ¯ ¯ ¯ ⎪ Vq = Zqq Iq = Zqq ⎪ ⎭

(4.12)

From the Fig. 4.10, the voltages V¯p and V¯q can be related as

V¯q = V¯p − v¯pq = Z¯pq I¯q − z¯pq¯ipq = Z¯pq + z¯pq

(4.13)

¯pq and V¯q = Z¯qq . Because, from the Fig. 4.10 , ¯ipq = −I¯q = −1 pu and from equation (4.12) V¯p = Z Thus, Z¯qq = Z¯pq + z¯pq (4.14) ¯qi one can write For calculating Z V¯q Z¯qi = ∣ I¯i ¯

Ik = 0; ∀ k = 1, 2, ⋯ m, ≠ i

¯qi a current source of I¯1 = 1 p.u is connected to the ith bus, with all the others Hence, to compute Z 107

¯qq Figure 4.9: Calculation of Z

Figure 4.10: Relation between V¯p and V¯q buses open circuited, and the bus voltage V¯i is computed for all the buses, as shown in Fig. 4.11. From equation (4.11) one gets

V¯1 = Z¯1i I¯i = Z¯1i ⎫ ⎪ ⎪ ⎪ ⎪ ¯ ¯ ¯ ¯ ⎪ V2 = Z2i Ii = Z2i ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⋮ ⎪ ⎪ ⎪ ¯ ¯ ¯ ¯ Vp = Zpi Ii = Zpi ⎬ ⎪ ⎪ ⎪ ⎪ ⋮ ⎪ ⎪ ⎪ ¯ ¯ ¯ ⎪ Vm = Zmi Ii = Zmi ⎪ ⎪ ⎪ ⎪ ⎪ ¯ ¯ ¯ ¯ ⎪ Vq = Zqi Ii = Zqi ⎪ ⎭ 108

¯qi for case 2 Figure 4.11: Calculation of Z From Fig. 4.11, V¯q = V¯p as the current in the branch p − q is zero. Hence, from the above equations one gets (4.15) Z¯qi = Z¯pi ; ∀ i = 1, 2, ⋯ m

¯ Bus matrix after addition of an element between an existing bus ‘p’ the Hence, the modified Z new bus ‘q’ is given as, ⎡¯ ⎢ Z11 ⎢ ⎢ Z¯21 ⎢ ⎢ ⎢ ⋮ ⎢ ¯ BUS = ⎢⎢ Z¯p1 Z ⎢ ⎢ ⋮ ⎢ ⎢¯ ⎢Zm1 ⎢ ⎢¯ ⎢ Zp1 ⎣

Z¯12 ⋯ Z¯1p ⋯ Z¯1m Z¯22 ⋯ Z¯2p ⋯ Z¯2m Z¯p2 ⋯ Z¯pp ⋯ Z¯pm Z¯m2 ⋯ Z¯mp ⋯ Z¯mm Z¯p2 ⋯ Z¯pp ⋯ Z¯pm

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ Z¯pp ⎥⎥ ⎥ ⎥ ⎥ ¯ Zmp ⎥⎥ ⎥ Z¯pq + z¯qp ⎥⎦ Z¯1p Z¯2p

(4.16)

So far in this lecture, we have considered the cases of addition of branches only. In the next lecture we will consider the case of addition of links.

109

4.1.3

Addition of a link between an existing node and the reference node (Case 3):

When an element is connected between an existing node and the reference, it creates a loop and thus, the addition of this element is equivalent to the addition of a link. This will not generate any ¯ Bus matrix remains unchanged. However, all the elements are new node and the size of modified Z modified and need to be recalculated. Let the added element,with an impedance of z¯qo , be connected between an existing node ‘q’ and the reference node ‘0’ as shown in the Fig. 4.12. I¯` is the current

Figure 4.12: Addition of a link between an existing node ‘q’ and the reference through the link as shown in the Fig. 4.12. This current modifies the current injected into q th bus from I¯q to I¯q − I¯` . The modified network equations can be written as,

⎫ V¯1 = Z¯11 I¯1 + Z¯12 I¯2 + ⋯ + Z¯1q (I¯q − I¯` ) + ⋯ + Z¯1m I¯m ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⋮ ⋮ ⎪ ⎪ ⎪ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ Vq = Zq1 I1 + Zq2 I2 + ⋯ + Zqq (Iq − I` ) + ⋯ + Zqm Im ⎬ ⎪ ⎪ ⎪ ⎪ ⋮ ⋮ ⎪ ⎪ ⎪ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ⎪ Vm = Zm1 I1 + Zm2 I2 + ⋯ + Zmq (Iq − I` ) + ⋯ + Zmm Im ⎪ ⎪ ⎭

(4.17)

V¯q = z¯qo I¯`

(4.18)

Also Substuting V¯q from the equation (4.17) into the equation (4.18) one can write

Z¯q1 I¯1 + Z¯q2 I¯2 + ⋯ + Z¯qq (I¯q − I¯` ) + ⋯ + Z¯qm I¯m = z¯qo I¯` 110

or

0 = −Z¯q1 I¯1 − Z¯q2 I¯2 − ⋯ − Z¯qq I¯q + ⋯ − Z¯qm I¯m + (Z¯qq + z¯qo )I¯`

(4.19)

Equations equation (4.17) and equation (4.19) together form the set of (m + 1) simultaneous network equations which can be expressed in matrix form as:

⎡ V¯1 ⎤ ⎢ ⎥ ⎢¯ ⎥ ⎢ V2 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⋮ ⎥ ⎢ ⎥ ⎢ V¯ ⎥ ⎢ q⎥ ⎢ ⎥= ⎢ ⋮ ⎥ ⎢ ⎥ ⎢¯ ⎥ ⎢Vm ⎥ ⎢ ⎥ ⎢⋯⎥ ⎢ ⎥ ⎢ ⎥ ⎢0⎥ ⎣ ⎦

⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

Z¯11 Z¯12 Z¯21 Z¯22 ⋮ ¯ Zq1 Z¯q2 ⋮ ¯ Zm1 Z¯m2 ⋯ ⋯ ¯ −Zq1 −Z¯q2

⋯ ⋯

Z¯1q Z¯2q

⋯ ⋯

Z¯1m Z¯2m

⋯

Z¯qq

⋯

Z¯qm

⋯ Z¯mq ⋯ Z¯mm ⋯ ⋯ ⋯ ⋯ ⋯ −Z¯qq ⋯ −Z¯qm

⋮ −Z¯1q ⋮ −Z¯2q ⋮ ⋮ −Z¯qq ⋮ ⋮ −Z¯mq ⋯ ⋯ ¯ ⋮ Zqq + z¯qo

⎤ ⎡ I¯1 ⎤ ⎥⎢ ⎥ ⎥⎢ ¯ ⎥ ⎥ ⎢ I2 ⎥ ⎥⎢ ⎥ ⎥⎢ ⎥ ⎥⎢ ⋮ ⎥ ⎥⎢ ⎥ ⎥ ⎢ I¯ ⎥ ⎥⎢ q ⎥ ⎥⎢ ⎥ ⎥⎢ ⋮ ⎥ ⎥⎢ ⎥ ⎥⎢¯ ⎥ ⎥ ⎢Im ⎥ ⎥⎢ ⎥ ⎥⎢⋯⎥ ⎥⎢ ⎥ ⎥⎢ ⎥ ⎥ ⎢ I¯` ⎥ ⎦⎣ ⎦

(4.20)

¯ Bus matrix The link current I¯` has to be eliminated and hence, the last row and column of modified Z have to be eliminated. The partitioned matrix relation of equation (4.20) can be written in compact form as: ⎡ ¯m ⎤ (m) ⎢[VBus ]⎥ ⎢ ⎥ ⎢ ⎥ (m) ¯m Z ⎢ ⎥ Bus ⎢ ⎥= [ ¯ T ⎢ ⎥ (1) [∆Z] ⎢ ⎥ ⎢ ⎥ ⎢ 0 ⎥ ⎣ ⎦

⎡¯m ⎤ ⎢IBus ⎥ ⎢ ⎥ ⎥ ¯ ⎢⎢ [∆Z] ⎥ ⎥ ]⎢ ¯ ⎢ ⎥ Z`` ⎢ ⎥ ⎢ ¯ ⎥ ⎢ I` ⎥ ⎣ ⎦ (1)

(4.21)

where,

¯ = [−Z¯1q − Z¯2q ⋯ − Z¯qq ⋯ − Z¯mm ]T [∆Z]

From equation (4.20) one can write

¯ T [¯Im ¯ ¯ 0 = [∆Z] Bus ] + Z`` I` or,

¯ T [¯Im ] [∆Z] Bus I¯` = − ¯ Z``

(4.22)

From equation (4.21) one can also write, m ¯ ¯ ¯ Bus ¯m ¯ [V ] = [Z Bus ][IBus ] + [∆Z]I`

(4.23)

Substituting I¯` from equation (4.22) into equation (4.23) one obtains,

¯ ¯ T [∆Z][ ∆Z] m ¯ Bus ¯m ] [¯Im [V ] = [[Z ] − Bus Bus ] Z¯`` 111

(4.24)

Hence, m ¯ Bus ¯ Bus ][¯Im [V ] = [Z Bus ]

(4.25)

¯ ¯ T [∆Z][∆ Z] ¯ Bus ] = [[Z ¯m [Z ] − ] Bus Z¯``

(4.26)

where

¯ Bus ] matrix is an m × m matrix i.e. the size of the [Z ¯ Bus ] matrix It is worth observing that the [Z does not increase when a link is added to the partial network of ‘m’ buses as no new node is created.

4.1.4

Addition of a link between two existing nodes (Case 4):

Let an element with impedance z¯pq be connected between two existing nodes ‘p’ and ‘q’. This is an addition of a link as it forms a loop encompassing nodes ‘p’ and ‘q’ as shown in Fig. 4.13. Let I¯`

Figure 4.13: Addition of a link between two existing nodes ‘p’ and ‘q’

be the current through the link as shown in the Fig. 4.13. This link current changes the injected current at pth node from I¯p to (I¯p − I¯` ) , while the injected current at node q th is modified from I¯q to (I¯q + I¯` ). The modified network equations can be written as: 112

⎪ V¯1 = Z¯11 I¯1 + Z¯12 I¯2 + ⋯ + Z¯1p (I¯p − I¯l ) + ⋯ + Z¯1q (I¯q + I¯` ) + ⋯ + Z¯1m I¯m ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⋮ ⎪ ⎪ ⎪ ⎪ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ⎪ Vp = Zp1 I1 + Zp2 I2 + ⋯ + Zpp (Ip − Il ) + ⋯ + Zpq (Iq + I` ) + ⋯ + Zpm Im ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⋮ ⎪ ⎬ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ Vq = Zq1 I1 + Zq2 I2 + ⋯ + Zqp (Ip − Il ) + ⋯ + Zqq (Iq + I` ) + ⋯ + Zqm Im ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⋮ ⎪ ⎪ ⎪ ⎪ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ⎪ Vm = Zm1 I1 + Zm2 I2 + ⋯ + Zmp (Ip − Il ) + ⋯ + Zmq (Iq + I` ) + ⋯ + Zmm Im ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

(4.27)

Also from Fig. 4.13, the relation between V¯p and V¯q in terms of I¯` and z¯pq can be written as

V¯p − V¯q = z¯pq I¯`

(4.28)

0 = −V¯p + V¯q + z¯pq I¯`

(4.29)

or Substituting V¯p and V¯q from the equation (4.27) into the equation (4.29) in the following relation is obtained:

0 =(Z¯q1 − Z¯p1 )I¯1 + (Z¯q2 − Z¯p2 )I¯2 + ⋯ + (Z¯qp − Z¯pp )I¯p + ⋯(Z¯qq − Zpq )I¯q + ⋯ + (Z¯qm − Z¯pm )I¯m + (Z¯pp + Z¯qq − 2Z¯pq + z¯pq )I¯` (4.30) Equations (4.27) and (4.30) form a set of (m + 1) simultaneous equation which can be written in matrix form as:

⎡ V¯1 ⎤ ⎢ ⎥ ⎢¯ ⎥ ⎢ V2 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⋮ ⎥ ⎢ ⎥ ⎢ V¯ ⎥ ⎢ q⎥ ⎢ ⎥= ⎢ ⋮ ⎥ ⎢ ⎥ ⎢¯ ⎥ ⎢Vm ⎥ ⎢ ⎥ ⎢⋯⎥ ⎢ ⎥ ⎢ ⎥ ⎢0⎥ ⎣ ⎦

⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

Z¯11 Z¯21 ⋮ ¯ Zq1 ⋮ ¯ Zm1 ⋯ ¯ Z`1

Z¯12 ⋯ Z¯1q ⋯ Z¯1m Z¯22 ⋯ Z¯2q ⋯ Z¯2m Z¯q2 ⋯ Z¯qq ⋯ Z¯qm Z¯m2 ⋯ Z¯mq ⋯ Z¯mm ⋯ ⋯ ⋯ ⋯ ⋯ ¯ Z`2 ⋯ Z¯`q ⋯ Z¯`m

Where,

Z¯1` = Z¯`1 = (Z¯1q − Z¯1p ) ; Z¯2` = Z¯`2 = (Z¯2q − Z¯2p ) Z¯q` = Z¯`q = (Z¯qq − Z¯qp ) ; Z¯m` = Z¯`m = (Z¯mq − Z¯mp ) also and 113

⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋯ ⋮

⎤ ⎡ I¯1 ⎤ ⎥⎢ ⎥ ⎥⎢ ¯ ⎥ ⎥ ⎢ I2 ⎥ ⎥⎢ ⎥ ⎥⎢ ⎥ ⎥⎢ ⋮ ⎥ ⎥⎢ ⎥ ¯ Zq` ⎥⎥ ⎢⎢ I¯q ⎥⎥ ⎥⎢ ⎥ ⎥⎢ ⋮ ⎥ ⎥⎢ ⎥ ⎥⎢ ⎥ Z¯m` ⎥⎥ ⎢⎢I¯m ⎥⎥ ⋯ ⎥⎥ ⎢⎢ ⋯ ⎥⎥ ⎥⎢ ⎥ Z¯`` ⎥⎦ ⎢⎣ I¯` ⎥⎦ Z¯1` Z¯2`

(4.31)

Z¯`` = Z¯qq + Z¯pp − 2Z¯pq + z¯pq .

For eliminating the link current I¯` equation (4.31) can be written in the compact form as:

⎡¯m ⎤ (m) ⎢VBus ⎥ ⎢ ⎥ ⎢ ⎥ (m) ¯m Z ⎢ ⎥ ⎥= ⎢ [ ¯ BusT ⎢ ⎥ (1) [∆Z] ⎢ ⎥ ⎢ ⎥ ⎢ 0 ⎥ ⎦ ⎣

⎡¯m ⎤ ⎢IBus ⎥ ⎥ ⎢ ⎥ ⎢ ¯ [∆Z] ⎥ ⎢ ⎥ ⎢ ] ⎥ ⎢ Z¯`` ⎥ ⎢ ⎢ ¯ ⎥ ⎢ I` ⎥ ⎦ ⎣ (1)

(4.32)

where,

¯ = [(Z¯1q − Z¯1p )⋯(Z¯pq − Z¯pp )⋯(Z¯qq − Z¯qp )⋯(Z¯mq − Z¯mp )]T [∆Z] ¯ Bus ] matrix, after the elemination of the link current I¯` , can be Now using equation (4.26), the [Z ¯ Bus ] is still (m × m) in size as no new node has been determined. It is worth noting that the [Z created. ¯ Bus ] as follows: Summarizing the step-by-step procedure for building the [Z Step 1: Draw the graph of the network and select a tree of the graph. Identify the branches and the links of the graph. A tree of a graph with branches and links is shown in Fig. 4.14.

Figure 4.14: Tree of a graph

¯ Bus ] matrix building Step 2: Select a branch connected to the reference node to initiate the [Z process. From Fig. 4.14, it is evident that the first branch selected could be either 1 or 3 or 4 as these are the only branches connected to the reference node. Let the branch 1 be selected as the starting branch and z¯po be the impedance of the branch then (m) Bus

¯ Z

(1)

= (1) [ z¯p0 ]

Step 3: Pick up another element from the graph. It should either be connected to an existing node or the reference node. Never select an element connected to two new nodes as it will be isolated from ¯ Bus ] matrix becoming infinite. the existing partial network and this will result in the elements of [Z ¯ Bus ] matrix building process, if For instance, with reference to Fig. 4.14, in the next step of [Z 114

element 5 is next added to the partial network as shown in Fig. 4.15, the resultant network is disjointed. This is an incorrect choice. The proper choice could be any one of the elements 2 or 3 or 4 or 6.

Figure 4.15: Selecting a wrong element in the step-by-step process

¯m If the bus impedance matrix of a partial network with m-nodes, [Z Bus ], is known, then depending on whether the added (m + 1)th element is a branch or a link, the following steps are to be followed ¯ Bus ] matrix: to obtain the new [Z (a) If the added element is a branch between a new node ‘q’ and the reference node with an impedance ¯ Bus ] matrix will increase by one and the new matrix is given as : z¯qo , then the size of the new [Z

(1)

¯ Bus = Z

⋮ (m) (q)

⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

(1)

(⋯)

(m)

(q)

Z¯11 ⋯ Z¯1m 0 ⋮ ⋮ ⋮ ⋮ Z¯m1 ⋯ Z¯mm 0 0 0 0 z¯q0

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

(b) If the added element is a branch between an existing node ‘p’ and a new node ‘q’ with an impedance z¯pq , then a new row and column corresponding to the new node ‘q’ is added to the ¯m existing [Z Bus ] matrix. The new matrix is calculated as follows:

(1) ⋮ (p)

¯ Bus = Z

⋮ (m)

(q)

⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

(1)

⋯

(p)

⋯

(m)

Z¯11 ⋯ Z¯1p ⋯ Z¯1m ⋮ ⋮ ⋮ ¯ ¯ ¯ Zp1 ⋯ Zpp ⋯ Zpm ⋮ ⋮ ⋮ ¯ ¯ ¯ Zm1 ⋯ Zmp ⋯ Zmm Z¯p1 ⋯ Z¯pp ⋯ Z¯pm 115

(q)

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ Z¯pp + z¯pq ⎥⎦ Z¯1p ⋮ ¯ Zpp ⋮ ¯ Zmp

(c) If the added element is a link between an existing node ‘q’ and the reference node with an impedance z¯qo , then no new node is added to the network. A two-step procedure has to be followed to find the new bus impedance matrix.

¯m In the first step, a column and a row will be temporarily added to existing [Z Bus ] matrix as:

(1) ⋮

¯ (temp) Z = Bus

(q) ⋮ (m) (`)

⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

(1)

⋯

Z¯11 ⋮ ¯ Zq1 ⋮ ¯ Zm1 −Z¯q1

⋯

(q)

Z¯1q ⋮ ¯ ⋯ Zqq ⋮ ¯ ⋯ Zmq ⋯ −Z¯qq

⋯

(m)

(`)

⋯

−Z¯1q ⋮ ¯ −Zqq ⋮ ¯ −Zmq Z¯``

Z¯1m ⋮ ¯ ⋯ Zqm ⋮ ¯ ⋯ Zmm ⋯ −Z¯qm

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

where,

Z¯`` = Z¯qq + z¯q0 The additional row and column have to be deleted so that [ZBus ] matrix is (m × m) in size. The elimination process is carried out using

¯ Bus ] = [[Z ¯m [Z Bus ] −

¯ ¯ T [∆Z][∆ Z] ]] Z¯``

where,

¯ = [−Z¯1q − Z¯2q ⋯ − Z¯qq ⋯ − Z¯mq ]T [∆Z] (d) If the added element is a link between two existing nodes ‘p’ and ‘q’ with an impedance z¯pq , then again the two step procedure as outlined in (step c) is to be followed. The temporary ¯ (temp) impedance matrix Z is calculated as: Bus (1)

(1)

(p)

(q)

(m) (`)

⎡ Z¯11 ⎢ ⎢ ⎢ ⋮ ⎢ ⎢ ⎢ Z¯p1 ⎢ ⎢ ⋮ ⎢ ⎢ ¯ ⎢ Zq1 ⎢ ⎢ ⎢ ⎢ ⎢ Z¯m1 ⎢ ⎢ ⎢ (Z¯q1 − Z¯p1 ) ⎣

(p)

(q)

(m)

⋯

Z¯1p

Z¯1q

⋯

Z¯1m

⋯

Z¯pp

Z¯pq

⋯

Z¯pm

⋯

Z¯qp

Z¯qq

⋯

Z¯qm

⋯ Z¯mp Z¯mq ⋯ Z¯mm ⋯ (Z¯qp − Z¯pp ) (Z¯qq − Z¯pq ) ⋯ (Z¯qm − Z¯pm )

where,

Z¯`` = Z¯pp + Z¯qq − 2Z¯pq + z¯pq 116

(`)

(Z¯1q − Z¯1p ) ⎤⎥ ⎥ ⎥ ⎥ ⎥ (Z¯pq − Z¯pp ) ⎥ ⎥ ⎥ ⎥ ⎥ ¯ ¯ (Zqq − Zqp ) ⎥⎥ ⎥ ⎥ ⎥ ¯ ¯ (Zmq − Zmp ) ⎥⎥ ⎥ ⎥ Z¯`` ⎦

Next eliminate the added row and column ‘`’ using the expression:

¯ Bus ] = [[Z ¯m [Z Bus ] −

¯ ¯ T [∆Z][∆ Z] ] Z¯``

where,

¯ = [(Z¯1q − Z¯1p )⋯(Z¯pq − Z¯pp )⋯(Z¯qq − Z¯qp )⋯(Z¯mq − Z¯mp )] [∆Z]

T

Step 4: Repeat Step 3 till all the elements are considered. In the next lecture, we will be looking at an example of [ZBus ] matrix building algorithm.

117

4.2

Example of [ZBus] matrix building algorithm

The single line diagram of a power system is shown in the Fig. 4.16. The line impedances in pu are ¯ Bus ] matrix formulation is explained as given below: also given. The step-by-step procedure for [Z

Figure 4.16: Single Line Diagram of the Power System for the example Preliminary Step: The graph of the network and a tree is shown in Fig. 4.17. Elements 1,2,4 and 5 are the tree branches while 3, 6 and 7 are the links.

Figure 4.17: Graph and a tree of the network of Fig. 4.17

¯ Bus ] matrix building algorithm starts with element 1, which is Step 1: The step-by-step [Z a tree branch connected between nodes 1 and the reference node 0 and has an impedance of z¯10 = j0.10 pu. This is shown in the accompanying figure ,Fig. 4.18. 118

Figure 4.18: Partial network of Step 1

¯ Bus ] matrix is The resulting [Z (1)

(1)

¯ Bus = (1) [ z10 ] = (1) [ j0.10 ] Z Step 2: Next, the element 2 connected between node 2 (q = 2) and the reference node ‘0’ is selected. This element has an impedance of z¯20 = j0.10 p.u. As this is the addition of a tree branch ¯ Bus ] matrix. This addition is illustrated in Fig. 4.19. it will add a new node ‘2’ to the existing [Z

Figure 4.19: Partial network of Step 2

The new bus impedance matrix is given by : (1)

(2)

(1)

(2)

0 ¯ Bus = (1) [ j0.10 0 ] = (1) [ j0.1 Z ] (2) 0 z20 (2) 0 j0.10 Step 3: Element 3 connected between existing nodes, node 1 (p = 1) and node 2 (q = 2), having an impedance of z¯12 = j0.20 p.u. is added to the partial network, as shown in Fig. 4.20. Since this is an addition of a link to the network a two step procedure is to be followed. In the 119

Figure 4.20: Partial network of Step 3

first step a new row and column is added to the matrix as given below : (1)

(1)

¯ (temp) Z = (2) Bus (`)

(`)

(2)

⎡ j0.10 0.0 (Z¯12 − Z¯11 ) ⎢ ⎢ ⎢ 0.0 j0.10 (Z¯22 − Z¯21 ) ⎢ ⎢ ¯ ⎢ (Z21 − Z¯11 ) (Z¯22 − Z¯12 ) Z¯`` ⎣

⎤ (1) ⎥ ⎥ ⎥ = (2) ⎥ ⎥ ⎥ (`) ⎦

(1)

(2)

(`)

⎡ j0.10 0.0 −j0.10 ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ 0.0 j0.10 j0.10 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ −j0.10 j0.10 j0.40 ⎥ ⎣ ⎦

where,

Z¯`` = Z¯11 + Z¯22 − 2Z¯12 + z¯20 = j0.10 + j0.10 − 0.0 + j0.20 = j0.40 p.u. ¯ Bus ] matrix as given below: Next this new row and column is eliminated to restore the size of [Z

¯ Bus ] = [j0.10 0.0 ] − [Z 0.0 j0.10

−j0.10 [ ] [−j0.10 j0.10] j0.10 j0.40

Hence, the impedance matrix after the addition of element 3 is found out to be :

(1)

¯ Bus ] = [Z

(1) (2)

[

(2)

j0.075 j0.025 ] j0.025 j0.075

Step 4: The element 4 , which is added next, is connected between an existing node, node 2 (p = 2) and a new node, node 3 (q = 3). The impedance of this element is z¯23 = j0.30 p.u. and it is a tree branch hence, a new node, node 3 is added to the partial network. This addition, shown ¯ Bus ] to (3 × 3). in Fig. 4.21, thus increases the size of [Z 120

Figure 4.21: Partial network of Step 4

The new impedance matrix can be calculated as: (1)

(1)

¯ Bus = (2) Z (3)

(2)

(3)

⎡ j0.075 j0.025 Z¯12 ⎢ ⎢ ⎢ j0.025 j0.0.075 Z¯22 ⎢ ⎢ ¯ ⎢ Z21 Z¯22 Z¯22 + z¯23 ⎣

⎤ (1) ⎥ ⎥ ⎥ = (2) ⎥ ⎥ ⎥ (3) ⎦

(1)

(2)

(3)

⎡ j0.075 j0.025 j0.025 ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ j0.025 j0.0.075 j0.075 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ j0.025 j0.075 j0.375 ⎥ ⎣ ⎦

Step 5: Element 5 is added next to the existing partial network. This is a tree branch connected between an existing node, node 3 (p = 3) and a new node, node 4 (q = 4). This is illustrated in Fig. 4.22.

¯ Bus ] increases to (4 × 4). The Since a new node is added to the partial network, the size of [Z impedance of the new element is z¯34 = j0.15 p.u. The new bus impedance matrix is : (1)

(1)

¯ Bus = (2) Z (3)

(4)

(2)

(3)

(4)

⎡ ⎢ j0.075 j0.025 j0.025 Z¯31 ⎢ ⎢ j0.025 j0.075 j0.075 Z¯32 ⎢ ⎢ ⎢ j0.025 j0.075 j0.375 Z¯33 ⎢ ⎢ ¯ ⎢ Z13 Z¯23 Z¯33 Z¯33 + z¯34 ⎣

⎤ ⎥ (1) ⎥ ⎥ (2) ⎥ ⎥= ⎥ (3) ⎥ ⎥ ⎥ (4) ⎦

⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

(1)

(2)

(3)

(4)

j0.075 j0.025 j0.025 j0.025

j0.025 j0.075 j0.075 j0.075

j0.025 j0.075 j0.375 j0.375

j0.025 j0.075 j0.375 j0.525

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

Step 6: Next,the element 6 connected between two existing nodes node 1 (p = 1) and node 4 (q = 4) is added to the network, as shown in the Fig. 4.23. The impedance of this element is z¯23 = j0.25 p.u. As this is a link addition, the two step procedure is used. The bus impedance 121

Figure 4.22: Partial network of Step 5

Figure 4.23: Partial network of Step 6 matrix is modified by adding a new row and column as given below: (1)

(1) (temp) Bus

¯ Z

(2)

= (3) (4) (`)

(2)

(3)

(4)

⎡ j0.075 j0.025 j0.025 j0.025 ⎢ ⎢ ⎢ j0.025 j0.075 j0.075 j0.075 ⎢ ⎢ ⎢ j0.025 j0.075 j0.375 j0.375 ⎢ ⎢ j0.025 j0.075 j0.375 j0.525 ⎢ ⎢ ¯ ⎢ (Z41 − Z¯11 ) (Z¯42 − Z¯12 ) (Z¯43 − Z¯13 ) (Z¯44 − Z¯14 ) ⎣ 122

(`)

(Z¯14 − Z¯11 ) (Z¯24 − Z¯21 ) (Z¯34 − Z¯31 ) (Z¯44 − Z¯41 ) Z¯``

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

¯ Bus ] matrix elements in the last row and column the Substituting the values of appropriate [Z intermediate impedance matrix is:

(1) (temp) Bus

¯ Z

(2)

= (3) (4) (`)

⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

(1)

(2)

(3)

(4)

j0.075 j0.025 j0.25 j0.025 −j0.05

j0.025 j0.075 j0.075 j0.075 0.05

j0.025 j0.075 j0.375 j0.375 j0.35

(`)

j0.025 −j0.05 ⎤⎥ ⎥ j0.075 j0.05 ⎥⎥ ⎥ j0.375 j0.35 ⎥⎥ j0.525 j0.50 ⎥⎥ ⎥ j0.50 j0.80 ⎥⎦

where,

Z¯`` = Z¯44 + Z¯11 − 2Z¯14 + z¯14 = j0.075 + j0.525 − 2 × j0.025 + j0.25 = j0.80 p.u. The additional row and column ‘`’ are to be eliminated to restore the impedance matrix size to ¯ Bus ] matrix after the addition of element 6 is calculated as: (m × m), and the [Z

⎡ ⎢j0.075 ⎢ ⎢ ¯ Bus ] = ⎢⎢j0.025 [Z ⎢ j0.25 ⎢ ⎢ ⎢j0.025 ⎣

j0.025 j0.075 j0.075 j0.075

j0.025 j0.075 j0.375 j0.375

⎡ ⎤ ⎢−j0.05⎥ ⎢ ⎥ ⎢ j0.50 ⎥ ⎢ ⎥ ⎤ ⎢⎢ j0.35 ⎥⎥ [−j0.05 j0.05 j0.35 j0.50] j0.025⎥ ⎢ ⎥ ⎥ ⎢ ⎥ ⎥ ⎢ ⎥ j0.50 j0.075⎥ ⎣ ⎦ ⎥− ⎥ j0.80 j0.375⎥ ⎥ j0.525⎥⎦

Hence, (1)

¯ Bus = (2) Z (3) (4)

⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

(1)

(2)

(3)

(4)

j0.0719 j0.0281 j0.0469 j0.0563

j0.0281 j0.0719 j0.0531 j0.0437

j0.0469 j0.0531 j0.2219 j0.1562

j0.0563 j0.0437 j0.1562 j0.2125

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

Step 7: Finally the element 7 connected between two existing nodes node 2 (p = 2) and node 4 (q = 4) is added to the partial network of step 6. The impedance of this element is is z¯23 = j0.40 pu. This is also a link addition, as shown in Fig. 4.24 and hence the two step precedure will be ¯ Bus ] matrix. In the first step the Z ¯ (temp) is calculated after a row and a followed to obtain the [Z Bus

123

Figure 4.24: Partial network of Step 7

¯ Bus as follows: column are added to the exiting Z (1)

(1) (2)

¯ (temp) Z = (3) Bus (4) (`)

(2)

(3)

(4)

⎡ j0.0719 j0.0281 j0.0469 j0.0563 ⎢ ⎢ ⎢ j0.0281 j0.0719 j0.0531 j0.0437 ⎢ ⎢ ⎢ j0.0469 j0.0531 j0.2219 j0.1562 ⎢ ⎢ j0.0563 j0.0437 j0.1562 j0.2125 ⎢ ⎢ ¯ ⎢ (Z41 − Z¯21 ) (Z¯42 − Z¯22 ) (Z¯43 − Z¯23 ) (Z¯44 − Z¯24 ) ⎣

(`)

(Z¯14 − Z¯12 ) (Z¯24 − Z¯22 ) (Z¯34 − Z¯32 ) (Z¯44 − Z¯42 ) Z¯``

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

Substituing the values of the elements of impedance matrix one gets:

(1) (temp) Bus

¯ Z

(2)

= (3) (4) (`)

⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

(1)

(2)

(3)

j0.0719 j0.02810 j0.0469 j0.0563 j0.281

j0.0281 j0.0719 j0.0531 j0.0437 −j0.281

j0.0469 j0.0531 j0.2219 j0.1562 j0.1031

(4)

(`)

j0.0563 j0.281 ⎤⎥ ⎥ j0.0437 −j0.281 ⎥⎥ ⎥ j0.1562 j1031 ⎥⎥ j0.2125 j0.1688 ⎥⎥ ⎥ j0.1688 j0.5969 ⎥⎦

where,

Z¯`` = Z¯22 + Z¯44 − 2Z¯24 + z¯24 = j0.0719 + j0.2125 − 2 × j0.0563 + j0.40 = j0.5969 p.u. The additional row and column ‘`’ are to be eliminated to restore the impedance matrix size to 124

¯ Bus ] after the addition of element 7 is calculated as: (m × m), and [Z

⎡ ⎢j0.0719 ⎢ ⎢ ¯ Bus ] = ⎢⎢j0.0281 [Z ⎢ j0.469 ⎢ ⎢ ⎢j0.0563 ⎣

j0.0281 j0.0719 j0.0531 j0.0437

j0.0469 j0.0531 j0.2219 j0.1562

⎤ ⎡ ⎢ j0.0281 ⎥ ⎥ ⎢ ⎢−j0.0281⎥ ⎥ ⎢ ⎤ ⎢⎢ j0.1031 ⎥⎥ [j0.0281 −j0.0281 j0.1031 j0.1688] j0.0563⎥ ⎢ ⎥ ⎥ ⎢ ⎥ ⎥ ⎥ ⎢ j0.1688 j0.0437⎥ ⎣ ⎦ ⎥− ⎥ j0.5969 j0.1562⎥ ⎥ j0.2125⎥⎦

Hence, (1)

¯ Bus = (2) Z (3) (4)

⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

(1)

(2)

(3)

(4)

j0.0705 j0.0295 j0.0420 j0.0483

j0.0295 j0.0705 j0.0580 j0.0517

j0.0420 j0.0580 j0.2041 j0.1271

j0.0483 j0.0517 j0.1271 j0.1648

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

¯ Bus ] matrix is a (4 × 4) matrix, as the network has 4 nodes and As can be seen that the final [Z a reference node. As there are 7 elements is the network, 7 steps are required for the formation of ¯ Bus ] matrix. [Z

4.2.1

¯ Bus ] : Modifications in the existing [Z

¯ Bus ] matrix is known, some modification such as line If in an existing network, for which the [Z ¯ Bus ] matrix can be easily modified removal or line impedance alteration is carried out then the [Z without any need of reconstructing the matrix from scratch. ¯ Bus matrix be the final bus impedance matrix given for the network of As an example, let the Z Fig. 4.16. Next, let the element 7 connecting nodes 2 and 4 be removed from the network and it ¯ Bus . is required to find the modified Z Removal of element 7 is equivalent to setting its impedance z¯24 to infinite. This can be obtained org add by connecting a fictitious element z¯24 in parallel to the existing element z¯24 such that the resultant result impedance z¯24 is infinite i.e.

1 result 24

z¯

=

1 1 1 + add = =0 z¯ z¯24 ∞ org 24

or org add z¯24 = −¯ z24 = −j0.40 p.u. org add Hence, by adding an element z¯24 = −j0.4 p.u. in parallel to z¯24 the removal of line between nodes 2 and 4 can be simulated. The new added fictitious element is a link addition between the two nodes, p = 2 and q = 4 and is shown in Fig. 4.25 . Hence, this will require a two-step procedure. The addition of the fictitious element 8 , which is a link, will introduce a temporary row and column.

125

Figure 4.25: Adding a link to simulate the removal of element 7

¯ (temp) The Z is given as: Bus (1)

(1) (temp) Bus

¯ Z

(2)

= (3) (4) (`)

(3)

(2)

(`)

(4)

⎡ j0.0705 j0.0295 j0.0420 j0.0483 ⎢ ⎢ ⎢ j0.0295 j0.0705 j0.0580 j0.0517 ⎢ ⎢ ⎢ j0.0420 j0.0580 j0.2041 j0.1271 ⎢ ⎢ j0.0483 j0.0517 j0.1271 j0.1648 ⎢ ⎢ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ⎢ (Z41 − Z21 ) (Z42 − Z22 ) (Z43 − Z23 ) (Z44 − Z¯24 ) ⎣

(Z¯14 − Z¯12 ) (Z¯24 − Z¯22 ) (Z¯34 − Z¯32 ) (Z¯44 − Z¯42 ) Z¯``

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

Substituting the appropriate values one gets:

(1) (2)

¯ (temp) Z = (3) Bus (4) (`)

⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

(1)

(2)

(3)

j0.0705 j0.0295 j0.0295 j0.0705 j0.0420 j0.0580 j0.0483 j0.0517 j0.0188 −j0.0188

j0.0420 j0.0580 j0.2041 j0.1271 j0.0691

(4)

(`)

j0.0483 j0.0188 ⎤⎥ ⎥ j0.0517 −j0.0188 ⎥⎥ ⎥ j0.1271 j0.0691 ⎥⎥ j0.1648 j0.1131 ⎥⎥ ⎥ j0.1131 −j0.2681 ⎥⎦

where, add Z¯`` = Z¯22 + Z¯44 − 2Z¯24 + z¯24 = j0.0705 + j0.1648 − 2 × j0.0483 + (−j0.40) = −j0.2681 p.u.

126

The additional row and column is eliminated in the following step:

⎡ ⎢j0.0705 ⎢ ⎢ ¯ Bus ] = ⎢⎢j0.0295 [Z ⎢j0.0420 ⎢ ⎢ ⎢j0.0483 ⎣

j0.0295 j0.0705 j0.0580 j0.0517

j0.0420 j0.0580 j0.2041 j0.1271

⎤ ⎡ ⎢ j0.0188 ⎥ ⎥ ⎢ ⎢−j0.0188⎥ ⎥ ⎢ ⎤ ⎢⎢ j0.0691 ⎥⎥ [j0.0188 −j0.0188 j0.0691 j0.1131] j0.0483⎥ ⎢ ⎥ ⎥ ⎢ ⎥ ⎥ ⎥ ⎢ j0.1131 j0.0517⎥ ⎣ ⎦ ⎥− −j0.2681 j0.1271⎥⎥ ⎥ j0.1648⎥⎦

Thus, the final impedance matrix after the removal of element 7 is :

(1)

¯ Bus = (2) Z (3) (4)

⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

(1)

(2)

(3)

(4)

j0.0719 j0.0281 j0.0469 j0.0563

j0.0281 j0.0719 j0.0531 j0.0437

j0.0469 j0.0531 j0.2219 j0.1562

j0.0563 j0.0437 j0.1562 j0.2125

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

¯ Bus matrix is identical to the Z ¯ Bus matrix obtained in step 6 of the previous The obtained Z example, which is the impdance matrix of the network before the addition of element 7. ¯ Bus matrix building algorithm without any presence of mutually So far we have considered the Z coupled elements. In the next lecture, we will take into account the presence of mutually coupled ¯ Bus matrix. elements while forming the Z

127

4.3

¯ BUS formation considering mutual coupling between Z elements

¯m Assume that the bus impedance matrix [Z BUS ] is known for a partial network of ‘m’ nodes and a reference node. The bus voltage and bus current relation for the partial network, shown in Fig. 4.26, can be expressed as:

Figure 4.26: Partial Network with m-buses

¯ BUS ] = [Z ¯m ¯ [V BUS ] [IBUS ]

(4.33)

In equation (4.33),

¯ BUS is (m × 1) bus voltage vector V ¯IBUS is (m × 1) bus current vector ¯m Z BUS is (m × m) bus impedance matrix The new added element ‘p-q’ may be a branch or may be a link as discussed in the previous algorithm. 128

4.3.1

Addition of a branch to this partial network:

The performance equation of the network with an added branch ‘p-q’ is

⎡¯ ⎤ ⎡¯ ⎢ V1 ⎥ ⎢ Z11 ⎢ ⎥ ⎢ ⎢ V¯2 ⎥ ⎢ Z¯21 ⎢ ⎥ ⎢ ⎢ ⎥ ⎢ ⎢ ⋮ ⎥ ⎢ ⋮ ⎢ ⎥ ⎢ ⎢¯ ⎥ ⎢¯ ⎢ Vp ⎥ = ⎢ Zp1 ⎢ ⎥ ⎢ ⎢ ⋮ ⎥ ⎢ ⋮ ⎢ ⎥ ⎢ ⎢¯ ⎥ ⎢ ¯ ⎢Vm ⎥ ⎢Zm1 ⎢ ⎥ ⎢ ⎢¯ ⎥ ⎢¯ ⎢ Vq ⎥ ⎢ Zq1 ⎣ ⎦ ⎣

Z¯12 ⋯ Z¯1p ⋯ Z¯1m Z¯22 ⋯ Z¯2p ⋯ Z¯2m Z¯p2 ⋯ Z¯pp ⋯ Z¯pm Z¯m2 ⋯ Z¯mp ⋯ Z¯mm Z¯q2 ⋯ Z¯qp ⋯ Z¯qm

⎤⎡ ⎤ Z¯1q ⎥ ⎢ I¯1 ⎥ ⎥⎢ ⎥ Z¯2q ⎥⎥ ⎢⎢ I¯2 ⎥⎥ ⎥⎢ ⎥ ⎥⎢ ⋮ ⎥ ⎥⎢ ⎥ ⎥⎢ ⎥ ¯ Zpq ⎥⎥ ⎢⎢ I¯p ⎥⎥ ⎥⎢ ⋮ ⎥ ⎥⎢ ⎥ ⎥⎢ ⎥ ¯ Zmq ⎥⎥ ⎢⎢I¯m ⎥⎥ ⎥⎢ ⎥ Z¯qq ⎥⎦ ⎢⎣ I¯q ⎥⎦

(4.34)

¯qi = Z¯iq for i = The network is assumed to contain bilateral passive elements and hence, Z 1, 2, ⋯m, i ≠ q . The added branch ‘p-q’ is assumed to be mutually coupled with one or more elements of the partial network. ¯qi , inject a current at ith node and calculate the voltage at q th node with To determine element Z respect to reference, as shown in Fig. 4.27.

¯ qi for the addition of branch Figure 4.27: Calculation of Z

¯qi Calculation of Z As all other bus currents are zero, bus voltages can be written as, 129

V¯1 = Z¯1i I¯i ⎫ ⎪ ⎪ ⎪ ⎪ ¯ ¯ ¯ ⎪ V2 = Z2i Ii ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⋮ ⎪ ⎪ ⎪ ¯ ¯ ¯ Vp = Zpi Ii ⎬ ⎪ ⎪ ⎪ ⎪ ⋮ ⎪ ⎪ ⎪ ¯ ¯ ⎪ Vm = Zmi Ii ⎪ ⎪ ⎪ ⎪ ⎪ ¯ ¯ ¯ ⎪ Vq = Zqi Ii ⎪ ⎭

(4.35)

Also from Fig. 4.27, V¯p and V¯q can be related as,

V¯q = V¯p − v¯pq

(4.36)

Where v¯pq is the voltage across the added element ‘p − q ’. Also, the currents in the elements of the network can be related to the voltages across the elements as,

¯ipq y¯pq,pq y ¯pq,ρσ v¯pq ][ ] [¯ ] = [ ¯ρσ y ¯ρσ,pq y ¯ρσ,ρσ v iρσ

(4.37)

Where,

¯ipq = the current through the added element ‘p − q ’. ¯iρσ = (m × 1) current vector of the elements of the partial network. v ¯ρσ = (m × 1) voltage vector of the elements of the partial network. y¯pq,pq = Self-admittance of the added element. y ¯pq,ρσ = (m × 1) vector of mutual admittances between the added element ‘p − q ’ and the elements ‘ρ − σ ’ of the partial network. y ¯ρσ,ρσ = (m × m) primitive admittance matrix of the partial network. y ¯ρσ,pq = [¯ ypq,ρσ ]

T

The diagonal elements of primitive impedance matrix [¯ z] are the self impedance of the individual elements while the off-diagonal elements are mutual impedances between the elements. The inverse of primitive impedance matrix is the primitive admittance matrix [¯ y]. This can be explained with the help of an illustrative example. A single line diagram of a power system is shown in Fig. 4.28. The self-impedances of lines are written by the side of the line and are in p.u.. The two lines between nodes 1 and 3 are mutually coupled with a mutual impedance of j0.10. The primitive impedance matrix for the network can be written as,

⎡j0.60 0 0 0 0 ⎤⎥ ⎢ ⎢ ⎥ ⎢ 0 j0.50 0 0 0 ⎥⎥ ⎢ ⎢ ⎥ [¯ z] = ⎢⎢ 0 0 j0.50 0 0 ⎥⎥ ⎢ 0 0 0 j0.25 j1.0 ⎥⎥ ⎢ ⎢ ⎥ ⎢ 0 ⎥ 0 0 j1.0 j0.20 ⎣ ⎦ 130

Figure 4.28: Sample Power System

The inverse of [¯ z] is [¯ y], the primitive admittance matrix.

⎡−j1.67 0 0 0 0 ⎤⎥ ⎢ ⎢ ⎥ ⎢ 0 −j2.0 0 0 0 ⎥⎥ ⎢ ⎢ ⎥ [¯ y] = ⎢⎢ 0 0 −j2.0 0 0 ⎥⎥ ⎢ 0 0 0 −j5.0 j2.5 ⎥⎥ ⎢ ⎢ ⎥ ⎢ 0 ⎥ 0 0 j2.5 −j6.25 ⎣ ⎦ The current ¯ipq in the added branch ‘p − q ’ equal to zero as node ‘q ’ is open.

¯ipq = 0

(4.38)

The voltage v¯pq , however, is not zero as the added branch is mutually coupled to one or more elements of the partial network. Thus, the voltage across other elements of the network can be expressed as,

¯ρ − V ¯σ v ¯ρσ = V

(4.39)

¯ ρ and V ¯ σ are the voltages of the nodes of the partial network. With ¯ipq = 0 from equation where V (4.37) one can write, y¯pq,pq v¯pq + y ¯pq,ρσ v ¯ρσ = 0

Hence,

v¯pq = −

y ¯pq,ρσ v ¯ρσ y¯pq,pq

131

(4.40)

Substituting v ¯ρσ from equation (4.39) and v¯pq equation (4.36) in equation (4.40) one gets,

¯ρ − V ¯ σ) y ¯pq,ρσ (V V¯q = V¯p + y¯pq,pq Substitution of I¯i = 1 pu in equation (4.35) results in V¯p , V¯q , V¯ρ and V¯σ being replaced by their corresponding impedances and hence,

¯ ρi − Z ¯ σi ) y ¯pq,ρσ (Z Z¯qi = Z¯pi + y¯pq,pq

(4.41)

∀i = 1, 2, ⋯, m, i ≠ q ¯qq , a current I¯q = 1 p.u. is injected into q th node with all For calculating the self impedance Z other currents equal to zero as shown in Fig. 4.29. Then the voltages of the nodes are calculated from equation (4.35) , as V¯1 = Z¯1q I¯q ⎫ ⎪ ⎪ ⎪ ⎪ ¯ ¯ ¯ ⎪ V2 = Z2q Iq ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⋮ ⎪ ⎪ ⎪ ¯ ¯ ¯ Vp = Zpq Iq ⎬ ⎪ ⎪ ⎪ ⎪ ⋮ ⎪ ⎪ ⎪ ¯ ¯ ⎪ Vm = Zmq Iq ⎪ ⎪ ⎪ ⎪ ⎪ ¯ ¯ ¯ ⎪ Vq = Zqq Iq ⎪ ⎭

(4.42)

¯qq Can be calculated directly by calculating V¯q . With I¯q = 1 p.u., Z also,

V¯q = V¯p − v¯pq

(4.43)

¯ipq = −I¯q = −1

(4.44)

¯ipq = −1 = y¯pq,pq v¯pq + y ¯pq,ρσ v ¯ρσ

(4.45)

and

Hence, from equation (4.37) one gets

And thus v¯pq can be written as,

v¯pq = −

1+y ¯pq,ρσ v ¯ρσ y¯pq,pq

(4.46)

Substituting v¯pq and v ¯ρσ , the above equation can be rewritten as,

¯ρ − V ¯ σ) 1+y ¯pq,ρσ (V V¯q = V¯p + ypq,pq 132

(4.47)

Figure 4.29: Calculation of Zqq for the addition of a branch

¯ ρ and V ¯ σ can be replaced by respective transfer With I¯q = 1 p.u., from equation (4.35) V¯p , V¯q , V impedances, ¯ ρq − Z ¯ σq ) 1+y ¯pq,ρσ (Z Z¯qq = Z¯pq + ypq,pq

4.3.2

(4.48)

Addition of a link to this partial network:

If the added element p − q is a link, then a fictitious node ` is created by connecting an ideal voltage in series with the added element, as shown in Fig. 4.30. The value of the source voltage is selected such that the current I¯` through the added link is zero. If e¯` is the voltage of node ` with respect to node q and I¯` is the current injected into node ` from node q . The performance equation of the partial network with the added link p − ` and ideal series voltage source e` is,

⎡ V¯ ⎤ ⎡ Z¯ ⎢ 1 ⎥ ⎢ 11 ⎢ ⎥ ⎢ ⎢ ⋮ ⎥ ⎢ ⋮ ⎢ ⎥ ⎢ ⎢¯ ⎥ ⎢¯ ⎢ Vp ⎥ ⎢ Zp1 ⎢ ⎥=⎢ ⎢ ⋮ ⎥ ⎢ ⋮ ⎢ ⎥ ⎢ ⎢ ⎥ ⎢ ⎢V¯m ⎥ ⎢Z¯m1 ⎢ ⎥ ⎢ ⎢ ⎥ ⎢¯ ⎢ e¯` ⎥ ⎢ Z`1 ⎣ ⎦ ⎣

. . . Z¯1p . . . Z¯1m . . . Z¯pp . . . Z¯pm . . . Z¯mp . . . Z¯mm . . . Z¯`p . . . Z¯`m 133

Z¯1` ⎤⎥ ⎡⎢ I¯1 ⎤⎥ ⎥⎢ ⎥ ⎥⎢ ⋮ ⎥ ⎥⎢ ⎥ ⎥⎢ ⎥ Z¯p` ⎥⎥ ⎢⎢ I¯p ⎥⎥ ⎥⎢ ⋮ ⎥ ⎥⎢ ⎥ ⎥⎢ ⎥ ¯ Zm` ⎥⎥ ⎢⎢I¯m ⎥⎥ ⎥⎢ ⎥ Z¯`` ⎥⎦ ⎢⎣ I¯` ⎥⎦

(4.49)

¯ `i for the addition of a link Figure 4.30: Calculation of Z Here, Z¯li represents the transfer impedance relating the current, Ii , injected into the ith bus and the voltage of the added source e¯` , connected between nodes ` and q . They are conceptually ¯ij of Z ¯ BUS matrix which relate the current injected into the j th bus different from the elements Z and the voltage of the ith bus with repect to the reference. As

e¯` = V¯` − V¯q , ¯`i , 1 = 1, 2, ⋯m, i ≠ `, of the added row and column, can be calculated by The elements Z injecting a current I¯i into the ith node and determining the voltage of `th node with repect to q th node. Hence,

e¯` Z¯`i = ∣ , I¯k = 0, k = 1, 2, ⋯m, k ≠ ` I¯i Also,

e¯` = V¯p − V¯q − v¯p`

(4.50)

The current ¯ip` through the link can be written as,

¯ip` = y¯p`,p` v¯p` + y ¯p`,pσ v ¯ρσ

(4.51)

Since the current through the link is zero, ¯ip` = ¯ipq = 0 Hence,

v¯p` = −

y ¯p`,ρσ v ¯ρσ y¯p`,p`

134

(4.52)

Since the voltage source is ideal source, one can write,

y ¯p`,ρσ = y ¯pq,ρσ and

y¯p`,p` = y¯pq,pq So,

v¯p` = −

¯ρ − V ¯ σ) y ¯pq,ρσ (V y ¯pq,ρσ v ¯ρσ =− y¯pq,pq y¯pq,pq

(4.53)

(4.54)

With Ii = 1 p.u., substituting V¯p , V¯q , V¯p and V¯σ from equation (4.35) and v¯p` from equation (4.52) in equation (4.50) one gets

¯ pi − Z ¯ σi ) y ¯pq,ρσ (Z Z¯`i = Z¯pi − Z¯qi + y¯pq,pq

(4.55)

∀i = 1, 2, ⋯ m, i ≠ ` ¯`` , a current is injected at the ‘`th ’ node with respect to node ‘q ’, as shown in Fig. To calculate Z 4.31. As all other node currents are zero, the node voltages can be written as,

¯`` for the addition of a link Figure 4.31: Calculation of Z

V¯k = Z¯k` I¯` , ∀k = 1, 2, ⋯m 135

e¯` = Z¯`` I¯`

(4.56)

¯`` can be directly computed by calculating e¯` . The current in the element p − ` is With I¯` = 1 p.u., Z ¯ip` = −I¯` = 1 p.u.

(4.57)

From equation (4.37) one gets

¯ip` = y¯p`,p` v¯p` + y ¯p`,pσ v ¯pσ = −1 Further as y¯p`,p` = y¯pq,pq and y ¯p`,pσ = y ¯pq,pσ , hence v¯p` can be expressed as

v¯p` = −

1+y ¯pq,pσ v ¯pσ y¯pq,pq

(4.58)

Substituting v¯p` from equation (4.50) , one can write

1 + y¯pq,pσ (V¯p − V¯σ ) e¯` = V¯p − V¯q + y¯pq,pq ¯ ρ, V ¯ σ and e¯` from equation (4.56), Z¯`` is obtained as With I¯` = 1 p.u., substituting V¯p , V¯q , V ¯ p` − Z ¯ σ` ) 1+y ¯pq,ρσ (Z Z¯`` = Z¯pi − Z¯qi + y¯pq,pq

(4.59)

∀i = 1, 2, ⋯m, i ≠ ` In the case of link addition the additional row and column corresponding to fictitious node ` are to be eliminated. For this the fictitious series voltage source e¯` is short circuited. From equation (4.49) the bus voltages can be written in compact from as m ¯ ¯ ¯m ¯ Bus ¯m ] = [Z [V Bus ][IBus ] + [∆Z][I` ]

(4.60)

Where,

¯ = [Z¯1` Z¯2` ⋯ Z¯m` ]T is an (m × 1) vector comprising of the entries of the column added [∆Z] ¯m to the Z Bus matrix ¯m [¯Im Bus ], [VBus ] = (m × 1) bus current and voltage vectors respectively, of the partial network before

the addition of element p − `. ¯m ¯ [Z Bus ] = (m × m) [ZBus ] matrix of the partial network before the addition of element p − `. I¯` = current injected in the link. Also,

¯ T [¯IBus ] + Z¯`` I¯` = 0 [¯ e` ] = [∆Z] 136

(4.61)

¯ Bus ] can be written as, On substituting I¯` from equation (4.61) into equation (4.60) [V T ¯ ∆Z.∆Z m ¯ ¯ ][I¯Bus ] [VBus ] = [ZBus ] − [ Z¯``

(4.62)

¯ Bus ] which is (m × m) in size can be written as Hence, the final [Z T ¯ ∆Z.∆Z m ¯ Final ¯ [Z ] = [ Z ] − [ ] Bus Bus Z¯``

¯ Bus ] building procedure will be discussed in the next lecture. An example illustrating the [Z

137

(4.63)

4.4

¯ Bus] matrix formulation in the presence of Example of [Z mutual impedances

Consider the network shown in Fig. 4.32.

¯ Bus ] example Figure 4.32: The power system for [Z A tree for the network is shown in Fig. 4.32. The system data is given in Table 4.1.

Figure 4.33: Tree of the network

Table 4.1: System data Self Element no. Bus code Impedance p-q z¯pq,pq (p.u.) 1 0 - 1(1) j0.4 2 0 - 1(2) j0.5 3 0-2 j0.5 4 2-3 j0.4 5 1-3 j0.6

138

Mutual Bus code Impedance r-s z¯pq,rs (p.u.) 0 - 1(2) j0.2 0 - 1(1)

j0.1

¯ Bus ] matrix element by element. To initiate the Step 1: The algorithm starts building the [Z process, start with element 1 connected between nodes p = 0 and q = 1, shown in Fig. 4.34. The ¯ Bus ] matrix of the partial network is given as, [Z

Figure 4.34: Partial network in Step 1

(1)

¯ Bus ] = (1) [ j0.4 ] [Z Step 2: Next add element 2 connected between p = 0 and q = 1 which is mutually coupled to the existing element 1, connected between ρ = 0 and σ = 1. This new element is a link as it does not ¯ (temp) create a new node, the partial network for this step is shown in Fig. 4.35. The augmented [Z ] Bus matrix after the addition of this element, is given by

Figure 4.35: Partial network in Step 2

(1) (1) ¯ (temp) Z = Bus (`)

(`)

j0.4 Z¯1` [ ¯ ] Z`1 Z¯``

y¯0−1(2),0−1(1) (Z¯01 − Z¯11 ) Z¯`1 = Z¯01 − Z¯11 + y¯0−1(2),0−1(2) 1 + y¯0−1(2),0−1(1) (Z¯0` − Z¯1` ) Z¯`` = Z¯0` − Z¯1` + y¯0−1(2),0−1(2) ¯01 and Z¯0` are the elements of [Z ¯ Bus ] matrix associated with the reference node. where, Z The primitive impedance matrix [¯ z] for the partial network is 0−1(1)

[¯ z] =

0−1(1) 0−1(2)

[

0−1(2)

j0.4 j0.2 ] j0.2 j0.5

The primitive admittance matrix [¯ y] for the partial network in nothing but the inverse of primitive 139

impedance matrix [¯ z] and is given by 0−1(1)

[¯ y] = [¯ z]−1 =

0−1(1) 0−1(2)

[

0−1(2)

−j3.125 j1.25 ] j1.25 −j2.5

¯01 = 0, since 0 is the reference node, Z¯`1 is evaluated as With Z j1.25(−j0.4) = −j0.2 = Z¯1` Z¯`1 = −j0.4 + −j2.5 ¯0` = 0, since 0 is the reference node, and hence, Z¯`` is calculated as Also as Z 1 + j1.25(j0.2) Z¯`` = j0.2 + = j0.50 −j2.5 (temp)

The augmented Zbus

matrix is given as (1)

(`)

(1) j0.4 −j0.2 ¯ (temp) Z = [ ] Bus (`) −j0.2 j0.5

The row and column corresponding to the `th row and column corresponding to a link addition,(shown in red in the above matrix), need to be eliminated as the link addition does not create ¯ Bus ] matrix, after the addition of second element to the partial network, is a new node. The [Z calculated using the following expression

¯ ¯ ¯ Bus ] = [Z ¯ Bus ] − Z1` Z`1 [Z Z¯`` = j0.4 −

(−j0.2)(−j0.2) j0.5 (1)

¯ Bus = (1) [ j0.32 ] Z ¯ Bus matrix is still (1×1) as no new node has been added to the partial network Note that the size of Z as yet. Step 3: Next add element 3, which is connected between the nodes p = 0 and q = 2. This is a branch addition as a new node, node 2 is created. This element is mutually coupled to the existing element 1. Hence, the primitive [¯ z] matrix of the partial network, shown in Fig. 4.36, is

140

Figure 4.36: Partial network in Step 3

0−1(1)

0−1(1)

[¯ z] = 0−1(2) 0−2

The primitive [¯ y] matrix is calculated as [¯ z]

−1

[¯ y] = 0−1(2) 0−2

0−2

and is equal to

0−1(1)

0−1(1)

0−1(2)

⎡ j0.4 j0.2 j0.1 ⎤ ⎥ ⎢ ⎥ ⎢ ⎢ j0.2 j0.5 0 ⎥ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ j0.1 0 j0.5 ⎦ ⎣

0−1(2)

0−2

⎡ j3.333 j1.333 j0.667 ⎤⎥ ⎢ ⎢ ⎥ ⎢ j1.333 −j2.533 −j0.2667 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ j0.667 −j0.2667 j2.133 ⎥ ⎣ ⎦

¯ Bus ] matrix is expressed as The modified [Z (1)

¯ Bus ] = [Z

(1) (2)

[

j0.32 Z¯21

(2)

Z¯12 ] Z¯22

For this element p = 0 and q = 2 and the set of elements [¯ ρσ ¯ ] mutually coupled to this element

is [0 − 1(1) 0 − 1(2)]

Z¯21 = Z¯01 +

Z¯02 − Z¯11 [¯ ] y0−2,0−1(1) y¯0−2,0−1(2) ] [ ¯ Z01 − Z¯11 y¯0−2,0−2

Z¯01 and Z¯02 are the transfer impedances associated with the reference node and are equal to zero.

Z¯21 =

−j0.32 [j0.667 −j0.2667] ∗ [ ] −j0.32 j2.133 141

= j0.06

¯12 = Z¯21 = j0.06 Hence, Z

Z¯22 = Z¯02 +

Z¯02 − Z¯12 1 + [¯ ] y0−2,0−1(1) y¯0−2,0−1(2) ] [ ¯ Z01 − Z¯12 y¯0−2,0−2

1 + [j0.667 −j0.2667] ∗ [ Z¯21 =

−j0.32 ] −j0.32

j2.133

= j0.48

¯ Bus ] matrix is The modified [Z (1)

¯ Bus ] = [Z

(1) (2)

[

(2)

j0.32 j0.06 ] j0.06 j0.48

Step 4: On adding element 4 between p = 2 and q = 3, a new node, node 3 is created. Hence, this ¯ Bus ] matrix can be written as is a branch addition and is shown in Fig. 4.37. The modified [Z

Figure 4.37: Partial network in Step 4

(1)

(1)

¯ Bus = (2) Z (3)

(2)

(3)

⎡ j0.32 j0.06 Z¯ ⎢ 13 ⎢ ⎢ j0.06 j0.48 Z¯23 ⎢ ⎢ ¯ ⎢ Z31 Z¯32 Z¯33 ⎣

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

As this element is not mutually coupled to other elements the elements of vector y ¯pq,ρσ are zero. ¯ Bus ] matrix can be calculated, using the expression given in (4.41), Hence, the new elements of [Z as : Off-diagonal elements

Z¯qi = Z¯pi ∀ i = 1, 2, 3 i ≠ q 142

Z¯31 = Z¯21 = j0.06 Z¯32 = Z¯22 = j0.48 Z¯13 = Z¯31 = j0.06 Z¯23 = Z¯32 = j0.48 Diagonal element Using the expression of (4.48) with no mutual coupling, the diagonal element can be written as:

Z¯qq = Z¯pq + z¯pq,pq hence,

Z¯33 = Z¯23 + z¯23,23 = j0.48 + j0.4 = j0.88 (1)

(2)

(3)

⎡ j0.32 j0.06 j0.06 ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ j0.06 j0.48 j0.48 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ j0.06 j0.48 j0.88 ⎥ ⎣ ⎦

(1)

¯ Bus = (2) Z (3)

Step 5: Finally add element 5 between nodes p = 1 and q = 3. This is an addition of a link hence a temporary row and column are added. Fig. 4.38 showns the final network after the addition of this ¯ (temp) element. The modified Z matrix can be written as Bus

Figure 4.38: The complete network after the addition of link in step 5

(1) (2) ¯ (temp) Z = Bus

(3) (`)

⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

(1)

(2)

(3)

(`)

⎤ j0.32 j0.06 j0.06 Z¯1` ⎥ ⎥ j0.06 j0.48 j0.48 Z¯2` ⎥⎥ ⎥ j0.06 j0.48 j0.88 Z¯3` ⎥⎥ ⎥ Z¯`1 Z¯`2 Z¯`3 Z¯`` ⎥⎦ 143

¯ (temp) Since this element is not mutually coupled to other elements, the new elements of [Z ] Bus matrix can be calculated, using the expression of (4.55), as : Off-diagonal elements Z¯`i = Z¯pi − Z¯qi ∀ i = 1, 2, 3 Z¯1` = Z¯11 − Z¯13 = j0.32 − j0.06 = j0.26 = Z¯`1 Z¯2` = Z¯21 − Z¯23 = j0.06 − j0.48 = −j0.42 = Z¯`2 Z¯3` = Z¯31 − Z¯33 = j0.06 − j0.88 = −j0.82 = Z¯`3 Diagonal element For calculating the diagonal element, the expression given in (4.59) is used. Hence,

Z¯`` = Z¯p` − Z¯q` + z¯pq,pq Z¯`` = Z¯1` − Z¯3` + z¯13,13 = j0.26 + j0.82 + j0.6 = j1.68

(temp)

Hence, the temporary [ZBus

] matrix can be written as

(1)

¯ (temp) [Z ]= Bus

(2) (3) (`)

⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

(1)

(2)

(3)

(`)

⎤ j0.32 j0.06 j0.06 j0.26 ⎥ ⎥ j0.06 j0.48 j0.06 −j0.42 ⎥⎥ ⎥ j0.06 j0.48 j0.88 −j0.82 ⎥⎥ ⎥ j0.26 −j0.42 −j0.82 j1.68 ⎥⎦

¯ Bus to 3 × 3. The elimination The `th row and `th column are to be eleminated to restore the size of Z is done using the relation ¯ ∗ ∆Z ¯T ∆Z (temp) ¯ Bus = Z ¯ Bus − Z Z¯`` ¯ T = [j0.26 −j0.42 −j0.82] ∆Z ⎡ j0.23 ⎤ ⎢ ⎥ ⎢ ⎥ ⎢−j0.42⎥ ∗ [−j0.26 −j0.42 −j0.82] ⎥ ⎡j0.32 j0.06 j0.06⎤ ⎢⎢ ⎢ ⎥ ⎢−j0.82⎥⎥ ⎢ ⎥ ⎦ ¯ Bus = ⎢⎢j0.06 j0.48 j0.48⎥⎥ − ⎣ Z j1.68 ⎢ ⎥ ⎢j0.06 j0.48 j0.88⎥ ⎣ ⎦ 144

¯ Bus ] is Hence, the final matrix [Z (1)

(1)

¯ Bus ] = (2) [Z (3)

(2)

(3)

⎡ j0.2798 j0.1250 j0.1869 ⎤ ⎥ ⎢ ⎥ ⎢ ⎢ j0.1250 j0.3750 j0.2750 ⎥ ⎥ ⎢ ⎥ ⎢ ⎢ j0.1869 j0.2750 j0.4798 ⎥ ⎦ ⎣

¯ Bus ] matrix, we are now ready to discuss fault analysis, After this discussion of formulation of [Z which we will start from the next lecture.

145

4.5

Fault Analysis:

Under normal conditions, a power system operates under balanced conditions with all equipments carrying normal load currents and the bus voltages within the prescribed limits. This condition can be disrupted due to a fault in the system. A fault in a circuit is a failure that interferes with the normal flow of current. A short circuit fault occurs when the insulation of the system fails resulting in low impedance path either between phases or phase(s) to ground. This causes excessively high currents to flow in the circuit, requiring the operation of protective equipments to prevent damage to equipment. The short circuit faults can be classified as: • Symmetrical faults • Unsymmetrical faults

4.6

Symmetrical faults:

¯f to the A three phase symmetrical fault is caused by application of three equal fault impedances Z ¯f = 0 the fault is called a solid or a bolted fault. These faults three phases, as shown in Fig. 4.39. If Z can be of two types: (a) line to line to line to ground fault (LLLG fault) or (b) line to line to line fault (LLL fault). Since the three phases are equally affected, the system remains balanced. That is why, this fault is called a symmetrical or a balanced fault and the fault analysis is done on per phase basis. The behaviour of LLLG fault and LLL fault is identical due to the balanced nature ¯f = 0, this is usually the of the fault. This is a very severe fault that can occur in a system and if Z most severe fault that can occur in a system. Fortunately, such faults occur infrequently and only about 5% of the system faults are three phase faults.

Figure 4.39: Symmetrical Fault

146

4.7

Unsymmetrical faults:

Faults in which the balanced state of the network is disturbed are called unsymmetrical or unbalanced faults. The most common type of unbalanced fault in a system is a single line to ground fault (LG fault). Almost 60 to 75% of faults in a system are LG faults. The other types of unbalanced faults are line to line faults (LL faults) and double line to ground faults (LLG faults). About 15 to 25% faults are LLG faults and 5 to 15% are LL faults. These faults are shown in Fig. 4.40.

Figure 4.40: Unsymmetrical Fault Majority of the faults occur on transmission lines as they are exposed to external elements. Lightening strokes may cause line insulators to flashover, high velocity winds may cause tower failure, ice loading and wind may result in mechanical failure of line or insulator and tree branches may cause short circuit. Much less common are the faults on cables, circuit breakers, generators, motors and transformers. Fault analysis is necessary for selecting proper circuit breaker rating and for relay settings and coordination.The symmetrical faults are analysed on per phase basis while the unsymmetrical faults ¯ BUS matrix is very usefull for short are analyzed using symmetrical components. Further, the Z circuit studies .

4.8

Symmetrical or Balanced three phase fault analysis:

In this type of faults all three phases are simultaneously short circuited. Since the network remains balanced, it is analyzed on per phase basis. The other two phases carry identical currents but with a phase shift of 120○ . A fault in the network is simulated by connecting impedances in the network at the fault location. The faulted network is then solved using Thevenin’s equivalent network as seen from the fault point. The bus impedance matrix is convenient to use for fault studies as its diagonal elements are the Thevenin’s impedance of the network as seen from different buses. Prior to the occurrence of fault, the system is assumed to be in a balanced steady state and hence per phase network model is used. The generators are represented by a constant voltage source behind a suitable reactance which may be sub-transient, transient or normal d-axis reactance. The transmission lines are represented by their π -models with all impedances referred to a common base. A typical bus 147

Figure 4.41: Fault at kth bus of a power system of an n- bus power system network is shown in Fig. 4.41. Further, a balanced three phase fault, ¯f is assumed to occur at k th bus as shown in the figure. A pre-fault through a fault impedance Z load flow provides the information about the pre-fault bus voltage. ¯ BUS (0)] be the prefault bus voltage vector =[V¯1 (0) . . . Vk (0) . . . Vn (0)]T p.u. The fault Let [V ¯f will cause a change in the voltage of all the buses [∆V ¯ BUS ] at k th bus through an impedance Z due to the flow of heavy currents through the transmission lines. This change can be calculated by applying a voltage V¯k (0) at k th bus and short circuiting all other voltage sources. The sources and ¯i loads are replaced by their equivalent impedances. This is shown in Fig. 4.42. In Fig. 4.42, Z

¯ BUS ] Figure 4.42: Network representation for calculating [∆V ¯k are the equivalent load impedances as bus i and k respectively, z¯ik is the impedance of line and Z 148

between ith and k th buses. x ¯di is the appropriate generator reactance, Z¯f is the fault impedance, I¯k (F ) is the fault current and V¯k (0) is the prefault voltage at k th bus. From the superposition theorem, the bus voltages due to a fault can be obtained as the sum of prefault bus voltages and the change in bus voltages due to fault,i.e.,

¯ BUS (F)] = [V ¯ BUS (0)] + [∆V ¯ BUS ] [V

(4.64)

where, T ¯ BUS (F)] = Vector of bus voltages during fault =[V¯1 (F ) . . . V¯i (F ) . . . V¯n (F )] [V T ¯ BUS (0)] = Vector of pre-fault bus voltages =[V¯1 (0) . . . V¯i (0) . . . V¯n (0)] [V T ¯ BUS ] = Vector of change in bus voltages due to fault= [∆V¯1 . . . ∆V¯k . . . ∆V¯n ] [∆V

Also the bus injected current [¯ IBUS ] can be expressed as,

¯ BUS ] [V ¯ BUS ] [¯IBUS ] = [Y

(4.65)

¯ BUS ] is the bus voltage vector and [Y ¯ BUS ] is the bus admittance matrix. where, [V With all the bus currents, except of the faulted bus k , equal to zero, the node equation for the network of Fig. 4.42 can be written as

⎡ 0 ⎤ ⎡ Y¯11 ⋯ Y¯1k ⋯ Y¯1n ⎤ ⎡ ∆V¯1 ⎤ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⋮ ⎥ ⎢ ⋮ ⎥⎢ ⋮ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ¯ ⎥ ⎢ ⎥⎢ ⎥ ⎢−Ik (F )⎥ = ⎢Y¯k1 ⋯ Y¯kk ⋯ Y¯kn ⎥ ⎢ ∆V¯k ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⋮ ⎥ ⎢ ⋮ ⎥⎢ ⋮ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢¯ ⎥ ⎢ ¯ n ⎥⎥ ⎢ 0 ⎥ ⎢Yn1 ⋯ Y¯nk ⋯ Y¯nn ⎥ ⎢ ∆V ⎣ ⎦ ⎣ ⎦⎣ ⎦

(4.66)

As the fault current I¯k (F ) is leaving the bus it is taken as a negative current entering the bus. Hence, ¯ BUS ] [∆V ¯ BUS ] [¯IBUS (F)] = [Y (4.67)

¯ BUS ] can be calculated as: [∆V −1

¯ BUS ] = [Y ¯ BUS ] [¯IBUS (F)] = [Z ¯ BUS ] [¯IBUS (F)] [∆V

(4.68)

¯ BUS ] is the bus impedance matrix = [Y ¯ BUS ]−1 . where, [Z ¯ BUS ] from equation (4.68) in equation (4.64) one can write, Substituting the expression of [∆V ¯ BUS (F)] = [V ¯ BUS (0)] + [Z ¯ BUS (F)] [¯IBus (F)] [V Expanding the above equation one can write, 149

(4.69)

⎡ V¯1 (F ) ⎤ ⎡ V¯1 (0) ⎤ ⎡ Z¯11 ⋯ Z¯1k ⋯ Z¯1n ⎤ ⎡ (0) ⎤ ⎥ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎢ ⎥ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎢ ⎥⎢ ⋮ ⎥ ⎢ ⋮ ⎥ ⎢ ⋮ ⎥ ⎢ ⋮ ⎥ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎢ ⎥ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎢¯ ⎢Vk (F )⎥ = ⎢V¯k (0)⎥ + ⎢Z¯k1 ⋯ Z¯kk ⋯ Z¯kn ⎥ ⎢−I¯k (F )⎥ ⎥ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎢ ⎥⎢ ⋮ ⎥ ⎢ ⋮ ⎥ ⎢ ⋮ ⎥ ⎢ ⋮ ⎥ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎢ ⎥ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎢¯ ⎢Vn (F )⎥ ⎢V¯n (0)⎥ ⎢Z¯n1 ⋯ Z¯nk ⋯ Z¯nn ⎥ ⎢ 0 ⎥ ⎦ ⎦⎣ ⎦ ⎣ ⎦ ⎣ ⎣

(4.70)

The bus voltage of kth bus can be expressed as:

V¯k (F ) = V¯k (0) − Z¯kk I¯k (F )

(4.71)

V¯k (F ) = Z¯F I¯k (F )

(4.72)

Also from Fig. 4.41

¯f = 0 and hence, V¯k (F ) = 0. Thus the fault current I¯k (F ) for bolted fault can For a bolted fault Z be expressed using equation (4.71) as, V¯k (0) I¯k (F ) = Z¯kk

(4.73)

¯f , the fault current can be calculated as: For faulty with non-zero fault impedance Z V¯k (0) I¯k (F ) = Z¯kk + Z¯f

(4.74)

¯kk in equation (4.73) and equation (4.74) is the Thevenin’s impedance or openThe quantity Z circuit impedance of the network as seen from the faulted bus k. From equation (4.70), the bus voltage after fault for the unfaulted or healthy buses can be written as: V¯i (F ) = V¯i (0) − Z¯ik I¯k (F ) ∀i = 1, 2, ⋯n, i ≠ k

(4.75)

Substituting I¯k (F ) from equation (4.73) , V¯i (F ) can be expressed as:

V¯i (F ) = V¯i (0) −

Z¯ik ¯ Vk (0) Z¯kk + Z¯f

(4.76)

The fault current I¯ij (F ) flowing in the line connecting ith and j th bus can be calculated as

V¯i (F ) − V¯j (F ) I¯ij (F ) = z¯ij where z¯ij is the impedance of line connecting buses i and j. 150

(4.77)

4.9

Unsymmetrical or Unbalanced fault analysis:

For the analysis of unsymmetrical or unbalanced faults, symmetrical component method is used. The use of symmetrical components simplifies the analysis procedure of unbalanced system and also helps in improving the understanding of the system behavior during fault conditions. A review of symmetrical components is presented next.

4.9.1

Symmetrical components:

Any unbalanced set of three phase voltage or current phasors can be replaced by three balanced sets of three phase voltage or current phasors. These three balanced set of voltage or current phasors are called symmetrical components of voltages or currents. Let I¯a , I¯b , and I¯c be an arbitrary set of three current phasors representing phase currents. Then using symmetrical components they can be expressed as:

⎡I¯ ⎤ ⎡I¯ ⎤ ⎡I¯ ⎤ ⎡I¯ ⎤ ⎢ a ⎥ ⎢ a0 ⎥ ⎢ a1 ⎥ ⎢ a2 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ I¯b ⎥ = ⎢ I¯b0 ⎥ + ⎢ I¯b1 ⎥ + ⎢ I¯b2 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢¯⎥ ⎢¯ ⎥ ⎢¯ ⎥ ⎢¯ ⎥ ⎢ Ic ⎥ ⎢ Ic0 ⎥ ⎢ Ic1 ⎥ ⎢ Ic2 ⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

(4.78)

Or,

[¯I]abc = [¯I0 ] + [¯I1 ] + [¯I2 ] where, T [¯I]abc = [I¯a I¯b I¯c ] is the arbitrary set of three current phasors of phase currents. T [¯I0 ] = [I¯a0 I¯b0 I¯c0 ] is the set of zero sequence components.The magnitudes of the three zero sequence components are equal i.e.∣I¯a0 ∣ = ∣I¯b0 ∣ = ∣I¯c0 ∣ and they are co-phasors. T [¯I1 ] = [I¯a1 I¯b1 I¯c1 ] is the set of positive sequence components, with I¯a1 = ∣I¯a1 ∣ ∠0○ , I¯b1 = ∣I¯b1 ∣ ∠ − 120○ , and I¯c1 = ∣I¯c1 ∣ ∠120○ , with ∣I¯a1 ∣ = ∣I¯b1 ∣ = ∣I¯c1 ∣. T [¯I2 ] = [I¯a2 I¯b2 I¯c2 ] is the set of negative sequence components, with I¯a2 = ∣I¯a2 ∣ ∠0○ , I¯b2 = ○ , and I ¯c2 = ∣I¯c2 ∣ ∠ − 120○ ,with ∣I¯a2 ∣ = ∣I¯b2 ∣ = ∣I¯c2 ∣ ¯ ∣I¯b2 ∣ ∠120

The graphical representation of the sequence components is shown in Fig. 4.43. Let an operator ‘a’ be defined such that a = ∠120○ . Any phasor multiplied by ‘a’ undergoes a counter clockwise rotation of 120○ without any change in the magnitude. Further,

a = 1∠120○ a2 = 1∠240○ a3 = 1∠360○ also 1 + a + a2 = 0

I¯a1 = Ia1 ∠θ1

where, ∠θ1 is the angle of phase ‘a’ positive sequence current. 151

Figure 4.43: Representation of Symmetrical Components

I¯b1 = a2 I¯a1 I¯c1 = aI¯a1 The phase sequence of the positive component set is ‘abc’. Similarly the negative sequence set can be written as:

I¯a2 = Ia2 ∠θ2 where, ∠θ2 is the angle of phase ‘a’ negative sequence current.

I¯b2 = aI¯a2 I¯c2 = a2 I¯a2 152

The phase sequence of the negative component set is ‘acb’. The zero-sequence component set can be written as:

I¯a0 = I¯a0 ∠θ0 = I¯b0 = I¯c0 where, ∠θ0 is the angle of phase ‘a’ zero sequence current. Hence, equation (4.78) can be simplified as:

⎡I¯ ⎤ ⎡1 1 1 ⎤ ⎡I¯ ⎤ ⎥ ⎢ a0 ⎥ ⎢ a⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎢ I¯b ⎥ = ⎢1 a2 a ⎥ ⎢I¯a1 ⎥ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢¯⎥ ⎢ ⎢ Ic ⎥ ⎢1 a a2 ⎥ ⎢I¯a2 ⎥ ⎦⎣ ⎦ ⎣ ⎦ ⎣

(4.79)

It can also expressed in a compact form as:

¯ [¯I] [¯I]abc = [A] 012 where, [¯ I]

abc

= set of phase quantities = [I¯a I¯b I¯c ]

(4.80)

T

T [¯I]012 = set of sequence quantities = [I¯a0 I¯a1 I¯a2 ]

⎡1 1 1 ⎤ ⎢ ⎥ ⎢ ⎥ 2 ⎢ ¯ A = ⎢1 a a ⎥⎥ is the symmetrical component transformation matrix. ⎢ ⎥ ⎢1 a a2 ⎥ ⎣ ⎦ The symmetrical components [¯ I]012 can be written in terms of phase quantities [¯I]abc as: −1

⎡1 1 1 ⎤ ⎢ ⎥ ⎢ ⎥ 1 −1 ¯ = ⎢⎢1 a a2 ⎥⎥ thus, where, A 3⎢ ⎥ ⎢1 a2 a ⎥ ⎣ ⎦

¯ [¯I] [¯I]012 = [A] abc

(4.81)

1 I¯a0 = [I¯a + I¯b + I¯c ] 3 1 I¯a1 = [I¯a + aI¯b + a2 I¯c ] 3 1 I¯a2 = [I¯a + a2 I¯b + aI¯c ] 3

(4.82)

¯ ¯ ¯ [V] [V] = [A] abc 012

(4.83)

To summarize:

• For voltage:

153

−1

¯ ¯ [V] ¯ [V] = [A] 012 abc

(4.84)

¯ abc is the set of phase voltages, and [V] ¯ 012 is the set of sequence voltages. where, [V] • For current:

¯ [¯I] [¯I]abc = [A] 012 −1

¯ [¯I] [¯I]012 = [A] abc

(4.85) (4.86)

where, [¯ I]abc is the set of phase voltages, and [¯I]012 is the set of sequence voltages. Before starting unbalanced fault analysis, it is necessary to learn about the sequence networks of different power system components, which we will discuss in the next lecture.

154

4.9.2

Sequence Networks of a loaded Synchronous Generator:

¯s per phase, with its A three-pahse synchronous generator, having a synchronous impedance of Z ¯n is shown in Fig.4.44. The generator is supplying a neutral grounded through a impedance Z ¯a , E¯b and E¯c are balanced and hence treated balanced three phase load. The generator voltages E

Figure 4.44: Three phase synchronous generator supplying a balanced load

as positive sequence set of voltage phasors and can be expressed as:

⎡1⎤ ⎢ ⎥ ⎢ ⎥ ¯ ¯ [E]abc = ⎢⎢ a2 ⎥⎥ [Ea] ⎢ ⎥ ⎢a⎥ ⎣ ⎦

(4.87)

As the generator is supplying a three-phase balanced load, the following KVL equations can be written for each phase :

V¯a = E¯a − Z¯s I¯a − Z¯n I¯n V¯b = E¯b − Z¯s I¯b − Z¯n I¯n V¯c = E¯c − Z¯s I¯c − Z¯n I¯n 155

(4.88)

Substituting the neutral current I¯n = I¯a + I¯b + I¯c in equation (4.88), and writing the resulting equation in matrix form, we get:

⎡V¯ ⎤ ⎡E¯ ⎤ ⎡Z¯ + Z¯ Z¯n Z¯n ⎤⎥ ⎡⎢I¯a ⎤⎥ ⎢ a⎥ ⎢ a⎥ ⎢ s n ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎢ V¯b ⎥ = ⎢ E¯b ⎥ − ⎢ Z¯n Z¯s + Z¯n Z¯n ⎥⎥ ⎢⎢ I¯b ⎥⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢¯ ⎥ ⎢ ¯ ⎥ ⎢ ¯ ⎢ Vc ⎥ ⎢ Ec ⎥ ⎢ Zn Z¯n Z¯s + Z¯n ⎥⎦ ⎢⎣ I¯c ⎥⎦ ⎣ ⎦ ⎣ ⎦ ⎣

(4.89)

The above matrix equation can be expressed in a compact form as:

¯ ¯ ¯ [V] [¯I]abc = [E] − [Z] abc abc abc

(4.90)

where, T ¯ abc = [V¯a V¯b V¯c ] is the vector of terminal phase voltages. [V] T [¯I]abc = [I¯a I¯b I¯c ] is the vector of terminal phase currents. ¯ abc is the impedance matrix which can be easily identified from equation (4.89). [Z]

Replacing the phase quantities of equation (4.90) by corresponding sequence quantities, using the transformation equation (4.83) and equation (4.85) one can write:

¯ [¯I] ¯ ¯ ¯ [E] ¯ ¯ [V] [A] [A] − [Z] = [A] 012 abc 012 012 ¯ Premultiplying bothsides of the equation (4.91) by [A]

−1

and after simplifications one gets:

¯ ¯ ¯ [V] [¯I]012 = [E] − [Z] 012 012 012 ¯ where,[Z]

012

(4.91)

(4.92)

is Generator Sequence Impedance Matrix and is defined as:

⎡Z¯ + 3Z¯ 0 0 ⎤ ⎢ s ⎥ n ⎢ ⎥ −1 ⎢ ¯ ¯ ¯ ¯ ¯ [Z]012 = [A] [Z]abc [A] = ⎢ 0 Zs 0 ⎥⎥ ⎢ ⎥ ⎢ 0 0 Z¯s ⎥⎦ ⎣ ¯ [E] 012

⎡0⎤ ⎢ ⎥ ⎢ ⎥ ⎥ is the generated sequence voltage vector and is defined as ⎢ ⎢E¯a ⎥ since the generated voltages ⎢ ⎥ ⎢0⎥ ⎣ ⎦

are always balanced and contain only the positive sequence component.

¯ Substituting [E]

012

¯ and [Z]

012

in equation (4.92) we get:

⎡V¯ ⎤ ⎡ 0 ⎤ ⎡Z¯ 0 0 ⎤ ⎡I¯ ⎤ ⎢ a0 ⎥ ⎢ ⎥ ⎢ 0 ⎥ ⎢ a0 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢V¯a1 ⎥ = ⎢E¯a ⎥ − ⎢ 0 Z¯1 0 ⎥ ⎢I¯a1 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢¯ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢Va2 ⎥ ⎢ 0 ⎥ ⎢ 0 0 Z¯2 ⎥ ⎢I¯a2 ⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎣ ⎦

(4.93)

¯1 = Z¯s is the positive sequence generator impedance, Z¯2 = Z¯s is the negative sequence generwhere, Z 156

¯0 = Z¯s + 3Z¯n is the zero sequence generator impedance. Expanding the above ator impedance and Z equation, one can write seperate equation for each of the sequence components as : V¯a0 = −Z¯0 I¯a0 V¯a1 = E¯a − Z¯1 I¯a1

(4.94)

V¯a2 = −Z¯2 I¯a2 From equation (4.94), it is evident that the three sequence components are independent of each other.The current of a particular sequence produces a voltage drop of that sequence only, hence the three sequences are decoupled from each other. The three sequence networks of a synchronous generator are shown in Fig. 4.45.

Figure 4.45: The sequence networks of a synchronous generator

4.9.3

Sequence networks of a transmission line:

For a static device such as a transmission line, the phase sequence of voltages and currents have no effect on the impedance offered by the line as both positive and negative phase sequences encounter identical line geometry. Hence, the positive and negative sequence impedances offered by a line are ¯1 = Z¯2 . identical i.e. Z The zero sequence currents, however, are in phase and flow through the conductors and return through grounded neutral and/or ground wires. As a result, the ground or ground wire are to be ¯0 is, therefore, included in the path of the zero sequence currents. The zero sequence impedance Z ¯1 and Z¯2 due to the inclusion of the ground return path. Z¯0 is usually more than different from Z ¯1 or Z¯2 . The three sequence networks of the transmission lines are shown in Fig. three times of Z 4.46.

4.9.4

Sequence networks of a tranformer:

For short circuit studies, the shunt magnetizing branch of transformer is neglected as the current through it is negiligible as compared to short circuit current. The transformer is, therefore, modelled with an equivalent series leakage impedance. Since the transformer is also a static device like a 157

Figure 4.46: The sequence networks of a transmission line transmission line, the series leakage impedance will not change if the phase sequence of applied voltage is reversed. Therefore, the positive and negative sequence impedances offered by a transformer are equal. The zero sequece current flows through a transformer if paths for it to flow exist on the primary as well the secondary sides. For such transformers the zero sequence impedance is equal to ¯0 = Z¯1 = Z¯2 . The positive and negative sequence networks the leakage impedance, as a consequence Z of a transformer are identical to the positive and negative sequence networks of a transmission line as shown in Fig. 4.46 (a) and (b). The sequence networks for zero sequence depends on the winding connections and whether or not the neutrals are grounded. To derive these circuits for different transformer connections, one has to keep in mind that an open circuit will exist on the primary (secondary) side if there is no ground return for primary (secondary) currents or if there is no corresponding path for secondary(primary) zero-sequence currents. The different three-phase transformer connections and their equivalent zero-sequence networks are discussed next. It is assumed that the neutrals, if grounded, are solidly grounded.

Figure 4.47: The zero-sequence equivalent circuit of a Star-Star transformer with both neutrals grounded (a) Star-Star connections with both neutrals grounded: Since both the neutrals are grounded,the phasor sum of three unbalanced phase currents is equal to three times the zero sequence current I¯a0 (equation (4.82)). Hence, the zero sequence currents can flow in the primary and secondary 158

windings and the transhomer, therefore, can be represented by the equivalent zero-sequence leakage impedance. The equivalent circuit is shown in Fig. 4.47. (b) Star-Star connections with only one neutral grounded: When the neutral of only one winding is grounded, the phase currents of the ungrounded winding must add up to zero. This implies that the zero sequence currents can not exist in the ungrounded winding and hence the zero sequence currents can not exist even in the transformer side with neutral grounded. The transformer in this case, is represented as an open circuit between primary and secondary windings and the equivalent circuit is shown in Fig. 4.48.

Figure 4.48: The zero-sequence equivalent circuit of a Star-Star transformer with only one neutral grounded

(c) Star-Star connections with only no neutral grounded: In this case also the phasor sum of the phase curents of both the windings is zero and hence the zero sequence currents can not exist on any winding in this case also. The zero sequence equivalent network is represented as an open circuit between the two windings and is shown in Fig. 4.49.

Figure 4.49: The zero-sequence equivalent circuit of a Star-Star transformer with both neutrals ungrounded

159

(d) Star-delta connections with neutral grounded: A zero sequence current on the grounded starwinding will cause a circulating zero-sequence current in the closed delta-winding. However, the zero-sequence current on the delta-winding can not exist on line side of the winding and is confined only to the closed delta-winding. As a result, an open circuit exists between the star and the delta sides. But, as the zero-sequence currents can exist on the line-side of the grounded star winding, the zero-sequence leakage impedance of the transformer is connected to ground on the star side of the transformer and an open circuit exists between the two windings. The equivalent circuit for this connection is shown in Fig. 4.50.

Figure 4.50: The zero-sequence equivalent circuit of a Star-Delta transformer with Star-side neutral grounded

(e) Star-delta connections with ungrounded neutral grounded: Since the neutral is isolated, no zerosequence current can exist in the star side of the transformer and as a consequence zero-sequence currents can not exist in the delta side. The transformer is, therefore, represented as an opencircuit and the equivalent circuit is shown in Fig. 4.51.

Figure 4.51: The zero-sequence equivalent circuit of a Star-Delta trnasformer with Star-side neutral ungrounded

(f) Delta-delta connections with ungrounded neutral grounded: In this case, the zero-sequence currents can only circulate within the closed delta windings and can not exit on line sides of both 160

the windings. Hence, an open circuit exists between the two windings as far as zero-sequence currents are concerned. To permit the circulating zero-sequence current to exist, the zero-sequence leakage impedance is represented as a closed path with the ground. The equivalent circuit is shown in Fig. 4.52.

Figure 4.52: The zero-sequence equivalent circuit of a Delta-Delta transformer

Figure 4.53: The zero-sequence equivalent circuit of a Star-Delta transformer with neutral grounded through impedance Point to remember: If the neutral of a transformer is grounded through a grounding impedance Z¯n , as shown in Fig. 4.53, then, the total zero-sequence equivalent impdance to be used in the equivalent circuit is Z¯0total = Z¯0 + 3Z¯n (4.95) This is due to the fact that the neutral current is 3 times the zero-sequence current per phase. Next, the concepts of unsymmetrical fault analysis are developed with help of Thevenin’s equivalent circuit of sequence networks and symmetrical components in the next lecture.

161

4.9.5

Single line to ground (LG) fault analysis :

An unloaded balanced three-phase synchronous generator with neutral grounded through an impedance Z¯n is shown in Fig. 4.54. Suppose a single line to ground fault (LG) occurs on phase ‘a’ though ¯f . an impedance Z

Figure 4.54: LG fault on phase ‘a’ of an unloaded generator

Since the generator is unloaded, the following terminal conditions exist at the fault point:

V¯a = Z¯f I¯a I¯b = 0

(4.96)

I¯c = 0 Substituting I¯b = I¯c = 0 in equation (4.86), the symmetrical components of currents can be calculated as:

⎡1 1 1 ⎤ ⎡I¯ ⎤ ⎡I¯ ⎤ ⎢ ⎥ ⎢ a⎥ ⎢ a0 ⎥ ⎢ ⎥ 1⎢ ⎥⎢ ⎥ ⎢I¯a1 ⎥ = ⎢1 a a2 ⎥ ⎢ 0 ⎥ ⎢ ⎥ 3⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢¯ ⎥ ⎢Ia2 ⎥ ⎢1 a 2 a ⎥ ⎢ 0 ⎥ ⎣ ⎦⎣ ⎦ ⎣ ⎦ 162

(4.97)

Solving the above equation, the values of the symmetrical components of fault current I¯a are:

1 I¯a0 = I¯a1 = I¯a2 = I¯a 3

(4.98)

The voltage of phase a can be expressed in terms of symmetrical components from equation (4.83), as V¯a = V¯a0 + V¯a1 + V¯a2 (4.99) Substituing in the equation the values of V¯a0 , V¯a1 and V¯a2 from equation (4.94) into equation (4.99), V¯a can be written as (with I¯a0 = I¯a1 = I¯a2 from equation (4.98)):

V¯a = E¯a − (Z¯0 + Z¯1 + Z¯2 )I¯a0

(4.100)

¯f I¯a = 3Z¯f I¯a0 . Hence, equation (4.100) can be expressed as: From equations (4.96) and (4.98), V¯a = Z 3Z¯f I¯a0 = E¯a − (Z¯0 + Z¯1 + Z¯2 )I¯a0 or,

I¯a0 =

E¯a Z¯0 + Z¯1 + Z¯2 + 3Z¯f

(4.101)

The fault current, therefore, is:

I¯f = I¯a = 3I¯a0 =

3E¯a Z¯0 + Z¯1 + Z¯2 + 3Z¯f

(4.102)

From equations (4.98) and (4.101), it be easily interpreted that the three sequence networks are connected in series as shown in Fig. 4.55.

Figure 4.55: Connection of sequence networks for LG fault

¯n = 0 and for bolted fault Z¯f = 0. Note that, for solidly grounded generator, Z Extending the above concept to the analysis of LG fault in a power system, the Thevenin’s equiv163

alent circuit (as seen from the fault point) is obtained, individually for the three sequence networks. ¯1th , For the positive sequence network V¯th , the open circuit pre-fault voltage at the fault point, and Z the positive sequence Thevenin’s equivalent impedance as seen from the fault point are determined. ¯2th and Z¯0th , For negative and zero sequence networks, only the Thevenin’s equivalent impedances Z respectively are calculated. The three Thevenin’s equivalent networks are then connected in series.

4.9.6

Line to Line (LL) fault analysis :

¯f , on Fig. 4.56 shows a line to line fault (LL) between phases ‘b’ and ‘c’ through an impedance Z an unloaded three phase generator. The terminal conditions at the fault point are:

Figure 4.56: LL fault between phases ‘b’ and ‘c’ of an unloaded generator

V¯b − V¯c = Z¯f I¯b I¯b + I¯c = 0 164

(4.103)

I¯a = 0 Substituting I¯a = 0 and I¯b = −I¯c in equation (4.86), the symmetrical components of cuurents can be calculated as:

⎡I¯ ⎤ ⎡1 1 1 ⎤ ⎡ 0 ⎤ ⎥⎢ ⎥ ⎢ a0 ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ 1⎢ ⎢I¯a1 ⎥ = ⎢1 a a2 ⎥ ⎢ I¯b ⎥ ⎥⎢ ⎥ ⎢ ⎥ 3⎢ ⎥⎢ ⎥ ⎢¯ ⎥ ⎢ 2 ⎢Ia2 ⎥ ⎢1 a a ⎥ ⎢−I¯b ⎥ ⎦⎣ ⎦ ⎣ ⎦ ⎣

(4.104)

Solving the above equation, the values of the symmetrical components of the current I¯a are:

I¯a0 = 0 1 I¯a1 = (a − a2 )I¯b 3 1 I¯a2 = (a2 − a)I¯b = −I¯a1 3

(4.105)

From equation (4.83), we have

V¯b − V¯c = (a2 − a)(V¯a1 − V¯a2 ) = Z¯f I¯b

(4.106)

Substituting V¯a1 and V¯a2 from equation (4.94) and noting that I¯a1 = −I¯a2 , one can write:

(a2 − a) [E¯a − (Z¯1 + Z¯2 )I¯a1 ] = Z¯f I¯b

(4.107)

Also from equation (4.105),

I¯b =

3I¯a1 (a − a2 )

(4.108)

Substituting this value of I¯b in equation (4.107), we get:

[E¯a − (Z¯1 + Z¯2 )I¯a1 ] =

3Z¯f I¯a1 (a − a2 )(a2 − a)

Since, (a − a2 )(a2 − a) = 3, the above expression can be simplified and written as:

I¯a1 =

E¯a (Z¯1 + Z¯2 + Z¯f )

(4.109)

The phase currents during fault can be calculated as:

⎡I¯ ⎤ ⎡1 1 1 ⎤ ⎡ 0 ⎤ ⎢ a⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ I¯b ⎥ = ⎢1 a2 a ⎥ ⎢ I¯a1 ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢¯⎥ ⎢ ⎥⎢ ⎥ ⎢ Ic ⎥ ⎢1 a a2 ⎥ ⎢−I¯a1 ⎥ ⎣ ⎦ ⎣ ⎦⎣ ⎦ Solving for the phase currents, the expressions for I¯b and I¯c can be written as:

165

(4.110)

I¯b = −I¯c = (a2 − a)I¯a1

(4.111)

Substituting I¯b from equation (4.111) in equation (4.106) one gets:

(V¯a1 − V¯a2 ) = Z¯f I¯a1 The equivalent circuit of the fault in terms of the sequence networks is shown in Fig. 4.57. The circuit has been drawn on the basis of equation (4.105) and the above equation. It shows that the positive sequence and negative sequence networks are connected in phase opposition bridged by the ¯f . Also, since I¯a0 = 0, the zero sequence network is open circuited and hence is fault impedance Z not shown in the diagram.

Figure 4.57: Connection of sequence networks for an LL fault between phases ‘b’ and ‘c’ of an unloaded generator Extending the above concept to LL fault calculations in a power system, it can be concluded that, the Thevenin’s equivalent positive and negative sequence networks, as seen from the fault point, can be connected in phase opposition through the fault impdedance for calculating fault current.

4.9.7

Double Line to ground (LLG) fault analysis :

Fig. 4.58 shows a double line to ground (LLG) fault on phases ‘b’ and ‘c’ through an impedance Z¯f on an unloaded three phase generator. The terminal conditions at the fault point are:

V¯b = V¯c = Z¯f I¯f = Z¯f (I¯b + I¯c ) I¯a = I¯a1 + I¯a2 + I¯a0 = 0

(4.112)

¯ b and V ¯ c can be written as: From equation (4.83), V V¯b = V¯a0 + a2 V¯a1 + aV¯a2 V¯c = V¯a0 + aV¯a1 + a2 V¯a2 166

(4.113)

Figure 4.58: LLG fault between phases ‘b’ and ‘c’ of an unloaded generator

Since V¯b = V¯c , from equation (4.113), one can write

V¯a1 = V¯a2

(4.114)

Substituting I¯b and I¯c in terms of their sequence components from equation (4.85), voltage of phase ’b’ can be expressed as:

V¯b = Z¯f (I¯a0 + a2 I¯a1 + aI¯a2 + I¯a0 + aI¯a1 + a2 I¯a2 ) 167

V¯b = Z¯f (I¯a0 + a2 I¯a1 + aI¯a2 + I¯a0 + aI¯a1 + a2 I¯a2 ) = Z¯f (2I¯a0 + (a2 + a)(I¯a1 + I¯a2 )) = Z¯f (2I¯a0 − (I¯a1 + I¯a2 )) Since 1 + a + a2 = 0 and I¯a = I¯a1 + I¯a2 + I¯a0 = 0, hence

V¯b = 3Z¯f I¯a0

(4.115)

Further substituting V¯b from equation (4.115) and the condition of equation (4.114) in equation (4.113), we get:

3Z¯f I¯a0 = V¯a0 + (a2 + a)V¯a1 = V¯a0 − V¯a1

(4.116)

Substituting V¯a0 and V¯a1 from equation (4.94) in equation (4.116), the zero sequence component of

Figure 4.59: Connection of sequence networks for an LLG fault between phases ‘b’ and ‘c’ of an unloaded generator

current I¯a0 is given by:

(E¯a − Z¯1 I¯a1 ) I¯a0 = − (Z¯0 + 3Z¯f )

(4.117)

For calculating the negative sequence component of current, I¯a2 , substitute V¯a1 and V¯a2 from equation (4.94) in equation (4.114). The expression for I¯a2 is:

(E¯a − Z¯1 I¯a1 ) I¯a2 = − Z¯2

(4.118)

Finally, by substituting I¯a0 and I¯a2 from equations (4.117) and (4.118) in equation (4.112), the value 168

of the positive sequence component of current I¯a1 is found out as:

I¯a1 =

E¯a Z¯2 (Z¯0 + 3Z¯f ) Z¯1 + (Z¯0 + Z¯2 + 3Z¯f )

(4.119)

¯f I¯f , from equation (4.115) one can conclude that Since V¯b = Z I¯f = 3I¯a0

(4.120)

The equivalent circuit for the fault in terms of the sequence networks is shown in Fig. 4.59. The circuit shown in Fig. 4.59 is based on equations (4.114) and (4.116). For LLG fault calculations in a power system, the Thevenin’s equivalent of the three sequence networks, as seen from the fault point, are found out. The positive and negative sequence equivalents are connected in parallel and ¯f . In the next lecture, the combination is then connected to the zero sequence network through 3Z ¯ BUS matrix. we will study the procedure of unbalanced fault analysis using Z

169

4.10

¯ BUS matrix: Unbalanced fault analysis using Z

In the previous section, it is observed that, for fault calculations the Thevenin’s equivalent networks, at the fault point, are needed for the three sequence networks. Since the three sequence networks are ¯ BUS matrices of these sequence networks can be found seperately. The diagonal independent, the Z ¯ BUS matrices infact, are the Thevenin’s equivalent impedances elements of the three sequence Z of the sequence networks as seen from the various buses. Let, the three sequence bus impedance matrices for zero sequence, positive sequence and negative sequence networks be represented as th bus, then Z(0) , Z(1) and Z(2) ¯ (0) ¯ (1) ¯ (2) [Z BUS ], [ZBUS ] and [ZBUS ] respectively. If the fault is at the k kk kk kk of the sequence bus impedance matrices are the zero, positive and negative Thevenin’s equivalent impedances, respectively, as seen from the faulted bus. Hence, the first step in the fault analysis ¯ BUS matrix is the determination of the three sequence networks and subsequently, finding using Z the bus impedance matrix for each sequence network. To illustrate this step, consider the single line diagram of the power system shown in Fig. 4.60.

Figure 4.60: Single line diagram of Power System The positive sequence equivalent network for the system is shown in Fig. 4.61. In this figure all the elements of the system have been represented by their positive sequence equivalents. Similarly by representing all elements by their negative sequence impedances, the negative sequence equivalent network can be obtained. The negative sequence network is shown in Fig. 4.62. For the zero sequence equivalent network, the generator neutral connections and transformer connections have to be considered. The zero sequence equivalent network is shown in Fig. 4.63. In ¯ Bus ] matrix for the three sequence networks is found using [Z ¯ Bus ] building the next step, the [Z (0) (1) (2) ¯ Bus ], [Z ¯ Bus ] and [Z ¯ Bus ] matrices are known, the following procedure is followed algorithm. Once [Z for the fault analysis of the given network. 170

Figure 4.61: The positive sequence equivalent network

Figure 4.62: The negative sequence equivalent network

¯f as shown in Fig. (a) LG fault: Let the fault be on phase ‘a’ of bus ‘k’ with a fault impedance Z 4.64. From equations (4.98) and (4.101), it can be seen that the three equivalent sequence networks are in series for calculating the sequence components of the fault currents. Hence, generalizing equation (4.101) for fault at kth bus, the expression for sequence component of fault current can be written as:

I¯k(0) (F ) = I¯k(1) (F ) = I¯k(2) (F ) =

V¯k (0) (1) (2) (0) Z¯(kk) + Z¯(kk) + Z¯(kk) + 3Z¯f

(4.121)

¯ (0) , Z¯ (1) and Z¯ (2) are the kth diagonal elements of [Z ¯ (0) ¯ (1) ¯ (2) • Z Bus ], [ZBus ] and [ZBus ] matrices kk kk kk respectively. • V¯k (0) is the prefault voltage of kth bus, usually taken as 1∠0○ pu. 171

Figure 4.63: The zero sequence equivalent network

Figure 4.64: LG fault on phase ‘a’ of k th bus The fault current is given by:

¯ [¯I(012) (F)] [¯I(abc) (F)] = [A] k k

(4.122)

(b) LL fault: Let the fault be between phases phase ‘b’ and phase ‘c’ of bus ‘k’ through an ¯f as shown in Fig. 4.65. From equation (4.109) and Fig. 4.57 it is observed that the impedance Z positive sequence and negative sequence equivalent networks are connected in phase opposition. Thus, the expression of equation (4.109) for the sequence components of fault current at bus k can be generalized as:

I¯k(0) (F ) = 0 172

Figure 4.65: LL fault between phase ‘b’ and phase ‘c’ of k th bus and

I¯k(1) (F ) =

V¯k (0) = −I¯k(2) (F ) (0) (1) (2) Z¯ + Z¯ + Z¯ kk

kk

(4.123)

kk

The phase components of fault current is the calculated from equation (4.122)

Ik (F ) = Ik(b) (F ) = −Ik(c) (F )

(4.124)

(c) LLG fault: Fig. 4.66 shows an LLG fault involving phases phase ‘b’ and phase ‘c’ of bus ‘k’ ¯f . Referring to equation (4.119) and Fig.4.66, the generalized expression through an impedance Z

Figure 4.66: LLG fault involving phase ‘b’ and phase ‘c’ of k th bus for sequence components of fault current at bus k can be written as

I¯k(1) (F ) =

V¯k (0)

(2) (0) Z¯kk (Zkk + 3Z¯f ) (1) ¯ Zkk + (2) Z¯ + Z¯ (0) + 3Z¯f kk

173

kk

(1) ¯(1) V¯k (0) − Z¯kk Ik (F ) (2) ¯ Ik (F ) = − (2) Z¯kk

(4.125)

(1) ¯(1) V¯k (0) − Z¯kk Ik (F ) (0) ¯ Ik (F ) = − (0) Z¯kk + 3Z¯f

The phase currents can be obtained from equation (4.122), the fault current is then calculated as

Ik (F ) = Ik(b) (F ) + Ik(c) (F )

4.10.1

(4.126)

Calculation of Bus voltages and Line currents during fault:

To calculate the voltages of buses during fault equation (4.94) can be generalized as:

V¯i(0) (F ) = −Z¯ik(0) I¯k(0) (F ) V¯i(1) (F ) = V¯i(1) (0) − Z¯ik(1) I¯k(1) (F )

(4.127)

V¯i(2) (F ) = −Z¯ik(2) I¯k(2) (F ) The pre fault voltage Vi1 (0)is usually set as 1.0 ∠00 pu. The bus phase voltage during fault is calculated from the following relation.

¯ [V¯i(012) (F )] V¯i(abc) (F ) = [A]

(4.128)

¯ is the symmetrical component transformation matrix. where [A] To calculate the symmetrical components of line currents in the line from bus i to bus j the following relation is used: (0) ij

I¯ (F ) = I¯ij(1) (F ) = (2) ij

I¯ (F ) = (0)

(1)

V¯i(0) (F ) − V¯j(0) (F ) z¯ij(0)

V¯i(1) (F ) − V¯j(1) (F ) z¯ij(1)

(4.129)

V¯i(2) (F ) − V¯j(2) (F ) z¯ij(2)

(2)

where z¯ij , z¯ij and z¯ij are the zero, positive and negative sequence impedance respectively of the line between bus i and bus j. The phase currents for the line can be calculated from the symmetrical components using the relation: 174

¯ I¯ij012 (F )] [I¯ijabc (F )] = [A][

(4.130)

The process of fault analysis of a power system network is illustrated in the next lecture with the help of an example.

175

4.11

Example of fault calculation for three phase and LG

faults in power system network A single line diagram of a power system is shown in Fig. 4.67 and the system data is as follows:-

¯1 = X ¯ 2 = 0.2 pu, X ¯ 0 = 0.05 pu • Generators G1 and G2 : X ¯1 = X ¯2 = X ¯0 = X ¯ ` = 0.05 pu • Transformers T1 and T2 : X ¯1 = X ¯ 2 = 0.1 pu, X ¯ 0 = 0.3 pu • Transmission Lines L1 , L2 and L3 : X

Figure 4.67: Single line diagram of the power System of the example Prefault voltage for all buses is taken as V¯i (0) = 1.0∠00 pu ∀ i = 1, 2, 3. We wish to carry out the complete short-circuit analysis of the system for: (a) three phase bolted fault at bus 5

¯f = 0.1 pu at bus 5 (b) LG fault with Z ¯f = 0.1 pu at bus 5 (c) LL fault with Z ¯f = 0.0 pu at bus 5 (d) LLG fault with Z Solution: (a) Three phase fault at bus 5 For the three phase bolted fault, only positive sequence network and the positive sequence bus ¯ (1) impedance matrix [Z Bus ] is required. The positive sequence network for the power system of Fig. 4.67 is shown in Fig. 4.68. In this diagram all the elements have been replaced by their per unit positive sequence impedances.

¯ (1) The [Z Bus ] matrix for the network of the Fig. 4.68 is given below: 176

Figure 4.68: Positive sequence equivalent network of Fig. 4.67

1 (1) Bus

¯ [Z

2

]=3 4 5

⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

1

2

3

j0.1294 j0.0706 j0.1118 j0.0882 j0.10

j0.0706 j0.1294 j0.0882 j0.1118 j0.10

j0.1118 j0.0882 j0.1397 j0.1103 j0.1250

4

5

j0.0882 j0.10 ⎤⎥ ⎥ j0.1118 j0.10 ⎥⎥ ⎥ j0.1103 j0.1250 ⎥⎥ pu j0.1397 j0.1250 ⎥⎥ ⎥ j0.1250 j0.1750 ⎥⎦

The sequence component of three phase fault current at bus 5 are given as, from equation (4.73):

⎡ 0 ⎤ ⎡ ⎥ ⎢ 0 ⎤⎥ ⎡ ⎤ ⎢ 0 ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ 1 ⎢ ⎥ ⎥ ⎢ ⎢ 1 ⎥ ⎢ (012) ⎥ ⎢ ¯ [I5 (F)] = ⎢ ¯ 1 ⎥ = ⎢ ⎥ = ⎢−j5.7143⎥⎥ pu ⎥ ⎢ Z55 ⎥ ⎢⎢ j0.1750 ⎥⎥ ⎢ ⎢ ⎥ ⎢ ⎥ ⎢ 0 ⎥ ⎦ ⎢ 0 ⎥ ⎣ 0 ⎦ ⎣ ⎦ ⎣ The phase components of the fault current are calculated using equation (4.122):

¯ ¯I5(012) (F)] [¯I5(abc) (F)] = [A][ ⎡1 1 1 ⎤ ⎡ ⎤ ⎡5.7143∠ − 900 ⎤ 0 ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ (abc) (abc) [¯Ifault ] = [¯I5 (F)] = ⎢⎢1 a2 a ⎥⎥ ⎢⎢−j5.7143⎥⎥ = ⎢⎢ 5.7143∠1500 ⎥⎥ pu ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢1 a a2 ⎥ ⎢ ⎥ ⎢ 5.7143∠300 ⎥ 0 ⎣ ⎦⎣ ⎦ ⎣ ⎦ Bus voltages during fault Bus 1: 177

(1) ¯(1) V¯1(1) (F ) = V¯1(1) (0) − Z¯15 I5 (F )

= 1.0 − j0.10 ∗ (−j5.7143) = 0.42857∠00 pu (a) (1) Since this is a balanced fault, V¯1 (F ) = V¯1 (F )

⎡ 0.42857∠00 ⎤ ⎥ ⎢ ⎥ ⎢ (abc) 0 ⎢ [V1 (F)] = ⎢0.42857∠ − 120 ⎥⎥ pu ⎥ ⎢ ⎢ 0.42857∠1200 ⎥ ⎦ ⎣ Bus 2:

(1) ¯(1) V¯2(1) (F ) = V¯2(1) (0) − Z¯25 I5 (F )

= 1.0 − j0.10 ∗ (−j5.7143) = 0.42857∠00 pu (a) (1) Since this is a balanced fault V¯2 (F ) = V¯2 (F )

⎡ 0.42857∠00 ⎤ ⎢ ⎥ ⎢ ⎥ (abc) [V2 (F)] = ⎢⎢0.42857∠ − 1200 ⎥⎥ pu ⎢ ⎥ ⎢ 0.42857∠1200 ⎥ ⎣ ⎦ Bus 3:

(1) ¯(1) V¯3(1) (F ) = V¯3(1) (0) − Z¯35 I5 (F )

= 1.0 − j0.125 ∗ (−j5.7143) = 0.28571∠00 pu (a)

(1)

Since this is a balanced fault V¯3 (F ) = V¯3 (F )

⎡ 0.28571∠00 ⎤ ⎢ ⎥ ⎢ ⎥ (abc) 0 ⎢ [V3 (F)] = ⎢0.28571∠ − 120 ⎥⎥ pu ⎢ ⎥ ⎢ 0.28571∠1200 ⎥ ⎣ ⎦ Bus 4: 178

(1) ¯(1) V¯4(1) (F ) = V¯4(1) (0) − Z¯45 I5 (F )

= 1.0 − j0.125 ∗ (−j5.7143) = 0.28571∠00 pu (a) (1) Since this is a balanced fault V¯4 (F ) = V¯4 (F )

⎡ 0.28571∠00 ⎤ ⎥ ⎢ ⎥ ⎢ (abc) [V4 (F)] = ⎢⎢0.28571∠ − 1200 ⎥⎥ pu ⎥ ⎢ ⎢ 0.28571∠1200 ⎥ ⎦ ⎣ ⎡0⎤ ⎢ ⎥ ⎢ ⎥ (abc) ¯ 5 (F) = ⎢⎢0⎥⎥ pu because the fault The bus voltage of bus 5 under faulted condition is V ⎢ ⎥ ⎢0⎥ ⎣ ⎦ impedance is zero.

Line Currents during fault (1)

For line L1 from bus 3 to bus 4 the positive sequence component for line current (I¯34 (F )) is calculated as:

V¯ (1) (F ) − V¯4(1) (F ) 0.28571 − 0.28571 (1) = =0 I¯34 (F ) = 3 (1) j0.1 Z¯34 Hence, the phase components of line current are

⎡0⎤ ⎢ ⎥ ⎢ ⎥ (abc) ¯I34 (F) = ⎢⎢0⎥⎥ pu ⎢ ⎥ ⎢0⎥ ⎣ ⎦ (1) For line L2 from bus 3 to bus 5 the positive sequence component for line current (I¯35 (F )) is calculated as:

V¯3(1) (F ) − V¯5(1) (F ) 0.28571 − 0.0 (1) ¯ I35 (F ) = = = 2.8571∠ − 900 pu (1) j0.1 Z¯35 Hence, the phase components of line current are

⎡2.8571∠ − 900 ⎤ ⎢ ⎥ ⎢ ⎥ (abc) 0 ⎢ [I35 (F)] = ⎢ 2.8571∠150 ⎥⎥ pu ⎢ ⎥ ⎢ 2.8571∠300 ⎥ ⎣ ⎦ 179

(1) For line L3 from bus 4 to bus 5 the positive sequence component for line current (I¯45 (F )) is calculated as:

V¯4(1) (F ) − V¯5(1) (F ) 0.28571 − 0.0 (1) ¯ = I45 (F ) = = 2.8571∠ − 900 pu (1) ¯ j0.1 Z45 Hence, the phase components of line current are

⎡2.8571∠ − 900 ⎤ ⎥ ⎢ ⎥ ⎢ (abc) 0 ⎢ [I45 (F)] = ⎢ 2.8571∠150 ⎥⎥ pu ⎥ ⎢ ⎢ 2.8571∠300 ⎥ ⎦ ⎣ Transformer Currents during fault For transformer T1 between bus 1 and bus 3 the positive sequence component fault current (1) (I¯13 (F )) is calculated as:

V¯1(1) (F ) − V¯3(1) (F ) 0.42857 − 0.28571 (1) ¯ I13 (F ) = = = 2.8571∠ − 900 pu (1) j0.05 z¯T1 The phase components of the transformer T1 current are:

⎡2.8571∠ − 900 ⎤ ⎢ ⎥ ⎢ ⎥ (abc) 0 ⎢ [¯I31 (F)] = ⎢ 2.8571∠150 ⎥⎥ pu ⎢ ⎥ ⎢ 2.8571∠300 ⎥ ⎣ ⎦ For transformer T2 between bus 2 and bus 4 the positive sequence component fault current (1) (I¯24 (F )) is calculated as:

V¯ (1) (F ) − V¯4(1) (F ) 0.42857 − 0.28571 (1) = 2.8571∠ − 900 pu I¯24 (F ) = 2 = (1) j0.05 z¯T2 The phase components of the transformer T2 current are:

⎡2.8571∠ − 900 ⎤ ⎢ ⎥ ⎢ ⎥ (abc) 0 ⎢ ¯ [I24 (F)] = ⎢ 2.8571∠150 ⎥⎥ pu ⎢ ⎥ ⎢ 2.8571∠300 ⎥ ⎣ ⎦ Generator Currents during fault (1) For generator G1 connected at bus 1 the positive sequence component fault current (I¯G1 (F )) is calculated as:

180

E¯a − V¯1(1) (F ) 1.0 − 0.42857 I¯G(1)1 (F ) = = = 2.8571∠ − 900 pu (1) j0.2 z¯G2 The phase components of the generator G1 current are:

⎡2.8571∠ − 900 ⎤ ⎥ ⎢ ⎥ ⎢ (abc) 0 ⎢ [¯IG1 (F)] = ⎢ 2.8571∠150 ⎥⎥ pu ⎥ ⎢ ⎢ 2.8571∠300 ⎥ ⎦ ⎣ (1) For Generator G2 connected at bus 2 the positive sequence component fault current (I¯G2 (F )) is calculated as:

Figure 4.69: Flow of fault current in the network

E¯a − V¯2(1) (F ) 1.0 − 0.42857 I¯G(1)2 (F ) = = = 2.8571∠ − 900 pu (1) j0.2 z¯G2 The phase components of the generator G2 current can be calculated as:

⎡2.8571∠ − 900 ⎤ ⎢ ⎥ ⎢ ⎥ (abc) 0 ⎢ [¯IG2 (F)] = ⎢ 2.8571∠150 ⎥⎥ pu ⎢ ⎥ ⎢ 2.8571∠300 ⎥ ⎣ ⎦ The flow of fault current in the system is shown in the single line diagram of Fig. 4.69. (b) Single line to ground fault at bus 5

181

In this case all the sequence networks are required. The positive sequence network is same as ¯ (1) the one shown in the Fig. 4.68 and [Z Bus ] is identical to the matrix used in three phase fault analysis. The negative sequence equivalent network for this network is as shown in Fig. 4.70. The network

Figure 4.70: Negative sequence equivalent network (2)

(1)

¯ Bus ] = [Z ¯ Bus ]. is identical to the network of Fig. 4.68 except for the voltage sources. Hence, [Z The zero sequence equivalent network is drawn next considering the transformer connections and grounding as well as generator grounding. The equivalent zero sequence networks is shown in Fig. 4.71.

Figure 4.71: Zero sequence equivalent network An explanation of the equivalent circuit will be in order. Generators G1 and G2 have their neutrals grounded, so their zero sequence impedances are connected to the reference. Transformer T1 has both the windings connected in star, with both neutrals solidly grounded. As a result, the zero sequence impedance of the transformer is directly connected between buses 1 and 2. 182

Transformer T2 has both the winding connected in delta, hence, no connection exists between the primary and secondary sides for zero sequence currents to flow. To represent circulating zero sequence currents in the delta connected transformer winding, it is represented as a short circuited winding.

¯ (0) [Z Bus ], the zero sequence bus impedance matrix is then calculated using the step-by-step ZBus building algorithm. The zero sequence bus impedance matrix is given as: 1 1 (0) Bus

¯ [Z

2

]=3 4 5

⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

2

3

4

5

j0.05 0.0 j0.05 j0.05 j0.05 ⎤⎥ ⎥ 0.0 j0.05 0.0 0.0 0.0 ⎥⎥ ⎥ j0.05 0.0 j0.10 j0.10 j0.10 ⎥⎥ pu j0.05 0.0 j0.10 j0.30 j0.20 ⎥⎥ ⎥ j0.05 0.0 j0.10 j0.20 j0.30 ⎥⎦

Fault current: The sequence component of the fault current at bus 5 are given as, from equation (4.121):

I¯5(0) (F ) = I¯5(1) (F ) = I¯5(2) (F ) =

V¯k (0) 1.0 = −j1.538 pu = (1) (2) (0) j0.175 + j0.175 + j0.30 Z¯55 + Z¯55 + Z¯55

⎡1 1 1 ⎤ ⎡−j1.538⎤ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ (abc) (abc) [¯Ifault ] = [¯I5 (F)] = ⎢⎢1 a2 a ⎥⎥ ⎢⎢−j1.538⎥⎥ ⎢ ⎥⎢ ⎥ ⎢1 a a2 ⎥ ⎢−j1.538⎥ ⎣ ⎦⎣ ⎦ ⎡4.6154∠ − 900 ⎤ ⎢ ⎥ ⎢ ⎥ (abc) ⎢ ⎥ pu ¯ [Ifault ] = ⎢ 0 ⎥ ⎢ ⎥ ⎢ ⎥ 0 ⎣ ⎦ Bus voltages: The bus voltage in sequence components, during fault, are calculated using equation (4.127) written in compact form as:

⎡V¯ (0) (F )⎤ ⎡ 0 ⎤ ⎡Z¯ (0) 0 0 ⎤⎥ ⎡⎢I¯k(0) (F )⎤⎥ ⎢ i ⎥ ⎢ ⎥ ⎢ ik ⎥ ⎢ ⎥ ⎢ ⎢ (1) ⎥⎢ ⎥ ⎢V¯ (F )⎥ = ⎢V¯i ⎥ − ⎢ 0 Z¯ (1) 0 ⎥ ⎢I¯(1) (F )⎥ ⎥ ⎢ ⎥ ⎢ ⎢ i ⎥ ⎢ ⎥ ik k ⎥ ⎢ ⎥ ⎢ ⎢ ¯ (2) ⎥ (2) ⎥ ⎢ ¯(2) ¯ ⎢Vi (F )⎥ ⎢ 0 ⎥ ⎢ 0 0 Zik ⎥⎦ ⎢⎣Ik (F )⎥⎦ ⎣ ⎦ ⎣ ⎦ ⎣

(4.131)

¯ (0) , Z¯ (1) and Z¯ (2) are the elements of the where, k represents the faulted bus number and Z ik ik ik respective sequence bus impedance matrices. I¯k(0) , I¯k(1) and I¯k(2) represent the sequence components of fault current at kth bus. V¯i (0) is the pre fault bus voltage of ith bus. Bus 1: The sequence voltages are: 183

⎡V¯ (0) (F )⎤ ⎡ 0 ⎤ ⎡j0.05 0 0 ⎤⎥ ⎡⎢−j1.538⎤⎥ ⎥ ⎢ ⎥ ⎢ ⎢ 1 ⎥ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎢ (1) ⎢V¯ (F )⎥ = ⎢1.0⎥ − ⎢ 0 j0.10 0 ⎥⎥ ⎢⎢−j1.538⎥⎥ ⎥ ⎢ ⎥ ⎢ ⎢ 1 ⎥ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎢ ¯ (2) ⎢V1 (F )⎥ ⎢ 0 ⎥ ⎢ 0 0 j0.10⎥⎦ ⎢⎣−j1.538⎥⎦ ⎦ ⎣ ⎦ ⎣ ⎣ Or,

⎡−0.0769⎤ ⎥ ⎢ ⎥ ⎢ (012) ⎢ ¯ [V1 (F)] = ⎢ 0.8462 ⎥⎥ pu ⎥ ⎢ ⎢−0.1538⎥ ⎦ ⎣

The bus voltage in the phase form is calculated using equation (4.128).

⎤ ⎡ 0.6154∠00 ⎥ ⎢ ⎥ ⎢ (abc) ¯ 1 (F)] = ⎢⎢0.9638∠ − 116.040 ⎥⎥ pu [V ⎥ ⎢ ⎢0.9638∠ − 116.040 ⎥ ⎦ ⎣ Bus 2: The sequence voltages are:

⎡V¯ (0) (F )⎤ ⎡ 0 ⎤ ⎡0.0 0 0 ⎤⎥ ⎡⎢−j1.538⎤⎥ ⎢ 2 ⎥ ⎢ ⎥ ⎢ ⎢ (1) ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢V¯ (F )⎥ = ⎢1.0⎥ − ⎢ 0 j0.10 0 ⎥⎥ ⎢⎢−j1.538⎥⎥ ⎢ 2 ⎥ ⎢ ⎥ ⎢ ⎢ ¯ (2) ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢V2 (F )⎥ ⎢ 0 ⎥ ⎢ 0 0 j0.10⎥⎦ ⎢⎣−j1.538⎥⎦ ⎣ ⎦ ⎣ ⎦ ⎣ Or,

⎡ 0.0 ⎤ ⎢ ⎥ ⎢ ⎥ (012) ⎢ ¯ 2 (F)] = ⎢ 0.8462 ⎥⎥ pu [V ⎢ ⎥ ⎢−0.1538⎥ ⎣ ⎦

The bus voltage in the phase form is calculated using equation (4.128).

⎡ ⎤ 0.6923∠00 ⎢ ⎥ ⎢ ⎥ (abc) ¯ 2 (F)] = ⎢⎢0.9326∠ − 111.790 ⎥⎥ pu [V ⎢ ⎥ ⎢0.9326∠ − 111.790 ⎥ ⎣ ⎦ Bus 3: The sequence voltages are:

⎡V¯ (0) (F )⎤ ⎡ 0 ⎤ ⎡j0.10 0 0 ⎤⎥ ⎡⎢−j1.538⎤⎥ ⎢ 3 ⎥ ⎢ ⎥ ⎢ ⎢ (1) ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢V¯ (F )⎥ = ⎢1.0⎥ − ⎢ 0 j0.125 0 ⎥⎥ ⎢⎢−j1.538⎥⎥ ⎢ 3 ⎥ ⎢ ⎥ ⎢ ⎢ ¯ (2) ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢V3 (F )⎥ ⎢ 0 ⎥ ⎢ 0 0 j0.125⎥⎦ ⎢⎣−j1.538⎥⎦ ⎣ ⎦ ⎣ ⎦ ⎣ Or,

⎡−0.1538⎤ ⎢ ⎥ ⎢ ⎥ (012) ⎢ ¯ 3 (F)] = ⎢ 0.8077 ⎥⎥ pu [V ⎢ ⎥ ⎢−0.1923⎥ ⎣ ⎦ 184

The bus voltage in the phase form is calculated using equation (4.128).

⎤ ⎡ 0.4615∠00 ⎥ ⎢ ⎥ ⎢ (abc) ¯ 3 (F)] = ⎢⎢0.9813∠ − 118.050 ⎥⎥ pu [V ⎥ ⎢ ⎢0.9813∠ − 118.050 ⎥ ⎦ ⎣ Bus 4: The sequence voltages are:

⎡V¯ (0) (F )⎤ ⎡ 0 ⎤ ⎡j0.20 0 0 ⎤⎥ ⎡⎢−j1.538⎤⎥ ⎥ ⎢ ⎥ ⎢ ⎢ 4 ⎥ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎢ (1) ⎥ ⎢−j1.538⎥ ⎢V¯ (F )⎥ = ⎢1.0⎥ − ⎢ 0 j0.125 0 ⎥ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎢ 4 ⎥ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎢ ¯ (2) ⎥ ⎢ ⎥ ⎢V4 (F )⎥ ⎢ 0 ⎥ ⎢ 0 −j1.538 0 j0.125 ⎦ ⎦⎣ ⎦ ⎣ ⎦ ⎣ ⎣ Or,

⎡−0.3076⎤ ⎥ ⎢ ⎥ ⎢ (012) ⎢ ¯ [V4 (F)] = ⎢ 0.8077 ⎥⎥ pu ⎥ ⎢ ⎢−0.1923⎥ ⎦ ⎣

The bus voltage in the phase form is calculated using equation (4.128).

⎡ ⎤ 0.3077∠00 ⎢ ⎥ ⎢ ⎥ (abc) 0 ⎢ ¯ 4 (F)] = ⎢1.0624∠ − 125.40 ⎥⎥ pu [V ⎢ ⎥ ⎢1.0624∠ − 125.400 ⎥ ⎣ ⎦ Bus 5: The sequence voltages are:

⎡V¯ (0) (F )⎤ ⎡ 0 ⎤ ⎡j0.30 0 0 ⎤⎥ ⎡⎢−j1.538⎤⎥ ⎢ 5 ⎥ ⎢ ⎥ ⎢ ⎢ (1) ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎥ ⎢−j1.538⎥ ⎢V¯ (F )⎥ = ⎢1.0⎥ − ⎢ 0 j0.175 0 ⎢ 5 ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ¯ (2) ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢V5 (F )⎥ ⎢ 0 ⎥ ⎢ 0 ⎥ ⎢−j1.538⎥ 0 j0.175 ⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎣ ⎦ Or,

⎡−0.4615⎤ ⎢ ⎥ ⎢ ⎥ (012) ⎢ ¯ [V5 (F)] = ⎢ 0.7308 ⎥⎥ pu ⎢ ⎥ ⎢−0.2692⎥ ⎣ ⎦

The bus voltage in the phase form is calculated using equation (4.128)

⎡ ⎤ 0.0∠00 ⎢ ⎥ ⎢ ⎥ (abc) 0 ⎢ ¯ 5 (F)] = ⎢1.087∠ − 128.64 ⎥⎥ pu [V ⎢ ⎥ ⎢1.087∠ − 128.640 ⎥ ⎣ ⎦ Observe that the phase voltage of the faulted phase ’a’ is zero due to a zero impedance fault. Line Currents 185

The sequence components of line currents during fault are calculated using equation (4.129), written here in compact form as

⎡ 1 ⎤ ⎢ (0) 0 0 ⎥⎥ ⎢ ⎥ ⎡V¯ (0) (F ) − V¯ (0) (F )⎤ ⎡I¯(0) (F )⎤ ⎢ z¯ij ⎥ ⎥ ⎢ ⎥⎢ i ⎢ ij j 1 ⎥ ⎥ ⎢ ⎥ ⎢ (1) ⎢ (1) (1) ⎢I¯ (F )⎥ = ⎢ 0 0 ⎥⎥ ⎢⎢V¯i (F ) − V¯j (F )⎥⎥ (1) ⎥ ⎢ ⎢ ij z¯ij ⎥ ⎥ ⎢ ⎥ ⎢ (2) ⎢ ¯(2) (2) ⎢Iij (F )⎥ ⎢ 1 ⎥⎥ ⎢⎣V¯i (F ) − V¯j (F )⎥⎦ ⎦ ⎢ ⎣ 0 ⎢ 0 ⎥ ⎢ z¯ij(2) ⎥⎦ ⎣

(4.132)

In equation (4.132), the line is between ith and jth buses.

z¯ij(0) , z¯ij(1) , z¯ij(2) represent the respective sequence impedances of the line i Ð→ j V¯i(0) (F ), V¯i(1) (F ), V¯i(2) (F ), V¯j(0) (F ), V¯j(1) (F ), V¯j(2) (F ) are the sequence components of voltages of ith and jth buses respectively during fault. Line 1: The sequence components of line current are

⎡ 1.0 ⎤ ⎢ 0 0 ⎥⎥ ⎡ ⎡I¯(0) (F )⎤ ⎢ j0.3 ⎤ ⎢ 34 ⎥ ⎢ ⎥ ⎢−0.1538 − (−0.3076)⎥ ⎢ (1) ⎥ ⎢ ⎥⎢ ⎥ 1.0 ⎢I¯ (F )⎥ = ⎢ 0 0 ⎥⎥ ⎢⎢ 0.8077 − 0.8077 ⎥⎥ ⎢ 34 ⎥ ⎢ j0.10 ⎢ ¯(2) ⎥ ⎢ ⎥⎢ ⎥ ⎢I34 (F )⎥ ⎢ 1.0 ⎥⎥ ⎢⎣−0.1923 − (−0.1923)⎥⎦ ⎣ ⎦ ⎢ 0 0 ⎢ ⎥ ⎣ j0.10 ⎦ Or,

⎡−j0.5128⎤ ⎢ ⎥ ⎢ ⎥ (012) ⎥ pu [¯I34 (F)] = ⎢⎢ 0 ⎥ ⎢ ⎥ ⎢ ⎥ 0 ⎣ ⎦

The line current in phase form is calculated as:

⎡0.5128∠ − 900 ⎤ ⎢ ⎥ ⎢ ⎥ (abc) [¯I34 (F)] = ⎢⎢0.5128∠ − 900 ⎥⎥ pu ⎢ ⎥ ⎢0.5128∠ − 900 ⎥ ⎣ ⎦ Line 2: The sequence components of line current are

⎡ 1.0 ⎤ ⎢ 0 0 ⎥⎥ ⎡ ⎤ ⎡I¯(0) (F )⎤ ⎢ j0.3 ⎢ 35 ⎥ ⎢ ⎥ ⎢−0.1538 − (−0.4615)⎥ ⎢ (1) ⎥ ⎢ ⎥⎢ ⎥ 1.0 ⎢I¯ (F )⎥ = ⎢ 0 0 ⎥⎥ ⎢⎢ 0.8077 − 0.7308 ⎥⎥ ⎥ ⎢ ⎢ 35 j0.10 ⎢ ¯(2) ⎥ ⎢ ⎥⎢ ⎥ ⎢I35 (F )⎥ ⎢ 1.0 ⎥⎥ ⎢⎣−0.1923 − (−0.2692)⎥⎦ ⎣ ⎦ ⎢ 0 0 ⎢ ⎥ ⎣ j0.10 ⎦ 186

Or,

⎡−j1.0256⎤ ⎥ ⎢ ⎥ ⎢ (012) ⎢ ¯ [I35 (F)] = ⎢−j0.7692⎥⎥ pu ⎥ ⎢ ⎢−j0.7692⎥ ⎦ ⎣

The line current in phase form is calculated as:

⎡2.5641∠ − 900 ⎤ ⎥ ⎢ ⎥ ⎢ (abc) [¯I35 (F)] = ⎢⎢0.2564∠ − 900 ⎥⎥ pu ⎥ ⎢ ⎢0.2564∠ − 900 ⎥ ⎦ ⎣ Line 3: The sequence components of line current are

⎤ ⎡ 1.0 ⎢ 0 0 ⎥⎥ ⎡ ⎡I¯(0) (F )⎤ ⎢ j0.3 ⎤ ⎢ 45 ⎥ ⎢−0.3077 − (−0.4615)⎥ ⎥ ⎢ ⎥ ⎢ (1) ⎥⎢ ⎥ ⎢ 1.0 ⎢I¯ (F )⎥ = ⎢ 0 0 ⎥⎥ ⎢⎢ 0.8077 − 0.7308 ⎥⎥ ⎢ 45 ⎥ ⎢ j0.10 ⎢ ¯(2) ⎥⎢ ⎥ ⎥ ⎢ ⎢I45 (F )⎥ ⎢ 1.0 ⎥⎥ ⎢⎣−0.1923 − (−0.2692)⎥⎦ ⎣ ⎦ ⎢ 0 0 ⎥ ⎢ ⎣ j0.10 ⎦ Or,

⎡−j0.5128⎤ ⎢ ⎥ ⎢ ⎥ (012) [¯I45 (F)] = ⎢⎢−j0.7692⎥⎥ pu ⎢ ⎥ ⎢−j0.7692⎥ ⎣ ⎦

The line current in phase form is calculated as:

⎡2.0513∠ − 900 ⎤ ⎢ ⎥ ⎢ ⎥ (abc) 0 ⎢ ¯ [I45 (F)] = ⎢0.2564∠ − 90 ⎥⎥ pu ⎢ ⎥ ⎢0.2564∠ − 900 ⎥ ⎣ ⎦ Transformer Currents

Transformer T1 ∶ The sequence components of line current are ⎤ ⎡ 1.0 ⎢ 0 0 ⎥⎥ ⎡ ⎡I¯(0) (F )⎤ ⎢ j0.05 ⎤ ⎢ 13 ⎥ ⎢ ⎥ ⎢−0.0769 − (−0.1538)⎥ ⎢ (1) ⎥ ⎥ ⎢ ⎢ ⎥ 1.0 ⎢I¯ (F )⎥ = ⎢ 0 ⎥ ⎢ 0.8462 − 0.8077 ⎥ 0 ⎢ 13 ⎥ ⎢ ⎥⎢ ⎥ j0.05 ⎢ ¯(2) ⎥ ⎢ ⎥⎢ ⎥ ⎢I13 (F )⎥ ⎢ ⎥ ⎢ ⎥ −0.1538 − (−0.1923) 1.0 ⎣ ⎥⎣ ⎦ ⎢ 0 ⎦ 0 ⎢ ⎥ ⎣ j0.05 ⎦ Or,

⎡ −j1.538 ⎤ ⎢ ⎥ ⎢ ⎥ (012) ⎢ [¯I13 (F)] = ⎢−j0.7692⎥⎥ pu ⎢ ⎥ ⎢−j0.7692⎥ ⎣ ⎦ 187

The line current in phase form is calculated as:

⎡3.0769∠ − 900 ⎤ ⎥ ⎢ ⎥ ⎢ (abc) [¯I13 (F)] = ⎢⎢0.7692∠ − 900 ⎥⎥ pu ⎥ ⎢ ⎢0.7692∠ − 900 ⎥ ⎦ ⎣ Transformer T2 ∶ The sequence components of line current are ⎡ 1.0 ⎤ ⎢ 0 0 ⎥⎥ ⎡ ⎤ ⎡I¯(0) (F )⎤ ⎢ ∞ ⎥ ⎢ ⎢ 24 ⎥ ⎢ 0 − (−0.3076) ⎥ 1.0 ⎥ ⎥ ⎢ ⎢ (1) ⎥⎢ ⎢I¯ (F )⎥ = ⎢ 0 0 ⎥⎥ ⎢⎢ 0.8462 − 0.8077 ⎥⎥ ⎥ ⎢ ⎢ 24 j0.05 ⎥ ⎥ ⎢ ⎢ ¯(2) ⎥⎢ ⎢I24 (F )⎥ ⎢ ⎥ ⎢−0.1538 − (−0.1923)⎥ 1.0 ⎦ ⎦ ⎢ 0 ⎣ ⎥⎣ 0 ⎢ j0.05 ⎥⎦ ⎣ ⎡ ⎤ 0 ⎢ ⎥ ⎢ ⎥ (012) ⎢ ¯ [I24 (F)] = ⎢−j0.7692⎥⎥ pu ⎢ ⎥ ⎢−j0.7692⎥ ⎣ ⎦ The line current in phase form is calculated as:

⎡ 1.538∠ − 900 ⎤ ⎢ ⎥ ⎢ ⎥ (abc) [¯I24 (F)] = ⎢⎢0.7692∠ − 900 ⎥⎥ pu ⎢ ⎥ ⎢0.7692∠ − 900 ⎥ ⎣ ⎦ Generator Currents The sequence components of generator currents during fault are calculated using the expression

⎡ 1 ⎤ ⎢ (0) 0 0 ⎥⎥ ⎢ ⎡I¯(0) (F )⎤ ⎢ z¯gi ⎥ ⎡E¯ (0) (F ) − V¯ (0) (F )⎤ ⎢ Gi ⎥ ⎢ ⎥ ⎢ Gi ⎥ ti 1 ⎢ (1) ⎥ ⎢ ⎥ ⎢ (1) ⎥ (1) ⎢I¯ (F )⎥ = ⎢ 0 ⎥ ⎢ ¯ ¯ 0 EGi (F ) − Vti (F )⎥⎥ (1) ⎢ Gi ⎥ ⎢ ⎥ ⎢ z¯gi ⎢ ¯(2) ⎥ ⎢ ⎥ ⎢ (2) ⎥ (2) ⎢IGi (F )⎥ ⎢ 1 ⎥⎥ ⎢⎣E¯Gi (F ) − V¯ti (F )⎥⎦ ⎣ ⎦ ⎢ 0 ⎢ 0 ⎥ (2) ⎥ ⎢ z¯gi ⎣ ⎦

(4.133)

where, (0) (1) (2) E¯Gi (F ), E¯Gi (F ), E¯Gi (F ) the zero, positive and negative sequence generated voltages respecth tively of i generator. V¯ti(0) (F ), V¯ti(1) (F ), V¯ti(2) (F ) are the zero, positive and negative sequence terminal voltages respectively of ith generator after fault. (0) (1) (2) z¯gi (F ), z¯gi (F ) and z¯gi (F ) are the sequence impedances of the ith generator.

Generator 1 : The sequence components of generator 1 current are 188

⎡ 1.0 ⎤ ⎢ ⎥ 0 0 ⎥ ⎡0 − (−0.0769)⎤ ⎡I¯(0) (F )⎤ ⎢ j0.05 ⎥ ⎥ ⎢ ⎢ G1 ⎥⎢ ⎥ ⎥ ⎢ ⎢ (1) ⎥⎢ 1.0 ⎢I¯ (F )⎥ = ⎢ 0 0 ⎥⎥ ⎢⎢ 1 − 0.8462 ⎥⎥ ⎥ ⎢ ⎢ G1 j0.20 ⎥ ⎥ ⎢ ⎢ ¯(2) ⎥⎢ ⎢IG1 (F )⎥ ⎢ 1.0 ⎥⎥ ⎢⎣0 − (−0.1538)⎥⎦ ⎦ ⎢ 0 ⎣ 0 ⎢ ⎥ ⎣ j0.20 ⎦ ⎡ −j1.538 ⎤ ⎥ ⎢ ⎥ ⎢ (012) [¯IG1 (F)] = ⎢⎢−j0.7692⎥⎥ pu ⎥ ⎢ ⎢−j0.7692⎥ ⎦ ⎣ The phase components generator current are calculated as:

⎡3.0769∠ − 900 ⎤ ⎥ ⎢ ⎥ ⎢ (abc) 0 ⎢ ¯ [IG1 (F)] = ⎢0.7692∠ − 90 ⎥⎥ pu ⎥ ⎢ ⎢0.7692∠ − 900 ⎥ ⎦ ⎣ Generator 2 : The sequence components of Generator 1 current are

⎡ 1.0 ⎤ ⎢ ⎥ 0 0 ⎡I¯(0) (F )⎤ ⎢ j0.05 ⎥⎡ ⎤ 0−0 ⎢ G2 ⎥ ⎢ ⎥⎢ ⎥ ⎢ (1) ⎥ ⎢ ⎥⎢ ⎥ 1.0 ⎢I¯ (F )⎥ = ⎢ 0 0 ⎥⎥ ⎢⎢ 1 − 0.8462 ⎥⎥ ⎢ G2 ⎥ ⎢ j0.20 ⎢ ¯(2) ⎥ ⎢ ⎥⎢ ⎥ ⎢IG2 (F )⎥ ⎢ 1.0 ⎥⎥ ⎢⎣0 − (−0.1538)⎥⎦ ⎣ ⎦ ⎢ 0 0 ⎢ ⎥ ⎣ j0.20 ⎦ ⎡ ⎤ 0 ⎢ ⎥ ⎢ ⎥ (012) [¯IG2 (F)] = ⎢⎢−j0.7692⎥⎥ pu ⎢ ⎥ ⎢−j0.7692⎥ ⎣ ⎦ The phase components generator current are calculated as:

⎡ 1.538∠ − 900 ⎤ ⎢ ⎥ ⎢ ⎥ (abc) 0 ⎢ ¯ [IG2 (F)] = ⎢0.7692∠ − 90 ⎥⎥ pu ⎢ ⎥ ⎢0.7692∠ − 900 ⎥ ⎣ ⎦ The flow of sequence currents in the sequence networks is shown next in the Fig. 4.72. From Fig. 4.72 the following points are worth observing: • Both generators contribute equal amount of positive and negative sequence currents as the network is symmetrical as seen from the fault point. • Since the positive and negative sequence fault voltages are equal for buses 3 and 4, the positive and negative sequence currents through line L1 between buses 3 and 4 are zero. 189

Figure 4.72: Flow of sequence currents for LG fault at bus 5 • The zero sequence circuit of generator G2 is open circuited due to ∆ − ∆ transformer T2 as a result, G2 does not contribute any zero sequence current to the fault. Generator G1 has to provide the entire zero sequence current. 190

• Moreover, the zero sequence network is not symmetrical, hence, zero sequence voltages of buses 3 and 4 are not equal and as a result a zero sequence current flows through line L1 . In the next lecture, we will look into the examples of short circuit fault calculation for LL and LLG faults.

191

4.12

Example of fault calculation for LL and LLG fault

in power system network (b) Double line (LL) fault between phases ‘b’ and ‘c’ at bus 5 (1)

(2)

¯ Bus ] and [Z ¯ Bus ] In this case only positive and negative sequence networks are required. Hence [Z ¯f = j0.1 pu. as calculated previously will be used. Let the fault impedance be Z Fault current calculations: The sequence components of fault current are calculated using equation (4.123).

I¯5(1) (F ) = 0 As zero sequence current can not flow without a ground path.

I¯5(1) (F ) =

1.0 V¯5(0) = = −j2.2222 pu (1) (2) Z¯55 + Z¯55 + Z¯f j0.175 + j0.175 + j0.1 I¯5(2) (F ) = −I¯5(1) (F ) = j2.222 pu

Hence

⎡ 0 ⎤ ⎢ ⎥ ⎢ ⎥ (012) ⎢ ¯ [I5 (F)] = ⎢−j2.222⎥⎥ pu ⎢ ⎥ ⎢ j2.222 ⎥ ⎣ ⎦ The phase components of the fault current are

⎡ 0 ⎤ ⎢ ⎥ ⎢ ⎥ (abc) ⎢ ¯ [I5 (F)] = ⎢−3.849⎥⎥ pu ⎢ ⎥ ⎢ 3.849 ⎥ ⎣ ⎦ Bus voltage calculations: Bus 1: The sequence components of bus 1 voltage are calculated as 192

Zero Sequence

V¯1(0) (F ) = 0.0 pu Positive Sequence (1) ¯(1) V¯1(1) (F ) = V¯1(1) (0) − Z¯15 I5 (F ) = 1.0 − j0.10 ∗ (−j2.222)

V¯1(1) (F ) = 0.7778 pu Negative Sequence (2) ¯(2) V¯1(2) (F ) = −Z¯15 I5 (F ) = −j0.10 ∗ (j2.222)

V¯1(2) (F ) = 0.2222 pu Hence, the sequence components of bus 1 voltage are:

⎡ 0 ⎤ ⎢ ⎥ ⎢ ⎥ 012 ⎢ ¯ 1 (F)] = ⎢0.7778⎥⎥ pu [V ⎢ ⎥ ⎢0.2222⎥ ⎣ ⎦ bus 1 voltage in phase form

⎡ ⎤ 1.0 ⎢ ⎥ ⎢ ⎥ (abc) 0 ⎢ ¯ [V(1) (F)] = ⎢0.6939∠ − 136 ⎥⎥ pu ⎢ ⎥ ⎢ 0.6939∠1360 ⎥ ⎣ ⎦ Bus 2: The sequence components of bus 2 voltage are calculated as Zero Sequence

V¯2(0) (F ) = 0.0 pu Positive Sequence (1) ¯(1) V¯2(1) (F ) = V¯2(1) (0) − Z¯25 I5 (F ) = 1.0 − j0.10 ∗ (−j2.222)

V¯2(1) (F ) = 0.7778 pu Negative Sequence

(2) ¯(2) V¯2(2) (F ) = −Z¯25 I5 (F ) = −j0.10 ∗ (j2.222)

193

V¯2(2) (F ) = 0.2222 pu Hence, the sequence components of bus 2 voltage are:

⎡ 0 ⎤ ⎥ ⎢ ⎥ ⎢ 012 ⎢ ¯ [V2 (F)] = ⎢0.7778⎥⎥ pu ⎥ ⎢ ⎢0.2222⎥ ⎦ ⎣ bus 2 voltage in phase form

⎤ ⎡ 1.0 ⎥ ⎢ ⎥ ⎢ (abc) 0 ⎥ ⎢ ¯ [V (2) (F)] = ⎢0.6939∠ − 136 ⎥ pu ⎥ ⎢ ⎢ 0.6939∠1360 ⎥ ⎦ ⎣ Bus 3: The sequence components of bus 3 voltage are calculated as Zero Sequence

V¯3(0) (F ) = 0.0 pu Positive Sequence (1) ¯(1) V¯3(1) (F ) = V¯3(1) (0) − Z¯35 I5 (F ) = 1.0 − j0.125 ∗ (−j2.222)

V¯3(1) (F ) = 0.7222 pu Negative Sequence (2) ¯(2) V¯3(2) (F ) = −Z¯35 I5 (F ) = −j0.125 ∗ (j2.222)

V¯3(2) (F ) = 0.2778 pu Hence, the sequence components of bus 3 voltage are:

⎡ 0 ⎤ ⎢ ⎥ ⎢ ⎥ 012 ¯ 3 (F)] = ⎢⎢0.7222⎥⎥ pu [V ⎢ ⎥ ⎢0.2778⎥ ⎣ ⎦ bus 3 voltage in phase form 194

⎤ ⎡ 1.0 ⎥ ⎢ ⎥ ⎢ (abc) ⎢0.6310∠ − 142.410 ⎥ pu ¯ [V = (F)] ⎥ ⎢ (3) ⎥ ⎢ 0 ⎢ 0.6310∠142.41 ⎥ ⎦ ⎣ Bus 4: The sequence components of bus 4 voltage are calculated as Zero Sequence

V¯4(0) (F ) = 0.0 pu Positive Sequence (1) ¯(1) V¯4(1) (F ) = V¯4(1) (0) − Z¯45 I5 (F ) = 1.0 − j0.125 ∗ (−j2.222)

V¯4(1) (F ) = 0.7222 pu Negative Sequence (2) ¯(2) V¯4(2) (F ) = −Z¯45 I5 (F ) = −j0.125 ∗ (j2.222)

V¯4(2) (F ) = 0.2778 pu Hence, the sequence components of bus 4 voltage are:

⎡ 0 ⎤ ⎢ ⎥ ⎢ ⎥ 012 ⎢ ¯ [V4 (F)] = ⎢0.7222⎥⎥ pu ⎢ ⎥ ⎢0.2778⎥ ⎣ ⎦ bus 4 voltage in phase form

⎡ ⎤ 1.0 ⎢ ⎥ ⎢ ⎥ (abc) 0 ⎢ ¯ [V(4) (F)] = ⎢0.6310∠ − 142.41 ⎥⎥ pu ⎢ ⎥ ⎢ 0.6310∠142.410 ⎥ ⎣ ⎦ Bus 5: The sequence components of bus 5 voltage are calculated as Zero Sequence

V¯5(0) (F ) = 0.0 pu Positive Sequence (1) ¯(1) V¯5(1) (F ) = V¯5(1) (0) − Z¯55 I5 (F ) = 1.0 − j0.175 ∗ (−j2.222)

195

V¯5(1) (F ) = 0.6111 pu Negative Sequence (2) ¯(2) V¯5(2) (F ) = −Z¯55 I5 (F ) = −j0.175 ∗ (j2.222)

V¯5(2) (F ) = 0.3889 pu Hence, the sequence components of bus 5 voltage are:

⎡ 0 ⎤ ⎥ ⎢ ⎥ ⎢ 012 ¯ 5 (F)] = ⎢⎢0.6111⎥⎥ pu [V ⎥ ⎢ ⎢0.3889⎥ ⎦ ⎣ bus 5 voltage in phase form

⎡ ⎤ 1.0 ⎢ ⎥ ⎥ ⎢ (abc) 0 ⎢ ¯ [V(5) (F)] = ⎢0.5358∠ − 158.95 ⎥⎥ pu ⎢ ⎥ ⎢ 0.5358∠158.950 ⎥ ⎣ ⎦ Line current calculations: Line 1: The sequence components of line current are

⎤ ⎡ 1.0 ⎥ ⎢ 0 0 ⎡I¯(0) (F )⎤ ⎢ j0.3 ⎥⎡ ⎤ 0 ⎢ 34 ⎥ ⎢ ⎥⎢ ⎥ ⎢ (1) ⎥⎢ ⎥ ⎥ ⎢ 1.0 ⎢I¯ (F )⎥ = ⎢ 0 0 ⎥⎥ ⎢⎢0.7222 − 0.7222⎥⎥ ⎢ 34 ⎥ ⎢ j0.10 ⎢ ¯(2) ⎥ ⎢ ⎥⎢ ⎥ ⎢I34 (F )⎥ ⎢ 1.0 ⎥⎥ ⎢⎣0.2778 − 0.2778⎥⎦ ⎣ ⎦ ⎢ 0 0 ⎥ ⎢ ⎣ j0.10 ⎦ ⎡0⎤ ⎢ ⎥ ⎢ ⎥ (012) [¯I34 (F)] = ⎢⎢0⎥⎥ pu ⎢ ⎥ ⎢0⎥ ⎣ ⎦ The line current in phase form is calculated as:

⎡0⎤ ⎢ ⎥ ⎢ ⎥ (abc) [¯I34 (F)] = ⎢⎢0⎥⎥ pu ⎢ ⎥ ⎢0⎥ ⎣ ⎦ Line 2: The sequence components of line current are 196

⎡ 1.0 ⎤ ⎢ ⎥ 0 0 ⎤ ⎥⎡ ⎡I¯(0) (F )⎤ ⎢ j0.3 0 ⎥ ⎢ 35 ⎥ ⎢ ⎥⎢ ⎥ ⎢ (1) ⎥ ⎢ ⎥⎢ 1.0 ⎢I¯ (F )⎥ = ⎢ 0 0 ⎥⎥ ⎢⎢0.7222 − 0.6111⎥⎥ ⎢ 35 ⎥ ⎢ j0.10 ⎥ ⎢ ¯(2) ⎥ ⎢ ⎥⎢ ⎢I35 (F )⎥ ⎢ 1.0 ⎥⎥ ⎢⎣0.2778 − 0.3889⎥⎦ ⎣ ⎦ ⎢ 0 0 ⎢ ⎥ ⎣ j0.10 ⎦ ⎤ ⎡ 0 ⎥ ⎢ ⎥ ⎢ (012) ⎢ [¯I35 (F)] = ⎢−j1.1111⎥⎥ pu ⎥ ⎢ ⎢ j1.1111 ⎥ ⎦ ⎣ The line current in phase form is calculated as:

⎡ 0 ⎤ ⎥ ⎢ ⎥ ⎢ (abc) [¯I35 (F)] = ⎢⎢−1.9245⎥⎥ pu ⎥ ⎢ ⎢ 1.9245 ⎥ ⎦ ⎣ Line 3: The sequence components of line current are

⎤ ⎡ 1.0 ⎢ 0 0 ⎥⎥ ⎡ ⎡I¯(0) (F )⎤ ⎢ j0.3 ⎤ 0 ⎥⎢ ⎢ 45 ⎥ ⎥ ⎢ ⎥⎢ ⎢ (1) ⎥ ⎥ ⎢ 1.0 ⎥ ⎢ ⎢I¯ (F )⎥ = ⎢ 0 0 ⎥ ⎢0.7222 − 0.6111⎥⎥ ⎢ 45 ⎥ ⎢ j0.10 ⎥⎢ ⎢ ¯(2) ⎥ ⎥ ⎢ ⎢I45 (F )⎥ ⎢ 1.0 ⎥⎥ ⎢⎣0.2778 − 0.3889⎥⎦ ⎣ ⎦ ⎢ 0 0 ⎥ ⎢ ⎣ j0.10 ⎦ ⎡ ⎤ 0 ⎢ ⎥ ⎢ ⎥ (012) [¯I45 (F)] = ⎢⎢−j1.1111⎥⎥ pu ⎢ ⎥ ⎢ j1.1111 ⎥ ⎣ ⎦ The line current in phase form is calculated as:

⎡ 0 ⎤ ⎢ ⎥ ⎢ ⎥ (abc) [¯I45 (F)] = ⎢⎢−1.9245⎥⎥ pu ⎢ ⎥ ⎢ 1.9245 ⎥ ⎣ ⎦ Transformer current calculations: Transformer 1: The sequence components of transformer 1 current are:

⎡ 1.0 ⎤ ⎢ 0 0 ⎥⎥ ⎡ ⎤ ⎡I¯(0) (F )⎤ ⎢ j0.05 0 ⎢ 13 ⎥ ⎢ ⎥⎢ ⎥ ⎢ (1) ⎥ ⎢ ⎥⎢ ⎥ 1.0 ⎢I¯ (F )⎥ = ⎢ 0 ⎥ ⎢ 0 ⎥ ⎢(0.7778 − 0.7222)⎥⎥ ⎢ 13 ⎥ ⎢ j0.05 ⎢ ¯(2) ⎥ ⎢ ⎥⎢ ⎥ ⎢I13 (F )⎥ ⎢ 1.0 ⎥⎥ ⎢⎣(0.2222 − 0.2778)⎥⎦ ⎣ ⎦ ⎢ 0 0 ⎢ ⎥ ⎣ j0.05 ⎦ 197

⎤ ⎡ 0 ⎥ ⎢ ⎥ ⎢ (012) [¯I13 (F)] = ⎢⎢−j1.1111⎥⎥ pu ⎥ ⎢ ⎢ j1.1111 ⎥ ⎦ ⎣ The transformer current in phase form is calculated as:

⎡ 0 ⎤ ⎥ ⎢ ⎥ ⎢ (abc) ⎢ ¯ [I13 (F)] = ⎢−1.9245⎥⎥ pu ⎥ ⎢ ⎢ 1.9245 ⎥ ⎦ ⎣ Transformer 2: The sequence components of transformer 2 current are:

⎤ ⎡ 1.0 ⎥ ⎢ 0 0 ⎥⎡ ⎡I¯ (F )⎤ ⎢ j0.05 ⎤ 0 ⎢ ⎥ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎢ ⎥ ⎥ ⎢ 1.0 ⎢I¯ (F )⎥ = ⎢ 0 0 ⎥⎥ ⎢⎢(0.7778 − 0.7222)⎥⎥ ⎢ ⎥ ⎢ j0.05 ⎥⎢ ⎢¯ ⎥ ⎥ ⎢ ⎢I (F )⎥ ⎢ 1.0 ⎥⎥ ⎢⎣(0.2222 − 0.2778)⎥⎦ ⎣ ⎦ ⎢ 0 0 ⎥ ⎢ ⎣ j0.05 ⎦ (0) 24 (1) 24 (2) 24

⎡ ⎤ 0 ⎢ ⎥ ⎢ ⎥ (012) ⎢ ¯ [I24 (F)] = ⎢−j1.1111⎥⎥ pu ⎢ ⎥ ⎢ j1.1111 ⎥ ⎣ ⎦ The transformer current in phase form is calculated as:

⎡ 0 ⎤ ⎢ ⎥ ⎢ ⎥ (abc) ⎢ ¯ [I24 (F)] = ⎢−1.9245⎥⎥ pu ⎢ ⎥ ⎢ 1.9245 ⎥ ⎣ ⎦ Generator current calculations: Generator 1: The sequence components of generator 1 current are

⎤ ⎡ 1.0 ⎢ 0 0 ⎥⎥ ⎡ ⎡I¯(0) (F )⎤ ⎢ j0.05 ⎤ 0 ⎢ G1 ⎥ ⎢ ⎥⎢ ⎥ ⎥⎢ ⎢ (1) ⎥ ⎢ ⎥ 1.0 ⎢I¯ (F )⎥ = ⎢ 0 ⎥ ⎢ 0 ⎥ ⎢(1 − 0.7778)⎥⎥ ⎢ G1 ⎥ ⎢ j0.20 ⎢ ¯(2) ⎥ ⎢ ⎥⎢ ⎥ ⎢IG1 (F )⎥ ⎢ ⎥ ⎢(0 − 0.2222)⎥ 1.0 ⎥⎣ ⎣ ⎦ ⎢ 0 ⎦ 0 ⎢ ⎥ ⎣ j0.20 ⎦ ⎡ ⎤ 0 ⎢ ⎥ ⎢ ⎥ (012) ⎢ [¯IG1 (F)] = ⎢−j1.1111⎥⎥ pu ⎢ ⎥ ⎢ j1.1111 ⎥ ⎣ ⎦ 198

The generator current in phase form is calculated as:

⎡ 0 ⎤ ⎥ ⎢ ⎥ ⎢ (abc) ⎢ [¯IG1 (F)] = ⎢−1.9245⎥⎥ pu ⎥ ⎢ ⎢ 1.9245 ⎥ ⎦ ⎣ Generator 2: The sequence components of generator 2 current are

⎡ 1.0 ⎤ ⎢ 0 0 ⎥⎥ ⎡ ⎤ ⎡I¯(0) (F )⎤ ⎢ j0.05 0 ⎥ ⎥ ⎢ ⎥⎢ ⎢ G2 ⎥ ⎥ ⎢ ⎥⎢ ⎢ (1) 1.0 ⎢ ⎥ ⎢I¯ (F )⎥ = ⎢ 0 0 ⎥ ⎢(1 − 0.7778)⎥⎥ ⎥ ⎢ ⎢ G2 j0.20 ⎥ ⎥ ⎢ ⎥⎢ ⎢ ¯(2) ⎢IG2 (F )⎥ ⎢ 1.0 ⎥⎥ ⎢⎣(0 − 0.2222)⎥⎦ ⎦ ⎢ 0 ⎣ 0 ⎢ ⎥ ⎣ j0.20 ⎦ ⎡ ⎤ 0 ⎢ ⎥ ⎢ ⎥ (012) ⎢ [¯IG2 (F)] = ⎢−j1.1111⎥⎥ pu ⎢ ⎥ ⎢ j1.1111 ⎥ ⎣ ⎦ The generator current in phase form is calculated as:

⎡ 0 ⎤ ⎢ ⎥ ⎢ ⎥ (abc) ⎢ ¯ [IG2 (F)] = ⎢−1.9245⎥⎥ pu ⎢ ⎥ ⎢ 1.9245 ⎥ ⎣ ⎦ From Fig. 5.54 it can be seen that:

V¯5(1) (F ) = V¯5(2) (F ) + Z¯f I¯5(1) (F ) = 0.3889 + j0.1 ∗ −j2.222 V¯5(1) (F ) = 0.6111pu This value is same as the one calculated earlier. The flow of sequence currents in the sequence networks is shown next in the Fig.4.73. (c) Double line to ground fault (LLG) fault involving phases ‘b’ and ‘c’ at bus 5

¯ (2) ¯ (0) ¯ (1) In this case all the three sequence networks are required. Hence, [Z Bus ],[ZBus ],and [ZBus ] as ¯f = 0. calculated earlier will be used. It is assumed that the fault impedance Z Fault current calculations: The sequence components of fault current are calculated using equation (4.125) as follows: Positive Sequence Current 199

Figure 4.73: Flow of sequence currents for LL fault at bus 5

I¯5(1) (F ) =

I¯5(1) (F ) =

(1) Z¯55 +

1.0 (2) ¯ (0) Z¯55 (Z55 + 3Z¯f )

(2) (0) Z¯55 + Z¯55 + 3Z¯f

1.0 j0.175(j0.175 + 3 ∗ 0) j0.175 + j0.175 + j0.3 + 3 ∗ 0 I¯5(1) (F ) = −j3.5023 pu

The negative and zero sequence currents are calculated using current division as: Negative Sequence Current

(0) Z¯55 ∗ I¯5(1) (F ) (0) (2) ¯ ¯ Z55 + Z55 j0.3 = − ∗ (−j3.5023) (j0.3 + j0.175)

I¯5(2) (F ) = −

I¯5(2) (F ) = j2.212 pu Zero Sequence Current 200

(2) Z¯55 ∗ I¯5(1) (F ) (0) (2) ¯ ¯ Z55 + Z55 j0.175 ∗ (−j3.5023) = − (j0.3 + j0.175)

I¯5(0) (F ) = −

I¯5(0) (F ) = j1.2903 pu Hence, the fault current in sequence components is :

⎡ j1.290 ⎤ ⎥ ⎢ ⎥ ⎢ (012) [¯I5 (F)] = ⎢⎢−j3.503⎥⎥ pu ⎥ ⎢ ⎢ j2.212 ⎥ ⎦ ⎣ The phase components of the fault current are :

⎡ ⎤ 0 ⎢ ⎥ ⎢ ⎥ (abc) 0 ⎢ [¯I5 (F)] = ⎢5.3137∠158.64 ⎥⎥ pu ⎢ ⎥ ⎢ 5.3137∠21.360 ⎥ ⎣ ⎦ The total fault current

¯ ) = I¯5(b) (F ) + I¯5(c) (F ) I(F = 5.3137∠158.640 + 5.3137∠21.360

¯ ) = 3.871∠900 pu I(F Bus Voltage Calculations: Bus 1: The sequence components of bus 1 voltage are calculated as

⎡V¯ (0) (F )⎤ ⎡ 0 ⎤ ⎡j0.05 0 0 ⎤⎥ ⎡⎢ j1.290 ⎤⎥ ⎢ 1 ⎥ ⎢ ⎥ ⎢ ⎢ (1) ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢V¯ (F )⎥ = ⎢1.0⎥ − ⎢ 0 j0.10 0 ⎥⎥ ⎢⎢−j3.503⎥⎥ ⎢ 1 ⎥ ⎢ ⎥ ⎢ ⎢ ¯ (2) ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢V1 (F )⎥ ⎢ 0 ⎥ ⎢ 0 0 j0.10⎥⎦ ⎢⎣ j2.212 ⎥⎦ ⎣ ⎦ ⎣ ⎦ ⎣ 201

⎡0.0645⎤ ⎥ ⎢ ⎥ ⎢ (012) ¯ 1 (F)] = ⎢⎢0.6497⎥⎥ pu [V ⎥ ⎢ ⎢0.2212⎥ ⎦ ⎣ The bus voltage in the phase form is calculated as:

⎤ ⎡ 0.9355∠00 ⎥ ⎢ ⎥ ⎢ (abc) 0 ⎢ ¯ [V1 (F)] = ⎢0.5248∠ − 134.99 ⎥⎥ pu ⎥ ⎢ ⎢ 0.5248∠134.990 ⎥ ⎦ ⎣ Bus 2: The sequence components of bus 2 voltage are calculated as

⎡V¯ (0) (F )⎤ ⎡ 0 ⎤ ⎡0 0 0 ⎤⎥ ⎡⎢ j1.290 ⎤⎥ ⎢ 2 ⎥ ⎢ ⎥ ⎢ ⎥ ⎥⎢ ⎢ (1) ⎥ ⎢ ⎥ ⎢ ⎥ ⎢−j3.503⎥ ⎢V¯ (F )⎥ = ⎢1.0⎥ − ⎢0 j0.10 0 ⎥ ⎥⎢ ⎢ 2 ⎥ ⎢ ⎥ ⎢ ⎥ ⎥⎢ ⎢ ¯ (2) ⎥ ⎢ ⎥ ⎢ ⎢ ⎥ ⎥ ⎢V2 (F )⎥ ⎢ 0 ⎥ ⎢0 j2.212 0 j0.10 ⎦ ⎦⎣ ⎣ ⎦ ⎣ ⎦ ⎣ ⎡ 0 ⎤ ⎢ ⎥ ⎢ ⎥ (012) ⎢ ¯ 2 (F)] = ⎢0.6497⎥⎥ pu [V ⎢ ⎥ ⎢0.2212⎥ ⎣ ⎦ The bus voltage in the phase form is calculated as:

⎡ ⎤ 0.8710∠00 ⎢ ⎥ ⎢ ⎥ (abc) 0 ⎢ ¯ [V2 (F)] = ⎢0.5722∠ − 139.56 ⎥⎥ pu ⎢ ⎥ ⎢ 0.5722∠139.560 ⎥ ⎣ ⎦ Bus 3: The sequence components of bus 3 voltage are calculated as

⎡V¯ (0) (F )⎤ ⎡ 0 ⎤ ⎡j0.10 0 0 ⎤⎥ ⎡⎢ j1.290 ⎤⎥ ⎢ 3 ⎥ ⎢ ⎥ ⎢ ⎢ (1) ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢V¯ (F )⎥ = ⎢1.0⎥ − ⎢ 0 j0.125 0 ⎥⎥ ⎢⎢−j3.503⎥⎥ ⎢ 3 ⎥ ⎢ ⎥ ⎢ ⎢ ¯ (2) ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢V3 (F )⎥ ⎢ 0 ⎥ ⎢ 0 0 j0.125⎥⎦ ⎢⎣ j2.212 ⎥⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎡0.1290⎤ ⎢ ⎥ ⎢ ⎥ (012) ⎢ ¯ [V3 (F)] = ⎢0.5622⎥⎥ pu ⎢ ⎥ ⎢0.2765⎥ ⎣ ⎦ The bus voltage in the phase form is calculated as: 202

⎤ ⎡ 0.9766∠00 ⎥ ⎢ ⎥ ⎢ (abc) ¯ 3 (F)] = ⎢⎢0.3815∠ − 139.560 ⎥⎥ pu [V ⎥ ⎢ ⎢ 0.3815∠139.560 ⎥ ⎦ ⎣ Bus 4: The sequence components of bus 5 voltage are calculated as

⎡V¯ (0) (F )⎤ ⎡ 0 ⎤ ⎡j0.20 0 0 ⎤⎥ ⎡⎢ j1.290 ⎤⎥ ⎢ 4 ⎥ ⎢ ⎥ ⎢ ⎥ ⎥⎢ ⎢ (1) ⎥ ⎢ ⎥ ⎢ ⎢V¯ (F )⎥ = ⎢1.0⎥ − ⎢ 0 j0.125 0 ⎥⎥ ⎢⎢−j3.503⎥⎥ ⎢ 4 ⎥ ⎢ ⎥ ⎢ ⎥ ⎥⎢ ⎢ ¯ (2) ⎥ ⎢ ⎥ ⎢ ⎢V4 (F )⎥ ⎢ 0 ⎥ ⎢ 0 0 j0.125⎥⎦ ⎢⎣ j2.212 ⎥⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎡0.2581⎤ ⎥ ⎢ ⎥ ⎢ (012) ⎢ ¯ [V4 (F)] = ⎢0.5622⎥⎥ pu ⎥ ⎢ ⎢0.2765⎥ ⎦ ⎣ The bus voltage in the phase form is calculated as:

⎡ ⎤ 1.0968∠00 ⎢ ⎥ ⎢ ⎥ (abc) ¯ 4 (F)] = ⎢⎢0.2954∠ − 123.100 ⎥⎥ pu [V ⎢ ⎥ ⎢ 0.2954∠123.100 ⎥ ⎣ ⎦ Bus 5: The sequence components of bus 5 voltage are calculated as

⎡V¯ (0) (F )⎤ ⎡ 0 ⎤ ⎡j0.30 0 0 ⎤⎥ ⎡⎢ j1.290 ⎤⎥ ⎢ 5 ⎥ ⎢ ⎥ ⎢ ⎢ (1) ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢V¯ (F )⎥ = ⎢1.0⎥ − ⎢ 0 j0.175 0 ⎥⎥ ⎢⎢−j3.503⎥⎥ ⎢ 5 ⎥ ⎢ ⎥ ⎢ ⎢ ¯ (2) ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢V5 (F )⎥ ⎢ 0 ⎥ ⎢ 0 0 j0.175⎥⎦ ⎢⎣ j2.212 ⎥⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎡0.3871⎤ ⎢ ⎥ ⎢ ⎥ (012) ⎢ ¯ 5 (F)] = ⎢0.3871⎥⎥ pu [V ⎢ ⎥ ⎢0.3871⎥ ⎣ ⎦ The bus voltage in the phase form is calculated as:

⎡1.1613∠00 ⎤ ⎢ ⎥ ⎢ ⎥ (abc) ⎢ ⎥ pu ¯ [V5 (F)] = ⎢ 0 ⎥ ⎢ ⎥ ⎢ ⎥ 0 ⎣ ⎦ Line Current Calculations:

Line L1 ∶ 203

The sequence components of Line 1 current are calculated as

⎡ 1 ⎤ ⎢ 0 0 ⎥⎥ ⎡ ⎤ ⎡I¯(0) (F )⎤ ⎢ j0.30 ⎢ 34 ⎥ ⎢ ⎥ ⎢(0.1290 − 0.2581)⎥ ⎥ ⎢ ⎢ (1) ⎥ ⎢ ⎥ 1 ⎢I¯ (F )⎥ = ⎢ 0 ⎥ ⎢(0.5622 − 0.5622)⎥ 0 ⎥ ⎢ 34 ⎥ ⎢ ⎥⎢ j0.10 ⎥ ⎢ ¯(2) ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎢I34 (F )⎥ ⎢ ⎥ (0.2765 − 0.2765) 1 ⎦ ⎣ ⎦ ⎢ 0 ⎥⎣ 0 ⎢ ⎥ ⎣ j0.10 ⎦ Hence,

⎡j0.4301⎤ ⎥ ⎢ ⎥ ⎢ (012) ⎢ [¯I34 (F)] = ⎢ 0 ⎥⎥ pu ⎥ ⎢ ⎢ 0 ⎥ ⎦ ⎣ The phase components of line 1 fault current are

⎡j0.4301⎤ ⎢ ⎥ ⎢ ⎥ (abc) ⎢ ¯ [I34 (F)] = ⎢j0.4301⎥⎥ pu ⎢ ⎥ ⎢j0.4301⎥ ⎣ ⎦ Line L2 ∶ The sequence components of Line 2 current are calculated as

⎤ ⎡ 1 ⎢ 0 0 ⎥⎥ ⎡ ⎤ ⎡I¯(0) (F )⎤ ⎢ j0.30 ⎥ ⎢(0.1290 − 0.3871)⎥ ⎢ 35 ⎥ ⎢ ⎥⎢ ⎥ ⎢ (1) ⎥ ⎢ 1 ⎢I¯ (F )⎥ = ⎢ 0 0 ⎥⎥ ⎢⎢(0.5622 − 0.3871)⎥⎥ ⎢ 35 ⎥ ⎢ j0.10 ⎥⎢ ⎢ ¯(2) ⎥ ⎥ ⎢ ⎢I35 (F )⎥ ⎢ 1 ⎥⎥ ⎢⎣(0.2765 − 0.3871)⎥⎦ ⎣ ⎦ ⎢ 0 0 ⎥ ⎢ ⎣ j0.10 ⎦ Hence,

⎡ j0.8602 ⎤ ⎢ ⎥ ⎢ ⎥ (012) [¯I35 (F)] = ⎢⎢−j1.7511⎥⎥ pu ⎢ ⎥ ⎢ j1.1060 ⎥ ⎣ ⎦ The phase components of line 2 fault current are

⎡ 0.2151∠900 ⎤ ⎢ ⎥ ⎢ ⎥ (abc) 0 ⎢ ¯ [I35 (F)] = ⎢2.7425∠154.25 ⎥⎥ pu ⎢ ⎥ ⎢ 0.2954∠25.550 ⎥ ⎣ ⎦ Line L3 ∶ 204

The sequence components of Line 3 current are calculated as

⎡ 1 ⎤ ⎢ 0 0 ⎥⎥ ⎡ ⎤ ⎡I¯(0) (F )⎤ ⎢ j0.30 ⎢ 45 ⎥ ⎢ ⎥ ⎢(0.2581 − 0.3871)⎥ ⎥ ⎢ ⎢ (1) ⎥ ⎢ ⎥ 1 ⎢I¯ (F )⎥ = ⎢ 0 ⎥ ⎢(0.5622 − 0.3871)⎥ 0 ⎥ ⎢ 45 ⎥ ⎢ ⎥⎢ j0.10 ⎥ ⎢ ¯(2) ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎢I45 (F )⎥ ⎢ ⎥ (0.2765 − 0.3871) 1 ⎦ ⎣ ⎦ ⎢ 0 ⎥⎣ 0 ⎢ ⎥ ⎣ j0.10 ⎦ Hence,

⎡ j0.4301 ⎤ ⎥ ⎢ ⎥ ⎢ (012) [¯I45 (F)] = ⎢⎢−j1.7511⎥⎥ pu ⎥ ⎢ ⎢ j1.1060 ⎥ ⎦ ⎣ The phase components of line 3 fault current are

⎡ 0.2151∠ − 900 ⎤ ⎢ ⎥ ⎢ ⎥ (abc) [¯I45 (F)] = ⎢⎢2.5863∠163.080 ⎥⎥ pu ⎢ ⎥ ⎢ 2.5863∠16.920 ⎥ ⎣ ⎦ Transformer Current Calculations:

Transformer T1 ∶ The sequence components of Transformer 1 current are calculated as:

⎤ ⎡ 1 ⎢ 0 0 ⎥⎥ ⎡ ⎡I¯(0) (F )⎤ ⎢ j0.05 ⎤ ⎥ ⎢(0.0645 − 0.1290)⎥ ⎢ 13 ⎥ ⎢ ⎥⎢ ⎢ (1) ⎥ ⎢ ⎥ 1 ⎢I¯ (F )⎥ = ⎢ 0 0 ⎥⎥ ⎢⎢(0.6497 − 0.5622)⎥⎥ ⎢ 13 ⎥ ⎢ j0.05 ⎥⎢ ⎢ ¯(2) ⎥ ⎥ ⎢ ⎢I13 (F )⎥ ⎢ 1 ⎥⎥ ⎢⎣(0.2212 − 0.2765)⎥⎦ ⎣ ⎦ ⎢ 0 0 ⎥ ⎢ ⎣ j0.05 ⎦ Hence,

⎡ j1.290 ⎤ ⎢ ⎥ ⎢ ⎥ (012) ⎢ [¯I13 (F)] = ⎢−j1.750⎥⎥ pu ⎢ ⎥ ⎢ j1.106 ⎥ ⎣ ⎦ The phase components of Transformer 1 fault current are

⎡ 0.6452∠900 ⎤ ⎢ ⎥ ⎢ ⎥ (abc) 0 ⎢ [¯I13 (F)] = ⎢2.9536∠146.90 ⎥⎥ pu ⎢ ⎥ ⎢ 2.9536∠33.100 ⎥ ⎣ ⎦ 205

Transformer T2 ∶ The sequence components of Transformer 2 current are calculated as

⎡1 ⎤ ⎢ 0 0 ⎥⎥ ⎡ ⎡I¯(0) (F )⎤ ⎢ ∞ (0.0 − 0.2581) ⎤⎥ ⎢ ⎥ ⎢ 24 ⎥ ⎢ 1 ⎥ ⎥⎢ ⎢ (1) ⎥ ⎢ ⎢I¯ (F )⎥ = ⎢ 0 0 ⎥⎥ ⎢⎢(0.6497 − 0.5622)⎥⎥ ⎢ 24 ⎥ ⎢ j0.05 ⎥ ⎥⎢ ⎢ ¯(2) ⎥ ⎢ ⎢I24 (F )⎥ ⎢ 1 ⎥⎥ ⎢⎣(0.2212 − 0.2765)⎥⎦ ⎣ ⎦ ⎢0 0 ⎢ j0.05 ⎥⎦ ⎣ Hence,

⎡ 0 ⎤ ⎥ ⎢ ⎥ ⎢ (012) [¯I24 (F)] = ⎢⎢−j1.750⎥⎥ pu ⎥ ⎢ ⎢ j1.106 ⎥ ⎦ ⎣ The phase components of Transformer 2 fault current are

⎡ 0.6452∠ − 900 ⎤ ⎢ ⎥ ⎢ ⎥ (abc) 0 ⎢ [¯I24 (F)] = ⎢2.4953∠172.57 ⎥⎥ pu ⎢ ⎥ ⎢ 2.4953∠7.430 ⎥ ⎣ ⎦

Figure 4.74: Flow of sequence currents for LLG fault at bus 5 Generator Current Calculations:

Line G1 ∶ The sequence components of Generator 1 current are calculated as: ⎤ ⎡ 1 ⎢ 0 0 ⎥⎥ ⎡ ⎤ ⎡I¯(0) (F )⎤ ⎢ j0.05 ⎢ G1 ⎥ ⎢ ⎥ ⎢(0.0 − 0.0645)⎥ ⎢ (1) ⎥ ⎢ ⎥⎢ ⎥ 1 ⎢I¯ (F )⎥ = ⎢ 0 0 ⎥⎥ ⎢⎢(1.0 − 0.6497)⎥⎥ ⎢ G1 ⎥ ⎢ j0.20 ⎥ ⎢ ⎥⎢ ⎥ ⎢ ¯(2) ⎢IG1 (F )⎥ ⎢ 1 ⎥⎥ ⎢⎣(0.0 − 0.2212)⎥⎦ ⎣ ⎦ ⎢ 0 0 ⎢ ⎥ ⎣ j0.20 ⎦ 206

⎡ j1.290 ⎤ ⎥ ⎢ ⎥ ⎢ (012) [¯IG1 (F)] = ⎢⎢−j1.750⎥⎥ pu ⎥ ⎢ ⎢ j1.106 ⎥ ⎦ ⎣ ⎡ 0.6452∠00 ⎤ ⎥ ⎢ ⎥ ⎢ (abc) [¯IG1 (F)] = ⎢⎢2.9536∠146.900 ⎥⎥ pu ⎥ ⎢ ⎢ 2.9536∠33.100 ⎥ ⎦ ⎣ Line G2 ∶ The sequence components of Generator 2 current are calculated as: ⎡ 1 ⎤ ⎢ 0 0 ⎥⎥ ⎡ ⎤ ⎡I¯(0) (F )⎤ ⎢ j0.05 ⎥ ⎢ ⎢ G2 ⎥ ⎢ (0.0 − 0.0) ⎥ ⎥ ⎥ ⎢ ⎢ (1) ⎥⎢ 1 ⎢I¯ (F )⎥ = ⎢ 0 0 ⎥⎥ ⎢⎢(1.0 − 0.6497)⎥⎥ ⎥ ⎢ ⎢ G2 j0.20 ⎥ ⎥ ⎢ ⎢ ¯(2) ⎥⎢ ⎢IG2 (F )⎥ ⎢ 1 ⎥⎥ ⎢⎣(0.0 − 0.2212)⎥⎦ ⎦ ⎢ 0 ⎣ 0 ⎢ ⎥ ⎣ j0.20 ⎦ ⎡ 0.0 ⎤ ⎢ ⎥ ⎢ ⎥ (012) ⎢ ¯ [IG2 (F)] = ⎢−j1.750⎥⎥ pu ⎢ ⎥ ⎢ j1.106 ⎥ ⎣ ⎦ ⎡ 0.6452∠00 ⎤ ⎢ ⎥ ⎢ ⎥ (abc) 0 ⎢ ¯ [IG2 (F)] = ⎢2.4953∠172.57 ⎥⎥ pu ⎢ ⎥ ⎢ 2.4953∠7.430 ⎥ ⎣ ⎦ The flow of sequence currents in the sequence networks is shown in the Fig.4.74. We will be discussing open-conductor faults in the next lecture.

207

4.13

Open Conductor Faults:

When one or two phases of a balanced three-phase line opens it creates an unbalance in the system and results in the flow of unbalanced currents. Such conditions occur in the system when one or two conductors of a trnansmission line are broken due to storm or if fuses, isolators or circuit breakers operate only on one or two phases leaving others connected. Such open conductor ¯ Bus ] matrices of sequence networks. faults can also be analysed with the help of [Z

Figure 4.75: Open Conductor faults on a section of three phase system In Fig. 4.75, a section of a three phase system between buses i and j is shown. Fig. 4.75 (a) shows one conductor open while Fig. 4.75 (b) shows two conductors open between points k and k′ . The positive direction of currents I¯a , I¯b and I¯c are shown in the figure. For the analysis of such faults, the Thevenin’s impedance between two buses i and j is required and the relationship ¯ Bus ] and Thevenin’s impedances at each bus of the network needs between the elements of [Z to be established.

¯ 0 ] be the vector of open-circuit bus voltages corresponding to the initial (pre-fault) value Let [V ¯ Bus ]. We can of bus current vector [¯ I0 ] injected in a network with bus impedance matrix [Z then write 208

¯ 0 ] = [Z ¯ Bus ][¯I0 ] [V

(4.134)

¯ can be If the bus currents are changed to a new value, [¯ I0 + ∆¯I], the new bus voltage [V] expressed as:

¯ = [Z ¯ Bus ] [¯I0 + ∆¯I] [V] ¯ Bus ][¯I0 ] + [Z ¯ Bus ][∆¯I] = [Z

(4.135)

¯ 0 ] + [∆V] ¯ = [V ¯ represents the change in the values of the original bus voltage [V ¯ 0 ]. where,[∆V] Fig. 4.76 represents a power system with buses i and j taken out along with the reference node. ¯ 0 ] and [¯I0 ] are zero. Currents [∆I¯i ] and [∆I¯j ] are The circuit is not energised so that [V injected into the ith and jth buses respectively, through current sources connected between the node and the reference node.

¯ due to current [∆I¯i ] and [∆I¯j ] Figure 4.76: Change in bus voltage [∆V] 209

¯ can be calculated from equation (4.135) as The changes in bus voltage [∆V] ⎡ ∆V¯ ⎤ 1 ⎢ 1⎥ ⎥ ⎢ ⎢ ⋮ ⎥ ⎥ ⎢ ⎢ ¯⎥ ⎢ ∆ Vi ⎥ i ⎥ ⎢ ⎢ ∆V¯ ⎥ = j ⎢ j⎥ ⎥ ⎢ ⎢ ⋮ ⎥ ⎥ ⎢ ⎢ ¯ ⎥ ⎢∆VN ⎥ N ⎦ ⎣

⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

1

i

Z¯11 ⋯ Z¯1i ⋮ ⋮ Z¯i1 ⋯ Z¯ii Z¯j1 ⋯ Z¯ji ⋮ ⋮ Z¯N 1 ⋯ Z¯N i

j

N

Z¯1j ⋯ Z¯1N ⋮ ⋮ Z¯ij ⋯ Z¯iN Z¯jj ⋯ Z¯jN ⋮ ⋮ Z¯N j ⋯ Z¯N N

⎤ ⎡ ∆I¯ ⎤ ⎥⎢ 1⎥ ⎥ ⎥⎢ ⎥⎢ ⋮ ⎥ ⎥ ⎥⎢ ⎥⎢ ¯ ⎥ ⎥ ⎢ ∆Ii ⎥ ⎥ ⎥⎢ ⎥ ⎢ ∆I¯ ⎥ ⎥⎢ j ⎥ ⎥ ⎥⎢ ⎥⎢ ⋮ ⎥ ⎥ ⎥⎢ ⎥⎢ ¯ ⎥ ⎥ ⎢∆IN ⎥ ⎦ ⎦⎣

⎡ ∆V¯ ⎤ ⎡ Z¯ ∆I¯ + Z¯ ∆I¯ ⎤ ⎢ ⎢ 1j j ⎥ 1⎥ ⎥ ⎥ ⎢ 1i i ⎢ ⎥ ⎢ ⋮ ⎥ ⎢ ⋮ ⎥ ⎥ ⎢ ⎢ ⎢ ¯⎥ ⎢ ¯ ¯ ¯ ¯ ⎥ ⎢ ∆Vi ⎥ ⎢ Zii ∆Ii + Zij ∆Ij ⎥ ⎥ ⎥ ⎢ ⎢ ⎢ ∆V¯ ⎥ = ⎢ Z¯ ∆I¯ + Z¯ ∆I¯ ⎥ ⎢ ⎢ jj j ⎥ j⎥ ⎥ ⎥ ⎢ ji i ⎢ ⎥ ⎢ ⋮ ⎥ ⎢ ⋮ ⎥ ⎥ ⎢ ⎢ ⎥ ⎢ ¯ ⎥ ⎢¯ ⎢∆VN ⎥ ⎢ZN i ∆I¯i + Z¯N j ∆I¯j ⎥ ⎦ ⎦ ⎣ ⎣

(4.136)

The modified voltage at ith bus can be written as :

V¯i = V¯i0 + ∆V¯i = V¯i0 + Z¯ii ∆I¯i + Z¯ij ∆I¯j

(4.137)

¯ij ∆I¯i in equation (4.137), one obtains adding and subtracting Z V¯i = V¯i0 + (Z¯ii − Z¯ij )∆I¯i + Z¯ij (∆I¯i + ∆I¯j )

(4.138)

Similarly the modified voltage at jth bus can be written as

¯ j = V¯j0 + Z¯ji ∆I¯i + Z¯jj ∆I¯j V¯j = V¯j0 + ∆V

(4.139)

¯ji ∆I¯j in equation (4.139) , one obtains adding and subtracting Z V¯j = V¯j0 + (Z¯jj − Z¯ji )∆I¯j + Z¯ji (∆I¯i + ∆I¯j )

(4.140)

Since the network is symmetrical

Z¯ji = Z¯ij

(4.141)

Thus the equations (4.138) and (4.140) can be represented by an equivalent circuit shown in Fig. 4.77, which is also the Thevenin’s Equivalent circuit of the network as seen from the ith and jth buses. From the figure it can be observed that the Thevenin’s open circuit voltage between ith and jth buses is (V¯i0 − V¯j0 ). 210

Figure 4.77: The Thevenin’s Equivalent of the original network To calculate the open circuit impedance between ith and jth buses, the initial voltages V¯i0 and V¯j0 are set equal to zero and an ideal current source I¯ is connected between the two busses. Next, the resulting voltages V¯i and V¯j are calculated. Note that ∆I¯i = I¯ and ∆I¯j = −I¯.

V¯i = (Z¯ii − Z¯ij )I¯

(4.142)

¯ V¯j = (Z¯jj − Z¯ji )(−I)

(4.143)

Next, calculate the voltage difference ∆V¯ij between ith and jth buses as:

∆V¯ij = V¯i − V¯j = (Z¯ii + Z¯jj − 2Z¯ij )I¯

(4.144)

Hence,

∆Z¯T hevenin,ij =

∆V¯ij = (Z¯ii + Z¯jj − 2Z¯ij ) I¯

(4.145)

Once the Thevenin’s equivalent is established, the analysis of open-conductor faults can proceed further. The opening of all the three phases is equivalent to the removal of the line i → j totally (0) (1) (2) from the network. If z¯ij , z¯ij and z¯ij are the the three sequence impedance of the line i → j, then the removal of this line from the network can be simulated by adding −¯ zij(0) , −¯ zij(1) and −¯ zij(2) to the corresponding Thevenin’s equivalent network of the three sequence networks of the original network as seen from ith and jth buses. Let x represents the fractional length of the broken line i → j from ith bus to the break point ‘k’, where 0 ≤ x ≤ 1. 211

The positive sequence impedance of the conductor segment between the ith bus and the point of break k is x¯ zij(1) , and the positive sequence impedance of the remaining conductor from point k to j th bus is (1 − x)¯ zij(1) . These two impedances are then added to represent the broken conductor. This is illustrated in Fig. 4.78

Figure 4.78: Positive sequence equivalent network with line open between buses k and k’ (a)

(b)

(c)

If V¯kk′ , V¯kk′ and V¯kk′ represent the phase component of voltage drops between points k and (0) (1) (2) k’, then V¯kk′ , V¯kk′ and V¯kk′ represent the sequence components of the voltage drops between points k and k’. These sequence voltages have different values depending on the type of open conductor fault. (1) To further simplify the circuit, the voltage V¯kk′ and the total series impedance [x¯ zij(1) + (1 −

x)¯ zij(1) ] = z¯ij(1) is replaced by a current source

V¯kk(1)′ (1) ij

z¯

(1)

and a parallel impedance z¯ij as shown in

Fig. 4.79. (1)

(1)

Further, the parallel combination of z¯ij and −¯ zij is ∞ and hence, is replaced by an open circuit. The final simplified positive sequence impedance equivalent circuit is shown in Fig. 4.80 Similarly, the negative sequence and zero sequence equivalent networks are shown in Fig.4.81 (a) and (b) repectively. These equivalent networks are identical to the positive sequence equivalent network but do not contain any internal voltage sources. The equivalent currents

V¯kk(1)′ V¯kk(2)′ , (1)

z¯ij

z¯ij(2)

and

V¯kk(0)′ z¯ij(0)

are due to open conductor fault between k and

k’. If no conductor is open then the sequence voltages are all zero and the current sources are not present in the equivalent circuit. Further, the current sources can be regarded as current injections into the buses i and j of the original sequence networks. All through the calculation 212

Figure 4.79: Thevenin’s Equivalent with transformed current source

Figure 4.80: Final positive sequence Thevenin’s Equivalent circuit repesenting the opening of line i → j between buses k and k’

(0)

(1)

(2)

¯ Bus ], [Z ¯ Bus ] and [Z ¯ Bus ] of the original network are process, the bus impedance matrices [Z used.

The current injections at the buses i and j can be tabulated as: 213

Figure 4.81: Final (a) negative sequence (b) zero sequence Thevenin’s Equivalent circuit repesenting the opening of line i → j between k and k’ Positive Sequence Negative Sequence Zero sequence at ith bus at j th bus

V¯kk(1)′

V¯kk(2)′

z¯ij(1) V¯kk(1)′ − (1) z¯ij

V¯kk(0)′

z¯ij(2) V¯kk(2)′ − (2) z¯ij (0)

(1)

z¯ij(0) V¯kk(0)′ − (0) z¯ij (2)

The sequence voltage drops ∆V¯n , ∆V¯n and ∆V¯n at any bus ‘n’ due to the current injections at the buses ‘i’ and ‘j’ can be calculates from equation (4.136) as:

(0) n

∆V¯

=

∆V¯n(1) = ∆V¯n(2) =

(0) (0) ¯ (0) (Z¯ni − Z¯nj )Vkk′

z¯ij(0) (1) (1) ¯ (1) (Z¯ni − Z¯nj )Vkk′

z¯ij(1) (2) (2) ¯ (2) (Z¯ni − Z¯nj )V ′ (2) ij

(4.146)

kk

z¯

Next, the Thevenin’s equivalent impedances for each sequence network, as seen from the busesk and k’, are calculated as follows:

¯ (1)′ is found out as : From Fig.4.78, the positive sequence equivalent impdeance Z kk (1) Z¯kk zij(1) + ′ = x¯

(1) Z¯th,ij (−¯ zij(1) ) + (1 − x)¯ zij(1) (1) (1) Z¯ + (−¯ zij ) th,ij

214

(1) kk′

Z¯

−(¯ zij(1) )2 = (1) Z¯ − z¯ij(1)

(4.147)

th,ij

Similarly from Fig. 4.81 (a) and (b), the negative sequence and zero sequence Thevenin’s equivalent impedances can be expressed as:

(2) Z¯kk = ′

−(¯ zij(2) )2 Z¯ (2) − z¯ij(2) th,ij

(4.148) (0) Z¯kk = ′

−(¯ zij(0) )2 Z¯ (0) − z¯ij(0) th,ij

The open-circuit voltage from point k to k’ can be calculated as: (1) th,kk′

V¯

(−¯ zij(1) )2 (V¯i(1) − V¯j(1) ) = (1) (1) ¯ Z − z¯ij

(4.149)

th,ij

substituting

(1) Z¯kk ′

z¯ij(1)

=−

(1) V¯th,kk ′ is obtained as:

z¯ij(1)

(1) Z¯th,ij − z¯ij(1)

from equation (4.147) in equation (4.149), the final value of

(1) Z¯kk ′ (1) V¯th,kk = (V¯i(1) − V¯j(1) ) ′ (1) z¯ij

(4.150)

(1) Also prior to the occurance of open-conductor fault on any conductor, the current I¯ij flowing in phase a is the positive sequence component and is given by the relation:

I¯ij(1) =

(V¯i(1) − V¯j(1) ) z¯ij(1)

(4.151)

Substituting equation (4.151) in equation (4.150), one gets (1) ¯ (1) ¯(1) V¯th,kk ′ = Zkk ′ Iij

(4.152)

The Thevenin’s equivalent network as seen from points k and k’ for the three sequence networks are shown in Fig. 4.82 We are now ready to discuss the two possible cases of open-circuit fault i.e. (a) open phase open (b) two phases open. We will be discussing them in the next lecture.

215

Figure 4.82: Thevenin’s Equivalent networks as seen from k and k’

4.13.1

One Phase open:

Consider that phase a conductor is open as shown in Fig. 4.75(a), hence phase a current I¯a = 0. As a result:

I¯a(1) + I¯a(2) + I¯a(0) = 0 (1)

(2)

(4.153)

(0)

where, I¯a , I¯a and I¯a are the symmetrical components of phase a current. Since phases b and c are closed , the voltage drops .

V¯kk′ ,b = 0 V¯kk′ ,c = 0 216

(4.154)

The symmetrical components of voltage drops across the fault point can be calculated as :

⎡V¯ (0) ⎤ ⎡1 1 1 ⎤ ⎡V¯ ′ ⎤ ⎡V¯ ′ ⎤ ⎥ ⎢ kk ,a ⎥ ⎢ a ⎥ ⎢ ⎢ kk ,a ⎥ ⎥ ⎢ ⎥ ⎥ ⎢ (1) ⎥ 1 ⎢ ⎢ 1 ⎢V¯a ⎥ = ⎢1 a a2 ⎥ ⎢ 0 ⎥ = ⎢Vkk′ ,a ⎥ ⎥ 3⎢ ⎥ ⎥ 3⎢ ⎥⎢ ⎢ ⎥ ⎥ ⎥⎢ ⎢ ¯ (2) ⎥ ⎢ ⎢ 2 ⎢Va ⎥ ⎢1 a a ⎥ ⎢ 0 ⎥ ⎢Vkk′ ,a ⎥ ⎦ ⎦ ⎦ ⎦⎣ ⎣ ⎣ ⎣

(4.155)

1 V¯a(0) = V¯a(1) = V¯a(2) = V¯kk′ ,a 3

(4.156)

Hence,

It implies that open conductor in phase a causes equal voltages to appear across points k and k’ of each sequence network. Hence, the three equivalent sequence networks can be connected in parallel across points k and k’ as shown in Fig. 4.83.

Figure 4.83: Connection of Equivalent sequence networks to represent open phase a between k and k’ (1) The current I¯a is given as :

I¯a(1) = I¯ij

(1) Z¯kk ′ +

(1) Z¯kk ′ (2) Z¯ ′ Z¯ (0)′ kk

kk

(2) ¯ (0) Z¯kk ′ + Zkk ′

simplifying

I¯a(1) = I¯ij

(1) ¯ (2) ¯ (0) Z¯kk ′ [Zkk ′ + Zkk ′ ] Z¯ (0)′ Z¯ (1)′ + Z¯ (1)′ Z¯ (2)′ + Z¯ (2)′ Z¯ (0)′ kk

kk

kk

kk

kk

(4.157)

kk

(1) (2) (0) The sequence voltage drops V¯kk′ , V¯kk′ and V¯kk′ can be calculated with reference to Fig.4.83 as:

¯ (2) ¯ (0) (1) (1) Zkk′ Zkk′ ¯ ¯ Vkk′ = Ia (2) ¯ (0) Z¯kk ′ + Zkk ′ (1) Substituting I¯a and simplifying we get:

217

V¯kk(1)′ = V¯kk(2)′ = V¯kk(0)′ = I¯ij

(1) ¯ (2) ¯ (0) Z¯kk ′ Zkk ′ Zkk ′ (0) (1) Z¯ ′ Z¯ ′ + Z¯ (1)′ Z¯ (2)′ + Z¯ (2)′ Z¯ (0)′ kk

(1)

(2)

kk

kk

kk

kk

(4.158)

kk

(0)

¯ ′ , Z¯ ′ , and Z¯ ′ are obtained from the impedance parameters of the sequence networks • Z kk kk kk [equation (4.147) and equation (4.148)]. • I¯ij is the pre-fault current or load current in phase a of the line i → j Next, the equivalent injected currents (0)

(1)

(2)

V¯kk(1)′ V¯kk(2)′ , (1)

z¯ij

z¯ij(2)

and

V¯kk(0)′ z¯ij(0)

are calculated.

Further, ∆V¯i ,∆V¯i and ∆V¯i representing the changes in the symmetrical components of bus voltage are calculated using equation (4.146). Finally, the bus voltages after fault are calculated using superposition principle as :

V¯i(1) (F ) = V¯i(1) (0) + ∆V¯i(1) V¯i(2) (F ) = ∆V¯i(1) V¯i(0) (F ) = ∆V¯i(2)

4.13.2

(4.159)

Two Phases open:

When two phases b and c are open then,

V¯kk(1)′ ,a = V¯a(0) + V¯a(1) + V¯a(2) = 0 I¯b = 0 I¯c = 0

(4.160)

The sequence components of line current are:

⎡I¯(0) ⎤ ⎡1 1 1 ⎤ ⎡I¯ ⎤ ⎢ ⎥ ⎢ a⎥ ⎢ a ⎥ ⎥⎢ ⎥ ⎢ (1) ⎥ 1 ⎢ ⎢I¯a ⎥ = ⎢1 a a2 ⎥ ⎢ 0 ⎥ ⎥⎢ ⎥ ⎢ ⎥ 3⎢ ⎢ ⎥⎢ ⎥ ⎢ ¯(2) ⎥ 2 ⎢1 a a ⎥ ⎢ 0 ⎥ ⎢Ia ⎥ ⎣ ⎣ ⎦ ⎦⎣ ⎦ Simplifying one gets the condition:

1 I¯a(0) = I¯a(1) = I¯a(2) = I¯a 3

(4.161)

equation (4.161) indicates that the three equivalent sequence networks are in series and to ensure V¯ + V¯a(1) + V¯a(2) = 0 the circuit should be closed.The interconnection of the sequence networks is shown in Fig.4.84. From the equivalent circuit of Fig.4.84, the sequence currents can be calculated as : (0) a

218

Figure 4.84: Connection of Equivalent sequence networks to represent open phases b and c between k and k’

I¯a(1) = I¯a(2) = I¯a(0) = I¯ij

(1) Z¯kk ′ (0) (1) ¯ ¯ Z ′ + Z ′ + Z¯ (2)′ kk

kk

(4.162)

kk

I¯ij is the pre-fault current in phase a. The sequence voltage can be calculated as: (1) ¯ (0) ¯ (2) Z¯kk ′ (Zkk ′ + Zkk ′ ) (1) (2) (1) ¯ (0) ¯ ¯ ¯ ¯ Vkk′ = Ia (Zkk′ + Zkk′ ) = Iij (0) (1) ¯ (2) Z¯kk′ + Z¯kk ′ + Zkk ′ (1) ¯ (2) Z¯kk ′ Zkk ′ (2) ¯ V¯kk(2)′ = −I¯a(2) Z¯kk = − I ′ ij (0) (1) ¯ (2) Z¯kk′ + Z¯kk ′ + Zkk ′ (1) ¯ (0) Z¯kk ′ Zkk ′ (0) ¯ V¯kk(0)′ = −I¯a(0) Z¯kk = − I ′ ij (0) (1) ¯ (2) Z¯kk′ + Z¯kk ′ + Zkk ′

(4.163)

Remaining calculations are similar to those performed for single conductor open case. In the next lecture we will be looking at an example of open conductor fault analysis. 219

4.14

Example of calculations for open conductor fault in power system:

The power system described in Fig. 4.67 is considered again. The prefault bus voltages for buses 3, 4 and 5 are :

⎡V (0)⎤ ⎡ 0.9165∠ − 8.7580 ⎤ ⎥ ⎢ 3 ⎥ ⎢ ⎥ ⎥ ⎢ ⎢ ⎢V4 (0)⎥ = ⎢0.9152∠ − 10.0980 ⎥ pu ⎥ ⎥ ⎢ ⎢ ⎥ ⎥ ⎢ ⎢ ⎢V5 (0)⎥ ⎢0.8858∠ − 12.9610 ⎥ ⎦ ⎦ ⎣ ⎣ ONE CONDUCTOR OPEN Let one conductor of line 7 between buses 4 and 5 be open. The pre-fault current in phase a of ¯ (0) ¯ (1) the faulted line is I¯45 = (0.3821 − j0.3779) pu. The sequence impedance matrices [Z Bus ], [ZBus ] ¯ (2) and [Z Bus ] are same as used in the example for calculation of short circuit faults on the power system of Fig. 4.67. The Thevenin’s impedance of the network as seen from buses 4 and 5 is calculated using equation (4.145) with i=4 and i=5 as: (1) (1) (1) (1) Z¯th,45 = Z¯44 + Z¯55 − 2Z¯45 = j0.1397 + j0.1750 − 2 ∗ 0.1250 = j0.0647 pu

(2) (1) Z¯th,45 = Z¯th,45 = j0.0647 pu

(0) (0) (0) (0) Z¯th,45 = Z¯44 + Z¯55 − 2Z¯45 = j0.30 + j0.30 − 2 ∗ 0.20 = j0.20 pu (1)

(2)

(0)

¯ ′ , Z¯ ′ andZ¯ ′ are calculated from equation (4.147) and equation (4.148). Next,Z kk kk kk (1) 2 (−¯ z45 ) −(j0.1)2 (1) ¯ Zkk′ = (1) = = j0.2833 pu (1) j0.0647 − j0.10 Z¯th,45 − z¯45 (2) ¯ (1) Z¯kk ′ = Zkk ′ = j0.2833 pu (0) 2 (−¯ z45 ) −(j0.30)2 (0) ¯ Zkk′ = (0) = = j0.90 pu (0) j0.20 − j0.30 Z¯th,45 − z¯45 (1) The current I¯a is calculated using equation (4.157)

I¯a(1) = I¯ij

(1) ¯ (2) ¯ (0) Z¯kk ′ [Zkk ′ + Zkk ′ ] Z¯ (0)′ Z¯ (1)′ + Z¯ (1)′ Z¯ (2)′ + Z¯ (2)′ Z¯ (0)′ kk

I¯a(1) = (0.3821 − j0.3779)

kk

kk

kk

kk

kk

j0.2833(j0.2833 + j0.3) = (0.2170 − j0.2146) pu j0.9 ∗ j0.2833 + j0.2833 ∗ j0.2833 + j0.2833 ∗ j0.9 220

(1) (2) (0) The sequence voltage drops V¯kk′ , V¯kk′ and V¯kk′ are then calculated using the equation:

¯ (2) ¯ (0) (1) (1) Zkk′ Zkk′ ¯ ¯ Vkk′ = Ia (2) ¯ (0) Z¯kk ′ + Zkk ′ j0.2833 ∗ j0.9 V¯kk(1)′ = (0.2170 − j0.2146) = (0.0463 + j0.0468) pu j0.2833 + j0.9 V¯kk(2)′ = V¯kk(0)′ = V¯kk(1)′ = (0.0463 + j0.0468) pu As a check calculate

I¯a(2)

=−

V¯kk(2)′ Z¯ (2)′

=−

0.0463 + j0.0468 = (−0.1651 + j0.1633) pu j0.2833

I¯a(0)

V¯ (0)′ = − kk Z¯ (0)′

=−

0.0463 + j0.0468 = (−0.052 + j0.0514) pu j0.90

kk

kk

I¯a = I¯a(0) + I¯a(1) + I¯a(2) = (0.2170 − j0.2146) + (−0.1651 + j0.1633) + (−0.052 + j0.0514) =0

Q.E.D.

Then we calculate the changes in bus voltages using equation (4.146). Bus 3 (0) (0) (Z¯34 − Z¯35 )Vkk(0)′ j0.10 − j0.10 (0) ¯ ∆ V3 = = ∗ (0.0463 + j0.0468) = 0 pu (0) j0.30 z¯45

(1) (1) (Z¯34 − Z¯35 )Vkk(1)′ j0.1103 − j0.1250 (1) ¯ ∗ (0.0463 + j0.0468) = (−0.0068 − j0.0069) pu ∆ V3 = = (1) j0.1 z¯45

∆V¯3(2) = ∆V¯3(1) = (−0.0068 − j0.0069) pu Hence,

∆V¯3 = ∆V¯3(0) + ∆V¯3(1) + ∆V¯3(2) = (−0.0136 − j0.0138) pu Bus 4

(0) (0) (Z¯44 − Z¯45 )Vkk(0)′ j0.30 − j0.20 (0) ¯ ∆V4 = = ∗ (0.0463 + j0.0468) = (0.0154 + j0.0156) pu (0) j0.3 z¯45

221

(1) (1) (Z¯44 − Z¯45 )Vkk(1)′ j0.01397 − j0.125 (1) ¯ ∆ V4 = = ∗ (0.0463 + j0.0468) = (0.0068 + j0.0069) pu (1) j0.10 z¯45

∆V¯4(2) = ∆V¯4(1) = (0.0068 + j0.0069) pu ∆V¯4 = ∆V¯4(0) + ∆V¯4(1) + ∆V¯4(2) = (0.0290 − j0.0293) pu Bus 5

(0) (0) (Z¯54 − Z¯55 )Vkk(0)′ j0.20 − j0.30 (0) ¯ ∆V5 = = ∗ (0.0463 + j0.0468) = (−0.0154 − j0.0156) pu (0) j0.3 z¯45

(1) (1) (Z¯54 − Z¯45 )Vkk(1)′ j0.1250 − j0.1750 (1) ¯ = ∗ (0.0463 + j0.0468) = (−0.0231 − j0.0234) pu ∆ V5 = (1) j0.3 z¯45

∆V¯5(2) = ∆V¯5(1) = (−0.0231 − j0.0234) pu ∆V¯5 = ∆V¯5(0) + ∆V¯5(1) + ∆V¯5(2) = (−0.0617 − j0.0624) pu The bus voltages during fault are

V¯3 (F ) = V¯30 + ∆V¯3 = (0.9058 − j0.1395) + (−0.0136 − j0.138) = (0.8922 − j0.1533) pu

V¯3 (F ) = 0.9053∠ − 9.750 pu

V¯4 (F ) = V¯40 + ∆V¯4 = (0.9010 − j0.1605) + (0.0290 + j0.0293) = (0.9300 − j0.1311) pu

V¯4 (F ) = 0.9392∠ − 8.020 pu

V¯5 (F ) = V¯50 + ∆V¯5 = (0.8632 − j0.1987) + (−0.0617 − j0.0624) = (0.8016 − j0.2610) pu 222

V¯5 (F ) = 0.8430∠ − 18.040 pu The phase components of current in line 4-5 are:

⎤ ⎡I¯(a) ⎤ ⎡1 1 1 ⎤ ⎡ (−0.052 + j0.0514) ⎤ ⎡ 0 ⎥ ⎥ ⎢ ⎥⎢ ⎢ 45 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ (b) ⎥ ⎢ ⎢ I¯ ⎥ = ⎢1 a2 a ⎥ ⎢ (0.2170 − j0.2146) ⎥ = ⎢0.4782∠ − 147.840 ⎥ pu ⎥ ⎥ ⎢ ⎥⎢ ⎢ 45 ⎥ ⎢ ⎥ ⎥ ⎢ ⎥⎢ ⎢ ¯(c) ⎥ ⎢ 0 2 ⎢ I45 ⎥ ⎢1 a a ⎥ ⎢(−0.1651 + j0.1633)⎥ ⎢ 0.4782∠58.56 ⎥ ⎦ ⎦ ⎣ ⎦⎣ ⎣ ⎦ ⎣ TWO CONDUCTORS OPEN The sequence currents can be calculated with the help of equation (4.162) as:

I¯a(1) = I¯ij

(1) Z¯kk (0.3821 − j0.3779) ∗ j0.2833 ′ = (0) (1) (2) j0.9 + j0.2833 + j0.2833 Z¯kk′ + Z¯kk′ + Z¯kk′

I¯a(1) = I¯a(2) = I¯a(0) = (0.0738 − j0.073) pu The sequence voltages of Thevenin’s equivalent sequence network is calculated using equation (4.163)

(0) ¯ (2) V¯kk(1)′ = I¯a(1) (Z¯kk ′ + Zkk ′ ) = (0.0738 − j0.073) ∗ (j0.9 + j0.2833) = (0.0864 + j0.0873) pu

(2) V¯kk(2)′ = −I¯a(2) Z¯kk ′ = −(0.0738 − j0.073)(j0.2833) = −(0.0207 + j0.0209) pu

(0) V¯kk(0)′ = −I¯a(0) Z¯kk ′ = −(0.0738 − j0.073)(j0.9) = −(0.0657 + j0.0664) pu

The changes in bus voltages are calculated using equation (4.146). Bus 3 (0) (0) (Z¯34 − Z¯35 )Vkk(0)′ j0.1 − j0.10 (0) ¯ ∆ V3 = = ∗ (−0.0657 − j0.0664) = 0 pu (0) j0.3 z¯45

(1) (1) (Z¯34 − Z¯35 )Vkk(1)′ j0.1103 − j0.1250 (1) ¯ ∆ V3 = = ∗ (0.0864 + j0.0873) = (−0.0127 − j0.0128) pu (1) j0.1 z¯45

(2) (2) (Z¯34 − Z¯35 )Vkk(2)′ j0.1103 − j0.1250 (2) ¯ ∆ V3 = = ∗ (−0.0207 − j0.0208) = (0.0030 + j0.0030) pu (2) j0.1 z¯45

∆V¯3 = ∆V¯3(0) + ∆V¯3(1) + ∆V¯3(2) = (−0.0097 − j0.0098) pu 223

Hence, the voltage of bus 3 during fault is:

V¯3 (F ) = V¯30 + ∆V¯3 = (0.9058 − j0.1395) + (−0.0097 − j0.0098) = (0.8962 − j0.1493) pu

V¯3 (F ) = 0.9085∠ − 9.460 pu Bus 4

(0) (0) (Z¯44 − Z¯45 )Vkk(0)′ j0.30 − j0.20 (0) ¯ = ∗ (−0.0657 − j0.0664) = (−0.0219 − j0.0221) pu ∆ V4 = (0) j0.30 z¯45

(1) (1) (Z¯44 − Z¯45 )Vkk(1)′ j0.1397 − j0.1250 (1) ¯ = ∗ (0.0864 + j0.0873) = (0.0127 + j0.0128) pu ∆V4 = (1) j0.1 z¯45

(2) (2) (Z¯44 − Z¯45 )Vkk(1)′ j0.1397 − j0.1250 (2) ¯ ∆ V4 = = ∗ (0.0207 − j0.0209) = (−0.0030 − j0.0031) pu (2) j0.1 z¯45

∆V¯4 = ∆V¯4(0) + ∆V¯4(1) + ∆V¯4(2) = −(0.0122 + j0.0124) pu Hence, the voltage of bus 4 during fault is:

V¯4 (F ) = V¯40 + ∆V¯4 = (0.9010 − j0.1605) + (−0.0122 − j0.0124) = (0.8888 − j0.1728) pu

V¯4 (F ) = 0.9054∠ − 11.00 pu Bus 5

(0) (0) (Z¯54 − Z¯55 )Vkk(0)′ j0.20 − j0.30 (0) ¯ = ∗ (−0.0657 − j0.0664) = (0.0219 + j0.0221) pu ∆V5 = (0) j0.30 z¯45

(1) (1) (Z¯54 − Z¯55 )Vkk(1)′ j0.1250 − j0.1750 (1) ¯ ∆ V5 = = ∗ (0.0864 + j0.0873) = (−0.0432 − j0.0437) pu (1) j0.10 z¯45

224

(2) (2) (Z¯54 − Z¯55 )Vkk(2)′ j0.1250 − j0.1750 (2) ¯ ∆ V5 = = ∗ (−0.0207 − j0.0209) = (0.0103 + j0.0105) pu (2) j0.10 z¯45

∆V¯5 = ∆V¯5(0) + ∆V¯5(1) + ∆V¯5(2) = −(0.0111 + j0.01111) pu Hence, the voltage of bus 5 during fault is:

V¯5 (F ) = V¯50 + ∆V¯5 = (0.8632 − j0.1987) + (−0.0111 − j0.01111) = (0.8523 − j0.2097) pu V¯5 (F ) = 0.8777∠ − 13.830 pu The phase components of current in line 4-5 are:

⎡I¯(a) ⎤ ⎡1 1 1 ⎤ ⎡(0.0738 − j0.073)⎤ ⎡0.3114∠ − 44.690 ⎤ ⎢ 45 ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ (b) ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ I¯ ⎥ = ⎢1 a2 a ⎥ ⎢(0.0738 − j0.073)⎥ = ⎢ ⎥ pu 0 ⎢ 45 ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ¯(c) ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ 2 ⎢ I45 ⎥ ⎢1 a a ⎥ ⎢(0.0738 − j0.073)⎥ ⎢ ⎥ 0 ⎣ ⎦ ⎣ ⎦⎣ ⎦ ⎣ ⎦

225

Module 5 Contingency Analysis 5.1

Introduction and concept of linear sensitivity factors

In Chapter 2, we have discussed about the methods for analyzing the power system performance at a particular operating point. However, for practical system operation, apart from ensuring the satisfactory operation of the system at a particular operating condition, it is also equally important to make sure that the system operates with adequate level of security. Broadly, the term ‘security’ implies the ability of the system to operate within system constraints (on bus voltage magnitudes, current and power flow over the lines) in the event of outage (contingency) of any component (generator or transmission line). Now, if the system is operating at high loading (light loading) conditions, then the post-contingency system condition would be highly stressed (lightly stressed). Therefore, the post contingency values of different quantities (voltages, current/power flow) depend on the present operating condition. In case the post-outage (post contingency) does not involve any violation of any operating constraints, the system is said to be operating securely. Otherwise, the system is said to enter an emergency operating condition. Therefore, for detecting the possibility of appearance of emergency operating conditions, analysis of the post-contingency scenario (henceforth termed as contingency analysis) of the system needs to be carried out. Now, in any practical sized power system, there is a very large number of elements. Hence, for carrying out contingency analysis, outages of all these elements (preferably) need to be carried out one-by-one corresponding to any particular operating condition. However, in any power system, the operating point of the system changes quite frequently with change is loading/generating conditions. With the change in system operating conditions, the contingency analysis exercise needs to be carried out again at the new operating point. Thus, for proper monitoring of system security, a large number of outage cases need to be simulated repeatedly over a short span of time. Ideally, these outage cases should be studied with the help of full AC load flow solutions. However, analysis of thousands of outage cases with full AC power flow technique will involve a significant amount of computation time and as a result, it might not be possible to complete this entire exercise before the new operating condition emerges. Therefore, instead of using full non-linear AC power flow analysis, approximate, but much faster techniques based on linear sensitivity factors are used to estimate the post contingency values of different quantities of interest. The basic concept of sensitivity factors 226

is described below. Essestially, the linear sensitivity factors approximately estimate the changes in different line flows for any particular outage condition without the need of full AC power flow solution. Basically, there are two types of sensitivity factors and these are: a. Generation outage sensitivity factor (GOSF) b. Line outage sensitivity factor (LOSF) GOSF relates the approximate change in power flow in line ‘i-j’ (i.e. between bus ‘i’ and ‘j’) due to the outage of generator at bus ‘k’, whereas LOSF helps to calculate the approximate change in power flow in line ‘i-j’ due to outage of line ‘m-n’. The generation outage sensitivity factor is defined by,

αijk =

∆fij ∆Pk

(5.1)

k where, αij → GOSF of line ‘i-j’ for generation change at bus ‘k’

∆fij → Change in power flow in line ‘i-j’ ∆Pk → Change in generation at bus ‘k’

k The factor αij denotes the sensitivity of the line flow on line ‘i-j’ due to change in generation at bus ‘k’. In equation (5.1), it is assumed that the generation lost at bus ‘k’ would be exactly compensated by the reference or slack bus. Now, if the generation at bus ‘k’ was generating an amount of power equal to Pk0 , then to represent the outage condition, ∆Pk = −Pk0 .

Hence, the new power flow over the line ‘i-j’ would be given as,

fijn = fij0 + ∆fij = fij0 + αijk ∆Pk = fij0 − αijk Pk0

(5.2)

k The factor αij would be pre-calculated and stored in the memory. As we will see later, the values k of αij depend only on the network parameters and therefore, are constant. However, it should be k m noted that for any particular line ‘i-j’, the factors αij and αij (for generation outage at bus ‘m’) are different and therefore need to be pre-calculated separately. Once these factors are pre-calculated and stored, the new values of line flow over any line can easily be estimated very quickly from equation (5.2). If the new power flow over any line is found to be more than the corresponding limit, then the operator can be alerted for taking an appropriate pre-emptive action.

In equation (5.2), it is assumed that the lost generation at bus ’k’ would be taken up by the slack bus. However, it is also quite possible that the lost generation would be compensated by all the remaining ‘on-line’ generators combinedly, in which, each of the ‘on-line’ generators would take up some fraction of the lost generation in some particular ratio. One of the most frequently used methods assumes that the ‘on-line’ generators share the lost generation in proportion to their maximum MW rating. Thus, the proportion of generation picked up by generation ‘g’ is given by 227

g ≠ k, γgk =

Pgmax

(5.3)

M

∑ Pamax

a=1 ≠k

Where, M → Total number of generators in the system

γgk → Proportionality factor for generation ‘g’ to pick up generation when unit ‘k’ fails Pamax → Maximum MW rating for generator ‘a’. Now, as the sensitivity factors shown in equation (5.1) are linear in nature, the effects of simultaneous generation change in several generators on a particular line can be obtained by following superposition principle. Hence, the new line flow in the line ‘i-j’ becomes, m

fij(n) = fij(0) + αijk ∆Pk − ∑ αija ∆Pa γak

(5.4)

a=1

In equation (5.4) it is assumed that no remaining ‘on-line’ generation hits the generation limit. The line outage distribution factors are also defined similarly. The LOSF is defined by,

βij, mn =

∆fij

(5.5)

(0) fmn

Where, βij, mn →Line outage distribution factor for line ‘i-j’ under outage of line ‘m-n’. (0) fmn → Power flow over line ‘m-n’ in the pre-outage condition.

Therefore, for the outage of line ‘m-n’, the new flow over line ‘i-j’ is given by, (0) fij(n) = fij(0) + βij, mn fmn

(5.6)

Again, as we will show later, the factors βij,mn are constant as they are dependent only on the line parameters. Therefore, they would be pre-calculated and stored in the memory. As a result, for the outage of any line ‘m-n’, the new power flows over all the other lines can be estimated very quickly.

5.2

DC load flow and generation outage distribution factor

We have already discussed the concepts of linear sensitivity factors. These factors are calculated based on the concept of DC power flow and hence, let us first have a look at DC power flow technique.

5.2.1

DC power flow

From FDLF method, we know, ∆P = [B ] ∆θ , where each elements of the matrix [B ] is negative ¯ BUS bus matrix. Now, in DC power of the imaginary parts of the corresponding elements of the Y ′

′

228

flow analysis, apart from using the above decoupled relation between ∆P and ∆θ , several other simplifying assumptions are also taken as follows: a. The system is lossless (i.e. line resistance is neglected) and therefore, each line is represented by its reactance only. b. The voltage of each bus is maintained at the rated voltage, i.e. 1.0 p.u. c. For any line ‘m-n’, the angular difference between its terminal buses is quite small, and hence, cos θm ≈ cos θn (as θm ≈ θn ) and sin(θm − θn ) ≈ (θm − θn ) rad (as θm − θn ≈ 0) . With these assumptions, the power flow over a line becomes,

Pij =

Vi Vj 1 sin(θi − θj ) ≈ (θi − θj ) (p.u) xij xij

(5.7)

In equation (5.7), the quantity xij denotes the reactance of the line ‘i-j’. From this equation it is observed that the line power flow is basically a linear combination of the terminal bus voltage angles. Moreover, the current flow over the line ‘i-j’ is given by,

V¯i − V¯j I¯ij = jxij (Vi cos θi − Vj cos θj ) + j(Vi sin θi − Vj sin θj ) = jxij 1 1 = (sin θi − sin θj ) ≈ (θi − θj ) xij xij

(5.8)

Equation (5.8) has been written under an added assumption that both the angles θi and θj are individually quite small in magnitude. From equations (5.7) and (5.8) it is observed that in DC power flow model, the expressions of line power flow and line current are same in per unit. We will utilize this fact for computing LOSF in future. In the next lecture, we will discuss a method for calculating GOSF.

229

5.2.2

Calculation of GOSF −1

From FDLF we know that ∆P = [B ] ∆θ or ∆θ = [X] ∆P where [X] = [B ] . Or, ′

′

⎤ ⎤⎡ ⎤ ⎡ ⎡ ⎢ ∆θ2 ⎥ ⎢ X22 X23 ⋯ X2n ⎥ ⎢ ∆P2 ⎥ ⎥ ⎥⎢ ⎥ ⎢ ⎢ ⎥ ⎥⎢ ⎢ ∆θ ⎥ ⎢ X ⎢ 3 ⎥ ⎢ 32 X33 ⋯ X3n ⎥ ⎢ ∆P3 ⎥ ⎥ ⎥⎢ ⎥=⎢ ⎢ ⎢ ⋮ ⎥ ⎢ ⋮ ⎥⎢ ⋮ ⎥ ⎥ ⎥⎢ ⎥ ⎢ ⎢ ⎥ ⎥⎢ ⎥ ⎢ ⎢ ⎢∆θn ⎥ ⎢Xn2 Xn3 ⋯ Xnn ⎥ ⎢∆Pn ⎥ ⎦ ⎦⎣ ⎦ ⎣ ⎣

(5.9)

In equation (5.9), the matrix [X] is of size (n − 1) × (n − 1) and any element Xij is actually ′ located in the location (i − 1), (j − 1). As the matrix [B ] is a constant real matrix dependent only on the line parameters, matrix [X] is also a constant real matrix dependent on the line parameters. Furthermore, in equation (5.9), the quantity ∆θ1 is not included in the vector ∆θ as the reference angle does not change with any outage in the system. Now, to calculate the GOSF for the outage of generator at bus ‘k’, the perturbation at bus ‘k’ is set to ∆Pk and the perturbation at the slack bus is set to −∆Pk (assuming that the entire lost generation is taken up by the slack bus) while keeping the perturbations at the other buses equal to zero. Therefore, the perturbed values of the bus angles can be given as,

∆θi = Xik ∆Pk for i = 2, ⋯⋯ n

(5.10)

Now, from equation (5.7), the change in power flow over line ‘i-j’ is given by,

∆Pij = Therefore,

1 1 (∆θi − ∆θj ) = (Xik − Xjk )∆Pk , where, xl = xij is the reactance of the line ‘i-j’. xl xl αijk =

∆Pij Xik − Xjk = ∆Pk xl

(5.11)

k As can be seen from equation (5.11), the factor αij is dependent only on the line parameters.

Let us now look at the application and utility of these sensitivity factors. For this purpose, let as consider the IEEE-14 bus system (the data of which are given in Tables A.3 and A.4). In this system, apart from the slack bus,there are two other generations at bus 2 and bus 6 (refer Table A.3). The real power flows in all the lines have been calculated with all the three generatiors operating in the system and are shown in column 2 of Table 5.1 (under the heading ‘Pline(ori)’). The GSOFs for these two generators have been calculated using equation (5.11) and are shown is columns 3 and 7 of Table 5.1 respectively (under the heading GSOF(2) and GSOF(6) respectively). From Table A.3 it can be seen that the specified real power generated by these two generations are 18.3 MW and 11.2 MW respectively. Hence, following the argument given for equation (5.2), the quantities ∆P2 and ∆P6 are equal to -0.183(p.u) and 0.112 (p.u) respectively (on a 100 MVA base) as shown in columns 4 and 8 of Table 5.1. With these information, the estimated line flows after the outage of generation 2 and 6 are calculated using equation (5.2) and are shown in columns 5 and 9 respectively. Lastly, full AC power flow studies have been carried out by removing generator 2 and 6 one by one (by modeling them as PQ bus after reducing their real power generation to zero) and the results of line 230

flows (obtained with full AC power flow) are shown in columns 6 and 10 respectively. From columns 5 and 6 as well as from columns 9 and 10 it is observed that the post-outage line-flows estimated by the sensitivity analysis technique match quite closely with those obtained by the full AC power flow method.

Table 5.1: Results of generator outage analysis in IEEE 14 bus system (all powers are given in p.u.) For outage of generator at bus 2 Line Pline GSOF Pline Pline ∆P2 no. (ori) (2) (cal) (ACLF) 1 1.6262 -0.9709 -0.183 1.803875 1.7891 2 0.7464 -0.1935 -0.183 0.781811 0.7776 3 0.8655 0.0089 -0.183 0.863871 0.8606 4 0.5314 0.0556 -0.183 0.521225 0.5207 5 0.3662 0.0821 -0.183 0.351176 0.3511 6 -0.358 0.047 -0.183 -0.3666 -0.3622 7 -0.7068 0.1064 -0.183 -0.72627 -0.7217 8 0.2689 -0.0013 -0.183 0.269138 0.2691 9 0.1063 -0.0004 -0.183 0.106373 0.1064 10 0.2893 -0.0078 -0.183 0.290727 0.2887 11 0.1156 -0.0059 -0.183 0.11668 0.1154 12 0.0852 -0.0007 -0.183 0.085328 0.0851 13 0.2005 -0.0031 -0.183 0.201067 0.2003 14 0.0025 -0.0065 -0.183 0.00369 0.0025 15 0.2664 0.0052 -0.183 0.265448 0.2667 16 0.0123 0.0055 -0.183 0.011294 0.0124 17 0.0655 0.0037 -0.183 0.064823 0.0657 18 -0.0777 0.0057 -0.183 -0.07874 -0.0776 19 0.0233 -0.0012 -0.183 0.02352 0.0232 20 0.0857 -0.0038 -0.183 0.086395 0.0854

For outage of GSOF ∆P6 (6) -0.7628 -0.112 -0.4344 -0.112 -0.1568 -0.112 -0.2928 -0.112 -0.2977 -0.112 -0.1203 -0.112 -0.0033 -0.112 -0.2562 -0.112 -0.1009 -0.112 -0.6729 -0.112 0.2459 -0.112 0.0277 -0.112 0.1312 -0.112 -0.0204 -0.112 -0.2358 -0.112 -0.2287 -0.112 -0.1548 -0.112 -0.2368 -0.112 0.05 -0.112 0.1574 -0.112

generator at bus 6 Pline Pline (cal) (ACLF) 1.711634 1.7203 0.795053 0.7753 0.883062 0.8888 0.564194 0.5678 0.399542 0.395 -0.34453 -0.3357 -0.70643 -0.7189 0.297594 0.3173 0.117601 0.1255 0.364665 0.3309 0.088059 0.0748 0.082098 0.0778 0.185806 0.1782 0.004785 0.0022 0.29281 0.3151 0.037914 0.0512 0.082838 0.0944 -0.05118 -0.039 0.0177 0.016 0.068071 0.0567

To calculate the LOSF, we first need to understand the concept of Thevenin equivalent impedance of a power system. Let as first have a look at that.

5.2.3

Determination of Thevenin’s equivalent impedance

¯V ¯ where ¯I, V ¯ and Y ¯ are the vector of bus injection currents, vector of bus voltages We know ¯ I=Y ¯ =Z ¯¯I where Z ¯ is the bus impedance matrix and and the admittance matrix respectively. Hence,V ¯=Y ¯ −1 . As the matrix Z ¯ is a constant matrix, the relation V ¯ =Z ¯¯I denotes a linear is given by Z ¯ and ¯I and hence, for incremental charges in V ¯ and ¯I the same relation also equation connecting V 231

¯ = Z∆ ¯ ¯I. Expanding this relation we get, holds good, i.e. ∆V ⎤ ⎤⎡ ⎡ ¯ ⎤ ⎡¯ ⎢ ∆V1 ⎥ ⎢ Z11 Z¯12 ⋯ Z¯1n ⎥ ⎢ ∆I¯1 ⎥ ⎥ ⎥⎢ ⎥ ⎢ ⎢ ⎥ ⎢ ∆I¯ ⎥ ⎢ ∆V¯ ⎥ ⎢ Z¯ ¯ ¯ Z ⋯ Z ⎢ 2 ⎥ ⎢ 21 2⎥ 22 2n ⎥ ⎢ ⎥ ⎥⎢ ⎥=⎢ ⎢ ⎥⎢ ⋮ ⎥ ⎢ ⋮ ⎥ ⎢ ⋮ ⎥ ⎥⎢ ⎥ ⎢ ⎢ ⎥ ⎥⎢ ⎢ ¯ ⎥ ⎢¯ ⎢∆Vn ⎥ ⎢Zn1 Z¯n2 ⋯ Z¯nn ⎥ ⎢∆I¯n ⎥ ⎦ ⎦⎣ ⎦ ⎣ ⎣

(5.12)

Now, suppose that there is an incremental charge is the current of bus ‘k’ only while the incremental changes at the other buses are zero. Hence, ∆Ik ≠ 0 and ∆Ii = 0 for i = 1, ⋯⋯n; ≠ k . Hence, ¯ik ∆I¯k for i = 1, ⋯⋯n. Hence, from equation (5.12), the changes in bus voltages are given by, ∆V¯i = Z (0) if the initial bus voltage is V¯k , then the final voltage after perturbation is given by,

V¯k = V¯k(0) + Z¯kk ∆I¯k

(5.13)

Equation (5.13) can be represented as an equivalent circuit as shown in Fig. 5.1, which shows the Thevenin’s equivalent circuit at bus ‘k’. From this circuit, the Thevenin’s equivalent impedance ¯kk . of the system (looking from bus ‘k’) in equal to Z

Figure 5.1: Thevenin equivalent circuit from bus ‘k’ In the next lecture, we will use the Thevenin’s equivalent impedance for calculating LOSF.

232

5.3

Calculation of LOSF

In the last lecture, we have looked at the Thevenin’s impedance of the system looking from a particular bus ‘k’. Let us now look at the Thevenin’s equivalent impedance of any system between bus ‘m’ and bus ‘n’. To achieve that let us assume that the incremental currents ∆I¯m and ∆I¯n are injected at bus ‘m’ and ‘n’ respectively (the incremental current at all the other buses are zero). Consequently, from equation (5.12), the changes in the bus voltages (∆V¯m and ∆V¯n ) can be given as, ∆V¯m = Z¯mm ∆I¯m + Z¯mm ∆I¯n (5.14)

∆V¯n = Z¯nm ∆I¯m + Z¯nn ∆I¯n

(5.15)

Therefore, the new bus voltages are

V¯m = V¯m(0) + Z¯mm ∆I¯m + Z¯mn ∆I¯n = V¯m(0) + (Z¯mm − Z¯mn )∆I¯m + Z¯mn (∆I¯m + ∆I¯n )

(5.16)

V¯n = V¯n(0) + Z¯nm ∆I¯m + Z¯nn ∆I¯n = V¯n(0) + Z¯nm (∆I¯m + ∆I¯n ) + (Z¯nn − Z¯nm )∆I¯n

(5.17)

From equations (5.16) and (5.17), the equivalent circuit of the power system looking from bus ‘m’ and bus ‘n’ can be drawn as shown in Fig. 5.2.

Figure 5.2: Equivalent circuit between bus ‘m’ and bus ‘n’ (0)

(0)

From Fig. 5.2 it is observed that the open-circuit voltage between bus ‘m’ and bus ‘n’ is V¯m −V¯n 233

and the Thevenin’s equivalent impedance between bus ‘m’ and bus ‘n’ is given by,

Z¯T h, mn = Z¯mm + Z¯nn − 2Z¯mn

(5.18)

From Fig. 5.2 it is also observed that the Thevenin’s equivalent impedance between bus ‘m’ and ¯mm − Z¯nm + Z¯nm ) = Z¯mm and that between bus ‘n’ and the reference is (Z¯nn − Z¯nm + reference is (Z Z¯mn ) = Z¯nn which is in complete agreement with the results already obtained in equation (5.13). ¯b is connected between Also, from Fig. 5.2 it can be readily seen that if a line having an impedance Z bus ‘m’ and ‘n’, then the current through this line is given by,

V¯m(0) − V¯n(0) ¯ Ib = Z¯T h, mn + Z¯b

(5.19)

¯b is removed Now, let as consider the case when a line or transformer having an impedance Z from the system. Let us assume that this branch is removed between bus ‘m’ and ‘n’. The removal ¯b between bus ‘m’ and ‘n’. The of this branch can be represented by addition of an impedance −Z situation is shown in Fig. 5.3.

Figure 5.3: Removal of a branch between bus ‘m’ and bus ‘n’

¯ b and this Now, in Fig. 5.3, it can be seen that the current I¯a flows in the added impedance −Z current is given by, V¯m(0) − V¯n(0) ¯ Ia = Z¯T h, mn − Z¯b

(5.20)

This current I¯a flows ‘out’ of bus ‘m’ and flows ‘into’ bus ‘n’. Therefore, because of the addition 234

¯b , the injected current changes at bus ‘m’ by ∆I¯m = −I¯a and that changes at bus of impedance −Z ‘n’ by ∆I¯n = I¯a . However, at all the other buses (not equal to ‘m’ or ‘n’), the current injections do not change. Because of these changed current injections at bus ‘m’ and ‘n’, the voltages at all the bues will get changed. The changes in voltages at any bus ‘p’ and ‘q’ can be given as, ∆V¯p = Z¯pm ∆I¯m + Z¯pn ∆I¯n = (Z¯pn − Z¯pm )I¯a

(5.21)

∆V¯q = Z¯qm ∆I¯m + Z¯qn ∆I¯n = (Z¯qn − Z¯qm )I¯a

(5.22)

and

Therefore, change in current flow is line ‘p-q’ can be given as,

∆I¯pq =

∆V¯p − ∆V¯q (Z¯pn − Z¯pm ) − (Z¯qn − Z¯qm ) ¯ = Ia Z¯c Z¯c

(5.23)

¯c is the impedance of the line ‘p-q’. Combining equations (5.20) In equation (5.23), the quantity Z and (5.23) we get, (Z¯pn − Z¯pm ) − (Z¯qn − Z¯qm ) V¯m(0) − V¯n(0) ¯ [ ] ∆Ipq = Z¯c Z¯T h, mn − Z¯b

(5.24)

¯b , then the pre-outage current in line ‘m-n’ given by, If the impedance of the line ‘m-n’ is Z V¯m(0) − V¯n(0) ¯ Imn = Z¯b

(5.25)

Therefore, from equation (5.24) we get,

Or,

Z¯b (Z¯pn − Z¯pm ) − (Z¯qn − Z¯qm ) V¯m(0) − V¯n(0) ¯ ∆Ipq = [ ] Z¯c Z¯T h, mn − Z¯b Z¯b

(5.26)

∆I¯pq Z¯b (Z¯pn − Z¯pm ) − (Z¯qn − Z¯qm ) = [ ] I¯mn Z¯c Z¯T h, mn − Z¯b

(5.27)

Now, in equations (5.7) and (5.8) we have seen that under the assumptions of DC power flow method, the expressions of power flow and current flow in a line are same in per unit. Therefore, if we impose the same assumptions in equation (5.27), then all the impedances would be replaced by the corresponding reactances and ∆I¯pq (I¯mn ) would be replaced by ∆Ppq (Pmn ). Therefore, the LOSF (βpq, mn ) is given by,

βpq, mn =

∆Ppq xb (Xpn − Xpm ) − (Xqn − Xqm ) = [ ] Pmn xc XT h, mn − Xb

(5.28)

Again, as an example, IEEE 14 bus system is chosen and the estimated and calculated line-power flows for two line outage cases are shown in Table 5.2. In this table, the pre-outage real power flows 235

in the line are shown in column 2. The values of LOSF are shown in columns 3 and 7. Using these LOSF, the estimated values of power flow in the lines are tabulated in columns 5 and 9. Finally, the actual power flow in the lines obtained with full AC power flow (after removing the line under consideration) are shown in columns 6 and 10 respectively. From this table, it is observed that the estimated and calculated values are reasonably close to each other, thereby establishing the usefulness of the line outage sensitivity factors. Table 5.2: Results of line outage analysis in IEEE 14 bus system (all powers are given in p.u.) For outage of line no. 18 Line Pline LSOF Pline Pline P18 no. (ori) (2) (cal) (ACLF) 1 1.6262 -0.0363 -0.0777 1.6290205 1.6309 2 0.7464 0.0351 -0.0777 0.7436727 0.742 3 0.8655 -0.0294 -0.0777 0.8677844 0.869 4 0.5314 -0.0682 -0.0777 0.5366991 0.5364 5 0.3662 0.0574 -0.0777 0.36174 0.362 6 -0.358 -0.0363 -0.0777 -0.355179 -0.3545 7 -0.7068 0.5224 -0.0777 -0.747390 -0.7442 8 0.2689 -0.4868 -0.0777 0.3067244 0.3014 9 0.1063 -0.1923 -0.0777 0.1212417 0.1192 10 0.2893 0.6604 -0.0777 0.2379869 0.2433 11 0.1156 0.9961 -0.0777 0.038203 0.0351 12 0.0852 -0.0857 -0.0777 0.0918589 0.0925 13 0.2005 -0.325 -0.0777 0.2257525 0.2277 14 0.0025 -0.0125 -0.0777 0.0034713 0.0024 15 0.2664 -0.4743 -0.0777 0.3032531 0.2991 16 0.0123 -1.0039 -0.0777 0.090303 0.0904 17 0.0655 0.4085 -0.0777 0.0337596 0.0329 18 -0.0777 — -0.0777 — — 19 0.0233 -0.1021 -0.0777 0.0312332 0.0305 20 0.0857 -0.4102 -0.0777 0.1175725 0.1192

For outage of line no. LSOF Pline P2 (6) (cal) 1.0113 0.7464 2.381034 — 0.7464 — 0.1721 0.7464 0.993955 0.3597 0.7464 0.799880 0.4845 0.7464 0.727830 0.1716 0.7464 -0.22991 0.4944 0.7464 -0.3377 0.0258 0.7464 0.288157 0.0101 0.7464 0.113838 -0.0288 0.7464 0.267803 -0.0166 0.7464 0.103209 -0.0024 0.7464 0.083408 -0.0086 0.7464 0.19408 0.0048 0.7464 0.006082 0.021 0.7464 0.282074 0.0168 0.7464 0.024839 0.011 0.7464 0.073710 0.0167 0.7464 -0.06523 -0.0025 0.7464 0.02143 -0.011 0.7464 0.077489

2 Pline (ACLF) 2.4451 — 0.9987 0.7986 0.7256 -0.2352 -0.3481 0.2815 0.1113 0.272 0.1046 0.0842 0.1952 0.0024 0.2791 0.0234 0.0719 -0.0666 0.0224 0.0794

So far, we have discussed the case of single contingency only. In the next lecture, we will start looking into the procedure of analysis of multiple contingencies.

236

5.4

Analysis of multiple contingencies

Let us now examine the case of outage of more than one elements. We will discuss here the case of outage of two transmission lines. Although theoretically the outage of any number of lines can be analyzed, but practically it is not feasible as outage of more than two lines would almost surely overload the other existing lines and hence trip the whole system. We will also not discuss the case of outage of two generations as outage of two generators would also most probably create a huge deficit of power supply in the system, which would necessitate shedding of loads to maintain the frequency of the grid within acceptable limits. Hence, we will confine ourselves to the case of outage of two lines only. Towards this goal, let us first analyze the situation where two transmission lines are simultaneously added to the power system, as the expressions derived would be directly applicable to the case of our interest, i.e. outage of two lines.

Let us consider an N-bus power system in which I¯1 , I¯2 , ⋯⋯I¯N are the bus injection currents and V¯1 , V¯2 , ⋯⋯V¯N are the bus voltages. The bus voltages and the bus injection currents are related by ¯ and the relation is given as, the bus impedance matrix Z

⎤⎡ ⎤ ⎡¯ ⎤ ⎡¯ ⎢ V1 ⎥ ⎢ Z11 Z¯12 ⋯ Z¯1N ⎥ ⎢ I¯1 ⎥ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ V¯ ⎥ ⎢ Z¯ ⎢ 2 ⎥ ⎢ 21 Z¯22 ⋯ Z¯2N ⎥ ⎢ I¯2 ⎥ ⎥⎢ ⎥ ⎢ ⎥=⎢ ⎥⎢ ⋮ ⎥ ⎢ ⋮ ⎥ ⎢ ⋮ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢¯ ⎥ ⎢ ¯ ¯ ¯ ⎢VN ⎥ ⎢ZN 1 ZN 2 ⋯ ZN N ⎥ ⎢I¯N ⎥ ⎦⎣ ⎦ ⎣ ⎦ ⎣

(5.29)

¯a in connected between buses ‘e-f’ while another Now, let as consider that a line of impedance Z ¯b in connected between buses ‘g-h’. It is to be again noted that the lines Z¯a and line of impedance Z Z¯b are not part of the original system described by equation (5.29) and our objective is to analyze the effect of these two line additions on the bus voltages and power flowers in the lines of the original system. To accomplish this, we assume that the bus injections currents I¯1 , I¯2 , ⋯⋯I¯N remain constant i.e. they are un-effected by the addition of these two lines. Now, due to the addition of these two ¯ ′ = [V¯1′ , V¯2′ , ⋯⋯V¯N′ ] be the vector of the lines, the bus voltages of the system would charge. Let V new bus voltages. Also, due to the addition of these two lines, currents I¯a and I¯b would flow through ¯a and Z¯b respectively as shown in Fig. 5.4. the impedance Z

¯a 0 I¯a ¯e′ − V¯ ′ Z V ¯a I¯a = V¯e − V¯ and Z¯b I¯b = V¯g − V¯ . Hence, [ From Fig. 5.4, Z ] [ ¯ ] = [ ¯ ′ ¯f′ ]. f h ¯ 0 Zb Ib Vg − Vh ′

′

′

237

′

Figure 5.4: Addition of two lines in the original system

Or,

Z¯a [ 0

⎡ V¯ ′ ⎤ ⎢ 1⎥ ⎢ ¯′⎥ ⎢ V2 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⋮ ⎥ ⎢ ′⎥ ⎢ V¯ ⎥ ⎢ e⎥ ⎢ ⎥ ⎢ ⋮ ⎥ e f g h ⎢ ⎥ ⎢ ′⎥ ¯ 0 Ia 0 ⋯ 0 1 ⋯ −1 0 ⋯ 0 0 ⋯ 0 ⋯ 0 ⎢⎢ V¯f ⎥⎥ ][ ] = [ ] Z¯b I¯b 0 ⋯ 0 0 ⋯ 0 0 ⋯ 1 0 ⋯ −1 ⋯ 0 ⎢⎢ ⋮ ⎥⎥ ⎢ ′⎥ ⎢ V¯ ⎥ ⎢ g⎥ ⎢ ⎥ ⎢ ⋮ ⎥ ⎢ ′⎥ ⎢¯ ⎥ ⎢ Vh ⎥ ⎢ ⎥ ⎢ ⋮ ⎥ ⎢ ⎥ ⎢¯′ ⎥ ⎢VN ⎥ ⎣ ⎦ ′ ¯ = AV 238

(5.30)

In equation (5.30), the matrix A can easily be identified. Now, from Fig. 5.4 it can be seen ¯a and Z¯b , extra currents I¯a and I¯b are being injected into buses ‘f’ that due to the addition of lines Z and ‘h’ respectively (which were not present before the addition of these two lines). Similarly, extra currents I¯a and I¯b are being extracted from the buses ‘e’ and ‘g’ respectively (which were also not present before the addition of these two lines). Therefore, it can be said that, due to the addition of these two lines, injection currents at these four buses have changed by ∆I¯e , ∆I¯f , ∆I¯g and ∆I¯h respectively, where ∆I¯e = −I¯a ; ∆I¯f = +I¯a ; ∆I¯g = −I¯b and ∆I¯h = +I¯b and due to these extra injection ¯ to V ¯ ′ . Therefore, the addition of these two lines can currents, the bus voltages have changed from V be indirectly represented by these extra four injection currents which will produce the same effect on the existing system. This situation is shown in Fig. 5.5. Now, the vector of change in injection currents can be expressed as; T ∆¯I = [0⋯0 ∆I¯e ⋯ ∆I¯f ⋯ ∆I¯g ⋯ ∆I¯h 0⋯0] . Or,

⎡ 0 ⎤ ⎡0 0⎤ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎢ ⋮ ⎥ ⎢ ⋮ ⋮ ⎥⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎢ 0 ⎥ ⎢0 0⎥ ⎥ ⎢ ⎥ ⎢ ⎢−I¯ ⎥ ⎢−1 0 ⎥ ⎥ ⎢ a⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⋮ ⎥ ⎢ ⋮ ⋮ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ¯ ⎥ ⎢ ¯ ⎢ Ia ⎥ ⎢ 1 0 ⎥ I¯a ⎥ [ ] = −AT [Ia ] ∆¯I = ⎢⎢ ⎥⎥ = ⎢⎢ ⎥ I¯b ⋮ ⎥ I¯b ⎢ ⋮ ⎥ ⎢ ⋮ ⎥ ⎢ ⎥ ⎢ ⎢ −I¯b ⎥ ⎢ 0 −1⎥ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎢ ⋮ ⎥ ⎢ ⋮ ⋮ ⎥⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ¯ ⎥ ⎢ ⎢ Ib ⎥ ⎢ 0 1 ⎥ ⎥ ⎢ ⎥ ⎢ ⎢ ⋮ ⎥ ⎢ ⋮ ⋮ ⎥⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎢ 0 ⎥ ⎢0 0⎥ ⎦ ⎣ ⎦ ⎣

(5.31)

¯ = Z∆ ¯ ¯I Due to the above change in current ∆¯ I, the bus voltages would undergo a change ∆V ′ ′ ¯ is the bus impedance matrix of the origin system. Now, ∆V ¯ =V ¯ −V ¯ . Or, V ¯ =V ¯ + ∆V ¯. Where Z Or,

¯ ¯′ = V ¯ + Z∆ ¯ ¯I = V ¯ − ZA ¯ T [Ia ] V I¯b I¯a I¯b

¯ = AV ¯ − AZA ¯ T [ ]; Or, AV ′

Or,

(5.32)

¯ Z¯a 0 I¯a ¯ − AZA ¯ T [Ia ] (from equation (5.30)) ] [ ] = A V 0 Z¯b I¯b I¯b

Or, [

−1 I¯a Z¯a 0 ¯ T } AV ¯ =Z ¯ −1 ¯ [ ¯ ] = {[ ] + AZA 1 AV Ib 0 Z¯b

(5.33)

Where,

¯ ¯ 1 = [Za 0 ] + AZA ¯ T Z 0 Z¯b

239

(5.34)

Figure 5.5: Equivalent representation of addition of lines Performing the matrix operation in equation (5.34) we get,

¯ ¯ ¯ ¯ ¯ (Z¯eg − Z¯eh ) − (Z¯f g − Z¯f h ) ¯ 1 = [(Zee − Zef ) − (Zf e − Zf f ) + Za Z ] (Z¯ge − Z¯gf ) − (Z¯he − Z¯hf ) (Z¯gg − Z¯gh ) − (Z¯hg − Z¯hh ) + Z¯b

(5.35)

Z¯T h,ef + Z¯a (Z¯eg − Z¯eh ) − (Z¯f g − Z¯f h ) ¯1 = [ Z ] (Z¯ge − Z¯gf ) − (Z¯he − Z¯hf ) Z¯T h,gh + Z¯b

(5.36)

Or,

¯T h,gh and Z¯T h,gf denote the thevenin equivalent impedance of the system as observed Where, Z from terminals ‘e-f’ and ‘g-h’ respectively (please see equation (5.18)). Reproducing equation (5.33) we have, ¯ ¯ I¯a −1 Ve − Vf ¯ −1 ¯ ¯ [¯] = Z A V = Z [ ] 1 1 Ib V¯g − V¯h

(5.37)

From equations (5.36) and (5.37), the vector [I¯a I¯b ]T is calculated and subsequently, from ¯ ′ can be calculated. equation (5.32), the new bus voltage vector V We are now ready to analyze the effect of outages of two lines, which we will do in the next lecture.

240

5.5

Analysis of multiple contingencies (contd..)

¯a and Z¯b are removed from the Let us now consider a case where two lines, having impedances Z ¯a terminals ‘e-f’ and ‘g-h’ respectively. Here it is to be noted that in the original system, impedance Z ¯b are already included and we are now considering the outages of these two lines. As has been and Z done previously, here also we assume that the bus injection currents do not change on account of outages of these two lines. Now, we have already seen that outage of any line having an impedance of Z¯b can be simulated by adding an impedance of −Z¯b between the same two terminals. Therefore, the ¯a and Z¯b can be simulated by connecting impedances −Z¯a and outages of lines having impedance Z −Z¯b between the terminals ‘e-f’ and ‘g-h’ respectively. Subsequently, following the same procedure ¯1 can be formed as; as discussed in the best lecture, the matrix Z Z¯T h, ef − Z¯a (Z¯eg − Z¯eh ) − (Z¯f g − Z¯f h ) ¯1 = [ Z ] (Z¯ge − Z¯gf ) − (Z¯he − Z¯hf ) Z¯T h, gh − Z¯b

(5.38)

¯a by −Z¯a and Z¯b by −Z¯b respectively Note that equation (5.38) has been obtained by replacing Z in equation (5.36). Hence, from equation (5.37) we have, Z¯T h, ef − Z¯a (Z¯eg − Z¯ef ) − (Z¯f g − Z¯f h ) I¯a V¯e − V¯f ] ] = [ [ ¯ ] [ V¯g − V¯h I¯b (Zge − Z¯gf ) − (Z¯he − Z¯hf ) Z¯T h, gh − Z¯b

(5.39)

Dividing each row of equation (5.39) by its diagonal element, we get,

⎡ V¯ − V¯ ⎤ ⎢ ⎥ e f ⎡ ⎤ ¯ ¯ ¯ ¯ ⎢ ⎥ ( Z − Z ) − ( Z − Z ) eg ef fg fh ⎥ ⎡ ¯ ⎤ ⎢ 1 ⎥ ⎢Ia ⎥ ⎢⎢ Z¯T h, ef − Z¯a ⎥⎥ ⎢ ⎢ ⎥⎢ ⎥ ⎢ Z¯T h, ef − Z¯a ⎥ ⎢ ⎥⎢ ⎥ = ⎢ ⎥ ⎢ (Z¯ − Z¯ ) − (Z¯ − Z¯ ) ⎥⎢ ⎥ ⎢ ⎥ ⎢ ¯ ⎥ ⎢ ¯ ¯ ⎥⎥ ⎢ ge gf he hf ⎥ ⎢ Ib ⎥ ⎢ Vg − Vh ⎥ ⎢ 1 ⎥⎣ ⎦ ⎢ ⎢ ⎥ Z¯T h, gh − Z¯b ⎣ ⎦ ⎢ Z¯T h, gh − Z¯b ⎥ ⎦ ⎣

(5.40)

Now, from equation (5.27), we have

∆I¯ef Z¯b (Z¯eh − Z¯eg ) − (Z¯f h − Z¯f g ) = [ ] = Lef, gh (say) I¯gh Z¯a Z¯T h, gh − Z¯b

(5.41)

∆I¯gh Z¯a (Z¯gf − Z¯ge ) − (Z¯hf − Z¯he ) = [ ] = Lgh, ef I¯ef Z¯b Z¯T h,ef − Z¯a

(5.42)

241

(say)

¯ is symmetrical (Z¯ij = Z¯ji ), from equations (5.40)-(5.42) we have, Nothing that the matrix Z ⎡ ⎢ 1 ⎢ ⎢ ⎢ ¯ ⎢ Za ⎢− Lef, gh ⎢ ¯ ⎣ Zb

⎡ V¯ − V¯ ⎤ ⎢ ⎥ e f ⎥ ⎤⎡¯ ⎤ ⎢ Z¯b ⎢ ⎥ − Lgh, ef ⎥ ⎢⎢Ia ⎥⎥ ⎢ Z¯T h, ef − Z¯a ⎥⎥ ⎥ ⎥⎢ ⎥ ⎢ Z¯a ⎥ ⎥⎢ ⎥ = ⎢ ⎥ ⎥⎢ ⎥ ⎢ ⎥ ⎢ I¯ ⎥ ⎢ V¯g − V¯h ⎥ 1 ⎥ ⎥ ⎣ b⎦ ⎢ ⎢¯ ⎥ ⎦ ⎢ ZT h, gh − Z¯b ⎥ ⎣ ⎦

Or,

⎡ ⎡I¯ ⎤ ⎢ ⎢ a⎥ 1 ⎢ ⎢ ⎥ 1 ⎢ ⎢ ⎥= ⎢ ⎢ ⎥ 1−L ⎢¯ ⎢¯⎥ gh, ef Lef, gh ⎢ Za ⎢ Ib ⎥ ⎢ ¯ Lef, gh ⎣ ⎦ ⎣ Zb

⎡ V¯ − V¯ ⎤ ⎢ ⎥ e f ⎥ ⎤⎢ Z¯b ⎢ ⎥ Lgh, ef ⎥ ⎢ Z¯T h, ef − Z¯a ⎥⎥ ⎥ ⎥⎢ Z¯a ⎥ ⎥⎢ ⎥ ⎥⎢ ⎥ ⎢ V¯g − V¯h ⎥ 1 ⎥ ⎥⎢ ⎥ ⎦⎢ ¯ ⎢ ZT h, gh − Z¯b ⎥ ⎣ ⎦

(5.43)

(5.44)

¯a and Z¯b are given by, Now, the pre-outage currents in lines Z V¯e − V¯f I¯ef = Z¯a

and I¯gh =

V¯g − V¯h Z¯b

(5.45)

Utilizes equation (5.45) in equation (5.44) we get,

I¯a =

I¯b =

1 1 − Lgh, ef Lef, gh 1 1 − Lgh, ef Lef, gh

[

Z¯ 2 Lgh, ef ¯ Z¯a I¯ef + b Igh ] Z¯T h, ef − Z¯a Z¯a Z¯T h, gh − Z¯b

(5.46)

[

Z¯b Z¯a2 Lef, gh ¯ Ief + I¯gh ] ¯ ¯ ¯ ¯ ¯ Zb ZT h, ef − Za ZT h, gh − Zb

(5.47)

¯c ) is given by, Now, the change in current in line ‘i-j’ (having an impedance Z ∆I¯ij =

∆Vi − ∆Vj Z¯c

(5.48)

Now,

Thus,

∆I¯ij =

∆Vi = Z¯ie ∆I¯e + Z¯if ∆I¯f + Z¯ig ∆I¯g + Z¯ih ∆I¯h = (Z¯if − Z¯ie )I¯a + (Z¯ih − Z¯ig )I¯b

(5.49)

∆V¯j = Z¯je ∆I¯e + Z¯jf ∆I¯f + Z¯jg ∆I¯g + Z¯jh ∆I¯h = (Z¯jf − Z¯je )I¯a + (Z¯jh − Z¯jg )I¯b

(5.50)

(Z¯if − Z¯ie ) − (Z¯jf − Z¯je ) ¯ (Z¯ih − Z¯ig ) − (Z¯jh − Z¯jg ) ¯ Ia + Ib Z¯c Z¯c 242

(5.51)

Substituting equations (5.46) and (5.47) into equation (5.51) we have,

∆I¯ij =

1 1 − Lgh, ef Lef, gh

¯ 1 I¯ef + X ¯ 2 I¯gh ] [X

(5.52)

Where,

¯ ¯ ¯ ¯ ¯ ¯2 Lef, gh ¯ 1 = Za (Zif − Zie ) − (Zjf − Zje ) + Za [(Z¯ih − Z¯ig ) − (Z¯jh − Z¯jg )] X ¯ ¯ ¯ ¯ ¯ ¯ ¯ Zc ZT h, ef − Za Zb Zc ZT h, ef − Za

(5.53)

Now, we recognize,

Lij, ef =

Z¯a (Z¯if − Z¯ie ) − (Z¯jf − Z¯je ) Z¯c Z¯T h, ef − Z¯a

(5.54)

Hence,

Lef, gh Z¯a2 ¯ [(Z¯ih − Z¯ig ) − (Z¯jh − Z¯jg )] X1 = Lij, ef + Z¯b Z¯c Z¯T h, ef − Z¯a

(5.55)

Now, expanding the second term of equation (5.55), we have [using equation (5.41)],

Lef, gh Z¯a2 [(Z¯ih − Z¯ig ) − (Z¯jh − Z¯jg )] Z¯b Z¯c Z¯T h, ef − Z¯a Z¯ 2 Z¯b (Z¯eh − Z¯eg ) − (Z¯f h − Zf g ) (Z¯ih − Z¯ig ) − (Z¯jh − Z¯jg ) = a Z¯b Z¯c Z¯a Z¯T h,gh − Z¯b Z¯T h,ef − Z¯a =[

Z¯b (Z¯ih − Z¯ig ) − (Z¯jh − Z¯jg ) Z¯a (Z¯eh − Z¯eg ) − (Z¯f h − Z¯f g ) ]×[ ] Z¯c Z¯T h, gh − Z¯b Z¯b Z¯T h, ef − Z¯a

Or,

Lef, gh Z¯a2 [(Z¯ih − Z¯ig ) − (Z¯jh − Z¯jg )] = Lij, gh Lgh, ef ¯ ¯ ¯ ¯ Zb Zc ZT h, ef − Za

(5.56)

Equation (5.56) has been written using equation (5.42) and by noting that,

Lij, gh =

Z¯b (Z¯ih − Z¯ig ) − (Z¯jh − Z¯jg ) Z¯c Z¯T h, gh − Z¯b

(5.57)

Hence,

¯ 1 = Lij, ef + Lij, gh Lgh, ef X

(5.58)

Again, from equations (5.52), (5.46), (5.47) and (5.51) we have,

¯2 Lgh, ef Z¯b (Z¯ih − Z¯ig ) − (Z¯jh − Z¯jg ) ¯ 2 = Zb [(Z¯if − Z¯ie ) − (Z¯jf − Z¯je )] + X Z¯a Z¯c Z¯T h, gh − Z¯b Z¯c Z¯T h, gh − Z¯b Again,

Lgh, ef Z¯b2 [(Z¯if − Z¯ie ) − (Z¯jf − Z¯je )] ¯ ¯ ¯ ¯ Za Zc ZT h, gh − Zb 243

(5.59)

Z¯b2 Z¯a (Z¯gf − Z¯ge ) − (Z¯hf − Zhe ) (Z¯if − Z¯ie ) − (Z¯jf − Z¯je ) Z¯a Z¯c Z¯b Z¯T h, ef − Za Z¯T h, gh − Z¯b Z¯a (Z¯if − Z¯ie ) − (Z¯jf − Z¯je ) Z¯b (Z¯gf − Z¯ge ) − (Z¯hf − Z¯he ) =[ ]×[ ] Z¯c Z¯T h, ef − Z¯a Z¯a Z¯T h, gh − Z¯b

=

Or,

Lgh, ef Z¯b2 [(Z¯if − Z¯ie ) − (Z¯jf − Z¯je )] = Lij, ef Lef, gh Z¯a Z¯c Z¯T h, gh − Z¯b

(5.60)

Please note that equation (5.60) has been obtained by using equations (5.41) and (5.54). Hence,

¯ 2 = Lij, gh + Lij, ef Lef, gh X

(5.61)

Therefore from equations (5.61), (5.58) and (5.52) we get,

∆I¯ij =

1 1 − Lef, gh Lgh, ef

[(Lij, ef + Lij, gh Lgh, ef )I¯ef + (Lij, gh + Lij, ef Lef, gh )I¯gh ]

(5.62)

Equation (5.62) gives the change in the current of line ‘i-j’ in terms of the original currents of the lines which have gone out of service. Now, in equation (5.62) all the factors are in terms of complex impedances. However, imposition of the conditions of the DC power flow makes all the factors being represented by the corresponding reactances (i.e. all complex impedances are replaced by their corresponding reactances). Now, from equation (5.28) we see that when these functions are represented by reactances only, they actually represent the different line outage sensitivity factors (LOSF). As already noted earlier, there LOSFs are already pre-calculated and stored. Also, as discussed earlier, under the assumption of DC power flow, for contingency calculation, the current flow quantities are replaced by corresponding power flow quantities. Therefore, the change in power flow in line ‘i-j’ can very easily be calculated from equation (5.62) using the information of pre-outage power flow in lines ‘e-f’ and ‘g-h’. In the next lecture, we will look into an example of contingency analysis for outages of two lines.

244

5.5.1

Example of contingency analysis for two line outages

As an example, let us consider outage of two lines in the IEEE-14 bus system. Towards this goal, let us assume that line no. 3 (let it be considered as line ‘e-f’) and 17 (let it be considered as line ‘g-h’) have been taken out of the system. Subsequently, for any line ‘i-j’ (in the set of lines which are still remaining in the system), the factors LOSFM1 and LOSFM2 are calculated as follows.

LOSF M 1 =

Lij,ef + Lij,gh Lgh,ef 1 − Lef,gh Lgh,ef

(5.63)

LOSF M 2 =

Lij,gh + Lij,ef Lef,gh 1 − Lef,gh Lgh,ef

(5.64)

Now, under the assumption of DC power flow, for contingency calculation, the impedances are replaced by corresponding reactances. Therefore, for calculating the quantities Lef, gh , Lgh, ef , Lij, ef and Lij, gh , the expressions given in equations (5.41), (5.42), (5.54) and (5.57) have been used respectively, with the impedances replaced by the corresponding reactances. After obtaining the values of these four quantities, the factors LOSFM1 and LOSFM2 have been calculated from equations (5.63) and (5.64). The calculated values of these two factors for all the lines remaining in the system are shown in columns 5 and 6 of Table 5.3. Please note that for lines 3 and 17, thes two factors have no relevance and therefore, these values have been indicated as zero in this table for these two lines. After obtaining these two factors, the change in power flow in line ‘i-j’ are calculated from equation (5.62) using the information of pre-outage power flow in line no. 3 and 17. For ready reference, the pre-outage power flows of line 3 and 17 are given in columns 3 and 4 of Table 5.3 respectively. After calculating the change in power flow of any line ‘i-j’, its post-outage power flow is obtained by adding its pre-outage power flow to the change in power flow. Again, for ready reference, the pre-outage line power flows for all the lines are given in column 2 of Table 5.3. The calculated post-outage power flows in all the lines are given in column 7 of this table. Please note that, as line no. 3 and 17 are now out of the circuit, the post-outage power flows in these two lines are zero. Finally, for the purpose of comparison, complete AC power flow solution of the IEEE-14 bus system has been obtained after removing line no. 3 and 17 from the system. The line power flows obtained through AC power flow study are shown in the last column of Table 5.3. From the results given in last two columns it is observed that there is some difference between the power flows calculated by complete AC power flow analysis and contingency analysis. Moreover, comparison of Tables 5.2 and 5.3 shows that this difference is more for double line contingencies as compared to that for single line contingency.

5.6

Contingency ranking and selection

From the results of the Tables 5.1 and 5.2 it can be observed that the sensitivity factors give reasonably close estimates of real power flows in the lines in the event of outage of a generator or a line. However, the sensitivity factors give the estimate of only the real power flows over the lines. On the other hand, in several situations, it is also equally important to consider the bus 245

Table 5.3: Contingency calculation for outage of line no. 3 and 17 (all powers are given in p.u.) Line Pline P3 P17 LOSFM1 no. (ori) 1 1.6262 0.8655 0.0655 -0.2053 2 0.7464 0.8655 0.0655 0.2177 3 0.8655 0.8655 0.0655 0 4 0.5314 0.8655 0.0655 0.4758 5 0.3662 0.8655 0.0655 0.3491 6 -0.358 0.8655 0.0655 -1.072 7 -0.7068 0.8655 0.0655 -0.5508 8 0.2689 0.8655 0.0655 -0.0274 9 0.1063 0.8655 0.0655 -0.0107 10 0.2893 0.8655 0.0655 0.0238 11 0.1156 0.8655 0.0655 0.0237 12 0.0852 0.8655 0.0655 0 13 0.2005 0.8655 0.0655 0 14 0.0025 0.8655 0.0655 -0.0097 15 0.2664 0.8655 0.0655 -0.0177 16 0.0123 0.8655 0.0655 -0.0237 17 0.0655 0.8655 0.0655 0 18 -0.0777 0.8655 0.0655 -0.0237 19 0.0233 0.8655 0.0655 0 20 0.0857 0.8655 0.0655 0

LOSFM2 -0.025 0.0237 0 -0.0678 0.0389 -0.004 0.4444 -0.3998 -0.1579 0.5427 -0.5152 0.2174 0.7805 -0.01 -0.3898 0.507 0 0.5111 0.2304 0.9991

Pline Pline (cal) (ACLF) 1.44687535 1.5449 0.9363717 0.9864 0 0 0.938764 0.9803 0.670894 0.7059 -1.286078 -1.1902 -1.1544092 -1.1585 0.2189984 0.218 0.0866967 0.0862 0.34544575 0.3668 0.10236675 0.1236 0.0994397 0.0994 0.25162275 0.2558 -0.00655035 0.0023 0.22554875 0.2157 0.02499615 0.0069 0 0 -0.0647353 -0.0833 0.0383912 0.0373 0.15114105 0.1533

voltage variations as well as the reactive power flows over the lines in the event of any outage. In these situations, the full AC power flow analysis needs to be carried out as the sensitivity factors are not able to estimate the changes in bus voltage and the reactive power flows in the lines. However, the full AC power flow analysis is considerably slower than the sensitivity based methods and these are not suitable for analyzing thousands of potential outage cases within the time frame required by on-line contingency analysis. Thus, we have a contradicting situation here. On one hand, for fast evaluation of contingencies, sensitivity based methods need to be used whereas, for accurate estimation of the effects of any outage, slower, full AC power flow analysis needs to be carried out. To break this dilemma, a middle path is followed. Initially, all the outage studies are carried out using the sensitivity factors. Based on the results of the sensitivity analysis, all the outage cases are ranked according to a suitably chosen performance index (PI). Once the outage cases are ranked and sequentially arranged in decreasing order of the performance index, the top few outage cases are analyzed further in detail using the AC power flow analysis. Therefore, for contingency ranking, the choice of the performance index is very important. The PI should be such that it should satisfy the following criteria. a. It should adequately reflect the severity of any particular contingency. b. The final list of contingencies for which full AC power flow analysis is to be carried out, should 246

not be too short or too long. c. The PI should consider both real and reactive power variation in the system. To achieve the above objectives, different performance indices are used. Below, some of the most prominent performance indices used are discussed. The PI can be categorized into two groups; i) MW ranking method (in which the changes in real power flows only are considered) and ii) reactive power or voltage security ranking (in which the variations of voltage magnitude or reactive power only are considered). Of course, for considering both real and reactive power variations, the performance indices from these two categories need to be suitably combined. Now, let us look at the various indices from these two categories.

5.6.1

MW ranking methods

i) The simplest form of the PI is; L

P I = ∑ Wj [ j=i

n

Pj Pj max

]

(5.65)

In equation (5.65), ‘L’ is the number of lines in the system, Pj and Pj max are MW flow and MW capacity of the line ‘j’ respectively and ‘n’ is a suitable index. However, the PI given in equation (5.65) is prone to masking phenomenon in which a contingency causing many lines to be heavily loaded with no lines being overloaded is ranked higher than a contingency causing few lines to be overloaded with the remaining lines being lightly loaded. However, this masking phenomenon can be removed if the summation in equation (5.65) is taken over only the set of overloaded lines as shown in equation (5.66).

P I = ∑ Wj [ j∈SL

n

Pj Pj max

]

(5.66)

In equation (5.66), SL denotes the set of overloaded lines. Even with the PI of equation (5.66), there is a chance of misranking. For example, if a contingency C1 causes many lines to be slightly overloaded and another contingency C2 causes some lines to be heavily overloaded, then C2 is more severe than C1 . However, equation (5.66) may identify C1 to be more severe than C2 . To prevent this, a two term PI can be used as shown in equation (5.67).

P I = ∣Hd1 ∣ + ∑ Wj [ n1

j∈SL

n2

Pj Pj max

]

(5.67)

In equation (5.67), ∣Hd1 ∣ is the change in power flow in the highest overloaded line while n1 and n2 are suitable indices. In case the highest overload in two cases of contingency are same, then the effect of second highest overloaded line is also taken into account as shown in equation (5.68).

P I = ∣Hd1 ∣ + ∣Hd2 ∣ + ∑ Wj [ n1

n2

j∈SL

Pj Pj max

n3

]

(5.68)

In equation (5.68), ∣Hd2 ∣ denotes the change in power flow in the second highest overloaded 247

line. Similarly, if both the highest and second highest overloading conditions are same for two contingencies, then the third highest overloaded line is taken into account separately.

P I = ∣Hd1 ∣ + ∣Hd2 ∣ + ∣Hd3 ∣ + ∑ Wj [ n1

n2

n3

j∈SL

Pj Pj max

n4

]

(5.69)

As before,∣Hd3 ∣ denotes the change in power flow in third highest overloaded line. Of course, this same philosophy can be extended further to include more number of lines for the calculation of PI, if necessary.

5.6.2

Voltage security/reactive power ranking methods

These are several PIs suggested for properly ranking the voltage /reactive power contingencies. Some of them are: i) 2 αi ∆Vi ] P Iv = ∑ [ lim i=1 2 ∆Vi N

Where,

(5.70)

1 ∆Vi = Vi − Visp ; ∆Vilim = (Vimax − Vimin ) ; 2 Vi = post-contingency voltage magnitude of bus ‘i’; Visp = Nominal or specified voltage of bus ‘i’; Vimax , Vimin = Maximum and minimum limit of voltage magnitudes of bus ‘i’ αi = User selected weighting factor P Iv = Performance index corresponding to voltage security N = number of buses in the system

ii)

∣Vi − Vilim ∣ P Iv = ∑ Wvi Vilim i∈S1

(5.71)

where,

Vilim = Vimax if Vi > Vimax = Vimin if Vi < Vimin

(5.72)

Wvi in the weighting factor for bus ‘i’ and S1 in the set of all buses at which the voltage limits have been violated. iii)

P Iv = ∑ Wvi (∆Vi )2 i∈S1

where,

∆Vi =

∆Vinom − 1; ∆Vilim

∆Vinom = Vi − Vinom ; 248

(5.73)

1 Vinom = (Vimax + Vimin ); 2

1 ∆Vilim = (Vimax − Vimin ) 2

iv) 2 N dmax dmin i P Iv = ∑ [ max ] + ∑ [ imin ] i=1 ai i=1 ai N

2

(5.74)

Where,

Vi − V amax if Vi > Viamax Vinom = 0 otherwise

(5.75)

Vimin − Vi if Vi < Viamin nom Vi = 0 otherwise

(5.76)

dmax = i

dmin = i

amax = i

1 V

nom i

(Vimax − Viamax );

amin = i

1 Vinom

(Viamin − Vimin )

Viamax and Viamin are the higher and lower volatage alarm limits for bus ‘i’. v)

⎡ ⎤2 ⎢ ⎥ N ⎢ o nom ⎥ V + ∆V − V ⎢ ⎥ i i ⎥ P Iv = ∑ ⎢ i max min ⎢ ⎥ V − V i=1 ⎢ i i ⎥ ⎢ ⎥ 2 ⎣ ⎦

(5.77)

∆Vi is the post-contingency change in bus voltage magnitude of bus ‘i’. vi) m

N

1

m ∆Vi ∣ ] P Iv = [∑ Wi ∣ ∆Vimax i=1

(5.78)

The value of ‘m’ is taken to be very large (≈ 20) to avoid the masking effect. One popular way of combining the above discussed real power and reactive power ranking method is to use the 1P1Q method. In this method, a decoupled power flow is used and after one iteration, (one P-Q computation and one Q-V computation), the bus voltages are noted and with these bus voltages, the line power flows are calculated. It generally appears that there is sufficient information available (in the bus voltages) to arrive at a reasonable values of the performance indices. After calculating the real power and reactive power performance indices separately, the combined PI is calculated by adding the appropriate real power and reactive power indices together. With these descriptions of contingency ranking methods, we conclude our discussion of contingency analysis. From the next lecture, we will start our discussion of stability analysis.

249

Module 6 Power system stability 6.1

Introduction

In general terms, power system stability refers to that property of the power system which enables the system to maintain an equilibrium operating point under normal conditions and to attain a state of equilibrium after being subjected to a disturbance. As primarily synchronous generators are used for generating power in grid, power system stability is generally implied by the ability of the synchronous generators to remain in ’synchronism’ or ’in step’. On the other hand, if the synchronous generators loose synchronism after a disturbance, then the system is called unstable. The basic concept of ’synchronism’ can be explained as follows. In the normal equilibrium condition, all the synchronous generators run at a constant speed and the difference between the rotor angles of any two generators is constant. Under any disturbance, the speed of the machines will deviate from the steady state values due to mismatch between mechanical and electrical powers (torque) and therefore, the difference of the rotor angles would also change. If these rotor angle differences (between any pair of generators) attain steady state values (not necessarily the same as in the pre-disturbance condition) after some finite time, then the synchronous generators are said to be in ’synchronism’. On the other hand, if the rotor angle differences keep on increasing indefinitely, then the machines are considered to have lost ’synchronism’. Under this ’out of step’ condition, the output power, voltage etc. of the generator continuously drift away from the corresponding pre-disturbance values until the protection system trips the machine. The above phenomenon of instability is essentially related with the instability of the rotor angles and hence, this form of instability is termed as ’rotor angle instability’. Now, as discussed above, this instability is triggered by the occurrence of a disturbance. Depending on the severity of the disturbance, the rotor angle instability can be classified into two categories: • Small signal instability: In this case, the disturbance occurring in the system is small. Such kind of small disturbances always take place in the system due to random variations of the loads and the generation. It will be shown later in this chapter that under small perturbation (or disturbance), the change in the electrical torque of a synchronous generator can be resolved into two components, namely, a) synchronizing torque (Ts ) - which is proportional to the change in the rotor angle and b) damping torque (Td ), which is proportional to the change in the speed 250

of the machine. As a result, depending on the amounts of synchronizing and damping torques, small signal instability can manifest itself in two forms. When there is insufficient amount of synchronizing torque, the rotor angle increases steadily. On the other hand, for inadequate amount of damping torque, the rotor angle undergoes oscillations with increasing amplitude. These two phenomena are illustrated in Fig. 6.1. In Fig. 6.1(a), both the synchronizing and damping torques are positive and sufficient and hence, the rotor angle comes back to a steady state value after undergoing oscillations with decreasing magnitude. In Fig. 6.1(b), the synchronizing torque is negative while the damping toque is positive and thus, the rotor angle envelope is increasing monotonically. Fig. 6.1(c) depicts the classic oscillatory instability in which Ts is positive while Td is negative.

Figure 6.1: Influence of synchronous and damping torque In an integrated power system, there can be different types of manifestation of the small signal instability. These are: 251

a. Local mode: In this type, the units within a generating station oscillate with respect to the rest of the system. The term ’local’ is used because the oscillations are localized in a particular generating station. b. Inter-area mode: In this case, the generators in one part of the system oscillate with respect to the machines in another part of the system. c. Control mode: This type of instability is excited due to poorly damped control systems such as exciter, speed governor, static var compensators, HVDC converters etc. d. Torsional mode: This type is associated with the rotating turbine-governor shaft. This type is more prominent in a series compensated transmission system in which the mechanical system resonates with the electrical system. • Transient instability: In this case, the disturbance on the system is quite severe and sudden and the machine is unable to maintain synchronism under the impact of this disturbance. In this case, there is a large excursion of the rotor angle (even if the generator is transiently stable). Fig. 6.2 shows various cases of stable and unstable behavior of the generator. In case 1, under the influence of the fault, the generator rotor angle increases to a maximum, subsequently decreases and settles to a steady state value following oscillations with decreasing magnitude. In case 2, the rotor angle decreases after attaining a maximum value. However, subsequently, it undergoes oscillations with increasing amplitude. This type of instability is not caused by the lack of synchronizing torque; rather it occurs due to lack of sufficient damping torque in the post fault system condition. In case 3, the rotor angle monotonically keeps on increasing due to insufficient synchronizing torque till the protective relay trips it. This type of instability, in which the rotor angle never decreases, is termed as ’first swing instability’.

Figure 6.2: Illustration of various stability phenomenon

Apart from rotor angle instability, instability can also occur even when the synchronous generators are maintaining synchronism. For example, when a synchronous generator is supplying power to an induction motor load over a transmission line, the voltage at the load terminal can progressively 252

reduce under some conditions of real and reactive power drawn by the load. In this case, loss of synchronism is not an issue but the challenge is to maintain a stable voltage. This type of instability is termed as voltage instability or voltage collapse. We will discuss about the voltage instability issue later in this course. Now, for analysing rotor angle stability, we have to first understand the basic equation of motion of a synchronous machine, which is our next topic.

6.2

Equation of motion of a synchronous machine

The equation of motion of a synchronous generator is based on the fact that the accelerating torque is the product of inertia and its angular acceleration. In the MKS system, this equation can be written as,

J

d2 θm = Ta = Tm − Te dt2

(6.1)

In equation (6.1),

J → The total moment of inertia of the rotor masses in Kg − m2 θm → The angular displacement of the rotor with respect to a stationary axis, in mechanical radians

t → Time in seconds Ta → The net accelerating torque, in N-m Tm → The mechanical or shaft torque supplied by the prime mover less retarding torque due to rotational losses, in N-m

Te → The net electrical or electromagnetic torque in N-m Under steady state operation of the generator, Tm and Te are equal and therefore, Ta is zero. In this case, there is no acceleration or deceleration of the rotor masses and the generator runs at constant synchronous speed. The electrical torque Te corresponds to the air gap power of the generator and is equal to the output power plus the real power loss of the armature winding. Now, the angle θm is measured with respect to a stationary reference axis on the stator and hence, it is an absolute measure of the rotor angle. Thus, it continuously increases with time even with constant synchronous speed. However, in stability studies, the rotor speed relative to the synchronous speed is of interest and hence, it is more convenient to measure the rotor angular position with respect to a reference axis which also rotates at synchronous speed. Hence, let us define, θm = ωsm t + δm (6.2) In equation (6.2), ωsm is the synchronous speed of the machine in mechanical radian/sec. and δm (in mechanical radian) is the angular displacement of the rotor from the synchronously rotating reference axis. From equation (6.2),

dδm dθm = ωsm + dt dt

or, 253

dδm dθm = − ωsm dt dt

(6.3)

d2 θm d2 δm = (6.4) dt2 dt2 dδm represents the deviation of the actual rotor speed Equation (6.3) shows that the quantity dt

from the synchronous speed in mechanical radian per second. Substituting equation (6.4) into equation (6.1) one gets,

J

d2 δm = Ta = Tm − Te dt2

(6.5)

Now. let us define the angular velocity of the rotor to be ωm =

Jωm

dθm . From equation (6.5) we get, dt

d2 δm = ωm Ta = ωm Tm − ωm Te dt2

Or,

Jωm

d2 δm = Pa = Pm − Pe dt2

(6.6)

In equation (6.6), Pa , Pe and Pm denote the accelerating power, electrical output power and the input mechanical power (less than the rotational power loss) respectively. The quantity Jωm is the angular momentum of the rotor and at synchronous speed, it is known as the inertia constant and is denoted by M . Strictly, the quantity Jωm is not constant at all operating conditions since ωm keeps on varying. However, when the machine is stable, ωm does not differ significantly from ωsm and hence, Jωm can be taken approximately equal to M . Hence, from equation (6.6) we obtain,

M

d2 δm = Pa = Pm − Pe dt2

(6.7)

Now, in machine data, another constant related to inertia, namely H-constant is often encountered. This is defined as;

H=

stored kinetic energy in megajoules at synchronous speed machine rating in MVA

Or,

1 Jωsm 2 1 M ωsm H= = 2 Smc 2 Smc

MJ/MVA =

1 M ωsm 2 Smc

sec.

(6.8)

In equation (6.8), the quantity Smc is the three phase MVA rating of the synchronous machine. Now, from equation (6.8),

M=

2HSmc ωsm

MJ/mech. rad

(6.9)

Substituting for M in equation (6.7), we get,

2H d2 δm Pa Pm − Pe = = ωsm dt2 Smc Smc

(6.10)

In equation (6.10), both δm and ωsm are in mechanical units. Now, the corresponding quantities 254

in electrical units are given as,

ωs =

P ωsm ; 2

δ=

P δm ; 2

(6.11)

In equation (6.11), P is the number of pole in the generator, ωs is the synchronous speed of the machine in electrical radian/sec. and δ (in electrical radian) is the angular displacement of the rotor from the synchronously rotating reference axis. Substituting equation (6.11) in equation (6.10) we get,

2H d2 δ = Pa = Pm − Pe per unit ωs dt2

(6.12)

Equation (6.12) is known as the swing equation of the synchronous machine. As this is a second order differential equation, it can be written as a set of two first order differential equations as below.

2H dω = Pm − Pe per unit ωs dt

(6.13)

dδ = ω − ωs dt

(6.14)

In equations (6.13) - (6.14), the quantity ω is the speed of the synchronous machine and is expressed in electrical radian per second. Now, in the above two equations, no damping of the machine is considered. If damping is considered (which opposes the motion of the machine), a term proportional to the deviation of the speed (from the synchronous speed) is introduced in equation (6.13). Therefore, the modified equation becomes;

2H dω = Pm − Pe − d(ω − ωs ) per unit ωs dt

(6.15)

In equation (6.15), d is called the damping co-efficient. However, in the presence of damping, equation (6.14) does not change. Therefore, in the presence of damping, this pair of equations ((6.14) and (6.15)) describe the motion of the synchronous machine. With this introduction of motion of synchronous machine, we are now ready to address the various stability issues. From the next lecture we will start with transient stability analysis.

255

6.3

Transient stability analysis

Before solving the differential equations to determine the transient stability or instability of the system, it is necessary to compute the initial conditions.

6.3.1

Computation of initial conditions

Let us consider an ‘n’ bus power system with ‘m’ generators (m < n). Without loss of generality, it is assumed that the ‘m’ generators are located at first ‘m’ buses of the system. Towards computation of initial conditions, initially the load flow solution of the system is computed. From the load flow solution, following quantities are availbale: a. Vi ∠θi

for i = 1, 2, ⋯⋯ n

b. PLi , QLi

for i = m + 1, m + 2, ⋯⋯ n

c. PLi , QLi , PGi , QGi

for i = 1, 2, ⋯⋯ m

In the above, PLi and QLi denote the real and reactive power load at bus ‘i’ respectively. Similarly, PGi and QGi denote the real and reactive power generation at bus ‘i’ respectively. Further, the quantities Vi and θi denote the voltage magnitude and angle of it h bus respectively. With these information, following calculations are carried out: (i) At any bus ‘i’, the loads are converted to equivalent admittance as;

y¯Li =

PLi − jQLi Vi 2

for i = 1, 2, ⋯⋯ n

(6.16)

(ii) Augment the YBUS matrix of the system as; old ¯ BUS (i, i) = Y ¯ BUS Y (i, i) + y¯Li

for i = 1, 2, ⋯⋯ n

(6.17)

old ¯ BUS ¯ BUS matrix of the system used in load flow calculation. where, Y is the original Y (iii) At the generator buses, the generators are represented as equivalent voltage sources behind the direct axis transient reactances as,

PGi − jQGi E¯i = V¯i + jx∕di I¯i = jx∕di = Ei ∠δi V¯i∗

for i = 1, 2, ⋯⋯ m

(6.18)

In equation (6.18), the quantity x∕di denotes the direct axis transient reactance of the ith machine. While performing the transient stability analysis, the magnitude ∣Ei ∣ is held constant. The equivalent diagram of the ‘n’ bus power system is shown in Fig. 6.3. With the initial conditions computed as above, we are now ready to solve the transient stability problem. Basically, the transient stability problem is solved by two techniques: i) partition explicit (PE) method and ii) simultaneous implicit (SI) method. In the PE method, the network algebraic equations and the generator differential equations are solved separately. In the SI method, these 256

Figure 6.3: Equivalent representation of generators for transient stability analysis

algebraic and differential equations are solved together. In this course, however, we are going to discuss PE method only. Before, discussing the PE method, it is necessary to describe the network algebraic equations.

6.3.2

Network algebraic equations

The network algebraic equations are represented as;

¯IBUS = Y ¯ BUS V ¯ BUS

(6.19)

¯ BUS is the bus impedance matrix of the system and In equation (6.19), Y ¯IBUS = [I¯1 I¯2 ⋯⋯ I¯m I¯m+1 ⋯⋯ I¯n ]T ¯ BUS = [V¯1 V¯2 ⋯⋯ V¯m V¯m+1 ⋯⋯ V¯n ]T V 257

(6.20)

The voltage current relationship of the generator reactance is given by,

v¯k = jx∕di¯ik

⇒

¯ik = y¯k v¯k

(6.21)

In equation (6.21), ¯ik and v¯k are the current through and voltage across the generator reactance

1 = −j/x∕di . Now, at the generator terminals, I¯k = ¯ik = y¯k v¯k ; or, I¯k = jx∕di y¯k (E¯k − V¯k ) as (v¯k = E¯k − V¯k ). Or, I¯k + y¯k V¯k = y¯k E¯k (6.22)

respectively and y¯k =

From equations (6.22) and (6.23) we get,

(Y¯11 + y¯1 ) V¯1 Y¯21 V¯1 ⋮ ¯ Ym1 V¯1 Y¯(m+1),1 V¯1 ⋮ Y¯n1 V¯1

+ Y¯12 V¯2 + (Y¯22 + y¯2 ) V¯2 + ⋮ ¯ + Ym2 V¯2 + Y¯(m+1),2 V¯2 + ⋮ + Y¯n2 V¯2

+ + + + + + +

⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯

+ Y¯1m V¯m + Y¯2m V¯m + ⋮ ¯ + (Ymm + y¯m ) V¯m + ⋯⋯⋯ + ⋮ + ⋯⋯⋯

+ + + + + + +

⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯

+ Y1n V¯n + Y2n V¯n + ⋮ + Ymn V¯n + Y¯(m+1),n V¯n + ⋮ + Y¯nn V¯n

= y¯1 E¯1 = y¯2 E¯2 = ⋮ = y¯m E¯m = 0 = ⋮ = 0

⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

(6.23)

¯1 , From equation (6.23), the voltages V¯1 , V¯2 ⋯⋯ V¯n can be solved, for known values of E E¯2 ⋯⋯ E¯n . With these known terminal voltages, the electrical power output of each generator can be calculated as; for i = 1, 2, ⋯ m Pei = Re (E¯I I¯i∗ ) (6.24) Where,

I¯i = y¯i (E¯i − V¯i )

for i = 1, 2, ⋯ m

(6.25)

With the above equations, we are now in a position to discuss the PE method.

6.3.3

Partition explicit solution scheme

In the PE method, the numerical integration of the generator differential equations are carried out separately from the solution of the network algebraic equations. For numerical integration of differential equations, the total simulation time (tT ) is divided into N intervals, each interval being

tT

of duration ∆t seconds. Thus, ∆t = . Now, the major steps for solving the transient stability N problem, with PE method, are as follows. 1. Obtain the load flow solution of the given system. Thereafter, compute the internal ¯i , for i = 1, 2, ⋯⋯ m) using equations (6.16) - (6.18). Please note voltages of all the generators (E that the magnitudes Ei would be kept constant at these calculated values throughout the simulation.

¯i obtained above, solve the equation set (6.23) to obtain the terminal 2. With the values of E voltages at all the buses (V¯i , for i = 1, 2, ⋯⋯ n). Subsequently, the electrical power output of all the generators (P¯ei , for i = 1, 2, ⋯⋯ m) are computed from equations (6.24) - (6.25). 258

3. Under steady state condition, mechanical power input to each generator is equal to its generator electrical output power (neglecting losses). Therefore, Pmi = Pei , for i = 1, 2, ⋯⋯ m. Also, under steady state, all the generators are assumed to operate at synchronous speed (ωs ). 4. Thus, after the above three steps, at t = 0, the variables pertaining to generators (Ei , δi , ωi , Pmi for i = 1, 2, ⋯⋯ m) and the network bus voltages (V¯i , for i = 1, 2, ⋯⋯ n) are all known.

5. The simulation process advances to t = ∆t. At this instant, first the network equations given in the equation set (6.23) are solved to compute the bus voltages and subsequently, the output electrical power of each generator is calculated by using equations (6.24) - (6.25). Now, if the steady state condition is still maintained, i.e. if there is no change in the network (as compared to the network condition at t = 0), then the calculated values of Pei would be again equal to the corresponding value of Pmi . 6. With the solution of the network equations at hand, we should now solve the solve the generator differential equations for calculating the values of δi and ωi at t = ∆t. Towards that end, let us first re-write the swing equations of ith machine for convenience in equations (6.26) - (6.27) below. 2Hi dωi = Pmi − Pei (6.26)

ωs dt

dδi = ωi − ωs (6.27) dt dωi Now, from equation (6.26), as Pmi = Pei , = 0. In other words, there is no change in the dt dδi speed of the generator and hence ωi = ωs . As a result, from equation (6.27), = 0 and hence, the dt angle δi would also be maintained at the value calculated at t = 0. Therefore, under steady state condition, at t = ∆t, both δi and ωi would be maintained at the values calculated at t = 0. 7. Assume that the steady state condition continues from t = 0 to t = (k − 1)∆t, where, k is a positive integer. Following the arguments described at steps 5 and 6, it can be easily seen that at the end of t = (k − 1)∆t sec., both δi and ωi would still be maintained at the values calculated at t = 0. 8. Now, let us assume that a three phase to ground short circuit fault occurs in the system at t = to = k∆t at the `th bus. To accommodate this fault condition in the network equations, the element Y¯`` is increased manyfold to reflect very high admittance from bus ‘`’ to ground. With this imposed condition, the network bus voltages are calculated from equation set (6.23). Subsequently, the output electrical power of each generator is calculated by using equations (6.24) - (6.25) corresponding to time t = to . 9. With the values of Pei calculated in step 8, the swing equations (6.26) - (6.27) are integrated to obtain the values of δi and ωi at t = to . Now, for integrating the swing equations, initial values of δi and ωi are required. These initial values are taken to be equal to the values of δi and ωi obtained at the end of t = (k − 1)∆t. For brevity, the value of δi (i = 1, 2, ⋯⋯ m) calculated at t = to is denoted as δi (to ). With this value of δi (to ), the voltage behind the transient reactance is updated ¯i = Ei ∠δi (to ), for i = 1, 2, ⋯⋯ m, where the magnitude Ei has been taken to be equal to the as E constant value calculated at step 1. 259

10. The simulation advances to t = to + ∆t. If the fault still persists, the values of Pei are ¯i in equation set (6.23) calculated as described in step 8. However, for this purpose, the values of E are taken to be equal to the latest values calculated at the end of t = to sec. 11. With the values of Pei calculated at t = to + ∆t, step 9 is repeated to update the variables ωi , δi and E¯i at the end of t = to + ∆t. Again, please note that for integrating the differential equations, the latest values of ωi and δi (calculated at t = to ) have been taken as the initial values. 12. Steps 10 and 11 are repeated to update the variables at t = to + 2∆t, to + 3∆t, to + 4∆t, ⋯⋯ till the fault clears. 13. At t = tcl (tcl = p∆t, p being a positive integer and p > k ), the fault clears. This condition is imposed in the network equations by restoring Y¯`` to its original pre-fault value and subsequently, steps 8 - 12 are repeated to obtain the variations of δi and ωi . By observing the variation of δi , the stability of the system is assessed. In step 9, the generator differential equations are numerically integrated. In the next lecture, we will look into two such numerical integration techniques, namely, i) modified Euler’s method and ii) 4th order Runga-Kutta technique.

260

6.3.4

Modified Euler’s method of integration

Before discussing the application of Euler’s method for solving the swing equations, let us first review the basic Euler’s method of numerical integration. Let the general from of a differential equation is given by;

dy = f (x, y); dx

y(xo ) = yo ;

(6.28)

In equation (6.28), x and y are independent and dependent quantities respectively and xo and yo are initial values of x and y respectively. For the purpose of numerical integration, the independent axis (x-axis) is divided into intervals of length ‘h’ such that discrete points on the independent axis are xo , xo + h, xo + 2h, ⋯⋯ etc. As indicated in equation (6.28), the value of y at x = xo is yo . The task is to calculate the values y1 , y2 , ⋯⋯ corresponding to the x co-ordinates xo + h, xo + 2h, ⋯⋯ respectively. Once these values are obtained, the smooth curve representing the solution of the differential equation given in equation (6.28) can be plotted. In the modified Euler’s method, the values y1 , y2 , ⋯⋯ are calculated in two steps: Predictor (1)

In this step, the approximate value of y1 (denoted as y1 ) as;

y1(1) = yo + h

dy ∣ = yo + hf (xo , yo ) dx x=xo

Corrector (1)

a) With the calculated value of y1 , calculate the approximate value of

dy ∣ = f (xo + h, y1(1) ) dx x=xo +h b) With this updated value of

(6.29)

dy at x = xo + h as; dx (6.30)

dy at x = x1 = xo + h, the final value of y1 is calculated as; dx dy h dy [ ∣ + ∣ ] 2 dx x=xo dx x=xo +h h = yo + [f (xo , yo ) + f (xo + h, y1(1) )] 2

y1 = yo + or,

y1

(6.31)

With this final value of y1 obtained at x = xo + h = x1 , the above two steps are repeated to calculate y2 at x = xo + 2h = x2 and subsequently, this process is repeated to obtain the complete solution of the differential equation. Now, for our application, let us note that the independent axis (x-axis) denotes time. Therefore, as already discussed, for solving the differential equations, the time axis is divided into intervals of duration ∆t sec. (i.e. h = ∆t). Further, let us also assume that the values of δi and ωi (i = 1, 2, ⋯⋯ m) have already been obtained at t = to and these values are denoted as δio and ωio respectively. More(o) over, the initial values of Pei (denoted as Pei ) are also assumed to be calculated utilising the values 261

of δio . With thse known values, the values of δi and ωi at t = to + ∆t are calculated as follows. Predictor step (1)

In this step, the approximate values of δi and ωi are calculated as (denoted as δi respectively);

dδi ∣ = δio + ∆t(ωio − ωs ) dt t=to

(6.32)

dωi ωs (Pmi − Pei(o) ) ∣ = ωio + ∆t dt t=to 2Hi

(6.33)

δi(1) = δio + ∆t ωi(1) = ωio + ∆t

(1)

and ωi

Corrector step (1)

With the new values of δi (i = 1, 2, ⋯⋯ m) obtained in the predictor step, the values of Pei (i = 1, 2, ⋯⋯ m) are updated using equations (6.23)-(6.25) (after appropriately incorporating the (1) network conditions in the equation set (6.23)). Let these updates values of Pei be denoted as Pei . Thereafter, the derivatives at the end of the present time step are calculated as follows:

dδi ∣ = ωi(1) − ωs dt t=to +∆t

(6.34)

dωi ωs ∣ = (Pmi − Pei(1) ) dt t=to +∆t 2Hi

(6.35)

With the above new derivative values obtained, the final values of δi and ωi at t = to +∆t (denoted as δi1 and ωi1 respectively) are calculated as;

δi1 = δio +

∆t dδi dδi [ ∣ + ∣ ] 2 dt t=to dt t=to +∆t

(6.36)

ωi1 = ωio +

∆t dωi dωi [ ∣ + ∣ ] 2 dt t=to dt t=to +∆t

(6.37)

Proceeding further, for calculating δi and ωi at t = to +2∆t, the quantities δio and ωio are replaced by δi1 and ωi1 respectively and equations (6.32)-(6.37) are followed again.

6.3.5

Runga Kutta 4th order method of integration

Let us again consider the same general form of a differential equation as in equation (6.28):

dy = f (x, y); dx

y(xo ) = yo ;

(6.38)

Again, the meanings of different notations used in equation (6.38) are same as those in equation 262

(6.28). In RK 4th order method, the value y1 (corresponding to x = xo + h) is calculated as;

h y1 = yo + (k1 + 2k2 + 2k3 + k4 ) 6

(6.39)

k1 = hf (xo , yo )

(6.40)

In equation (6.39),

k1 h k2 = hf (xo + , yo + ) 2 2 h k2 k3 = hf (xo + , yo + ) 2 2 k4 = hf (xo + h, yo + k3 )

(6.41) (6.42) (6.43)

Now, for solving the transient stability problem with RK 4th order method, let us again assume that the value of δi and ωi (i = 1, 2, ⋯⋯ m) are known at t = to (denoted as δio and ωio respectively). Moreover, the values of Pei are also assumed to be known (calculated utilising the values of δio ). From these initial values, the procedure of calculation of δi1 and ωi1 (values of δi and ωi at t = to + ∆t) is as follows.

Calculation of first estimate of the derivatives In this step, the first estimates of the derivatives for the ith machine (i = 1, 2, ⋯⋯ m) are calculated as:

dδi (1) ∣ = ωio − ωs dt

(6.44)

dωi (1) ωs [Pmi − Pei(o) ] ∣ = dt 2Hi

(6.45)

With these first estimates of the derivatives, the values of δi and ωi (i = 1, 2, ⋯⋯ m) are updated as: (1) i

1 dδi (1) = δio + ∆t ∣ 2 dt

(6.46)

(1) i

1 dωi (1) = ωio + ∆t ∣ 2 dt

(6.47)

δ

ω (1)

With these values of δi (i = 1, 2, ⋯⋯ m), the values of Pei (i = 1, 2, ⋯⋯ m) are updated (1) from equations (6.23) - (6.25). Let these newly calculated values of Pei be denoted as Pei (i = 1, 2, ⋯⋯ m). We now proceed to the next step.

Calculation of second estimate of the derivatives 263

The second estimates of the derivatives are calculated as:

dδi (2) ∣ = ωi(1) − ωs dt

(6.48)

dωi (2) ωs [Pmi − Pei(1) ] ∣ = dt 2Hi

(6.49)

With these second estimates of the derivatives, the values of δi and ωi (i = 1, 2, ⋯⋯ m) are updated as:

1 dδi (2) δi(2) = δio + ∆t ∣ 2 dt

(6.50)

1 dωi (2) = ωio + ∆t ∣ 2 dt

(6.51)

ω

(2) i

(2)

With these values of δi , the values of Pei are again updated from equations (6.23) - (6.25). Let (2) these newly calculated values of Pei be denoted as Pei (i = 1, 2, ⋯⋯ m). We now proceed to the next step.

Calculation of third estimate of the derivatives The third estimates of the derivatives are calculated as;

dδi (3) ∣ = ωi(2) − ωs dt

(6.52)

dωi (3) ωs [Pmi − Pei(2) ] ∣ = dt 2Hi

(6.53)

With these third estimates of the derivatives, the values of δi and ωi (i = 1, 2, ⋯⋯ m) are updated as; (3) i

dδi (3) = δio + ∆t ∣ dt

(6.54)

(3) i

dωi (3) = ωio + ∆t ∣ dt

(6.55)

δ

ω (3)

With these values of δi , the values of Pei are again updated from equations (6.23) - (6.25). Let (3) these newly calculated values of Pei be denoted as Pei (i = 1, 2, ⋯⋯ m). We now proceed to the next step.

Calculation of fourth estimate of the derivatives 264

The fourth estimates of the derivatives are calculated as:

dδi (4) ∣ = ωi(3) − ωs dt

(6.56)

dωi (4) ωs [Pmi − Pei(3) ] ∣ = dt 2Hi

(6.57)

After the fourth estimates are obtained, we are now in a position to calculate δi1 and ωi1 (i = 1, 2, ⋯⋯ m). Calculation of final values The final, updated values are calculated as:

∆t dδi (1) dδi (2) dδi (3) dδi (4) δi1 = δio + [ ∣ +2 ∣ +2 ∣ + ∣ ] 6 dt dt dt dt

(6.58)

dωi (2) dωi (3) dωi (4) ∆t dωi (1) [ ∣ +2 ∣ +2 ∣ + ∣ ] ωi1 = ωio + 6 dt dt dt dt

(6.59)

Proceeding further, for calculating δi and ωi at t = to +2∆t, the quantities δio and ωio are replaced by δi1 and ωi1 respectively and equations (6.44)-(6.59) are followed again. In the next lecture, we will illustrate the applicaion of Modified Euler’s method for transient stability calculation.

265

6.3.6

Example with Modified Euler’s method

As an illustration of the Modified Euler’s method, let us consider a three machine, 9 bus system. The schematic diagram of this system is shown in Fig. 6.4. The bus data of this system is given in Table A.7 while the line data are given in Table A.8. Further, the values of x∕di , di (damping constant) and Hi (inertia constant) for all the three generators are given in Table 6.1. With the data given in Tables A.7 and Table A.8, the load flow solution of this system has been carried out and the load flow results are given in Table 6.2. With the help of load flow results and ¯ and δ ) of the internal voltages of all the generators values of x∕di , the magnitudes and angles (∣E∣ have been calculated by utilising equations (6.16)-(6.18) and are also shown in Table 6.2. It is to ¯i ∣ (i = 1, 2, 3) are to be noted that, throughout the transient stability simulation, the values of ∣E be kept constant at the values given in Table 6.2. Also, at steady state, the speed of all generators are assumed to be equal to ωs (i.e. ωio = ωs ; i = 1, 2, 3). Now, following the arguments given in step 7 of sub-section 6.3.3, under steady state, the values of δi (i = 1, 2, 3) will remain constant at those values given in Table 6.2. Similarly, under steady state, the values of ωi (i = 1, 2, 3) will all be equal to ωs (= 376.9911184307752 rad./sec for a 60 Hz. system). Moreover, for transient stability simulation, a time step of 0.001 sec. (∆t = 0.001) has been taken. Further, to start with, the damping of the generators have been neglected.

Figure 6.4: 3 machine, 9 bus system Now, let us assume that a three phase to ground fault takes place at bus 7 at t = 0.5 sec. To simulate this fault, the element Y¯77 is increased 1000 times to represent very high admittance to ground. With this modification in Y¯77 , the equation set (6.23) are solved to calculate the faulted 266

Table 6.1: Machine data for 9 bus system Gen. x∕di di Hi no. (p.u) 1 0.0608 0.0254 23.64 2 0.1198 0.0066 6.40 3 0.1813 0.0026 3.01

Table 6.2: Load flow result for 9 bus system Gen. PG QG ∣V ∣ θ ∣E∣ δ no. (MW) (MVAR) (p.u) (deg.) (p.u) (deg.) 1 71.64102147 27.0459235334 1.04 0.0 1.0566418430 2.2716458404 2 163.0 6.6536603184 1.025 9.2800054816 1.0502010147 19.7315857693 3 85.0 -10.8597090709 1.025 4.6647513331 1.0169664112 13.1664110346

values of V¯i (i = 1, 2, 3). These values are shown in second column of Table 6.3. With these (o) calculated values of the generator terminal voltages, the values of Pei (i = 1, 2, 3) have been calculated using equations (6.24)-(6.25) and are shown in third column of Table 6.3. Now, from Fig. 6.4, bus 7 is the terminal bus of generator 2 (just after the transformer) and therefore, for any short circuit fault at bus 7, the real power output of generator 2 is expected to fall drastically. Indeed, (o) from Table 6.3, after the fault, Pe2 is indeed very low (0.0003691143 p.u.). Also, note that the values of δio and ωio (i = 1, 2, 3), which are to be used for calculation at this time instant, are all equal to the corresponding steady state values as there was no fault prior to this time instant. Table 6.3: Calculations with Euler’s method for predictor stage at t = 0.5 sec. (damping = 0) Gen. no. 1 2 3

V¯ (p.u) 0.8515307492 - 0.0053320553i 0.3391142785 + 0.1215867116i 0.6169489129 + 0.0743553183i

(o)

With these values of Pei

Pe(o) (p.u) 0.6791748939 0.0003691143 0.3821503052

and ωio , the initial estimates of

dδi dωi and (i = 1, 2, 3) are dt dt

calculated from equations (6.27) and (6.26) respectively and are shown in columns 2 and 3 of Table (1) (1) 6.4 respectively. Finally, the values of δi and ωi (i = 1, 2, 3) are calculated by using equations (6.32) - (6.33) and are shown in columns 4 and 5 of Table 6.4 respectively. From Tables 6.1 - 6.4 it is observed that the steady state values of δi (for i = 1, 2, 3; shown in the last column of Table 6.1) (1) are equal to the corresponding values of δi (for i = 1, 2, 3; shown in the fourth column of Table

dδi (i = 1, 2, 3) are all equal to zero. dt (1) However, due to fall in Pei (i = 1, 2, 3), from equation (6.33), the values of ωi (i = 1, 2, 3) are all 6.4). This is due to the fact that at this predictor stage,

267

greater than ωs . Table 6.4: Calculations with Euler’s method for predictor stage at t = 0.5 sec. (damping = 0) Gen. no. 1 2 3

dδ dt 0 0 0

δ (1)

dω dt

ω (1)

(deg.) (rad/sec.) 0.2968990107 2.2716458404 376.9914153297 47.9965914231 19.7315857693 377.0391150221 29.2982026019 13.1664110346 377.0204166333

Once the calculations pertaining to the predictor stage at t = 0.5 sec. are over, we move on to the calculations pertaining to the corrector stage. Towards this goal, initially the values of Pei (1) (i = 1, 2, 3) are updated using the values of δi (i = 1, 2, 3) in equations (6.23) - (6.25). As (1) (1) discussed earlier, the updated value of Pei is denoted as Pei . Now, as the values of δi (i = 1, 2, 3) (1) are equal to the corresponding steady state values, the values of Pei (i = 1, 2, 3) are also equal (1) (1) to the values given in Table 6.3. Subsequently, with the values of Pei and ωi (i = 1, 2, 3), the values of

dωi dδi and (i = 1, 2, 3) at the end of the present time step are calculated from equations dt dt

(6.34) - (6.35) and are shown in columns 2 and 3 of Table 6.5 respectively. Lastly, the final values of δi and ωi (i = 1, 2, 3) at t = 0.5 sec. are calculated by using equations (6.36) - (6.37), which are shown in columns 4 and 5 of Table 6.5 respectively. Table 6.5: Calculations with Euler’s method for corrector stage at t = 0.5 sec. (damping = 0) Gen. δ ω dδ dω no. (deg.) (rad/sec.) dt dt 1 0.0002968990 0.2968990107 2.2716543459 376.9914153297 2 0.0479965914 47.9965914231 19.7329607703 377.0391150221 3 0.0292982026 29.2982026019 13.1672503663 377.0204166333

With these final values of δi and ωi (i = 1, 2, 3) corresponding to t = 0.5 sec. at hand, we now increment the time by ∆t (= 0.001 sec.) and repeat the calculations for predictor and corrector stages for t = 0.501 sec. The detailed calculations are shown in Tables 6.6 - 6.9. Initially, with the values of δi and ωi (i = 1, 2, 3) just obtained, the generator terminal voltages and the generator output electrical powers are calculated and are shown in Table 6.6. With these newly calculated

dδi dωi (1) (1) , , δi and ωi (i = 1, 2, 3) corresponding to the dt dt (1) (1) predictor stage are calculated and are shown in Table 6.7. With these updated values of δi and ωi (i = 1, 2, 3), the values of the generator terminal voltages and Pei (i = 1, 2, 3) are re-calculated values of Pei (i = 1, 2, 3), the values of

corresponding to the corrector stage and are shown in Table 6.8. Lastly, using these updated values

dδi dωi and corresponding to the corrector stage are calculated and dt dt are shown in Table 6.9. Finally, the values of δi and ωi (i = 1, 2, 3) at t = 0.501 sec. are calculated

of Pei (i = 1, 2, 3), the values of

by using equations (6.36)-(6.37), which are shown in columns 4 and 5 of Table 6.9 respectively. 268

Table 6.6: Caculations with Euler’s method for predictor stage at t = 0.501 sec. (damping = 0) Gen. no. 1 2 3

V¯ (p.u) 0.8515306823 - 0.0053313633i 0.3391113605 + 0.1215948479i 0.6169476981 + 0.0743625981i

Pe (p.u) 0.6791650241 0.0003691288 0.3821597463

Table 6.7: Calculations with Euler’s method for predictor stage at t = 0.501 sec. (damping = 0) Gen. δ (1) ω (1) dδ dω no. (deg.) (rad/sec.) dt dt 1 0.0002968990 0.2969777087 2.2716713570 376.9917123074 2 0.0479965914 47.9965909948 19.7357107724 377.0871116131 3 0.0292982026 29.2976113769 13.1689290296 377.0497142447

Table 6.8: Caculations with Euler’s method for corrector stage at t = 0.501 sec. (damping = 0) Gen. no. 1 2 3

V¯ (p.u) 0.8515305484 - 0.0053299791i 0.3391055238 + 0.1216111204i 0.6169452683 + 0.0743771578i

Pe (p.u) 0.6791452844 0.0003691579 0.3821786281

Table 6.9: Calculations with Euler’s method for corrector stage at t = 0.501 sec. (damping = 0) Gen. δ ω dδ dω no. (deg.) (rad/sec.) dt dt 1 0.0005938767 0.2971351045 2.2716798648 376.9917123861 2 0.0959931824 47.9965901381 19.7370857735 377.0871116127 3 0.0585958139 29.2964289348 13.1697683443 377.0497136535

For subsequent time instants, the calculations proceed exactly in the same way as described above. The fault is assumed to be cleared at t = 0.6 sec. To simulate this event (clearing of fault), the value of Y¯77 is restored to its pre-fault value and subsequently, the values of Pei (i = 1, 2, 3) are calculated from equations (6.23)-(6.25). Please note that, while doing so, the latest values of δi (i = 1, 2, 3) obatained at t = 0.599 sec. are used in equation set (6.23). For subsequent instances (beyond t = 0.6 sec.), the calculations proceed in the identical manner and finally, the simulation study is stopped at t = 5.0 sec. The variations of δi (i = 1, 2, 3) with respect to the center of inertia (COI) are shown in Fig. 6.5. For calculating the value of δi with respect to the COI (denoted as 269

δiCOI ) at each time step, the following expression has been used: δiCOI = δi − δCOI ;

for i = 1, 2, 3;

(6.60)

where, m

δCOI =

∑ Hi δi i=1 m

(6.61)

∑ Hj j=1

In equation (6.61), ‘m’ denotes the number of generators in the system (in our present case, m = 3). At each time step, with the final calculated values of δi (i = 1, 2, 3) calculated at the end of corrector step, the value of δCOI is computed and thereafter, each value of δiCOI (i = 1, 2, 3) is computed with the help of equation (6.60). With the values of δiCOI thus obtained for all time steps, the plots shown in Fig. 6.5 are obtained. These plots show that the generators experience sustained oscillations. This is due to the fact that in this study, the damping of the generators have been neglected.

Figure 6.5: Variation of δ1COI (with no damping) obtained with Euler’s method Let us now consider the damping of the generators. As discussed earlier (in the context of equations (6.14) and (6.15)), when damping is considered, an extra term (representing damping) is introduced in the differential equation corresponding to rate of change of speed. However, there would be no change in the differential equation representing the rate of change of generator angle. Therefore, the set of differential equations for ith generator is given by;

dδi = ωi − ωs dt

(6.62)

2Hi dωi = Pmi − Pei − di (ωi − ωs ) ωs dt

(6.63)

In equation (6.63), the extra term di (ωi − ωs ) represnts the damping of the generator. There would be, of course, no change in the algebraic equations and the set of algebraic equations would 270

still be represented by equation set (6.23). With these sets of differential and algebraic equations, the calculations proceed in identically the same way as described above and the variations of δiCOI (i = 1, 2, 3) for the same fault considered above are shown in Fig. 6.6. Comparison of these three plots with those shown in Fig. 6.5 shows that when damping of the generators are taken into consideration, the oscillations in all the three generators reduce gradually with time, which, indeed should be the case. As the generator oscillations are decreasing with time, the generators will remain in synchronism and therefore, the system is stable.

Figure 6.6: Variation of δ1COI (with damping) obtained with Euler’s method We will now discuss the application of Runga Kutta (RK) 4th order method of integration for solving the transient stability problem in the next lecture.

271

6.3.7

Example with Runga Kutta 4th order method

Again, as an example, 3 machine, 9 bus system shown in Fig. 6.4 is again considered. Initially, the damping of the generators are neglected (i.e. di = 0 for i = 1, 2, 3). The load flow results and the initial values of the magnitudes and angles (∣E∣ and δ ) of the internal voltages of all the generators are same as those already given in Table 6.2. As before, it is again assumed that at t = 0.5 sec., a three phase to ground short circuit fault takes place at bus 7. The faulted generator terminal voltages and generator output powers are same as those shown in Table 6.3. With these values of

P

(o) ei

dδi (1) dωi (1) and ωio , the estimates ∣ and ∣ (i = 1, 2, 3) are calculated from equations (6.44) dt dt

and (6.45) respectively and are shown in columns 2 and 3 of Table 6.10 respectively. With these first (1) (1) estimates of the derivatives, the values of δi and ωi (i = 1, 2, 3) are calculated from equations (6.46) - (6.47) and are shown in columns 4 and 5 of Table 6.10 respectively. Table 6.10: Calculations with RK method for first estimate at t = 0.5 sec. (damping = 0)

Gen. no.

dδi (1) ∣ dt

1 2 3

0 0 0

dωi (1) ∣ dt

δ (1)

ω (1)

(deg.) (rad/sec.) 0.2968990107 2.2716458404 376.99126688 47.9965914231 19.7315857693 377.01511672 29.2982026019 13.1664110346 377.00576753

(1)

With the values of δi (i = 1, 2, 3) obtained above, the generator terminal voltages and output powers are updated from equations (6.23) - (6.25) and are shown in Table 6.11. Please note that (1) following the notations used earlier, the output powers calculated at this stage are denoted as Pei (i = 1, 2, 3). Table 6.11: Caculations with RK method for second estimate at t = 0.5 sec. (damping = 0) Gen. no. 1 2 3

(1)

With the values of ωi

V¯ (p.u) 0.8515307492 - 0.0053320553i 0.3391142785 + 0.1215867116i 0.6169489129 + 0.0743553183i

(1)

Pe(1) (p.u) 0.6791748939 0.0003691143 0.3821503052

and Pei (i = 1, 2, 3) calculated above, the quantities

dδi (2) dωi (2) ∣ and ∣ dt dt

(i = 1, 2, 3) are calculated from equations (6.48) - (6.49) and are shown in columns 2 and 3 of Table (2) (2) 6.12 respectively. Subsequently, the values of δi and ωi (i = 1, 2, 3) are calculated from equations (6.50) - (6.51) and are shown in columns 4 and 5 of Table 6.12 respectively. (2)

Again, with the values of δi (i = 1, 2, 3) obtained above, the generator terminal voltages and output powers are updated from equations (6.23) - (6.25) and are shown in Table 6.13. Please note 272

Table 6.12: Calculations with RK method for second estimate at t = 0.5 sec. (damping = 0)

Gen. no. 1 2 3

dδi (2) ∣ dt

dωi (2) ∣ dt

δ (2)

ω (2)

(deg.) (rad/sec.) 0.0001484495 0.2968990107 2.2716500931 376.9912668802 0.0239982957 47.9965914231 19.7322732698 377.0151167264 0.0146491013 29.2982026019 13.1668307004 377.0057675320

that following the notations used earlier, the output powers calculated at this stage are denoted as Pei(2) (i = 1, 2, 3). Table 6.13: Caculations with RK method for third estimate at t = 0.5 sec. (damping = 0) Gen. no. 1 2 3

Pe(2)

V¯ (p.u) 0.8515307158 - 0.0053317093i 0.3391128195 + 0.1215907798i 0.6169483055 + 0.0743589582i (2)

Proceeding further, with the values of ωi

(p.u) 0.6791699590 0.0003691216 0.3821550258

(2)

and Pei (i = 1, 2, 3) calculated above, the quantities

dδi (3) dωi (3) ∣ and ∣ (i = 1, 2, 3) are calculated from equations (6.52) - (6.53) and are shown in dt dt (3) (3) columns 2 and 3 of Table 6.14 respectively. Subsequently, the values of δi and ωi (i = 1, 2, 3) are calculated from equations (6.54) - (6.55) and are shown in columns 4 and 5 of Table 6.14 respectively.

Table 6.14: Calculations with RK method for third estimate at t = 0.5 sec. (damping = 0)

Gen. no. 1 2 3

dδi (3) ∣ dt

dωi (3) ∣ dt

δ (3)

ω (3)

(deg.) (rad/sec.) 0.0001484495 0.2969383597 2.2716543459 376.9914153691 0.0239982957 47.9965912089 19.7329607703 377.0391150219 0.0146491013 29.2979069891 13.1672503663 377.0204163377 (3)

Again, with the values of δi (i = 1, 2, 3) obtained above, the generator terminal voltages and output powers are updated from equations (6.23) - (6.25) and are shown in Table 6.15. Please note that following the notations used earlier, the output powers calculated at this stage are denoted as Pei(3) (i = 1, 2, 3). Lastly, with the values of ω and

(3) i

and P

(3) ei

dδi (4) (i = 1, 2, 3) calculated above, the quantities ∣ dt

dωi (4) ∣ (i = 1, 2, 3) are calculated from equations (6.56) - (6.57) and are shown in columns 2 dt 273

Table 6.15: Calculations with RK method for final estimate at t = 0.5 sec. (damping = 0) Gen. no. 1 2 3

V¯ (p.u) 0.8515306823 - 0.0053313633i 0.3391113605 + 0.1215948479i 0.6169476981 + 0.0743625981i

Pe(3) (p.u) 0.6791650241 0.0003691288 0.3821597463

and 3 of Table 6.16 respectively. Finally, the values of δi and ωi (i = 1, 2, 3) at the end of t = 0.5 sec. are calculated from equations (6.58) - (6.59) and are shown in columns 4 and 5 of Table 6.16 respectively. Table 6.16: Calculations with RK method for final estimate at t = 0.5 sec. (damping = 0)

dδi (4) ∣ dt

Gen. no. 1 2 3

dωi (4) ∣ dt

δi

ωi

(deg.) (rad/sec.) 0.0002969383 0.2969777087 2.2716543463 376.9914153560 0.0479965912 47.9965909948 19.7329607703 377.0391150220 0.0292979069 29.2976113769 13.1672503634 377.0204164363

After the final values of δi and ωi (i = 1, 2, 3) corresponding to t = 0.5 sec. are obtained, we increment the time by ∆t (= 0.001 sec.) and repeat the calculations for t = 0.501 sec. Towards this goal, the quantities δi and ωi (i = 1, 2, 3) shown in Table 6.16 are substituted for δio and ωio (i = 1, 2, 3) in equations (6.44) - (6.59). With these values of δio (i = 1, 2, 3), equations (6.23) (6.25) are solved to calculate the initial values of Pei (i = 1, 2, 3) at t = 0.501 sec. The results are shown in Table 6.17. Table 6.17: Calculations with RK method for initial estimate at t = 0.501 sec. (damping = 0) Gen. no. 1 2 3

V¯ (p.u) 0.8515306823 - 0.0053313633i 0.3391113605 + 0.1215948479i 0.6169476981 + 0.0743625981i

Pe(o) (p.u) 0.6791650241 0.0003691288 0.3821597462

(o)

With the values of Pei (i = 1, 2, 3) thus obtained, calculations pertaining to the first estimate are carried out by using equations (6.44) - (6.47) and the results are shown in Table 6.18. Sub(1) sequently, the values of Pei (i = 1, 2, 3) are calculated and the results are shown in Table 6.19. (1)

(1)

(1)

With the values of Pei , δi and ωi (i = 1, 2, 3) obtained as above, the calculations pertaining (2) (2) (2) to second estimate are performed to obtain Pei , δi and ωi (i = 1, 2, 3). The results are shown 274

Table 6.18: Calculations with RK method for first estimate at t = 0.501 sec. (damping = 0)

dδi (1) ∣ dt

Gen. no. 1 2 3

dωi (1) ∣ dt

δ (1)

ω (1)

(deg.) (rad/sec.) 0.0002969252 0.2969777084 2.2716628526 376.9915638448 0.0479965912 47.9965909948 19.7343357714 377.0631133175 0.0292980055 29.2976113792 13.1680896895 377.0350652419

Table 6.19: Calculations with RK method for first estimate at t = 0.501 sec. (damping = 0) Gen. no. 1 2 3

Pe(1)

V¯ (p.u) 0.8515306154 - 0.0053306712i 0.3391084422 + 0.1216029842i 0.6169464833 + 0.0743698779i

(p.u) 0.6791551544 0.0003691434 0.3821691871

in Tables 6.20 and 6.21. Table 6.20: Calculations with RK method for second estimate at t = 0.501 sec. (damping = 0)

Gen. no. 1 2 3

dδi (2) ∣ dt

dωi (2) ∣ dt

δ (2)

ω (2)

(deg.) (rad/sec.) 0.0004454140 0.2970564057 2.2716671065 376.9915638842 0.0719948867 47.9965905664 19.7350232719 377.0631133173 0.0439468112 29.2970201614 13.1685093468 377.0350649463

Table 6.21: Calculations with RK method for second estimate at t = 0.501 sec. (damping = 0) Gen. no. 1 2 3

Pe(2)

V¯ (p.u) 0.8515305819 - 0.0053303251i 0.3391069830 + 0.1216070523i 0.6169458758 + 0.0743735177i (2)

(2)

(2)

(p.u) 0.6791502196 0.0003691506 0.3821739075

Proceeding further, using the values of Pei , δi and ωi (i = 1, 2, 3), the calculations pertaining (3) (3) (3) to third estimate are performed to obtain Pei , δi and ωi (i = 1, 2, 3). The results are shown in Tables 6.22 and 6.23. (3) (3) (3) Using the values of Pei , δi and ωi (i = 1, 2, 3), the fourth estimates of the derivatives are computed and subsequently, the final values of δi and ωi (i = 1, 2, 3) corresponding to t = 0.501 sec. are obtained. The calculations are shown in Table 6.24. 275

Table 6.22: Calculations with RK method for third estimate at t = 0.501 sec. (damping = 0)

Gen. no. 1 2 3

dδi (3) ∣ dt

dωi (3) ∣ dt

δ (3)

ω (3)

(deg.) (rad/sec.) 0.0004454534 0.2970957537 2.2716798689 376.9917124517 0.0719948865 47.9965903523 19.7370857735 377.0871116124 0.0439465156 29.2967245580 13.1697683133 377.0497131608

Table 6.23: Calculations with RK method for third estimate at t = 0.501 sec. (damping = 0) Gen. no. 1 2 3

V¯ (p.u) 0.8515304815 - 0.0053292869i 0.3391026052 + 0.1216192565i 0.6169440533 + 0.0743844372i

Pe(3) (p.u) 0.6791354153 0.0003691725 0.3821880684

Table 6.24: Calculations with RK method for final estimate at t = 0.501 sec. (damping = 0)

Gen. no. 1 2 3

dδi (4) ∣ dt

dωi (4) ∣ dt

δi

ωi

(deg.) (rad/sec.) 0.0005940209 0.2972137971 2.2716798685 376.9917124386 0.0959931816 47.9965897098 19.7370857735 377.0871116124 0.0585947300 29.2958377565 13.1697683161 377.0497132593

For subsequent time instants, the calculations proceed exactly in the same way as described above. As in the case with Euler’s method, in this case also, the fault is assumed to be cleared at t = 0.6 sec. and finally, the simulation study is stopped at t = 5.0 sec. The variations of δi (i = 1, 2, 3) with respect to the center of inertia (COI) are shown in Fig. 6.7 below. Please note that in this figure, no damping of the generators has been considered. The simulation studies have also been carried out by considering the damping of the generators. The variations of δi (i = 1, 2, 3) with respect to the center of inertia (COI) for this case are shown in Fig. 6.8 below. Comaprison of Figs. 6.7 - 6.8 with Figs. 6.5 - 6.6 reveals that the responses obtained with these two methods are almost identical to each other. With this example, we are now at the end of discussion of transient stability analysis. From the next lecture, we will start the discussion of small signal stability analysis.

276

Figure 6.7: Variation of δ1COI (with no damping) obtained with Runga-Kutta method

Figure 6.8: Variation of δ1COI (with damping) obtained with Runga-Kutta method

277

6.4

Small signal analysis

For small signal analysis of a multimachine power system, we need to linearise the differential equations of the machines and form the system A matrix. Thereafter, by computing the eigenvalues of the A matrix, the system stability can be assessed. Now, let us consider an ‘n’ bus power system having ‘m’ generators. Each generator is represented by its classical model. Further, without any loss of generality, it is assumed that the generators are connected at the first ‘m’ buses of the system. Now, for linearising the differential equations, let us recall the swing equations of each generator here for ready reference.

dδi = ωi − ωs for i = 1, 2, ⋯⋯ m dt 2Hi dωi = Pmi − Pei for i = 1, 2, ⋯⋯ m ωs dt

(6.64) (6.65)

Linearising equation (6.64) for the ith generator, one can get,

d∆δi = ∆ωi dt

(6.66)

Now, let us define,

∆δ = [∆δ1 , ∆δ2 , ⋯⋯ ∆δm ]

T

∆ω = [∆ω1 , ∆ω2 , ⋯⋯ ∆ωm ]

T

(6.67) (6.68)

In equations (6.67) and (6.68), the vectors ∆δ and ∆ω denote the vectors of perturbed values of rotor angle and machine speed respectively. Please note that the size of each of these two vectors is (m × 1). Now, collecting equation (6.66) for all the ‘m’ generators and re-writing them in matrix notation we can obtain,

d ∆δ = F1 ∆ω dt

(6.69)

In equation (6.69), F1 is a (m × m) identity matrix. Now, linearising equation (6.65) for the ith generator, one can get,

2Hi d∆ωi = −∆Pei ωs dt

(6.70)

For performing linearisation of Pei , its expression its required. This expression can be derived as follows. From equation (6.25), one can write,

Ei ∠δi − Vi ∠θi Ei j(δi −π/2) Vi j(θi −π/2) I¯i = = ∕e − ∕e jx∕di xdi xdi 278

Or,

E i Vi Ei E¯i I¯i∗ = Ei ejδi I¯i∗ = ∕ ejπ/2 − ∕ ej(π/2+δi −θi ) xdi xdi Therefore,

E i Vi Pei = Re (E¯I I¯i∗ ) = ∕ sin(δi − θi ) xdi

(6.71)

Linearisation of equation (6.71) yields,

∆Pei = k1i ∆δi + k2i ∆θi + k3i ∆Vi

(6.72)

Where,

k1i =

Ei Vi cos(δi − θi ); x∕di

k2i = −

Ei Vi cos(δi − θi ); x∕di

k3i =

Ei sin(δi − θi ); x∕di

(6.73)

In equation (6.73), the constants k1i , k2i and k3i are evaluated using the values of Ei , Vi , δi and θi at the current operating point for i = 1, 2, ⋯⋯ m. Now, again let us define,

∆θg = [∆θ1 , ∆θ2 , ⋯⋯ ∆θm ]

T

∆Vg = [∆V1 , ∆V2 , ⋯⋯ ∆Vm ]

(6.74)

T

(6.75)

∆Pe = [∆Pe1 , ∆Pe2 , ⋯⋯ ∆Pem ]

T

(6.76)

In equations (6.74) and (6.75), the vectors ∆θg and ∆Vg denote the vectors of perturbed values of generator terminal voltage angles and magnitudes respectively. Similarly, the vector ∆Pe denotes the perturbed values of the generator real powers. Please note that the size of each of these three vectors is also (m × 1). Now, collecting equation (6.72) for all the ‘m’ generators and re-writing them in matrix notation we can obtain, ∆Pe = K1 ∆δ + K2 ∆θg + K3 ∆Vg (6.77) In equation (6.77), the size of each of the matrices K1 , K2 and K3 is (m × m). Moreover, all these three matrices are diagonal matrices and are given by;

K1 = diag (k11 , k12 , ⋯⋯ k1m ) K2 = diag (k21 , k22 , ⋯⋯ k2m ) K3 = diag (k31 , k32 , ⋯⋯ k3m ) 279

⎫ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎭

(6.78)

6.4.1

Linearisation of network equations (at the generator buses)

To illustrate the procedure for linearisation of the network equations, let us first consider the first equation of the equation set (6.23) below.

(Y¯11 + y¯1 ) V¯1 + Y¯12 V¯2 + ⋯ + Y¯1m V¯m + ⋯ + Y1n V¯n = y¯1 E¯1

(6.79)

Now, let,

Y¯ij = Yij exp (jαij );

V¯i = Vi exp (jθi );

for i, j = 1, 2, ⋯⋯ n;

(6.80)

y¯k = yk exp (jβk );

E¯k = Ek exp (jδk );

for k = 1, 2, ⋯⋯ m;

(6.81)

and

Utilising equations (6.80) and (6.81) in equation (6.79), we get,

Y11 V1 exp (j (α11 + θ1 )) + y1 V1 exp (j (β1 + θ1 )) + Y12 V2 exp (j (α12 + θ2 )) + ⋯+ Y1m Vm exp (j (α1m + θm )) + ⋯ + Y1n Vn exp (j (α1n + θn )) = y1 E1 exp (j (δ1 + β1 ))

(6.82)

Taking the real part of both the sides of equation (6.82), we get,

Y11 V1 cos (α11 + θ1 ) + y1 V1 cos (β1 + θ1 ) + Y12 V2 cos (α12 + θ2 ) + ⋯+ Y1m Vm cos (α1m + θm ) + ⋯ + Y1n Vn cos (α1n + θn ) = y1 E1 cos (δ1 + β1 )

(6.83)

Linearising equation (6.83), we get,

Y11 cos (α11 + θ1 )∆V1 − Y11 V1 sin (α11 + θ1 )∆θ1 + y1 cos (β1 + θ1 )∆V1 − y1 V1 sin (β1 + θ1 )∆θ1 + Y12 cos (α12 + θ2 )∆V2 − Y12 V2 sin (α12 + θ2 )∆θ2 + ⋯ + Y1m cos (α1m + θm )∆Vm − Y1m Vm sin (α1m + θm )∆θm + ⋯+ Y1n cos (α1n + θn )∆Vn − Y1n Vn sin (α1n + θn )∆θn = −y1 E1 sin (δ1 + β1 )∆δ1

(6.84)

Equation (6.84) can be re-written as,

a11 ∆V1 + a12 ∆V2 + ⋯ + a1m ∆Vm + ⋯ + a1n ∆Vn + b11 ∆θ1 + b12 ∆θ2 + ⋯ + b1m ∆θm + ⋯ + b1n ∆θn = g1 ∆δ1

(6.85)

In equation (6.85),

a11 = Y11 cos (α11 + θ1 ) + y1 cos (β1 + θ1 ); a12 = Y12 cos (α12 + θ2 ); a1m = Y1m cos (α1m + θm ); a1n = Y1n cos (α1n + θn ); b11 = −Y11 V1 sin (α11 + θ1 ) − y1 V1 sin (β1 + θ1 ); b12 = −Y12 V2 sin (α12 + θ2 ); b1m = −Y1m Vm sin (α1m + θm ); b1n = −Y1n Vn sin (α1n + θn ); g1 = −y1 E1 sin (δ1 + β1 );

(6.86)

Again, taking the real part of both sides of the second equation of the equation set (6.23) (after 280

substituting equations (6.80) and (6.81) into it), we get,

Y21 V1 cos (α21 + θ1 ) + y2 V2 cos (β2 + θ2 ) + Y22 V2 cos (α22 + θ2 ) + ⋯+ Y2m Vm cos (α2m + θm ) + ⋯ + Y2n Vn cos (α2n + θn ) = y2 E2 cos (δ2 + β2 )

(6.87)

Linearising equation (6.87), one can get,

a21 ∆V1 + a22 ∆V2 + ⋯ + a2m ∆Vm + ⋯ + a2n ∆Vn + b21 ∆θ1 + b22 ∆θ2 + ⋯ + b2m ∆θm + ⋯ + b2n ∆θn = g2 ∆δ2

(6.88)

In equation (6.88),

a21 = Y21 cos (α21 + θ1 ); a22 = Y22 cos (α22 + θ2 ) + y2 cos (β2 + θ2 ); a2m = Y2m cos (α2m + θm ); a2n = Y2n cos (α2n + θn ); b21 = −Y21 V1 sin (α21 + θ1 ); b22 = −Y22 V2 sin (α22 + θ2 ) − y2 V2 sin (β2 + θ2 ); b2m = −Y2m Vm sin (α2m + θm ); b2n = −Y2n Vn sin (α2n + θn ); g2 = −y2 E2 sin (δ2 + β2 );

(6.89)

Similarly, continuing with linearisation of the real parts of first ‘m’ equations (corresponding to the generator buses) of the equation set (6.23), we get,

[A1

⎤ ⎡ ⎢ ∆Vg ⎥ ⎥ ⎢ ⎢∆V ⎥ ⎢ L⎥ ⎥ = [G] [∆δ] B1 ] ⎢⎢ ⎥ ⎢ ∆θg ⎥ ⎥ ⎢ ⎢ ∆θL ⎥ ⎣ ⎦

(6.90)

In equation (6.90),

⎡ a11 a12 ⋯ a1m a1,(m+1) ⋯⋯ a1n ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ a21 a22 ⋯ a2m a2,(m+1) ⋯⋯ a2n ⎥ ⎢ ⎥ ⎢ ⎥ ⎥ ⇒ is a (m × n) matrix A1 = ⎢ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⎢ ⎥ ⎢ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⎥⎥ ⎢ ⎢ ⎥ ⎢am1 am2 ⋯ amm am,(m+1) ⋯⋯ amn ⎥ ⎣ ⎦

(6.91)

⎡ b11 b12 ⋯ b1m b1,(m+1) ⋯⋯ b1n ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ b21 b22 ⋯ b2m b2,(m+1) ⋯⋯ b2n ⎥ ⎢ ⎥ ⎢ ⎥ B1 = ⎢ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⎥⎥ ⇒ is a (m × n) matrix ⎢ ⎢ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⎥⎥ ⎢ ⎢ ⎥ ⎢bm1 bm2 ⋯ bmm bm,(m+1) ⋯⋯ bmn ⎥ ⎣ ⎦

(6.92)

281

⎡g1 0 ⋯ 0 ⋯⋯ 0 ⎤ ⎥ ⎢ ⎥ ⎢ ⎢ 0 g2 ⋯ 0 ⋯⋯ 0 ⎥ ⎥ ⎢ ⎥ ⎢ ⎥ ⇒ is a (m × m) diagonal matrix G=⎢⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⎥ ⎢ ⎢⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⎥⎥ ⎢ ⎥ ⎢ ⎢ 0 0 ⋯ 0 ⋯⋯ gm ⎥ ⎦ ⎣

(6.93)

The elements of the matrices A1 , B1 and G are given as;

aii = Yii cos (αii + θi ) + yi cos (βi + θi ); aij = Yij cos (αij + θj );

i = 1, 2, ⋯⋯ m; j = 1, 2, ⋯⋯ n; i ≠ j;

bii = −Yii Vi sin (αii + θi ) − yi Vi sin (βi + θi ); bij = −Yij Vj sin (αij + θj );

i = 1, 2, ⋯⋯ m;

i = 1, 2, ⋯⋯ m;

(6.94)

i = 1, 2, ⋯⋯ m; j = 1, 2, ⋯⋯ n; i ≠ j;

gi = −yi Ei sin (δi + βi );

i = 1, 2, ⋯⋯ m;

Further, the vectors ∆θL and ∆VL are defined as;

∆θL = [∆θm+1 , ∆θm+2 , ⋯⋯ ∆θn ]

T

∆VL = [∆Vm+1 , ∆Vm+2 , ⋯⋯ ∆Vn ]

T

(6.95) (6.96)

Please note that the size of each of the above two vectors is ((n − m) × 1). So far, we have considered only the algebraic equations at the generator buses. However, for completing the small signal model, the algebraic equations at the load buses all need to be linearised. We will discuss this issue in the next lecture.

282

6.4.2

Linearisation of network equations (at the load buses)

Now, let us consider the algebraic equations at the load buses. For this, please recollect that the loads have been assumed to be connected at the last (n−m) buses. The real part of the pth equation (p = (m + 1), (m + 2), ⋯⋯ n) of the equations set (6.23) is given by,

Yp1 V1 cos (αp1 + θ1 ) + Yp2 V2 cos (αp2 + θ2 ) + ⋯ + Ypn Vn cos (αpn + θn ) = 0

(6.97)

Linearising the above equation we get,

Yp1 cos (αp1 + θ1 )∆V1 − Yp1 V1 sin (αp1 + θ1 )∆θ1 + Yp2 cos (αp2 + θ2 )∆V2 −Yp2 V2 sin (αp2 + θ2 )∆θ2 + ⋯ + Ypn cos (αpn + θn )∆Vn − Ypn Vn sin (αpn + θn )∆θn = 0

(6.98)

Or,

ap1 ∆V1 + ap2 ∆V2 + ⋯ + apn ∆Vn + bp1 ∆θ1 + bp2 ∆θ2 + ⋯ + bpn ∆θn = 0

(6.99)

In equation (6.99),

apj = Ypj cos (αpj + θj ); j = 1, 2, ⋯⋯ n; bpj = −Ypj Vj sin (αpj + θj ); j = 1, 2, ⋯⋯ n;

(6.100)

Please note that equation (6.99) can be written for all the ‘n − m’ load buses (by varying ‘p’ from ‘m + 1’ to ‘n’). In that case, in equation (6.100) also, ‘p’ will vary from ‘m + 1’ to ‘n’. Collecting all these ‘n − m’ equations and putting them in a matrix form, we get,

[R1

⎡ ⎤ ⎢ ∆Vg ⎥ ⎢ ⎥ ⎢∆V ⎥ ⎢ L⎥ ⎥ = [Φ1 ] [∆δ] S1 ] ⎢⎢ ⎥ ∆θ g ⎢ ⎥ ⎢ ⎥ ⎢ ∆θL ⎥ ⎣ ⎦

(6.101)

In equation (6.101), the matrix Φ1 is a (n − m) × m null matrix and the matrices R1 and S1 are given by,

⎡ r11 r12 ⋯⋯ r1n ⎤⎥ ⎢ ⎢ ⎥ ⎢ r21 r22 ⋯⋯ r2n ⎥⎥ ⎢ ⎢ ⎥ R1 = ⎢ ⋮ ⋮ ⋮ ⋮ ⎥⎥ ⇒ is a ((n − m) × n) matrix ⎢ ⎢ ⋮ ⋮ ⋮ ⋮ ⎥⎥ ⎢ ⎢ ⎥ ⎢r(n−m),1 r(n−m),2 ⋯⋯ r(n−m),n ⎥ ⎣ ⎦ 283

(6.102)

⎡ s11 s12 ⋯⋯ s1n ⎤⎥ ⎢ ⎥ ⎢ ⎢ s21 ⎥ s ⋯⋯ s 22 2n ⎥ ⎢ ⎥ ⎢ ⎥ ⇒ is a ((n − m) × n) matrix S1 = ⎢ ⋮ ⋮ ⋮ ⋮ ⎥ ⎢ ⎢ ⋮ ⋮ ⋮ ⋮ ⎥⎥ ⎢ ⎥ ⎢ ⎢s(n−m),1 s(n−m),2 ⋯⋯ s(n−m),n ⎥ ⎦ ⎣

(6.103)

The elements of the matrices R1 and S1 are given by,

rij = a(i+m),j ; sij = b(i+m),j ;

i = 1, 2, ⋯⋯ (n − m); j = 1, 2, ⋯⋯ n; i = 1, 2, ⋯⋯ (n − m); j = 1, 2, ⋯⋯ n;

(6.104)

In equation (6.104), the expressions of the co-efficients a(i+m),j and b(i+m),j are given by equation (6.100). Now, taking the imaginary part of both the sides of equation (6.82), we get,

Y11 V1 sin (α11 + θ1 ) + y1 V1 sin (β1 + θ1 ) + Y12 V2 sin (α12 + θ2 ) + ⋯+ Y1m Vm sin (α1m + θm ) + ⋯ + Y1n Vn sin (α1n + θn ) = y1 E1 sin (δ1 + β1 )

(6.105)

Linearising equation (6.105), we get,

Y11 sin (α11 + θ1 )∆V1 + Y11 V1 cos (α11 + θ1 )∆θ1 + y1 sin (β1 + θ1 )∆V1 + y1 V1 cos (β1 + θ1 )∆θ1 + Y12 sin (α12 + θ2 )∆V2 + Y12 V2 cos (α12 + θ2 )∆θ2 + ⋯ + Y1m sin (α1m + θm )∆Vm + Y1m Vm cos (α1m + θm )∆θm + ⋯+ Y1n sin (α1n + θn )∆Vn + Y1n Vn cos (α1n + θn )∆θn = y1 E1 cos (δ1 + β1 )∆δ1

(6.106)

Equation (6.106) can be re-written as,

c11 ∆V1 + c12 ∆V2 + ⋯ + c1m ∆Vm + ⋯ + c1n ∆Vn + d11 ∆θ1 + d12 ∆θ2 + ⋯ + d1m ∆θm + ⋯ + d1n ∆θn = h1 ∆δ1

(6.107)

In equation (6.107),

c11 = Y11 sin (α11 + θ1 ) + y1 sin (β1 + θ1 ); c12 = Y12 sin (α12 + θ2 ); c1m = Y1m sin (α1m + θm ); c1n = Y1n sin (α1n + θn ); d11 = Y11 V1 cos (α11 + θ1 ) + y1 V1 cos (β1 + θ1 ); d12 = Y12 V2 cos (α12 + θ2 ); d1m = Y1m Vm cos (α1m + θm ); d1n = Y1n Vn cos (α1n + θn ); h1 = y1 E1 cos (δ1 + β1 );

(6.108)

Again, taking the imaginary part of both sides of the second equation of the equation set (6.23) (after substituting equations (6.80) and (6.81) into it), we get,

Y21 V1 sin (α21 + θ1 ) + y2 V2 sin (β2 + θ2 ) + Y22 V2 sin (α22 + θ2 ) + ⋯+ Y2m Vm sin (α2m + θm ) + ⋯ + Y2n Vn sin (α2n + θn ) = y2 E2 sin (δ2 + β2 ) 284

(6.109)

Linearising equation (6.109), one can get,

c21 ∆V1 + c22 ∆V2 + ⋯ + c2m ∆Vm + ⋯ + c2n ∆Vn + d21 ∆θ1 + d22 ∆θ2 + ⋯ + d2m ∆θm + ⋯ + d2n ∆θn = h2 ∆δ2

(6.110)

In equation (6.110),

c21 = Y21 sin (α21 + θ1 ); c22 = Y22 sin (α22 + θ2 ) + y2 sin (β2 + θ2 ); c2m = Y2m sin (α2m + θm ); c2n = Y2n sin (α2n + θn ); d21 = Y21 V1 cos (α21 + θ1 ); d22 = Y22 V2 cos (α22 + θ2 ) + y2 V2 cos (β2 + θ2 ); d2m = Y2m Vm cos (α2m + θm ); d2n = Y2n Vn cos (α2n + θn ); h2 = y2 E2 sin (δ2 + β2 );

(6.111)

Continuing with linearisation of the imaginary parts of first ‘m’ equations (corresponding to the generator buses) of the equation set (6.23), we get,

[C1

⎤ ⎡ ⎢ ∆Vg ⎥ ⎥ ⎢ ⎢∆V ⎥ ⎢ L⎥ ⎥ = [H] [∆δ] D1 ] ⎢⎢ ⎥ ⎢ ∆θg ⎥ ⎥ ⎢ ⎢ ∆θL ⎥ ⎣ ⎦

(6.112)

In equation (6.112),

⎡ c11 c12 ⋯ c1m c1,(m+1) ⋯⋯ c1n ⎤ ⎥ ⎢ ⎥ ⎢ ⎢ c21 c22 ⋯ c2m c2,(m+1) ⋯⋯ c2n ⎥ ⎥ ⎢ ⎢ ⎥ C1 = ⎢ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⎥⎥ ⇒ is a (m × n) matrix ⎢ ⎢ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⎥⎥ ⎢ ⎥ ⎢ ⎢cm1 cm2 ⋯ cmm cm,(m+1) ⋯⋯ cmn ⎥ ⎦ ⎣ ⎡ d11 d12 ⋯ d1m d1,(m+1) ⋯⋯ d1n ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ d21 d22 ⋯ d2m d2,(m+1) ⋯⋯ d2n ⎥ ⎢ ⎥ ⎢ ⎥ ⎥ ⇒ is a (m × n) matrix D1 = ⎢ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⎢ ⎥ ⎢ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⎥⎥ ⎢ ⎢ ⎥ ⎢dm1 dm2 ⋯ dmm dm,(m+1) ⋯⋯ dmn ⎥ ⎣ ⎦ ⎡h1 0 ⋯ 0 ⋯⋯ 0 ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ 0 h2 ⋯ 0 ⋯⋯ 0 ⎥ ⎢ ⎥ ⎢ ⎥ H=⎢ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⎥⎥ ⇒ is a (m × m) diagonal matrix ⎢ ⎢⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⎥⎥ ⎢ ⎢ ⎥ ⎢ 0 0 ⋯ 0 ⋯⋯ hm ⎥ ⎣ ⎦ 285

(6.113)

(6.114)

(6.115)

The elements of the matrices C1 , D1 and H are given as;

cii = Yii sin (αii + θi ) + yi sin (βi + θi ); cij = Yij sin (αij + θj );

i = 1, 2, ⋯⋯ m; j = 1, 2, ⋯⋯ n; i ≠ j;

dii = Yii Vi cos (αii + θi ) + yi Vi cos (βi + θi ); dij = Yij Vj cos (αij + θj );

i = 1, 2, ⋯⋯ m;

i = 1, 2, ⋯⋯ m;

(6.116)

i = 1, 2, ⋯⋯ m; j = 1, 2, ⋯⋯ n; i ≠ j;

hi = yi Ei cos (δi + βi );

i = 1, 2, ⋯⋯ m;

Now, let us consider the imaginary parts of the algebraic equations at the load buses. The real part of the pth equation (p = (m + 1), (m + 2), ⋯⋯ n) of the equations set (6.23) is given by,

Yp1 V1 sin (αp1 + θ1 ) + Yp2 V2 sin (αp2 + θ2 ) + ⋯ + Ypn Vn sin (αpn + θn ) = 0

(6.117)

Linearising the above equation we get,

Yp1 sin (αp1 + θ1 )∆V1 + Yp1 V1 cos (αp1 + θ1 )∆θ1 + Yp2 sin (αp2 + θ2 )∆V2 +Yp2 V2 cos (αp2 + θ2 )∆θ2 + ⋯ + Ypn sin (αpn + θn )∆Vn + Ypn Vn cos (αpn + θn )∆θn = 0

(6.118)

Or,

cp1 ∆V1 + cp2 ∆V2 + ⋯ + cpn ∆Vn + dp1 ∆θ1 + dp2 ∆θ2 + ⋯ + dpn ∆θn = 0

(6.119)

In equation (6.119),

cpj = Ypj sin (αpj + θj ); j = 1, 2, ⋯⋯ n; dpj = Ypj Vj cos (αpj + θj ); j = 1, 2, ⋯⋯ n;

(6.120)

Again, equation (6.119) can be written for all the ‘n − m’ load buses (by varying ‘p’ from ‘m + 1’ to ‘n’). In that case, in equation (6.120) also, ‘p’ will vary from ‘m + 1’ to ‘n’. Collecting all these ‘n − m’ equations and putting them in a matrix form, we get,

[U1

⎡ ⎤ ⎢ ∆Vg ⎥ ⎢ ⎥ ⎢∆V ⎥ ⎢ L⎥ ⎥ = [Φ2 ] [∆δ] V1 ] ⎢⎢ ⎥ ⎢ ∆θg ⎥ ⎢ ⎥ ⎢ ∆θL ⎥ ⎣ ⎦

(6.121)

In equation (6.121), the matrix Φ2 is a (n − m) × m null matrix and the matrices U1 and V1 286

are given by,

⎡ u11 u12 ⋯⋯ u1n ⎤⎥ ⎢ ⎥ ⎢ ⎢ u21 u22 ⋯⋯ u2n ⎥⎥ ⎢ ⎥ ⎢ U1 = ⎢ ⋮ ⋮ ⋮ ⋮ ⎥⎥ ⇒ is a ((n − m) × n) matrix ⎢ ⎢ ⋮ ⋮ ⋮ ⋮ ⎥⎥ ⎢ ⎥ ⎢ ⎢u(n−m),1 u(n−m),2 ⋯⋯ u(n−m),n ⎥ ⎦ ⎣

(6.122)

⎡ v11 v12 ⋯⋯ v1n ⎤⎥ ⎢ ⎥ ⎢ ⎢ v21 v22 ⋯⋯ v2n ⎥⎥ ⎢ ⎥ ⎢ V1 = ⎢ ⋮ ⋮ ⋮ ⋮ ⎥⎥ ⇒ is a ((n − m) × n) matrix ⎢ ⎢ ⋮ ⋮ ⋮ ⋮ ⎥⎥ ⎢ ⎥ ⎢ ⎢v(n−m),1 v(n−m),2 ⋯⋯ v(n−m),n ⎥ ⎦ ⎣

(6.123)

The elements of the matrices U1 and V1 are given by,

uij = c(i+m),j ; vij = d(i+m),j ;

i = 1, 2, ⋯⋯ (n − m); j = 1, 2, ⋯⋯ n; i = 1, 2, ⋯⋯ (n − m); j = 1, 2, ⋯⋯ n;

(6.124)

In equation (6.124), the expressions of the co-efficients c(i+m),j and d(i+m),j are given by equation (6.120). Now, combining equations (6.90), (6.101), (6.112) and (6.121), one can write,

⎡ ⎢A1 ⎢ ⎢R ⎢ 1 ⎢ ⎢C1 ⎢ ⎢ ⎢U1 ⎣ Or,

Where,

⎤ ⎡ ⎤ ⎤⎡ B1 ⎥ ⎢ ∆Vg ⎥ ⎢ G ⎥ ⎥ ⎢ ⎥ ⎥⎢ S1 ⎥⎥ ⎢⎢∆VL ⎥⎥ ⎢⎢Φ1 ⎥⎥ ⎥ = ⎢ ⎥ [∆δ] ⎥⎢ D1 ⎥⎥ ⎢⎢ ∆θg ⎥⎥ ⎢⎢ H ⎥⎥ ⎥ ⎢ ⎥ ⎥⎢ V1 ⎥⎦ ⎢⎣ ∆θL ⎥⎦ ⎢⎣Φ2 ⎥⎦

⎡ ⎤ ⎢ ∆Vg ⎥ ⎢ ⎥ ⎢∆V ⎥ ⎢ L⎥ [J1 ] ⎢ ⎥ = [J2 ] [∆δ] ⎢ ∆θg ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ∆θL ⎥ ⎣ ⎦ ⎡ ⎢A1 ⎢ ⎢R ⎢ 1 [J1 ] = ⎢ ⎢C1 ⎢ ⎢ ⎢U1 ⎣

⎤ B1 ⎥ ⎥ S1 ⎥⎥ ⎥; D1 ⎥⎥ ⎥ V1 ⎥⎦

⎡ ⎤ ⎢G⎥ ⎢ ⎥ ⎢Φ ⎥ ⎢ 1⎥ [J2 ] = ⎢ ⎥ ⎢H⎥ ⎢ ⎥ ⎢ ⎥ ⎢Φ2 ⎥ ⎣ ⎦

(6.125)

(6.126)

(6.127)

Now, let us recollect that the dimensions of various sub-matrices are as follows: A1 → (m × n), B1 → (m × n), G → (m × m), R1 → ((n − m) × n), S1 → ((n − m) × n), Φ1 → ((n − m) × m), C1 → (m × n), D1 → (m × n), H → (m × m), U1 → ((n − m) × n), V1 → ((n − m) × n), 287

Φ2 → ((n − m) × m). Therefore, the size of the matrix J1 is (2n × 2n) and that of the matrix J2 is (2n × m). Hence, matrix J1 is a square matrix and hence invertible. Thus, from equation (6.126), ⎡ ⎤ ⎤ ⎡ ⎢A2 ⎥ ⎢ ∆Vg ⎥ ⎢ ⎥ ⎥ ⎢ ⎢B ⎥ ⎢∆V ⎥ −1 ⎢ 2⎥ ⎢ L⎥ ⎥ = [J1 ] [J2 ] [∆δ] = ⎢ ⎥ [∆δ] ⎢ ⎢C2 ⎥ ⎢ ∆θg ⎥ ⎢ ⎥ ⎥ ⎢ ⎢ ⎥ ⎥ ⎢ ⎢D2 ⎥ ⎢ ∆θL ⎥ ⎣ ⎦ ⎦ ⎣

(6.128)

Hence, from equation (6.128),

[∆Vg ] = [A2 ] [∆δ] ;

[∆θg ] = [C2 ] [∆δ] ;

(6.129)

Please note that the dimension of both the matrices A2 and C2 is (m×m). We are now ready to form the system state matrix which we will discuss in the next lecture. Further, in the next lecture, we will also look at an example of small signal stability analysis.

288

6.4.3

Formation of system state matrix

From equation (6.70), let us recall the following linearised equation for ith generator,

2Hi d∆ωi = −∆Pei ωs dt

(6.130)

Collecting equation (6.130) for all the ‘m’ generators, the following matrix equation can be written;

d 2 HM ∆ω = −∆Pe ωs dt

(6.131)

In equation (6.131), the matrix HM is a (m × m) diagonal matrix and is given by,

HM = diag (H1 , H2 , ⋯⋯ Hm )

(6.132)

Substituting equation (6.77) into equation (6.131) we can get,

2 d HM ∆ω = − (K1 ∆δ + K2 ∆θg + K3 ∆Vg ) ωs dt

(6.133)

Further, substituting equation (6.129) into equation (6.133) one can write,

d 2 HM ∆ω = − (K1 + K2 C2 + K3 A2 ) ∆δ ωs dt

(6.134)

d ∆ω = K4 ∆δ dt

(6.135)

Or,

Where,

K4 = −

ωs −1 H (K1 + K2 C2 + K3 A2 ) 2 M

(6.136)

Again, recalling equation (6.69), we have,

d ∆δ = F1 ∆ω dt

(6.137)

Combining equations (6.135) and (6.137), we can write,

O F1 ∆δ d ∆δ [ ]=[ ][ ] dt ∆ω K4 O ∆ω

(6.138)

O F1 ] K4 O

Equation (6.138) is the required state space equation of the system and the matrix A = [

is the corresponding state matrix. The eigenvalues of the matrix A gives the necessary information about the small signal stability of the system under study. Now, let us consider the damping of the generators. Towards this goal, let us now linearize 289

equation (6.63) to obtain,

2Hi d∆ωi = −∆Pei − di ∆ωi ωs dt

(6.139)

Collecting equation (6.139) for all the ‘m’ generators, the following matrix equation can be written;

d 2 HM ∆ω = −∆Pe − DM ∆ω ωs dt

(6.140)

In equation (6.140), the matrix DM is a (m × m) diagonal matrix and is given by,

DM = diag (d1 , d2 , ⋯⋯ dm )

(6.141)

Following the same procedure as in equations (6.133) and (6.134), we can get,

d 2 HM ∆ω = − (K1 + K2 C2 + K3 A2 ) ∆δ − DM ∆ω ωs dt

(6.142)

d ∆ω = K4 ∆δ + K5 ∆ω dt

(6.143)

Or,

Where,

K5 = −

ωs −1 H DM 2 M

(6.144)

Combining equations (6.137) and (6.144), we can write,

O F1 ∆δ d ∆δ [ ]=[ ][ ] dt ∆ω K4 K5 ∆ω

(6.145)

Equation (6.145) is the required state space equation of the system when damping of the gener-

O F1 ] is the corresponding state matrix. K4 K5

ators are considered and the matrix A = [

6.5

Example of small signal stability

As an illustration of small signal stability, let us consider the three machine, 9 bus system (m = 3 and n = 9) shown in Fig. Fig. 6.4. The complete data of this system are given in Tables A.7, A.8 and 6.1. Further, the different quantities pertaining to the initial condition are given in Table 6.2. The load flow solution of the system is shown in Table 6.25. Initially, the damping of the machines are neglected (di = 0; i = 1, 2, 3). With these initial values, the different matrices are calculated as follows:

⎡18.0599 0 0 ⎤⎥ ⎢ ⎢ ⎥ K1 = ⎢⎢ 0 8.8364 0 ⎥⎥ ⎢ ⎥ ⎢ 0 0 5.6864⎥⎦ ⎣ 290

(6.146)

Table 6.25: Load flow solution of the 3 machine system Bus No. 1 2 3 4 5 6 7 8 9

Voltage (p.u.) 1.0400 1.0116 + 0.1653i 1.0216 + 0.0834i 1.0250 - 0.0397i 0.9932 - 0.0693i 1.0106 - 0.0651i 1.0236 + 0.0665i 1.0158 + 0.0129i 1.0317 + 0.0354i

⎡−18.0599 0 0 ⎤⎥ ⎢ ⎢ ⎥ K2 = ⎢⎢ 0 −8.8364 0 ⎥⎥ ⎢ ⎥ ⎢ 0 0 −5.6864⎥⎦ ⎣ ⎡0.6889 0 0 ⎤⎥ ⎢ ⎢ ⎥ K3 = ⎢⎢ 0 1.5902 0 ⎥⎥ ⎢ ⎥ ⎢ 0 0 0.8293⎥⎦ ⎣

(6.147)

(6.148)

⎡0.0000 0 0 0.6715 0 0 0 0 0 ⎤⎥ ⎢ ⎢ ⎥ A1 = ⎢⎢ 0 (6.149) 3.9262 0 0 0 0 −1.0380 0 0 ⎥⎥ ⎢ ⎥ ⎢ 0 0 1.8364 0 0 0 0 0 −0.5856⎥⎦ ⎣ ⎡35.1608 ⎤ 0 0 −17.7955 0 0 0 0 0 ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ B1 = ⎢ 0 (6.150) 24.6293 0 0 0 0 −16.3777 0 0 ⎥ ⎢ ⎥ ⎢ 0 0 23.0684 0 0 0 0 0 −17.6066⎥⎦ ⎣ ⎡17.3653 0 0 ⎤⎥ ⎢ ⎢ ⎥ G = ⎢⎢ 0 (6.151) 8.2516 0 ⎥⎥ ⎢ ⎥ ⎢ 0 0 5.4618⎥⎦ ⎣ ⎡0.0000 0 0 1.7844 −0.5547 −1.2622 0 0 0 ⎤⎥ ⎢ ⎢ ⎥ ⎢ 0 0 0 −0.9153 2.5634 0 −1.5727 0 0 ⎥⎥ ⎢ ⎢ ⎥ ⎢ 0 0 0 −1.5342 0 3.0558 0 0 −1.4730⎥⎥ ⎢ R1 = ⎢ −2.5802 0 0 −0.7691 0 5.0984 −1.7909 0 ⎥⎥ ⎢ 0 ⎢ ⎥ ⎢ 0 0 0 0 0 0 −2.5024 4.0411 −1.4902⎥⎥ ⎢ ⎢ ⎥ ⎢ 0 0 −1.3878 0 0 −0.9200 0 −1.2792 3.5391 ⎥⎦ ⎣

(6.152)

The matrix S1 is given in parts in equations (6.153) and (6.154) below, as the width of the page is not sufficient enough to accomodate the complete matrix S1 . 291

The matrix S1 (∶, 1 ∶ 5) (comprising of the elements in columns 1 to 5 for all rows) is given by;

⎡−18.0556 0 0 40.4237 −11.6200⎤⎥ ⎢ ⎥ ⎢ ⎢ 0 0 0 −11.9486 17.9858 ⎥⎥ ⎢ ⎥ ⎢ ⎥ ⎢ 0 0 0 −10.8507 0 ⎥ ⎢ S1 (∶, 1 ∶ 5) = ⎢ 0 −16.1854 0 0 −6.0169 ⎥⎥ ⎢ ⎥ ⎢ ⎥ ⎢ 0 0 0 0 0 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ 0 0 −17.4335 0 0 ⎦ ⎣

(6.153)

The matrix S1 (∶, 6 ∶ 9) (comprising of the elements in columns 6 to 9 for all rows) is given by;

⎤ ⎡−10.7481 0 0 0 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ 0 −6.0372 0 0 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ 16.5710 0 0 −5.7202 ⎥ S1 (∶, 6 ∶ 9) = ⎢⎢ ⎥ 0 36.0958 −13.8936 0 ⎥ ⎢ ⎥ ⎢ ⎢ 0 −13.9138 23.9677 −10.0539⎥⎥ ⎢ ⎥ ⎢ ⎢ −5.7307 0 −9.9240 33.0882 ⎥⎦ ⎣ ⎡−33.8085 0 0 17.3481 0 0 0 0 0 ⎤⎥ ⎢ ⎢ ⎥ C1 = ⎢⎢ 0 −24.0286 0 0 0 0 15.9663 0 0 ⎥⎥ ⎢ ⎥ ⎢ 0 0 −22.5058 0 0 0 0 0 17.0548⎥⎦ ⎣ ⎡0.0000 0 0 0.6889 0 0 0 0 0 ⎤⎥ ⎢ ⎢ ⎥ D1 = ⎢⎢ 0 4.0244 0 0 0 0 −1.0648 0 0 ⎥⎥ ⎢ ⎥ ⎢ 0 0 1.8823 0 0 0 0 0 −0.6046⎥⎦ ⎣ ⎡0.6889 0 0 ⎤⎥ ⎢ ⎢ ⎥ H = ⎢⎢ 0 2.9596 0 ⎥⎥ ⎢ ⎥ ⎢ 0 ⎥ 0 1.2777 ⎣ ⎦

(6.154)

(6.155)

(6.156)

(6.157)

The matrix U1 is given in parts in equations (6.158) and (6.159) below, as the width of the page is not sufficient enough to accomodate the complete matrix U1 . The matrix U1 (∶, 1 ∶ 5) (comprising of the elements in columns 1 to 5 for all rows) is given by;

⎡17.3611 0 0 −39.4074 11.6710 ⎤⎥ ⎢ ⎢ ⎥ ⎢ 0 ⎥ 0 0 11.6482 −18.0647 ⎢ ⎥ ⎢ ⎥ ⎢ 0 ⎥ 0 0 10.5779 0 ⎥ U1 (∶, 1 ∶ 5) = ⎢⎢ 15.7906 0 0 6.0433 ⎥⎥ ⎢ 0 ⎢ ⎥ ⎢ 0 ⎥ 0 0 0 0 ⎢ ⎥ ⎢ ⎥ ⎢ 0 ⎥ 0 17.0083 0 0 ⎣ ⎦ 292

(6.158)

The matrix U1 (∶, 6 ∶ 9) (comprising of the elements in columns 6 to 9 for all rows) is given by;

⎤ ⎡ 10.6138 0 0 0 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ 0 5.8855 0 0 ⎥ ⎢ ⎥ ⎢ ⎢−16.3639 0 0 5.5410 ⎥⎥ ⎢ U1 (∶, 6 ∶ 9) = ⎢ ⎥ 0 −35.1890 13.6763 0 ⎥ ⎢ ⎥ ⎢ ⎢ 0 13.5642 −23.5930 9.7389 ⎥⎥ ⎢ ⎥ ⎢ ⎢ 5.6591 0 9.7688 −32.0513⎥⎦ ⎣

(6.159)

⎡0.0000 0 0 1.8304 −0.5523 −1.2782 0 0 0 ⎤⎥ ⎢ ⎥ ⎢ ⎢ 0 0 0 −0.9389 2.5522 0 −1.6133 0 0 ⎥⎥ ⎢ ⎥ ⎢ ⎢ 0 0 0 −1.5737 0 3.0944 0 0 −1.5207⎥⎥ ⎢ V1 = ⎢ −2.6447 0 0 −0.7657 0 5.2297 −1.8194 0 ⎥⎥ ⎢ 0 ⎥ ⎢ ⎢ 0 0 0 0 0 0 −2.5669 4.1053 −1.5384⎥⎥ ⎢ ⎥ ⎢ ⎢ 0 0 −1.4225 0 0 −0.9316 0 −1.2995 3.6536 ⎥⎦ ⎣

(6.160)

The matrix J1 is given in parts in equations (6.161) - (6.163) below, as the width of the page is not sufficient enough to accomodate the complete matrix J1 . The matrix J1 (∶, 1 ∶ 6) (comprising of the elements in columns 1 to 6 for all rows) is given by;

⎡ 0.0000 ⎤ 0 0 0.6715 0 0 ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ 0 3.9262 0 0 0 0 ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ 0 0 1.8364 0 0 0 ⎢ ⎥ ⎢ 0.0000 ⎥ 0 0 1.7844 −0.5547 −1.2622 ⎥ ⎢ ⎢ ⎥ ⎢ ⎥ 0 0 0 −0.9153 2.5634 0 ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ 0 0 0 −1.5342 0 3.0558 ⎢ ⎥ ⎢ ⎥ 0 −2.5802 0 0 −0.7691 0 ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ 0 0 0 0 0 0 ⎢ ⎥ ⎢ ⎥ ⎢ 0 0 −1.3878 0 0 −0.9200 ⎥⎥ ⎢ J1 (∶, 1 ∶ 6) = ⎢ ⎥ 0 0 17.3481 0 0 ⎥ ⎢−33.8085 ⎢ ⎥ ⎢ ⎥ 0 −24.0286 0 0 0 0 ⎢ ⎥ ⎢ ⎥ ⎥ ⎢ 0 0 −22.5058 0 0 0 ⎢ ⎥ ⎢ ⎥ 0 0 −39.4074 11.6710 10.6138 ⎥ ⎢ 17.3611 ⎢ ⎥ ⎢ ⎥ 0 0 0 11.6482 −18.0647 0 ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ 0 0 0 10.5779 0 −16.3639 ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ 0 15.7906 0 0 6.0433 0 ⎢ ⎥ ⎢ ⎥ 0 0 0 0 0 0 ⎢ ⎥ ⎢ ⎥ ⎢ 0 0 17.0083 0 0 5.6591 ⎥⎦ ⎣

(6.161)

The matrices J1 (∶, 7 ∶ 12) (comprising of the elements in columns 7 to 12 for all rows) and 293

J1 (∶, 13 ∶ 18) (comprising of the elements in columns 13 to 18 for all rows) are given by; ⎤ ⎡ 0 0 0 35.1608 0 0 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ −1.0380 0 0 0 24.6293 0 ⎥ ⎢ ⎥ ⎢ ⎢ 0 0 −0.5856 0 0 23.0684 ⎥⎥ ⎢ ⎥ ⎢ 0 0 0 −18.0556 0 0 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ −1.5727 0 0 0 0 0 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ 0 0 −1.4730 0 0 0 ⎥ ⎢ ⎥ ⎢ 5.0984 −1.7909 0 0 −16.1854 0 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ −2.5024 4.0411 −1.4902 0 0 0 ⎥ ⎢ ⎥ ⎢ ⎢ 0 −1.2792 3.5391 0 0 −17.4335⎥⎥ ⎢ J1 (∶, 7 ∶ 12) = ⎢ ⎥ 0 0 0 0.0000 0 0 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ 15.9663 0 0 0 4.0244 0 ⎥ ⎢ ⎥ ⎢ ⎢ 0 0 17.0548 0 0 1.8823 ⎥⎥ ⎢ ⎥ ⎢ 0 0 0 0.0000 0 0 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ 5.8855 0 0 0 0 0 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ 0 0 5.5410 0 0 0 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢−35.1890 13.6763 0 0 −2.6447 0 ⎥ ⎢ ⎥ ⎢ 13.5642 −23.5930 9.7389 0 0 0 ⎥ ⎢ ⎥ ⎢ ⎢ 0 9.7688 −32.0513 0 0 −1.4225 ⎥⎦ ⎣

(6.162)

⎡−17.7955 ⎤ 0 0 0 0 0 ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ 0 0 0 −16.3777 0 0 ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ 0 0 0 0 0 −17.6066 ⎢ ⎥ ⎢ 40.4237 −11.6200 −10.7481 ⎥ 0 0 0 ⎢ ⎥ ⎢ ⎥ ⎢−11.9486 17.9858 ⎥ 0 −6.0372 0 0 ⎢ ⎥ ⎢ ⎥ ⎢−10.8507 0 16.5710 0 0 −5.7202 ⎥⎥ ⎢ ⎢ ⎥ 0 −6.0169 0 36.0958 −13.8936 0 ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ 0 0 0 −13.9138 23.9677 −10.0539 ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ 0 0 −5.7307 0 −9.9240 33.0882 ⎥ J1 (∶, 13 ∶ 18) = ⎢⎢ ⎥ 0 0 0 0 0 ⎢ 0.6889 ⎥ ⎢ ⎥ ⎢ ⎥ 0 0 0 −1.0648 0 0 ⎢ ⎥ ⎢ ⎥ ⎢ 0 0 0 0 0 −0.6046 ⎥⎥ ⎢ ⎢ ⎥ −0.5523 −1.2782 0 0 0 ⎢ 1.8304 ⎥ ⎢ ⎥ ⎢ −0.9389 ⎥ 2.5522 0 −1.6133 0 0 ⎥ ⎢ ⎢ ⎥ ⎢ −1.5737 0 3.0944 0 0 −1.5207 ⎥⎥ ⎢ ⎢ ⎥ ⎢ ⎥ 0 −0.7657 0 5.2297 −1.8194 0 ⎢ ⎥ ⎢ ⎥ 0 0 0 −2.5669 4.1053 −1.5384 ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ 0 0 −0.9316 0 −1.2995 3.6536 ⎣ ⎦

(6.163)

294

⎡17.3653 0 0 ⎤⎥ ⎢ ⎥ ⎢ ⎥ ⎢ 0 8.2516 0 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ 0 0 5.4618 ⎥ ⎢ ⎢ 0 0 0 ⎥⎥ ⎢ ⎥ ⎢ ⎢ 0 0 0 ⎥⎥ ⎢ ⎥ ⎢ ⎢ 0 0 0 ⎥⎥ ⎢ ⎢ 0 0 0 ⎥⎥ ⎢ ⎥ ⎢ ⎢ 0 0 0 ⎥⎥ ⎢ ⎥ ⎢ ⎢ 0 0 0 ⎥⎥ ⎢ J2 = ⎢ 0 0 ⎥⎥ ⎢ 0.6889 ⎥ ⎢ ⎥ ⎢ 0 2.9596 0 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ 0 0 1.2777 ⎥ ⎢ ⎥ ⎢ 0 0 ⎥ ⎢ 0 ⎥ ⎢ ⎢ 0 0 0 ⎥⎥ ⎢ ⎥ ⎢ ⎥ ⎢ 0 0 0 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ 0 0 0 ⎥ ⎢ ⎢ 0 0 0 ⎥⎥ ⎢ ⎥ ⎢ ⎢ 0 0 0 ⎥⎦ ⎣ ⎡0.0200 −0.0154 −0.0046⎤ ⎢ ⎥ ⎢ ⎥ A2 = ⎢⎢0.0594 −0.0764 0.0169 ⎥⎥ ⎢ ⎥ ⎢0.0499 −0.0146 −0.0353⎥ ⎣ ⎦ ⎡0.8328 0.0934 0.0738⎤ ⎢ ⎥ ⎢ ⎥ ⎢ C2 = ⎢0.1811 0.6881 0.1307⎥⎥ ⎢ ⎥ ⎢0.2272 0.2054 0.5673⎥ ⎣ ⎦ HM

⎡23.6400 0 0 ⎤⎥ ⎢ ⎢ ⎥ = ⎢⎢ 0 6.4000 0 ⎥⎥ ⎢ ⎥ ⎢ 0 ⎥ 0 3.0100 ⎣ ⎦

⎡−24.1837 13.5322 10.6514 ⎤⎥ ⎢ ⎢ ⎥ K4 = ⎢⎢ 44.3537 −77.5840 33.2302 ⎥⎥ ⎢ ⎥ ⎢ 78.3229 73.9174 −152.2403⎥ ⎣ ⎦ ⎡1 0 0⎤ ⎢ ⎥ ⎢ ⎥ ⎢ F1 = ⎢0 1 0⎥⎥ ⎢ ⎥ ⎢0 0 1⎥ ⎣ ⎦

295

(6.164)

(6.165)

(6.166)

(6.167)

(6.168)

(6.169)

With all the above calculated matrices, the system state matrix is finally computed as;

⎡ 0 0 0 1.0000 0 0 ⎤⎥ ⎢ ⎥ ⎢ ⎢ 0 0 0 0 1.0000 0 ⎥⎥ ⎢ ⎥ ⎢ ⎢ 0 0 0 0 0 1.0000⎥⎥ ⎢ A=⎢ 10.6514 0 0 0 ⎥⎥ ⎢−24.1837 13.5322 ⎥ ⎢ ⎢ 44.3537 −77.5840 33.2302 0 0 0 ⎥⎥ ⎢ ⎥ ⎢ ⎢ 78.3229 73.9174 −152.2403 0 0 0 ⎥⎦ ⎣

(6.170)

The eigenvalues of the system state matrix are lastly computed and are shown in Table 6.26. Table 6.26: Eigenvalues of the 3 machine system with damping neglected No. Eigenvalue 1 0.0000 +13.3602i 2 0.0000 -13.3602i 3 0.0000 + 8.6898i 4 0.0000 - 8.6898i 5 -0.0000 + 0.0000i 6 -0.0000 - 0.0000i

From Table 6.26, following points can be noted: • There are total 6 eigenvalues. This indeed should be the case as we have three machines in the system and each machine has two state variables. • There are two zero eigenvalues. One zero eigenvalue is due to the absence of damping in the system and the other eigenvalue is corresponding to δ1 (which is taken to be zero to provide the reference for load flow calculation). • All the non-zero eigenvalues appear in pairs. • The real parts of all the non-zero eigenvalues are zero. In other words, upon a disturbance, the oscillations in the system would be persisting (i.e. the magnitudes of the oscillations would not reduce with time). Indeed, from the plots shown in Figs. 6.5 and 6.7 it is observed that upon a fault, the oscillations are indeed sustaining with a constant amplitude. Now let us consider the damping of the generators. For this case, the matrices given above would remain the same. However, one extra matrix, K5 would appear (as shown in equation (6.145)) and this matrix is computed as;

⎡−0.2025 0 0 ⎤⎥ ⎢ ⎢ ⎥ K5 = ⎢⎢ 0 −0.1944 0 ⎥⎥ ⎢ ⎥ ⎢ 0 0 −0.1628⎥⎦ ⎣ 296

(6.171)

Subsequently, the system state matrix is finally computed as;

⎡ 0 0 0 1.0000 0 0 ⎤⎥ ⎢ ⎥ ⎢ ⎢ 0 0 0 0 1.0000 0 ⎥⎥ ⎢ ⎥ ⎢ ⎢ 0 0 0 0 0 1.0000 ⎥⎥ ⎢ A=⎢ 10.6514 −0.2025 0 0 ⎥⎥ ⎢−24.1837 13.5322 ⎥ ⎢ ⎢ 44.3537 −77.5840 33.2302 0 −0.1944 0 ⎥⎥ ⎢ ⎥ ⎢ ⎢ 78.3229 73.9174 −152.2403 0 0 −0.1628⎥⎦ ⎣

(6.172)

The eigenvalues of this state matrix are shown in Table 6.27. Table 6.27: Eigenvalues of the 3 machine system with damping neglected No. 1 2 3 4 5 6

Eigenvalue -0.0844 +13.3599i -0.0844 -13.3599i -0.0970 + 8.6893i -0.0970 - 8.6893i -0.0000 -0.1970

Comparison of Tables 6.26 and 6.27 reveals that: • When damping is considered, there is only one zero eigenvalue corresponding to δ1 . • The second zero eigenvalue in Table 6.26 is replaced with a real, negative eigenvalue. • The complex eigenvalues still appear in pairs. • The real parts of the complex eigenvalues are negative implying that upon a disturbance, the oscillations in the system will be damped (i.e. the amplitudes of the oscillations will decrease with time). In fact, Figs. 6.6 and 6.8 show that this is indeed the case. With this example, we are now at the end of discussion of small signal stability analysis. From the next lecture, we will start the discussion of voltage stability analysis.

297

6.6

Voltage stability

Voltage collapses usually occur on power system which are heavily loaded or faulted or have shortage of reactive power. Voltage collapse is a system instability involving many power system components. In fact, a voltage collapse may involve an entire power system. Voltage collapse is typically associated with reactive power demand of load not being met due to shortage in reactive power production and transmission. Voltage collapse is a manifestation of voltage instability in the system. The definition of voltage stability as proposed by IEEE/CIGRE task force is as follows: Voltage stability refer to the ability of power system to maintain steady voltages at all buses in the system after being subjected to a disturbance from a given initial operating point. The system state enters the voltage instability region when a disturbance or an increase in load demand or alteration in system state results in an uncontrollable and continuous drop in system voltage. A system is said to be in voltage stable state if at a given operating condition, for every bus in the system, the bus voltage magnitude increases as the reactive power injection at the same bus is increased. A system is voltage unstable if for at least one bus in the system, the bus voltage magnitude decreases as the reactive power injection at the same bus is increased. It implies that if, V-Q sensitivity is positive for every bus the system is voltage stable and if V-Q sensitivity is negative for at least one bus, the system is voltage unstable. The term voltage collapse is also often used for voltage instability conditions. It is the process, by which, the sequence of events following voltage instability leads to abnormally low voltages or even a black out in a large part of the system. The driving force for voltage instability is usually the loads and load characteristics, hence, voltage stability is sometimes also called load stability. In response to a disturbance, the power consumed by the loads tends to be restored by load dynamics. This in turn increases the stress on the high voltage network by increasing the reactive power consumption and further reducing the voltage. A major factor contributing to voltage instability is the voltage drop in the line impedances when active and reactive powers flow through it. As a result, the capability of the transmission network for power transfer and voltage support reduces. Voltage stability of a system is endangered when a disturbance increases the reactive power demand beyond the sustainable capacity of the available reactive power resources. The voltage stability has been further classified into four categories: Large disturbance voltage stability, small disturbance voltage stability, short term voltage satiability and long term voltage stability. A summary of these classifications is as follows: • Large disturbance voltage stability: It refers to the system’s ability to maintain steady voltage following large disturbances such as, system faults, loss of generation or circuit contingencies. This ability is determined by the system load characteristics and interaction of both continuous and discrete controls and protections. The study period of interest may be from few seconds to tens of minutes. This requires long term dynamic simulation study of the system to capture the interactions of under-load tap changer and generator field current limiter. 298

If following a large disturbance and subsequent system control actions, voltages at all the buses in the system settle down at acceptable levels, the system is said to be large disturbance voltage stable. • Small-disturbance voltage stability: This stability is concerned with the ability of the system to maintain acceptable level of steady voltages, when subjected to small perturbations such as incremental changes in system load. This form of stability is also influenced by the characteristics of loads, continuous controls, and discrete controls at a given instant of time. The basic processes contributing to small disturbance stability are essentially of a steady state nature. Therefore, static analysis can be effectively used to estimate stability margins. • Short term voltage satiability: It involves dynamics of fast acting load components such as induction motors, electronically controlled loads and HVDC converters. The study period of interest is in the order of several seconds and the analysis requires solution of appropriate system differential equations. • Long term voltage stability: The study of long term voltage stability involves the dynamics of slower acting equipment such as tap changing transformers, thermostatically controlled loads and generator current limiters. The study period of interest may extend to several or many minutes, and requires long term dynamics system simulation. Voltage instability may arise due many reasons, but some significant contributors are:

○ Increase in loading ○ Generators, synchronous condensers, or SVC reaching reactive power limits ○ Action of tap changing transformers ○ Load recovery dynamics ○ Line tripping or generator outages Most of these changes have a significant impact on the reactive power production, consumption and transmission in the system. Some counter measures to prevent voltage collapse are:

● Switching of shunt capacitors ● Blocking of tap-changing transformers ● Redispatch of generation ● Load shedding ● Temporary reactive power overloading of generators 299

Figure 6.9: Simple radial power system Voltage stability may occur in different ways. A simple case of voltage stability can be explained by considering the two terminal network of Fig. 6.9. In this system, the network is represented by an equivalent generator that can be modeled in the ¯ behind the equivalent impedance Z¯g .In general, the steady state by an equivalent voltage source E ¯L . The load generator, transformer and line impedances are combined together and represented as Z ¯D and V¯ is the receiving end or load voltage. The current I¯ in the circuit is given impedance is Z by:

I¯ = = =

E¯ Z¯L + Z¯D E¯ ZL ∠θ + ZD ∠φ

E¯ (ZL cos θ + ZD cos φ) + j(ZL sin θ + ZL sin φ)

(6.173)

The magnitude of current is

E I=√ (ZL cos θ + ZD cos φ)2 + (ZL sin θ + ZL sin φ)2 which may be written as:

I=

E √ ZL FL 300

(6.174)

where,

ZD ZD 2 FL = 1 + [ ] + 2 [ ] cos (θ − φ) ZL ZL Where The magnitude of the receiving and voltage is given by:

V = ZD I E ZD = √ [ ] FL ZL

(6.175)

The power supplied to the load is

PL = V Icos φ ZD E 2 PL = ( ) [ ] cos φ FL ZL The plots of I,V and PL are shown in Fig. 6.10 as a function of

θ and φ.

(6.176)

ZL ratio for a specific value of ZD

Figure 6.10: Receiving end voltage, Current and Power as a function of Load An explanation of the chracteristics of Fig. 6.10 is as follows: • As the load demand is increased by reducing ZD , the load power PL increases rapidly at first and then slowly, before reaching a maximum value and then starts decreasing. There is thus, a maximum value of active power that can be transmitted through an impedance from a constant voltage source. 301

• The transmitted power reaches a maximum when the voltage drop in the line is equal to the load voltage V. This occurs, when the load impedance ZD is equal to line impedance ZL . As ZD is gradually reduced, current in the line I increases and load voltage V decreases. Initially, for high values of ZD , the enhancement in I is more than the reduction in V, and hence load power PL increases rapidly with reduction in ZD . As ZD approaches ZL , the effect of the enhancement in I is only slightly greater than that of the reduction in V, hence increase in PL is slow. Finally, when ZD is quite less than ZL the reduction in V dominates over increase in I and hence, PL decreases. • The critical operating condition, corresponding to maximum power, is the limiting point of satisfactory operation. For higher load demand, control of power by varying load would be unstable, as a reduction in load impedance will reduce power. The load characteristics decides whether the system voltage decreases progressively and the system will become unstable. With a constant impedance static load characteristic, the system stabilizes at power and voltage levels lower than the desired values. For a constant power load characteristic, the system becomes unstable through collapse of load bus voltage. If the load is supplied by transformers with automatic under load-tap- changing (ULTC), the tap changer will try to raise the effective load impedance ZD as seen from the system. This will lower the load bus voltage still further and lead to a progressive reduction of voltage. This is the ease of simple voltage instability. From the study of voltage stability the relationship between PL and V is important and this will be discussed in the next lecture.

302

6.7

Relation between PL, QL and V

Neglecting the resistance of generator transformer and transmission line, the equivalent circuit of the system and its phasor diagram are shown in Fig. 6.11 (a) and (b).

Figure 6.11: Equivalent circuit of the system(a) and the phasor diagram (b)

From the phasor diagram of Fig. 6.11 (b), it can be observed that IX cos φ = E sin δ and

IX sin φ = E cos δ − V

IX cos φ EV = sin δ X X

PL (V ) = V I cos φ = V

(6.177)

QL (V ) = V I sin φ = V

2

IX sin φ EV V = cos δ − X X X

¯ and V¯ phasor can be eliminated using the identity sin2 δ + cos2 δ = 1 The angle δ between E and thus one can obtain: 303

EV 2 V2 2 2 [ ] = [PL (V )] + [QL (V ) + ] X X

(6.178)

This static power-voltage equation determines all the possible network solutions when the voltage characteristics PL (V ) and QL (V ) are taken into account. For an ideally stiff load, the power demand of the load is independent of voltage and is constant PL (V ) = PL and QL (V ) = QL , where, PL and QL are the real and reactive power demand of the load at the rated voltage V. For stiff load the equation (6.178) can now be written as:

EV 2 V2 2 2 [ ] = [PL ] + [QL + ] X X

(6.179)

Substituting QL = PL tan φ in equation (6.179) one can obtain,

EV 2 V2 2 V2 =[ ] −[ ] P + P tan φ + 2 PL tan φ X X X 2 L

Substituting tan φ =

2 L

2

(6.180)

sin φ and noting that sin2 φ + cos2 φ = 1 equation (6.180) can be further cos φ

simplified as

V2 V2 2 P + 2 PL sin φ cosφ = 2 (E − V 2 ) cos2 φ X X 2 L

(6.181)

2 V2 Adding and subtracting ( sin φ cos φ) to the right hand side of equation (6.181), gives X 2 V2 V2 2 V2 (PL + sin φ cos φ) − ( ) sin2 φ cos2 φ = ( 2 ) (E 2 − V 2 ) cos2 φ X X X

After simplifications the equation can be expressed as:

PL +

√ V V2 sin φ cos φ = cos φ E 2 − V 2 cos2 φ X X

(6.182)

The voltage at the load bus can be expressed in per unit as V/E. Equation (6.182) can be expressed as:

E2 V 2 E2 V PL = − ( ) ( ) sin φ cos φ + ( ) ( ) cos φ X E X E Or

Where p =

√

V 2 1 − ( ) cos2 φ E

√ p = −v 2 sin φ cos φ + v cos φ 1 − v 2 cos2 φ

(6.183)

PL V , and v = 2 (E /X) E

Equation (6.183) describes a family of curves with φ as a parameter. One such P-V curve is shown, for a particular value of power factor cos φ in Fig. 6.12. The Power-Voltage curve (PV-curve) presents load voltage as a function of load real power. For 304

Figure 6.12: PV curve for lagging power factor load static load (PL =constant) as shown in the figure, two operating points (A) and (B) are possible. Point (A) represents low current high voltage solution and is the desirable operating point, while point (B) represents high current low voltage solution. Operation at point B is possible, although, perhaps non-viable due to low voltage and high current condition. Further, with system initially at ‘point A’, if the load is increased then from the curve it can be seen that the voltage will drop. Increase in load will result in an increase in the current flowing in the transmission line, hence the voltage will reduce and this is a perfectly normal response of the system. Hence, point A and the upper portion of PV curve represent stable system operation region. At point B, however, as the load is increased the system voltage increases which is not possible at all. Hence, point B and the lower portion of PV curve represent unstable operating region. Power systems are operated in the upper part of the PV-curve. As the load increases point (A) and (B) come closer and coincide at the tip of P-V curve. This point is called the maximum loading point or critical point. Further increase in the load demand results in no intersection between the load-characteristic and PV curve and hence, represents voltage instability this is shown in Fig. 6.13. The impact of large a disturbance on voltage stability can also be explained with the help of PV curves. Suppose, a large disturbance causes the loss of a transmission line resulting in increases in reactance X or loss of generator resulting in reduction in E. The post-disturbance and pre-disturbance PV characteristics along with load characteristic are shown in Fig. 6.14. The large disturbance causes the network characteristic to shrink drastically, so that the post disturbance PV curve and the load characteristic do not intersect at all. This causes voltage instability leading to a voltage collapse. Assuming a smooth increase in load, the point where the load characteristic becomes tangent to the network PV characteristic defines the loadability limit of the system. Any increase in load beyond the loadability limit results in loss of voltage stability, and system can no longer function. In Fig. 6.13, the point where the load characteristic is tangent to network PV curve, coincides with the maximum deliverable power for a constant power load. However, a loadability limit need not necessarily coincide with the maximum deliverable power, as it is dependent on the load characteristic. 305

Figure 6.13: Changes in the operating point wih increasing load

Figure 6.14: Loss of voltage stability due to a large disturbance

Figure 6.15: Maximum deliverable power and loadabilty limit for polynomial load

V α This is shown in Fig. 6.15 for a polynomial load P = P0 ( ) , where, P0 represents the base V0 value of load active power at rated voltage V0 = 1 p.u. and α represents the voltage exponent. α = 0 represents a constant power load. Equation (6.183) when plotted for different values of φ gives a family of PV curves. Because of 306

Figure 6.16: PV curve drawn for different values of power factor their characteristic shape, these curves are referred to as nose curves. The following observations can be made regarding the curves shown in Fig. 6.16: • For a given load below the maximum, there are two possible solutions- one with higer voltage and lower current and the other with lower voltage and higher current. The former corresponds to ’normal’ operating conditions, with the load voltage V closer to the generator voltage. • As the load is more and more compensated (corresponding to smaller tan φ ), the maximum deliverable power increases, and the voltage at which this maximum occurs also increases. • For over-compensated loads (tan φ < 0 , leading power factor), there is a part of the upper PV curve along which the voltage increases, as the load power increases. This can be explained as follows: when tan φ is negative then, with more active power consumption more reactive power is produced by the load. At low values of load, the voltage drop due to increased active power is offset by the increase in voltage due to increased reactive power. The more negative tan φ is, the larger is the portion of PV curve where this voltage rise occurs. The usefulness of the nose curve is high in practice as the difference between a particular load and maximum load, determined by the peak of the characteristic, is equal to stability margin for a given power factor.

√

From equation (6.183), when QL = 0 ( and hence φ = 0), p = v 1 − v 2 . To find the value of PL

dp = 0 and the solution of the resulting equation gives the values dv 2 1 1 E E2 of v = ± and p = . Hence, PL = . Note that is the short circuit power at the load bus, 2 2 2X X at the peak of the nose curve set

as it is the product of no load voltage E and the short circuit current (E/X). The maximum power limit for a lossless line, with unity power power, thus corresponds to half the short circuit power. PV-curves (nose curves), illustrate the dependency of the voltage on real power of a composite load assuming that the power factor is a parameter. The QV curves discussed next are derived assuming that the voltage is a parameter. 307

For given value of V, equation (6.179) describes a circle in the (PL − QL ) plane as shown in Fig. 6.17 (a). The centre of the circle lies on the QL -axis and is shifted vertically down from the origin by (V 2 /X) , the radius of the curve is (EV /X) . Increasing the voltage V produces a family of circles of increasing radius and increasing downward shift, bounded by an envelope as shown in Fig. 6.17 (b).

Figure 6.17: QP curves for stiff load (a) one circle for a given V (b) family of curves for different voltages and their envelope For each point inside envelope, for example, point A, there are two possible solutions to equation (6.179), at voltages V1 and V2 , as it lies on both the circles. For any point on the envelope, say point B, there is only one value of V for which the equation (6.179) is satisfied. By determining the values of PL and VL for which only one solution to equation (6.179) exists, the equation of the envelope can be developed. Rearranging equation (6.179), one gets:

V2 2 E2 V 2 ( ) + (2QL − ) ( ) + (PL2 + Q2L ) = 0 X X X

(6.184)

This quadratic equation in (V 2 /X) has only one solution when

E2 2 D = (2QL − ) − 4 (PL2 + Q2L ) = 0 X

(6.185)

Solving for QL one gets

QL = (

P2 E2 )−[ 2L ] 4X (E /X)

(6.186)

This represents the equation of an inverted parabola that crosses PL axis at E 2 /2X and has its maximum at 308

PLmax = 0; QLmax =

E2 4X

(6.187)

Thus the maximum reactive power supplied to a load with PL = 0 i.e. a purely reactive load is equal to one fourth of the short-circuit power. Also a point with coordinates PL = E 2 /2X and QL = 0 corresponds to the peak of the nose curve for φ = 0 and QL = 0. The parabola as described equation (6.186) defines the shape of envelope in Fig. 6.17 (b) that encloses all the possible solutions to the network equation (6.179). Every point (PL , QL ) inside the parabola satisfies two possible network solutions corresponding to two distinct values of load voltage V, and each point on the parabola satisfies only one network solution corresponding to only one possible value of voltage. Further, there are no network solutions outside the parabola implying that it is not possible to deliver power for a (PL , QL ) point lying outside the parabola. Various stability criteria to assess the system voltage stability are discussed in the next lecture.

309

6.8

Criteria for assessing voltage stability

With reference to Fig. 6.17 (b), for each point inside the envelope of network solution, there are two voltage solutions, one with a higher value of voltage and the other with a lower value. It then becomes necessary, to find out which of these two solutions represents a stable operating point. This can be done by employing voltage stability criteria which are discussed next. (a) The d∆Q/dV criterion The classical voltage stability criterion is based on the capability of the system to supply reactive power for a given amount of load real power demand. For the explanation of this criterion, it is convenient to notionally separate the real and reactive power demands of the load as represented in Fig. 6.18.

Figure 6.18: Equivalent circuit for determining the reactive power characteristics of the system Let PL (V ) and QL (V ) be the load real and reactive power demands respectively.

Also, let PS (V ) and QS (V ) be the real and reactive powers supplied by the source to the load.

As the real power demand is always connected to the transmission link, PL (V ) = PS (V ) also during normal operation QL (V ) = QS (V ) , but for the purpose of stability analysis, this link between QL (V ) and QS (V ) is theoretically separated, hence, QS (V ) is treated as the reactive power supplied to the load and is assumed to be independent of the load reactive power demand QL (V ) . The real and reactive load powers are given as

PL (V ) = PS (V ) =

EV sin δ X

(6.188)

and

QS (V ) =

EV V2 cos δ − X X

(6.189)

Squaring and adding the above two equations and using the identity sin2 δ + cos2 δ = 1, and then solving for QS (V ) one can obtain, 310

⎡√ ⎤ ⎢ EV 2 V2 2 ⎥ ⎢ QS (V ) = ⎢ { } − {PL (V )} ⎥⎥ − ( ) X X ⎢ ⎥ ⎣ ⎦

(6.190)

This equation determines the reactive power-voltage characteristic of a system, and shows how much reactive power will be supplied by the source if the system is loaded only with the real power PL (V ) and the load voltage is treated as a variable. For a constant power load PL (V ) = PL = constant equation (6.186) takes the form of an inverted parabola as shown in Fig. 6.18. The first term of the equation (6.186) depends on the equivalent system reactance X and the load real power PL and has the effect of shifting the parabola downwards and towards the right.

Figure 6.19: QS − V Characteristics for PL = 0 and PL > 0

EV V2 )− and the parabola crosses the horizontal axis at V = E and X X dQS (V ) V = 0. For finding the maximum value Qmax , set the derivative = 0. On solving the dV E E2 resulting equation, the value of V at which Qmax occurs is and Qmax is equal to . 2 4X √ 2 E 2 PL (V )X • Similarly, for PL > 0,the maximum values of QS (V ) occurs at voltage V = ( ) +( ) 2 E E which is greater than 2 • For PL = 0, QS (V ) = (

Next, the QS (V ) and QL (V ) characteristics can be drawn on the same diagram as shown in Fig. 6.20 (a). At equilibrium the supply must equal the demand i.e., QL (V )=QS (V ) and the two possible equilibrium points VS and VU are obtained. This situation is similar to the one shown in Fig. 6.17 (b). The stability of the two equilibrium points can be evaluated using small perturbation method and the fact that an excess of reactive power produces an increase in voltage while a deficit of reactive power decreases the voltage. Now, consider the equilibrium points of Fig. 6.20(a), and assume a small reduction in voltage ∆V . At point ‘S’, this reduction will result in the supplied reactive power QS (V ) being greater than the reactive power demand QL (V ). This excess reactive power will try to increase the voltage 311

Figure 6.20: QS (V ) and QL (V ) Characteristics for (a) two equilibrium points ’S’ and ’U’ (b) the classical stability representation

and therefore, force the voltage to return to point ‘S’. If there is a small increase in ∆V , then at point ‘S’ the supplied reactive power QS (V ) becomes smaller than the load reactive power QL (V ). This deficit in reactive power brings the voltage back to point ‘S’ and thus, it can be concluded that the equilibrium point ‘S’ is a stable operating point. At point ‘U’, the other equilibrium point, a small reduction in voltage results in QL (V ) becoming greater than QS (V ). This deficit of reactive power, further reduces the voltage, and thus, the system fails to return to the equilibrium point ‘U’. Hence, the system is unstable at point ‘U’. Similarly, an increase in the voltage in the vicinity of ‘U’ results in QS (V ) becoming greater than QL (V ). This excess of reactive power increases the voltage further. Again the system fails to come back to equilibrium point, and hence, the equilibrium point ‘U’ is unstable. From Fig. 6.20(b), it can be observed that at the two equilibrium points ‘S’ and ‘U’ the derivative

d (QS − QL )

is of opposite sign. It is negative at the stable point ‘S’ of the surplus reactive power dV and positive at the unstable point ‘U’. Hence, it can be concluded that for stability:

d d∆Q = (QS − QL ) < 0 dV dV or

dQS dQL < dV dV From Fig. 6.20(b), it can be clear that at point ‘U’,

(6.191)

dQS dQL > , and hence, point ‘U’ is dV dV

not stable as per the criterion established in equation (6.191). While at point ‘S’ the condition of equation (6.191) is satisfied and hence, it is a stable point.

For the simple system of Fig. 6.18, the supplied real and reactive powers are functions of two variables V and δ 312

PS (V ) = PL (V ) = fP (V, δ) (6.192)

QS (V ) = fQ (V, δ) Hence, the incremental values of ∆PS and ∆QS can be expressed as :

∂PS ∂PS ∆V + ∆δ ∂V ∂δ

∆PL = ∆PS =

(6.193)

∆QS =

∂QS ∂QS ∆V + ∆δ ∂V ∂δ

∆δ can be written in terms of ∆PL from equation (6.193) as ∂PS −1 ∂PS ∆δ = [ ] [∆PL − ∆V ] ∂δ ∂V

(6.194)

Substituting ∆δ in the expression of ∆QS of equation (6.193) and dividing both sides of the resulting expression by ∆V one gets

∆QS ∂QS ∂QS ∂PS −1 ∆PL ∂PS = + [ ] [ − ] ∆V ∂V ∂δ ∂δ ∆V ∂V

(6.195)

For small values of ∆V one can write

dQS ∂QS ∂QS ∂PS −1 dPL ∂PL ≈ + [ ] [ − ] dV ∂V ∂δ ∂δ dV ∂V

(6.196)

The partial derivatives are calculated from equation (6.188) and equation (6.189)

∂PS ∂δ ∂PS ∂V ∂QS ∂δ ∂QS ∂V

EV cos δ X E = sin δ X EV = − sin δ X E V = cos δ − 2 X X

=

Substituting these partial derivatives into equation (6.196) gives

2V EV X dPL E dQS E ≈ cos δ − − sin δ [ − sin δ] dV X X X EV cos δ dV X 313

(6.197)

Figure 6.21: QS (V ) and QL (V ) Characteristics indicating stable and unstable operating points

E 2V dPL ∂QS ≈ −[ + tan δ] ∂V Xcos δ X dV

(6.198)

Hence, the stability criterion can be written as:

dQL E 2V dPL >[ −{ + tan δ}] dV Xcos δ X dV

(6.199)

dPL

dQL

The derivative of load active and reactive power w.r.t voltage and are calculated from dV dV the load characteristics expressed in terms of V. For a load reactive power characteristic QL (V ) , different QS (V ) ,the supplied reactive power can be plotted for different values of source voltage E. The resulting curves are shown in Fig. 6.21. The points of interaction of these curves correspond to the two possible equilibrium points ‘S’ and ‘U’. The curve QS4 is tangential to the QL (V ) characteristic and the point ‘S4′ represents the critical operating point. For any QS curve below QS4 , the system operation is not possible. In the figure Vcr represents the critical voltage and for normal stable operation it is necessary that V > Vcr . (b) The dE/dV criterion From equation (6.178) E can be expressed in terms of V as,

E(V ) =

¿ Á Á À

PL (V )X QL (V )X (V + ) +( ) V V 2

314

2

(6.200)

QL (V )X

In this equation, represents the voltage drop component in phase with V and V represents the voltage drop component in quadrature with V.

PL (V )X V

A typical E(V) characteristic is shown in Fig. 6.22 with the normal operating point of the load as ‘S’. V is large at this point and is much greater than both in phase and quadrature components of the voltage drop. In this case then, a drop in voltage V will result in a drop in the emf E(V). As V is reduced further, the in phase and quadrature components of the voltage drop become prominent and below a certain value of V, they will force E(V) to rise. As a result each value of E(V) may correspond to two possible network solutions of voltage V. The stability of these two solutions can be examined using small perturbation method.

Figure 6.22: The illustration of stability criterion

dE dV

Let us examine the system behavior at point ‘S’ as shown in Fig. 6.22 with the assumption that source emf is maintained at a constant value. A reduction in the load voltage by ∆V will cause a reduction in the required value of the emf E(V). As the available E is constant and greater than the required value of E(V), it will force the voltage to return to its initial value V. Thus, the system returns to the initial equilibrium point after the disturbance. Similarly, when the voltage is increased by ∆V , the required emf E(V) to maintain the enhanced voltage (V + ∆V ) is larger than the available source emf E. Hence, the voltage is again forced to return to its initial value V by the constant emf E. Thus, it can be safety established that the point ‘S’ is a stable equilibrium point. Next, consider the other equilibrium point ‘U’. As the voltage is reduced, a higher value of emf E(V) is required to maintain it. But as E is constant and less than the required value of emf E(V), this will result in further reduction in voltage. As a result, the voltage further reduces and moves away from the equilibrium point. Similarly, if the voltage is increased by ∆V , then, the required emf E(V) is smaller than the available source emf E. This larger available emf E will cause the voltage to increase further and move further away from the initial equilibrium point ‘U’. Hence, it can be concluded that point ‘U’ is an unstable equilibrium point. From the above arguments, it is apparent that the system is stable if the equilibrium point lies on the right hand side of the characteristics, that is when :

dE >0 dV 315

Figure 6.23: The

dE criterion of volatge stability dV

Vcr is the minimum system voltage at which the system can be operated stably. The usual system operation voltage is greater than Vcr . (c) The dQS /dQL criterion Let QS (V ) be the reactive power generation by the source and QL (V ) be the load demand. Then, QS (V ) = QL (V ) +line reactive power loss. From Fig. 6.11(b) one can write:

QS (V ) = EI sin (φ + δ) = EI sin φ cos δ + EI cos φ sin δ E E = IX sin φ cos δ + IX cos φ sin δ X X

(6.201)

Substituting IXsin φ = E cos δ − V and IX cos φ = E sin δ in equation (6.201), one gets:

E E (E cos δ − V ) cos δ + (E sin δ) sin δ X X E 2 EV QS (V ) = − cos δ X X QS (V ) =

(6.202)

Substituting the expression for QL (V ) from equation (6.177) in the above equation, we get:

QS (V ) =

E2 V 2 − − QL (V ) X X

Or

V 2 E2 = − QL (V ) − QS (V ) X X

(6.203)

Substituting this expression into equation (6.178) and rearranging the terms gives,

Q2S (V ) PL2 (V ) QL (V ) = − 2 + QS (V ) − (E /X) (E 2 /X) 316

(6.204)

Figure 6.24: Generation and load characteristics If the load is a constant real power load with PL (V ) = PL = constant, then equation (6.204) describes a horizontal parabola in the (QS , QL ) plane and is shown in Fig. 6.24(a). The vertex of the parabola is at a constant QS value equal to E 2 /2X while the minimum value of QL depends on PL and for PL = 0, the minimum is at E 2 /4X . An increase in PL shifts the parabola to the left along the QL axis with no corresponding shift with respect to the QS axis. Again a small perturbation in QL can be used to analyze the stability of the two equilibrium points. For a constant QL , less than the minimum value, the two equilibrium points are labeled as ‘S’ and ‘U’ as shown in Fig. 6.24. At the point ‘S’ a small increase ∆QL in the load reactive power is accompanied by an increase in generated reactive power QS and a small reduction in load reactive power causes a reduction in the generated reactive power. Thus, the balance between the load reactive demand and the generated reactive power is always maintained. Hence, the equilibrium point‘S’ is always stable. At the upper equilibrium point ‘U’, an increase in QL produce a reduction in QS , while a reduction in QL causes an increase in QS . Thus, the changes in reactive generation are now in opposite direction to the changes in demand and hence the equilibrium point ‘U’ is unstable. The stability criterion then can be stated as follows: A system is stable, if a small change in reactive load demand produces a change in the generation which has the same sign. In other words the derivative dQS /dQL is positive i.e. :

dQS >0 dQL

(6.205)

Further, at the maximum loading point at the nose of QS − QL characteristic of Fig. 6.24, the derivative dQG /dQL tends to infinity.

It is worth noting that the QG − QL characteristic is a parabola only for ideally stiff real power load PL (V ) = P =constant. For voltage dependant loads an explicit expression for QL (QG ) with PL (V ) cannot be obtained, and an iteractive procedure is required. 317

Methods of improving Voltage Stability Voltage stability can be improved by adopting the following means: • Enhancing the load reactive power support using shunt compensators. • Line length compensation using series compensation. • Load shedding during contingencies. • Constructing additional transmission lines. • Using FACTS controllers.

318

Appendix A The system data

Table A.1: Bus data for 5 bus system Bus no. 1 2 3 4 5

Type 1 2 2 3 3

∣V ∣

θ

PG

(p.u) (deg) (MW) 1 0 0 1 0 50 1 0 100 1 0 0 1 0 0

QG

PL

QL

(MVAR) (MW) (MVAR) 0 0 0 0 0 0 0 0 0 0 115 60 0 85 40 Base MVA = 100

QM IN

QM AX

(MVAR) 0 -500 -500 -500 -500

(MVAR) 0 500 500 500 500

Table A.2: Line data for 5 bus system Branch no. 1 2 3 4 5 6

From bus To bus R (p.u) 1 2 0.042 1 5 0.031 2 3 0.031 3 4 0.031 3 5 0.053 4 5 0.063

319

X (p.u) 0.168 0.126 0.126 0.126 0.210 0.252

B/2 (p.u) 0.041 0.031 0.031 0.031 0.051 0.061

Tx. Tap 0 0 0 0 0 0

Gsh

Bsh

(p.u) (p.u) 0 0 0 0 0 0 0 0 0 0

Table A.3: Bus data for 14 bus system Bus no. 1 2 3 4 5 6 7 8 9 10 11 12 13 14

Type 1 2 3 3 3 2 3 3 3 3 3 3 3 3

∣V ∣

θ

PG

QG

PL

QL

(p.u) (deg) (MW) (MVAR) (MW) (MVAR) 1.06 0 0 0 0 0 1.045 0 18.3 5.857 0 0 1 0 0 0 119 8.762 1 0 0 0 47.79 3.9 1 0 0 0 7.599 1.599 1.07 0 11.2 44.2 0 0 1 0 0 0 0 0 1 0 0 0 0 12.9 1 0 0 0 29.499 16.599 1 0 0 0 9 5.799 1 0 0 0 3.501 1.8 1 0 0 0 6.099 1.599 1 0 0 0 13.5 5.799 1 0 0 0 14.901 5.001 Base MVA = 100

QM IN

QM AX

(MVAR) 0 -500 0 0 0 -500 0 0 0 0 0 0 0 0

(MVAR) 0 500 0 500 0 500 0 0 0 0 500 0 0 0

Table A.4: Line data for 14 bus system Branch no. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

From bus To bus R (p.u) 1 2 0.0194 1 5 0.054 2 3 0.047 2 4 0.0581 2 5 0.0569 3 4 0.067 4 5 0.0134 4 7 0 4 9 0 5 6 0 6 11 0.095 6 12 0.1229 6 13 0.0661 7 8 0 7 9 0 9 10 0.0318 9 14 0.127 10 11 0.082 12 13 0.2209 13 14 0.1709

320

X (p.u) 0.0592 0.223 0.1979 0.1763 0.1738 0.171 0.0421 0.209 0.5562 0.2522 0.1989 0.2557 0.1302 0.1762 0.011 0.0845 0.2703 0.192 0.1999 0.3479

B/2 (p.u) 0.0528 0.0492 0.0438 0.0374 0.0339 0.0346 0.0128 0 0 0 0 0 0 0 0 0 0 0 0 0

Tx. Tap 0 0 0 0 0 0 0 1 1 1 0 0 0 1 1 0 0 0 0 0

Gsh

Bsh

(p.u) (p.u) 0 0 0 0 0.0002 0.502 0 0 0 0 0 0 0 0 0.0023 0.1325 0 0.0633 0 0 0 0 0 0 0 0 0 0

Table A.5: Bus data for 30 bus system Bus no. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

Type 1 2 3 3 2 3 3 2 3 3 2 3 2 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3

∣V ∣

θ

PG

QG

PL

QL

(p.u) (deg) (MW) (MVAR) (MW) (MVAR) 1.05 0 0 0 0 0 1.0338 0 57.56 2.47 21.7 12.7 1 0 0 0 2.4 1.2 1 0 0 0 7.6 1.6 1.0058 0 24.56 22.57 94.2 19 1 0 0 0 0 0 1 0 0 0 62.8 10.9 1.023 0 35 34.84 80 30 1 0 0 0 0 0 1 0 0 0 5.8 2 1.0913 0 17.93 30.78 0 0 1 0 0 0 11.2 7.5 1.0883 0 16.91 37.83 0 0 1 0 0 0 6.2 1.6 1 0 0 0 8.2 2.5 1 0 0 0 3.5 1.8 1 0 0 0 9 5.8 1 0 0 0 3.2 0.9 1 0 0 0 9.5 3.4 1 0 0 0 2.2 0.7 1 0 0 0 17.5 11.2 1 0 0 0 0 0 1 0 0 0 3.2 1.6 1 0 0 0 8.7 6.7 1 0 0 0 0 0 1 0 0 0 3.5 2.3 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 2.4 0.9 1 0 0 0 10.6 1.9 Base MVA = 100

321

QM IN

QM AX

(MVAR) 0 -500 0 0 -500 0 0 -500 0 0 -500 0 -500 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

(MVAR) 0 500 0 0 500 0 0 500 0 0 500 0 500 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

Gsh

Bsh

(p.u) (p.u) 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0.19 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0.04 0 0 0 0 0 0 0 0 0 0 0 0

Table A.6: Line data for 30 bus system Branch no. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41

From bus To bus R (p.u) 1 2 0.0192 1 3 0.0452 2 4 0.057 3 4 0.0132 2 5 0.0472 2 6 0.0581 4 6 0.0119 5 7 0.046 6 7 0.0267 6 8 0.012 6 9 0 6 10 0 9 11 0 9 10 0 4 12 0 12 13 0 12 14 0.1231 12 15 0.0662 12 16 0.0945 14 15 0.221 16 17 0.0824 15 18 0.107 18 19 0.0639 19 20 0.034 10 20 0.0936 10 17 0.0324 10 21 0.0348 10 22 0.0727 21 22 0.0116 15 23 0.1 22 24 0.115 23 24 0.132 24 25 0.1885 25 26 0.2544 25 27 0.1093 27 28 0 27 29 0.2198 27 30 0.3202 29 30 0.2399 8 28 0.0636 6 28 0.0169

322

X (p.u) 0.0575 0.1852 0.1737 0.0379 0.1983 0.1763 0.0414 0.116 0.082 0.042 0.208 0.556 0.208 0.11 0.256 0.14 0.2559 0.1304 0.1987 0.1997 0.1932 0.2185 0.1292 0.068 0.209 0.0845 0.0749 0.1499 0.0236 0.202 0.179 0.27 0.3292 0.38 0.2087 0.396 0.4153 0.6027 0.4533 0.2 0.0599

B/2 (p.u) 0.0528 0.0408 0.0368 0.0084 0.0418 0.0374 0.009 0.0204 0.017 0.009 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0.0418 0.013

Tx. Tap 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0

Table A.7: Bus data for 9 bus system Bus no. 1 2 3 4 5 6 7 8 9

Type 1 2 2 3 3 3 3 3 3

∣V ∣

θ

PG

QG

PL

QL

(p.u) (deg) (MW) (MVAR) (MW) (MVAR) 1.04 0 0 0 0 0 1.025 0 163 0 0 0 1.025 0 85 0 0 0 1 0 0 0 0 0 1 0 0 0 125 50 1 0 0 0 90 30 1 0 0 0 0 0 1 0 0 0 100 35 1 0 0 0 0 0 Base MVA = 100

QM IN

QM AX

(MVAR) 0 -500 -500 -500 -500 -500 -500 -500 -500

(MVAR) 0 500 500 500 500 500 500 500 500

Table A.8: Line data for 9 bus system Branch no. 1 2 3 4 5 6 7 8 9

From bus To bus R (p.u) 2 7 0.0 1 4 0.0 3 9 0.0 4 6 0.017 4 5 0.01 5 7 0.032 6 9 0.039 9 8 0.0119 8 7 0.0085

323

X (p.u) 0.0625 0.0576 0.0586 0.092 0.085 0.161 0.17 0.1008 0.072

B/2 (p.u) 0.0 0.0 0.0 0.079 0.088 0.153 0.179 0.1045 0.0745

Tx. Tap 1 1 1 0 0 0 0 0 0

Gsh

Bsh

(p.u) (p.u) 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0