Load Flow Matrices IIT Roorkee Notes NPTEL

Module 2

Load Flow Analysis

AC power flow analysis is basically a steady-state analysis of the AC transmission and distributiongrid. Essentially, AC power flow method computes the steady state values of bus voltages and linepower flows from the knowledge of electric loads and generations at di�erent buses of the systemunder study. In this module, we will look into the power flow solution of the AC transmission gridonly (the solution methodology of AC distribution grid will not be covered). Further, we will alsostudy the power flow solution technique when an HVDC link is embedded into an AC transmissiongrid. Also, we will be considering only a balanced system in which the transmission lines and loadsare balanced (the impedances are equal in all the three phases) and the generator produces balancedthree phase voltages (magnitudes are equal in all the phases while the angular di�erence betweenany 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) transformerand iv) transmission lines. For the purpose of power flow solution, synchronous generators are notrepresented explicitly, rather their presence in implicitly modeled. We will look into the implicitrepresentation of synchronous generators a little later. However, the other three components aremodeled 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 impedanceand iii) constant current. However, within the normal operating range of the voltage almost all theloads behave as constant power loads. As the objective of the AC power flow analysis is to computethe normal steady-state values of the bus voltages, the loads are always represented as constantpower loads. Hence, at any bus ‘k’ (say), the real and reactive power loads are specified as 100 MWand 50 MVAR (say) respectively. An important point needs to be mentioned here. As the loadsare always varying with time (the customers are always switching ‘ON’ and ‘OFF’ the loads), anyspecific 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. Aline of medium length is always represented by the nominal-fi model as shown in Fig. 2.1, wherez is the total series impedance of the line and Bc is the total shunt charging susceptance of theline. On the other hand, a long transmission line is most accurately represented by its distributedparameter model. However, for steady-state analysis, a long line can be accurately represented bythe equivalent-fi model, which predicts accurate behavior of the line with respect to its terminalmeasurements taken at its two ends. The equivalent-fi model is shown in Fig. 2.2.

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

Figure 2.2: Equivalent fi model of a long transmission line connected between buses ‘i’ and ‘j’

In Fig. 2.2,

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zc = z

yis the characteristic impedance of the line

“ =√zy is the propagation constantz = series impedance of the line per unit lengthy = shunt admittance of the line per unit lengthL = length of the line

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

2.1.3 Transformer

For power system steady-state and fault studies, generally the exciting current of the transformer isneglected 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 perunit 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 transformerwith 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 isas 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 ischanging only the voltage magnitude, not its angle). Further, in these two figures, the quantity y isthe per unit admittance of the transformer. Also, Fig. 2.5 can be derived from Fig. 2.4 by notingt = 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 studyof an ‘n’ bus power system. Towards that goal, we first must understand the concept of injectedpower and injected current, which is our next topic.

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

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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, thecorresponding injected current Ik is also taken as the negative of the load current. If both a generator

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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 tobe the di�erence of the generator current and the load current.

Figure 2.6: Illustration of injected power

To summarize, if Pk, Qk, and Ik denote the injected real power, reactive power and complexcurrent at bus ‘k’ respectively,

• Pk = PG ; Qk = QG and Ik = IG if only a generator is connected to the bus ‘k’.

• Pk = −PL ; Qk = −QL and Ik = −IL if only a load is connected to the bus ‘k’.

• Pk = PG − PL ; Qk = QG −QL and Ik = IG − IL if both generator and load are connected tothe bus ‘k’.

• Pk = 0 ; Qk = 0 ; Ik = 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 ofany general ‘n’ bus power system. The first step towards this goal is to derive the bus admittancematrix, which we will take up next.

2.3 Formation of bus admittance matrix (YBUS)Let us consider a 5 bus network as shown in Fig. 2.7. In this network, all the transmissions arerepresented by fi models. Therefore, the equivalent circuit of the above network is shown in Fig.2.8.

In Fig. 2.8, Ik; k = 1, 2, 3, 4, 5 are the injected currents at bus ‘k’. Further, the quantity yij

denotes the series admittance of the line ‘i-j’ whereas the quantity yijs denotes the half line chargingsusceptance of the line ‘i-j’. Now applying ‘KCL’ at each bus ‘k’ one obtains,

I1

= yT 1

(V1

− V2

) = yT 1

V1

− yT 1

V2

(2.1)

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Figure 2.7: A sample 5 bus network

Figure 2.8: Equivalent circuit of Fig. 2.7

I2

= yT 1

(V2

− V1

) + V2

y23s + (V2

− V3

)y23

+ V2

y24s + (V2

− V4

)y24

= −yT 1

V1

+ (yT 1

+ y23s + y

23

+ y24s + y

24

)V2

− y23

V3

− y24

V4

(2.2)

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I3

= (V3

− V2

)y23

+ V3

y23s + (V3

− V5

)tyT 2

+ t(t − 1)yT 2

V3

+ (V3

− V4

)y34

+ V3

y34s

= −V2

y23

+ {y23

+ y23s + tyT 2

+ t(t − 1)yT 2

+ y34

+ y34s} V

3

− y34

V4

− tyT 2

V5

(2.3)

I4

= (V4

− V2

)y24

+ y24sV4

+ (V4

− V3

)y34

+ y34sV4

= −V2

y24

− V3

y34

+ (y24

+ y24s + y

34

+ y34s)V4

(2.4)

I5

= (V5

− V3

)tyT 2

+ (1 − t)yT 2

V5

= −V3

tyT 2

+ {tyT 2

+ (1 − t)yT 2

} V5

(2.5)

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

��

I1

I2

I3

I4

I5

��=��

Y11

Y12

Y13

Y14

Y15

Y21

Y22

Y23

Y24

Y25

Y31

Y32

Y33

Y34

Y35

Y41

Y42

Y43

Y44

Y45

Y51

Y52

Y53

Y54

Y55

��

��

V1

V2

V3

V4

V5

��(2.6)

Where, Y11

= yT 1

; Y12

= −yT 1

; Y13

= Y14

= Y15

= 0; Y21

= −yT 1

;Y

22

= (yT 1

+ y23s + y

23

+ y24s + y

24

); Y23

= −y23

; Y24

= −y24

; Y25

= 0;Y

31

= 0; Y32

= −y23

; Y33

= {y23

+ y23s + tyT 2

+ t(t − 1)yT 2

+ y34

+ y34s} ;

Y34

= −y34

; Y35

= −tyT 2

; Y41

= 0; Y42

= −y24

; Y43

= −y34

;Y

44

= (y24

+ y24s + y

34

+ y34s); Y

45

= 0; Y51

= Y52

= 0;Y

53

= −tyT 2

; Y54

= 0; Y55

= {tyT 2

+ (1 − t)yT 2

}Equation (2.6) can be written as,

IBUS

= YBUS

VBUS

(2.7)

Where,I

BUS

= �I1

, I2

� I5

�T → (5 × 1) is the vector of bus injection currentsV

BUS

= �V1

, V2

� V5

�T → (5 × 1) is the vector of bus voltages measured with respect to theground

YBUS

→ (5 × 5) is the bus admittance matrix

Furthermore, from the elements of the YBUS

it can be observed that for i = 1, 2, �� 5;Yii = sum total of all the admittances connected at bus ‘i’Yij = negative of the admittance connected between bus ‘i’ and ‘j’ (if these two buses are physi-

cally connected with each other)Yij = 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,I

BUS

= �I1

, I2

� In�T → (n × 1) is the vector of bus injection currentsV

BUS

= �V1

, V2

� Vn�T → (n × 1) is the vector of bus voltagesY

BUS

→ (n × n) is the bus admittance matrixFurthermore, the elements of the Y

BUS

matrix are calculated in the same way as described above.

So far, we have considered the formation of the YBUS

matrix when there is no mutual couplingamong the elements of the network. In the next lecture, we will look into the procedure for formingthe Y

BUS

matrix in the presence of mutual coupling between the elements.

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2.4 Formation of YBUS matrix in the presence of mutuallycoupled elements

Let us consider Fig. 2.9. In this figure, the impedance Zc connected between nodes ‘u’ and ‘v’ ismutually coupled with the impedance Zd connected between nodes ‘x’ and ‘y’ through a mutualimpedance Zm. The currents through the impedances, the voltages across the impedances and theinjected 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 twoimpedances can be written as,

�Vc

Vd

� = � Zc Zm

Zm Zd

� �Ic

Id

�

Or, �Ic

Id

� = � Zc Zm

Zm Zd

�−1

�Vc

Vd

� = 1ZcZd − Z2

m

� Zd −Zm−Zm Zc

� �Vc

Vd

�Or,

�Ic

Id

� = � Yc Ym

Ym Yd

� �Vc

Vd

� (2.8)

Where,

Yc = Zd

ZcZd − Z2

m

; Yd = Zc

ZcZd − Z2

m

; and Ym = − Zm

ZcZd − Z2

m

(2.9)

Now from Fig. 2.9, �Vc

Vd

� = �Vu − Vv

Vx − Vy

� = � 1 −1 0 00 0 1 −1 �

��

Vu

Vv

Vx

Vy

��Or,

�Vc

Vd

� = �C��

Vu

Vv

Vx

Vy

��where, C = � 1 −1 0 0

0 0 1 −1 � (2.10)

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Again, from Fig. 2.9,

��

Iu

Iv

Ix

Iy

��=��

Ic−Ic

Id−Id

��=��

1 0−1 0

0 10 −1

��Ic

Id

� = �C�T �Ic

Id

� (2.11)

From equations (2.8) and (2.10),

�Ic

Id

� = � Yc Ym

Ym Yd

� �Vc

Vd

� = � Yc Ym

Ym Yd

� �C��

Vu

Vv

Vx

Vy

��Or,

�C�T �Ic

Id

� =��

Iu

Iv

Ix

Iy

��= ��C�T � � Yc Ym

Ym Yd

� �C��

Vu

Vv

Vx

Vy

��(2.12)

Now,

�C�T � Yc Ym

Ym Yd

� �C� =��

1 0−1 0

0 10 −1

�� Yc Ym

Ym Yd

� � 1 −1 0 00 0 1 −1 �

(2.13)

Or,

�C�T � Yc Ym

Ym Yd

� �C� =��

1 0−1 0

0 10 −1

�� Yc −Yc Ym −Ym

Ym −Ym Yd −Yd

�

(2.14)

Or,

�C�T � Yc Ym

Ym Yd

� �C� =��

Yc −Yc Ym −Ym−Yc Yc −Ym Ym

Ym −Ym Yd −Yd−Ym Ym −Yd Yd

��(2.15)

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Hence, from equations (2.12) and (2.15)

��

Iu

Iv

Ix

Iy

��=��

Yc −Yc Ym −Ym−Yc Yc −Ym Ym

Ym −Ym Yd −Yd−Ym Ym −Yd Yd

��

��

Vu

Vv

Vx

Vy

��(2.16)

From equation (2.16),

Iu = YcVu − YcVv + YmVx − YmVy

= YcVu − YcVv + YmVx − YmVy + YmVu − YmVu

= Yc(Vu − Vv) + (−Ym)(Vu − Vx) + Ym(Vu − Vy) (2.17)

Or,Iu = Iuv + Iux + Iuy (2.18)

Similarly,

Iv = −YcVu + YcVv − YmVx + YmVy

= −YcVu + YcVv − YmVx + YmVy + YmVv − YmVv

= Yc(Vu − Vv) + (−Ym)(Vv − Vy) + Ym(Vv − Vx) (2.19)

Or,Iv = Ivu + Ivy + Ivx (2.20)

Ix = YmVu − YmVv + YdVx − YdVy

= YmVu − YmVv + YdVx − YdVy + YmVx − YmVx

= Yd(Vx − Vy) + (−Ym)(Vx − Vu) + Ym(Vx − Vu) (2.21)

Or,Ix = Ixy + Ixu + Ixv (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),

Iy = −YmVu + YmVv − YdVx + YdVy

= −YmVu + YmVv − YdVx + YdVy + YmVy − YmVy

= Yd(Vy − Vx) + (−Ym)(Yy − Vv) + Ym(Yy − Vu) (2.23)

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

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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. 2.13can be considered as an equivalent circuit of Fig. 2.9. As Fig. 2.13 does not contain any mutualadmittance, usual method for Y

BUS

formulation can be adopted for this circuit also.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 nodebetween 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 shownin Fig. 2.14. Moreover, if the nodes ‘u’ and ‘x’ are also common (say ‘s’), then the equivalent circuitis shown in Fig. 2.15. Again, the usual method for Y

BUS

formulation can be adopted for these twocircuits also.

We are now in a position to write down the basic power flow equation, which we will take up inthe next lecture.

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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

I2⋮

In

��=��

Y11

Y12

� Y1n

Y21

Y22

� Y2n⋮ ⋮ ⋮ ⋮

Yn1

Yn2

� Ynn

��

��

V1

V2⋮

Vn

��(2.24)

Or,

Ii = n�j=1

YijVj (2.25)

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

Si = Pi + jQi = ViI∗i (2.26)

Now, Vi = Viej◊i; Vj = Vjej◊j ; Yij = Yijej–ij ;

Hence, Si = Pi + jQi = Viej◊i� n�j=1

YijVjej(◊j + –ij)�∗

Or,

Pi = n�j=1

ViVjYij cos(◊i − ◊j − –ij) (2.27)

Qi = n�j=1

ViVjYij sin(◊i − ◊j − –ij) (2.28)

Equations (2.27) and (2.28) are known as the basic load flow equations. It can be seen that forany 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, ◊i, Pi andQi) associated with the ith bus. Thus for the ‘n’-bus system, there are a total of ‘4n’ variables. Asthere are only ‘2n’ equations available, out of these ‘4n’ variables, ‘2n’ quantities need to be specifiedand remaining ‘2n’ quantities are solved from the ‘2n’ load-flow equations. As ‘2n’ variables are tobe 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 di�erent quantitiesare specified as described below.

1. PQ Bus: At these buses loads are connected and therefore, these buses are also termed as loadbuses. Generally the values of loads (real and reactive) connected at these buses are knownand hence, at these buses Pi and Qi are specified (or known). Consequently, Vi and ◊i need tobe calculated for these buses.

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2. PV Bus: Physically, these buses are the generator buses. Generally, the real power suppliedby the generator is known (as we say that the generation is supplying 100 MW) and also, themagnitude of the terminal voltage of the generator is maintained constant at a pre-specifiedvalue by the exciter (provided that the reactive power supplied or absorbed by the generatoris within the limits). Thus, at a PV bus, Pi and Vi are specified and consequently, Qi and ◊i

need to be calculatd.

3. Slack Bus: To calculate the angles ◊i (as discussed above), a reference angle (◊i = 0) needs tobe specified so that all the other bus voltage angles are calculated with respect to this referenceangle. Moreover, physically, total power supplied by all the generation must be equal to thesum of total load in the system and system power loss. However, as the system loss cannot becomputed before the load flow problem is solved, the real power output of all the generators inthe system cannot be pre-specified. There should be at least one generator in the system whichwould supply the loss (plus its share of the loads) and thus for this generator, the real poweroutput can’t be pre-specified. However, because of the exciter action, Vi for this generator canstill be specified. Hence for this generator, Vi and ◊i(= 0) are specified and the quantities Pi

and Qi are calculated. This generator bus is designated as the slack bus. Usually, the largestgenerator in the system is designated as the slack bus.

To summarise, the details of di�erent types of buses in a ‘n’ bus, ‘m’ generator power system areshown 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 Pi, Qi Vi, ◊i

PV m-1 Pi, Vi Qi, ◊i

Slack 1 Vi, ◊i Pi, Qi

PV while Pi and Qi at slack buses) are not directly solved. Only the quantities Vi and ◊i are directlysolved (Vi for all PQ buses while ◊i for all PV and PQ buses). This is because of the fact that onceVi and ◊i at all PV and PQ buses are solved, then the voltage magnitudes and angles at all the busesare known (Vi, ◊i 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’ ofthem) and ◊i (total ‘n-1’ of them). Therefore, total number of unknown quantities is ‘2n-m-1’. Onthe 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 isequal to the number of specified quantities, the load-flow problem is well-posed.

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Equations (2.27) and (2.28) represent a set of simultaneous, non-linear, algebraic equations. Asthe 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 theload 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 Gauss-Seidel 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) = 0f

2

(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 natureand no particular form of these equations is assumed. Now, with some algebraic manipulation, fromthe first equation of equation set (2.29), the variable x

1

can be represented in terms of the othervariables. Similarly from the second equation, the variable x

2

can be represented in terms of theother variables. Proceeding in the same way, from the nth equation, the variable xn can be expressedin terms of the other variables. Therefore, let,

x1

= g1

(x2

, x3

�� xn)x

2

= g2

(x1

, x3

�� xn)⋮ ⋮xk = gk (x1

, x2

�� xk−1

, xk+1

�� xn)⋮ ⋮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 toassume the initial values of these solution variables (x(0)

1

, x(0)2

�� x(0)n ). 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 = 1Step 2: Update the variables ;

x(k)1

= g1

(x(k−1)2

, x(k−1)3

,�� x(k−1)n );

x(k)2

= g2

(x(k)1

, x(k−1)3

,�� x(k−1)n );

⋮ ⋮x(k)p = gp(x(k)

1

, x(k)2

,�� x(k)p−1

, x(k−1)p+1

,�� x(k−1)n );

⋮ ⋮x(k)n = gn(x(k)

1

, x(k)2

,�� x(k)n−1

);Step 3: Compute e(k)i = �x(k)i − x(k−1)

i � for all i = 1, 2,�� n;Step 4: Compute er =max(e(k)

1

, e(k)2

,�� e(k)n ) ;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 ofx

1

, x2

,�� xp−1

(which are before xp in the sequence of the solution variables) are used while for thevariables xp+1

, xp+2

,�� xn (which are after xp in the sequence of the solution variables), the valuespertaining to previous iteration are used (as these variables have not been updated yet). Subse-quently, in steps 3 and 4, the maximum absolute error between the solutions of the current iterationand previous iteration is calculated. If this maximum absolute error is less then a pre-specifiedtolerance value, then the algorithm is considered to be converged. Otherwise, the solution variablesare 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),

Ii = n�k =1

YikVk = YiiVi + n�k =1≠ i

YikVk. Hence, Vi = 1Yii

��Ii − n�

k =1≠ i

YikVk

��. Now, from the relation

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Pi + jQi = ViI∗i we get, Ii = Pi − jQi

V ∗i. Thus,

Vi = 1Yii

��Pi − jQi

V ∗i− n�

k =1≠ i

YikVk

��(2.31)

Equation (2.31) is the basic equation for performing GSLF. It is to be noted that without lossof 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 thebasic GSLF procedure, let us assume that m = 1, i.e., there is only one generator (which is also theslack bus) and the rest ‘(n-1)’ buses are all load buses. To perform load-flow computation, initialguesses of the bus voltages are necessary. As any power system is generally expected to operate atthe normal steady-state operating condition (with the bus voltage magnitudes maintained between0.95 - 1.05 p.u.), all the unknown bus voltage are initialized to 1.0∠0o p.u (i.e. V (0)j = 1.0∠0o forj = 2, 3,�� n). This process of initializing all bus voltage to 1.0∠0o is called flat start. With theseinitial 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;

V (k)2

= 1Y

22

��P

2

− jQ2

�V (k−1)2

�∗ −n�

j =1≠2

Y2jV

(k−1)j

��⋮ ⋮V (k)p = 1

Ypp

��Pp − jQp

�V (k−1)p �∗ −

p−1�j=1

YpjV(k)

j − n�j=p+1

YpjV(k−1)

j

��⋮ ⋮V (k)n = 1

Ynn

��Pn − jQn

�V (k−1)n �∗ −

n−1�j=1

YnjV(k)

j

��Step 3: Compute e(k)i = �V (k)i − V (k−1)

i � for all i = 2,�� n;Step 4: Compute e(k) =max(e(k)

2

, e(k)3

,�� e(k)n ) ;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 GSLFprocedure for a system having multiple generators. Before we discuss the GSLF procedure, letus look into the procedure of initialisation of bus voltages (which is little di�erent than assuminga 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(i.e. V (0)j = 1.0∠0o for j = (m + 1), (m + 2),�� n) and ii) the magnitudes of the voltages of thePV buses are initialised with the corresponding specified voltage magnitudes while initialising allthese voltage angles to 0o (i.e. V (0)j = V sp

j ∠0o for j = 2, 3,�� m, where V spj is the specified bus

voltage magnitude of the jth generator). Now, as discussed earlier, the reactive power supplied orabsorbed by a generator (QG) is calculated by the load flow procedure. However any generator hasa maximum and minimum limit on QG. If the QG from the generator is within these limits, then thegenerator excitation system is able to maintain the terminal voltage at the specified value. On theother hand, if the generator reaches its limit on QG (either maximum or minimum), then becauseof the insu�cient amount of reactive power (either supplied or absorbed), the generator excitationsystem would not be able to maintain the terminal voltage magnitude at the specified value. In thatcase the generator bus would behave as a PQ bus (P being already specified for the generator andQ is set at either maximum or minimum limiting value of QG). In power system terminology, thisphenomenon (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 correspond-ing limits then this generator continues to behave as a PV bus. Hence �Vi� of this bus (at which thegenerator is connected ) is still maintained at the corresponding specified value and only the angleof this bus voltage is calculated in the present iteration. On the other hand if QG is found to exceedany limit (either maximum or minimum), then it is fixed at that limit and the bus is consideredto act like a PQ bus. Thus, both the magnitude and angle of the bus voltage are calculated inthe present iteration. With this background, the complete algorithm of GSLF involving multiplegenerator buses is as follows.

Complete GSLF algorithm

Step 1: Initialise V (0)j = V spj ∠0o for j = 2, 3,�� m and V (0)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,

Q(k)i = n�j=1

V (k−1)i V (k−1)

j Yij sin �◊(k−1)i − ◊(k−1)

j − –ij�b) If, Qmin

i ≤ Q(k)i ≤ Qmaxi ; then assign �V (k)i � = V sp

i and ◊(k)i = ∠ �A(k)i �. The quantity A(k)i

is given by,

A(k)i = 1Yii

��Pi − jQ(k)i�V (k−1)

i �∗ −i−1�j=1

YijV(k)

j − n�j=i+1

YijV(k−1)

j

��31

c) If Q(k)i ≥ Qmaxi , then calculate

V (k)i = 1Yii

��Pi − jQmax

i�V (k−1)i �∗ −

i−1�j=1

YijV(k)

j − n�j=i+1

YijV(k−1)

j

��d) If Q(k)i ≤ Qmin

i , then calculate

V (k)i = 1Yii

��Pi − jQmin

i�V (k−1)i �∗ −

i−1�j=1

YijV(k)

j − n�j=i+1

YijV(k−1)

j

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

V (k)i = 1Yii

��Pi − jQ(k)i�V (k−1)

i �∗ −i−1�j=1

YijV(k)

j − n�j=i+1

YijV(k−1)

j

��Step 4: Compute e(k)i = �V (k)i − V (k−1)

i � for all i = 2,�� n;Step 5: Compute e(k) =max(e(k)

2

, e(k)3

,�� e(k)n ) ;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 thissystem, n = 5 and m = 3. Moreover, bus 1 is taken to be the slack bus and thus, buses 2-3 areconsidered to be PV buses. The bus data and line data of this system are given in Tables A.1 andA.2 respectively. From Table A.1, the injected real and reactive powers at di�erent buses can beobtained as follows: P

2

= 0.5 p.u, P3

= 1.0 p.u, P4

= −1.15 p.u, P5

= −0.85 p.u, Q4

= −0.6 p.u, andQ

5

= −0.4 p.u. Moreover, the YBUS

of this system (computed from the line data given in TableA.2) is shown in equation (2.32). Note that in this equation, the real part (G) and the imaginarypart (B) of the Y

BUS

matrix (YBUS

=G + jB) are shown separately.

Figure 2.16: The example 5 bus system

G =��

3.2417 −1.4006 0 0 −1.8412−1.4006 3.2417 −1.8412 0 0

0 −1.8412 4.2294 −1.2584 −1.12980 0 −1.2584 2.1921 −0.9337

−1.8412 0 −1.1298 −0.9337 3.9047

��

B =��

−13.0138 5.6022 0 0 7.48355.6022 −13.0138 7.4835 0 0

0 7.4835 −18.9271 7.1309 4.47680 0 7.1309 −10.7227 3.7348

7.4835 0 4.4768 3.7348 −15.5521

��(2.32)

33

For applying GSLF, initially the flat start profile is assumed. Please note that the flat voltageprofile is followed for PQ buses. For PV buses, the initial voltage magnitude is taken to be equalto their corresponding specified voltage magnitude. However, the initial voltage angles are alwaysassumed to be zero. Therefore, from the data given in Table A.1, all the 5 bus voltages are initialisedto 1.0∠0o p.u. Now as the system contains both PV and PQ buses, we follow the ‘completeGSLF algorithm’. In step 2(a) of this algorithm, we first calculate the reactive power absorbedor generated by generators 2 and 3 (corresponding to i = 2 and i = 3). The calculated values of Q

2

and Q3

at iteration 1 (in p.u.) are shown in Table 2.2 (denoted as Qcal in this table). Now, the datain Table A.1 show that the minimum and maximum reactive power limits for both these generatorsare -5 p.u. and 5 p.u respectively. Hence, the calculated values of Q

2

and Q3

are well within thecorresponding reactive power limits. Therefore, both bus 2 and bus 3 are continued to operate as PVbuses and as a result, their voltage magnitudes are maintained at the corresponding specified valuesand only the voltage angles are calculated in step 2(b) (utilising the calculated values of Q

2

and Q3

).Subsequently in step 3, both the magnitude and angles of buses 4 and 5 are calculated. The resultsof iteration 1 are shown in Table 2.2. Finally in steps 4-5, the error is calculated, which is also shownin 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 iterationsare shown in Tables 2.2 and 2.3. Please observed from these two tables that because of high Qmax

G

and Qmin

G limits, the reactive powers supplied or observed by these two generators are always withinthese limits and thus bus 2 and 3 continue to act as PV buses from iteration to iteration. Also notefrom Tables 2.2 and 2.3 that the error reduces with iteration. Finally, the algorithm converges after69 iterations and the final solution is shown in Table 2.4. Please note from this table that the finalvalues of Q

2

and Q3

are -18.51 and 68.87 MVAR respectively. These values are well within theircorresponding reactive power limits and thus, the voltage magnitudes of bus 2 and 3 maintained at1.0 p.u (pre-specified values). The finally computed values of real and reactive power injection atall the buses are also shown in Table 2.4, which are also found to be exactly equal to the specifiedinjected values.

Table 2.2: GSLF results in 5 bus system without any generator Q limit for iterations 1-3

Bus no.Iteration = 1 Iteration = 2 Iteration = 3

Qcal �V � ◊ Qcal �V � ◊ Qcal �V � ◊(p.u) (p.u) (deg) (p.u) (p.u) (deg) (p.u) (p.u) (deg)

1 - 1.0 0 - 1.0 0 - 1.0 02 -0.0720 1.0 2.0533 -0.0663 1.0 4.1038 -0.2022 1.0 3.70913 -0.0932 1.0 3.5968 0.2937 1.0 2.6494 0.5142 1.0 1.62344 - 0.9394 -3.2379 - 0.9167 -5.0548 - 0.9101 -6.07765 - 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. Towardsthis 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



1 - 1.0 0 - 1.0 0 - 1.0 02 -0.2129 1.0 3.1318 -0.2071 1.0 2.6927 -0.2013 1.0 2.37823 0.5926 1.0 0.8641 0.6263 1.0 0.3228 0.6461 1.0 -0.05464 - 0.9082 -6.7844 - 0.9074 -7.266 - 0.9069 -7.59855 - 0.9452 -4.1735 - 0.9448 -4.435 - 0.9446 -4.6163


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

Bus no.Without generator Q limit�V � ◊ Pinj Qinj

(p.u) (deg) (p.u) (p.u)1 1.0 0 0.56743 0.265052 1.0 1.65757 0.5 -0.185193 1.0 -0.91206 1.0 0.688754 0.90594 -8.35088 -1.15 -0.65 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 iterationwise solutions for first 6 iterations of the load flow computation with this maximum limit are shownin Tables 2.5 and 2.6. Now, let us compare Tables 2.2 and 2.5. From these two tables it can beobserved that the load flow solution with generator Q limit proceeds in identical fashion for first twoiterations as in the case with no reactive power limit on the generators. However, from iteration 3onwards the solution changes. In iteration 3, Q

3

calculated is found to be equal to 51.42 MVAR. Asa result, Q

3

is limited to 50 MVAR and Bus 3 is converted to a PQ bus, and therefore its voltagemagnitude is calculated using the expression shown in step 2(c). Please observe that this voltagemagnitude is not maintained at 1.0 p.u (in fact, it becomes less than 1.0 p.u. because of insu�cientreactive power). In the subsequent iteration also, calculated Q

3

is always found to be more than themaximum limit and as a result, Q

3

is always maintained at 50 MVAR thereby making �V3

� < 1.0p.u. The algorithm finally converges with a tolerance of 1.0e−12 p.u. after 66 iterations and the finalsolution are shown in Table 2.7. In this table, the GSLF results without any reactive power limit (asshown in Table 2.4) are also reproduced for comparison. Please note from Table 2.7 that because ofcap on Q

3

, the overall voltage profile of the system is little lower than that obtained with no limiton generator reactive power.

As a second example, let us now consider the IEEE-14 bus system. The data of the IEEE 14bus 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



1 - 1.0 0 - 1.0 0 - 1.0 02 -0.0720 1.0 2.0533 -0.0663 1.0 4.1038 -0.2022 1.0 3.70913 -0.0932 1.0 3.5968 0.2937 1.0 2.6494 0.5142 0.9955 1.63184 - 0.9394 -3.2379 - 0.9167 -5.0548 - 0.9071 -6.09185 - 0.957 -2.5257 - 0.9482 -3.2562 - 0.944 -3.8036


Table 2.6: GSLF results in 5 bus system with generator Q limit on bus 3 for iterations 4-6



1 - 1.0 0 - 1.0 0 - 1.0 02 -0.1784 1.0 3.1067 -0.1365 1.0 2.6692 -0.1082 1.0 2.36563 0.5337 0.991 0.9288 0.5178 0.9882 0.4399 0.5109 0.9864 0.10254 - 0.9012 -6.8046 - 0.8973 -7.2918 - 0.8947 -7.62715 - 0.9409 -4.1739 - 0.9388 -4.4292 - 0.9374 -4.6044


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

Bus no.Without generator Q limit With generator Q limit�V � ◊ Pinj Qinj �V � ◊ Pinj Qinj

(p.u) (deg) (p.u) (p.u) (p.u) (deg) (p.u) (p.u)1 1.0 0 0.56743 0.26505 1.0 0 0.56979 0.339352 1.0 1.65757 0.5 -0.18519 1.0 1.69679 0.5 -0.047693 1.0 -0.91206 1.0 0.68875 0.9825 -0.63991 1.0 0.54 0.90594 -8.35088 -1.15 -0.6 0.88918 -8.35906 -1.15 -0.65 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 generatorsare maintained at their corresponding specified values. Also observe that the generator connectedat bus 6 supplies a reactive power of 37.27 MVAR. Now assume that generator 6 is constrained tosupply only 30 MVAR. With this Q limit, the power flow solution is also shown in Table 2.8. Fromthese 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



(p.u) (deg) (p.u) (p.u) (p.u) (deg) (p.u) (p.u)1 1.06 0 2.37259 -0.3308 1.06 0 2.37188 -0.312492 1.045 -5.17113 0.183 -0.166 1.045 -5.17845 0.183 -0.10663 1.04932 -14.54246 -1.19 -0.08762 1.04697 -14.55556 -1.19 -0.087624 1.03299 -10.39269 -0.4779 -0.039 1.02902 -10.35987 -0.4779 -0.0395 1.04015 -8.76418 -0.07599 -0.01599 1.03615 -8.71027 -0.07599 -0.015996 1.07 -12.52265 0.112 0.37278 1.05497 -12.45871 0.112 0.37 1.02076 -13.44781 0 0 1.01266 -13.49478 0 08 1.0224 -13.47154 0 -0.129 1.01391 -13.5185 0 -0.1299 1.0201 -13.60908 -0.29499 -0.16599 1.0118 -13.66101 -0.29499 -0.1659910 1.0211 -13.69541 -0.09 -0.05799 1.01154 -13.73679 -0.09 -0.0579911 1.04144 -13.22158 -0.03501 -0.018 1.02915 -13.21814 -0.03501 -0.01812 1.0526 -13.42868 -0.06099 -0.01599 1.03787 -13.39145 -0.06099 -0.0159913 1.04494 -13.50388 -0.135 -0.05799 1.03063 -13.48166 -0.135 -0.0579914 1.01249 -14.60128 -0.14901 -0.05001 1.00136 -14.64504 -0.14901 -0.05001


c. Because of the limit on generator reactive power, the overall voltage profile is on the lower sideas 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 TablesA.5 and A.6. Again initially the load flow solution has been computed without any limit on thegenerator reactive power and the result are shown in Table 2.9. Subsequently Q limits have beenimposed on both the generator connected at bus 11 (20 MVAR) and the generator connected at bus13 (30 MVAR). However, with a tolerance of 10e−12 p.u. GSLF algorithm fails to converge evenafter 10,000 iteration. When the tolerance is reduced to 10e−6 p.u., the algorithm converges in 348iteration and the result are again shown in Table 2.9. As can be seen from these results, for boththe generation, the reactive power supplied have been fixed at their corresponding limits and as aresult, the overall voltage profile of the system has gone down.

From these results it is observed that the convergence characteristics of the GSLF technique isquite poor. Usually the number of iteration taken by GSLF is quite large and moreover in manycases, GSLF even fails to converge. To overcome these di�culties of GSLF, Newton- Raphson(NR)techniques have been developed, which are our next topics of discussion. These are two versions ofNR techniques, namely, i) NR in polar co-ordinate and ii) NR in rectangular co-ordinate. We willstudy both these versions one by one and will start with the NR in polar co-ordinate from the nextlecture.

37



(p.u) (deg) (p.u) (p.u) (p.u) (deg) (p.u) (p.u)1 1.05 0 2.38673 -0.29842 1.05 0 2.3865 -0.293862 1.0338 -4.97945 0.3586 -0.05698 1.0338 -4.98084 0.35861 -0.045623 1.03128 -7.96653 -0.024 -0.012 1.03045 -7.95523 -0.02399 -0.0124 1.02578 -9.58235 -0.076 -0.016 1.02477 -9.56929 -0.07598 -0.0165 1.0058 -13.60103 -0.6964 0.05042 1.0058 -13.60836 -0.6964 0.055326 1.02178 -11.50296 0 0 1.02084 -11.49304 0.00003 07 1.00111 -13.9994 -0.628 -0.109 1.00055 -13.99792 -0.628 -0.1098 1.023 -12.56853 -0.45 0.12343 1.023 -12.57567 -0.45 0.151119 1.04608 -13.04088 0 0 1.04006 -13.02865 0.00001 010 1.03606 -14.88589 -0.058 -0.02 1.03117 -14.8866 -0.05797 -0.0200111 1.0913 -11.16876 0.1793 0.24018 1.07807 -11.12256 0.1793 0.212 1.04859 -13.74947 -0.112 -0.075 1.04456 -13.75755 -0.11198 -0.07513 1.0883 -12.56078 0.1691 0.31043 1.08311 -12.55854 0.1691 0.314 1.03346 -14.71704 -0.062 -0.016 1.02931 -14.73168 -0.062 -0.01615 1.02825 -14.86737 -0.082 -0.025 1.02403 -14.88097 -0.08199 -0.02516 1.0359 -14.50539 -0.035 -0.018 1.03148 -14.51198 -0.035 -0.01817 1.0306 -14.98291 -0.09 -0.058 1.02583 -14.98712 -0.09 -0.05818 1.01873 -15.58107 -0.032 -0.009 1.01422 -15.59618 -0.03199 -0.00919 1.01626 -15.81066 -0.095 -0.034 1.01159 -15.8251 -0.09499 -0.03420 1.02041 -15.63819 -0.022 -0.007 1.01568 -15.64961 -0.022 -0.00721 1.02305 -15.35955 -0.175 -0.112 1.01825 -15.36524 -0.17496 -0.1120122 1.02343 -15.35222 0 0 1.01867 -15.3581 0 023 1.0165 -15.41998 -0.032 -0.016 1.01226 -15.43637 -0.032 -0.01624 1.00939 -15.81043 -0.087 -0.067 1.00517 -15.8277 -0.087 -0.06725 1.00048 -15.84004 0 0 0.99748 -15.86849 0 026 0.9825 -16.27422 -0.035 -0.023 0.97944 -16.30534 -0.035 -0.02327 1.00379 -15.59587 0 0 1.00158 -15.62827 0 028 1.02049 -12.1474 0 0 1.01959 -12.14235 0 029 0.98353 -16.87497 -0.024 -0.009 0.98126 -16.91314 -0.024 -0.00930 0.97181 -17.79427 -0.106 -0.019 0.96951 -17.83675 -0.106 -0.019

Total iteration = 851 Total iteration = 348 (totlerance 10e−6)

38

2.8 Basic Newton - Raphson (NR) Techniques

Before discussing the application of NR technique in load flow solution, let us first review the basicprocedure of solving a set of non-linear algebraic equation by means of NR algorithm. Let there be‘n’ equations in ‘n’ unknown variables x

1

, x2

,�� xn as given below,

f1

(x1

, x2

,�� xn) = b1

f2

(x1

, x2

,�� xn) = b2⋮ ⋮ ⋮

⋮ ⋮ ⋮fn(x1

, x2

,�� xn) = 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 bedenoted as, x(0)

1

, x(0)2

,�� x(0)n . Subsequently, first order Taylor’s series expansion (neglecting thehigher order terms) is carried out for these equation around the initial guess of solution. Also letthe vector of initial guess be denoted as x(0) = �x(0)

1

, x(0)2

,�� x(0)n �T . Now, application of Taylor’sexpansion on the equations of set (2.33) yields,

f1

�x(0)1

, x(0)2

,�� x(0)n � + ˆf1

ˆx1

�x1

+ ˆf1

ˆx2

�x2

+�� + ˆf1

ˆxn

�xn = b1

f2

�x(0)1

, x(0)2

,�� x(0)n � + ˆf2

ˆx1

�x1

+ ˆf2

ˆx2

�x2

+�� + ˆf2

ˆxn

�xn = b2

⋮ ⋮ ⋮⋮ ⋮ ⋮

fn �x(0)1

, x(0)2

,�� x(0)n � + ˆfn

ˆx1

�x1

+ ˆfn

ˆx2

�x2

+�� + ˆfn

ˆxn

�xn = bn

��

(2.34)

Equation (2.34) can be written as,

��

f1

(x(0))f

2

(x(0))⋮

fn(x(0))

��+

��

ˆf1

ˆx1

ˆf1

ˆx2

� ˆf1

ˆxn

ˆf2

ˆx1

ˆf2

ˆx2

� ˆf2

ˆxn⋮ ⋮ ⋮ˆfn

ˆx1

ˆfn

ˆx2

� ˆfn

ˆxn

��

��

�x1

�x2⋮

�xn

��=��

b1

b2⋮

bn

��(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),

��

�x1

�x2⋮

�xn

��= [J]−1

��

b1

− f1

(x(0))b

2

− f2

(x(0))⋮

bn − fn(x(0))

��= [J]−1

��

�m1

�m2⋮

�mn

��(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 f

1

(x(k)), f2

(x(k)),�� fn(x(k)).Step 3: Compute �m

1

, �m2

,��mn.Step 4: Compute error =max [��m

1

� , ��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).Step 7: Calculate the correction vector �x = ��x

1

, �x2

,��xn�T 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 NRtechnique for load flow solution. We will first discuss the Newton Raphson load- flow (NRLF) inpolar co-ordinates.

2.9 Newton Raphson load flow (NRLF) in polar co-ordinatesFor NRLF techniques, the starting equations are same as those in equations (2.27) and (2.28), whichare reproduced below:

Pi = n�j=1


Qi = n�j=1


Now, as before let us again assume that in a ‘n’ bus, ‘m’ machine system, the first ‘m’ busesare the generator buses with bus 1 being the slack bus. Therefore, the unknown quantities are;◊

2

, ◊3

,�� ◊n (total ‘n-1’ quantities) and Vm+1

, Vm+2

,�� Vn (total ‘n-m’ quantities). Thus the totalnumber of unknown quantities is n−1+n−m = 2n−m−1. Against these unknown quantities, thespecified quantities are; P sp

2

, P sp3

,�� P spn (total ‘n-1’ quantities) and Qsp

m+1

, Qspm+2

,�� Qspn (total

‘n-m’ quantities). Hence, the total number of specified quantities is also (2n −m − 1). Let thevectors of unknown quantities be denoted as ◊ = [◊

2

, ◊3

,�� ◊n]T and V = [Vm+1

, Vm+2

,�� Vn]T .Similarly let the vector of the specified quantities be denoted as Psp = [P sp

2

, P sp3

,�� P spn ]T and

Qsp = [Qspm+1

, Qspm+2

,�� Qspn ]. Also note from equations (2.37) and (2.38) that the real and reactive

power injections at any bus are functions of ◊ and V. Thus, these injection quantities can be

40

written as Pi = Pi(◊, V) for i = 2, 3,�� n and Qi = Qi(◊, V) for i = (m + 1), (m + 2),�� n.For proceeding with NRLF, we assume initial guesses of the bus voltage angles �◊(0)� and the busvoltage 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),

��

ˆP2

ˆ◊2

� ˆP2

ˆ◊n

ˆP2

ˆVm+1

� ˆP2

ˆVn⋮ ⋮ ⋮ ⋮ˆPn

ˆ◊2

� ˆPn

ˆ◊n

ˆPn

ˆVm+1

� ˆPn

ˆVn

ˆQm+1

ˆ◊2

� ˆQm+1

ˆ◊n

ˆQm+1

ˆVm+1

� ˆQm+1

ˆVn⋮ ⋮ ⋮ ⋮ˆQn

ˆ◊2

� ˆQn

ˆ◊n

ˆQn

ˆVm+1

� ˆQn

ˆVn

��

��

�◊2⋮⋮

�◊n

�Vm+1⋮⋮

�Vm+1

��

=

��

P sp2

− P2

�◊(0), V(0)�⋮⋮

P spn − Pn �◊(0), V(0)�

Qspm+1

−Qm+1

�◊(0), V(0)�⋮⋮

Qspn −Qn �◊(0), V(0)�

��

(2.39)

In equation (2.39), the quantity Pi �◊(0), V(0)� is nothing but the calculated value of Pi withvectors ◊(0), V(0). As a result, commonly, the quantity Pi �◊(0), V(0)� is denoted as P cal

i . Withthese notations, equation (2.39) can be written as,

�J1

J2

J3

J4

� ��◊

�V� = �Psp −Pcal

Qsp −Qcal

� = ��P�Q� (2.40)

In equation (2.40) the vectors Pcal and Qcal are defined as; Pcal = [P cal2

, P cal3

,�� P caln ]T and

Qcal = [Qcalm+1

, Qcalm+2

,�� Qcaln ]. Also note that the vectors ◊ and Psp are of dimension (n − 1) × 1

each and the vectors V and Qsp are of dimension (n−m)×1 each. Therefore, from equations (2.39)and (2.40),

J1

= ˆP

ˆ◊=

��

ˆP2

ˆ◊2

ˆP2

ˆ◊3

� ˆP2

ˆ◊n

ˆP3

ˆ◊2

ˆP3

ˆ◊3

� ˆP3

ˆ◊n⋮ ⋮ ⋮ˆPn

ˆ◊2

ˆPn

ˆ◊3

� ˆPn

ˆ◊n

��

(2.41)

J2

= ˆP

ˆV=

��

ˆP2

ˆVm+1

ˆP2

ˆVm+2

� ˆP2

ˆVn

ˆP3

ˆVm+1

ˆP3

ˆVm+2

� ˆP3

ˆVn⋮ ⋮ ⋮ˆPn

ˆVm+1

ˆPn

ˆVm+2

� ˆPn

ˆVn

��

(2.42)

41

J3

= ˆQ

ˆ◊=

��

ˆQm+1

ˆ◊2

ˆQm+1

ˆ◊3

� ˆQm+1

ˆ◊n

ˆQm+2

ˆ◊2

ˆQm+2

ˆ◊3

� ˆQm+2

ˆ◊n⋮ ⋮ ⋮ˆQn

ˆ◊2

ˆQn

ˆ◊3

� ˆQn

ˆ◊n

��

(2.43)

J4

= ˆQ

ˆV=

��

ˆQm+1

ˆVm+1

ˆQm+1

ˆVm+2

� ˆQm+1

ˆVn

ˆQm+2

ˆVm+1

ˆQm+2

ˆVm+2

� ˆQm+2

ˆVn⋮ ⋮ ⋮ˆQn

ˆVm+1

ˆQn

ˆVm+2

� ˆQn

ˆVn

��

(2.44)

In equations (2.41) to (2.44) the sizes of the various matrices are as follows: J1

→ (n−1)×(n−1),J

2

→ (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)

In equation (2.45), the matrix J = �J1

J2

J3

J4

� is known as the Jacobian matrix, the vector �X =��◊

�V� is known as the correction vector and the vector �M = �Psp −Pcal

Qsp −Qcal

� is known as the

mismatch vector. Further, the size of the matrix J is (2n−m− 1)× (2n−m− 1) while the sizes ofboth the vectors �X and �M is (2n −m − 1) × 1.

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

Basic NRLF (polar) algorithm


2),�� n. Let the vectors of the initial voltage magnitudes and angles be denoted as V(0) and ◊(0)respectively.

Step 2: Set iteration counter k = 1.Step 3: Compute the vectors Pcal and Qcal with the vectors ◊(k−1) and V(k−1) thereby forming

the vector �M . Let this vector be represented as �M = [�M1

, �M2

,��M2n−m−1

]T .Step 4: Compute error =max (��M

1

� , ��M2

� ,�� M2n−m−1

�).Step 5: If error ≤ ‘ (pre - specified tolerance), then the final solution vectors are ◊(k−1) and

42

V(k−1) and print the results. Otherwise go to step 6.Step 6: Evaluate the Jacobian matrix with the vectors ◊(k−1) and V(k−1).Step 7: Compute the correction vector �X by solving equation (2.45).Step 8: Update the solution vectors ◊(k) = ◊(k−1) +�◊ 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).

Pi = n�j=1

ViVjYij cos(◊i − ◊j − –ij) = V 2

i Gii + n�j =1≠ i


Qi = n�j=1

ViVjYij sin(◊i − ◊j − –ij) = −V 2

i Bii + n�j =1≠ i


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

Matrix J1

�= ˆP

ˆ◊� (in this case, i = 2, 3,�� n, j = 2, 3,�� n)

ˆPi

ˆ◊j

= − n�k =1≠ i

ViVkYik sin(◊i − ◊k − –ik); j = i (2.48)

ˆPi

ˆ◊j

= ViVjYij sin(◊i − ◊j − –ij); j ≠ i (2.49)

Matrix J2

�= ˆP

ˆV� (in this case, i = 2, 3,�� n, j = (m + 1), (m + 2),�� n)

ˆPi

ˆVj

= 2ViGii + n�k =1≠ i

VkYik cos(◊i − ◊k − –ik); j = i (2.50)

ˆPi

ˆVj

= ViYij cos(◊i − ◊j − –ij); j ≠ i (2.51)

43

Matrix J3

�= ˆQ

ˆ◊� (in this case, i = (m + 1), (m + 2),�� n, j = 2, 3,�� n)

ˆQi

ˆ◊j

= n�k =1≠ i

ViVkYik cos(◊i − ◊k − –ik); j = i (2.52)

ˆQi

ˆ◊j

= −ViVjYij cos(◊i − ◊j − –ij); j ≠ i (2.53)

Matrix J4

�= ˆQ

ˆV� (in this case, i = (m + 1), (m + 2),�� n, j = (m + 1), (m + 2),�� n).

ˆQi

ˆVj

= −2ViBii + n�k =1≠ i

VkYik sin(◊i − ◊k − –ik); j = i (2.54)

ˆQi

ˆVj

= ViYij sin(◊i − ◊j − –ij); j ≠ i (2.55)

With these expressions of Jacobian elements given in equations (2.48)-(2.55), the Jacobian matrixcan be evaluated at each iteration as discussed earlier.

Now, in the basic NRLF (polar) algorithm described earlier, the generator Q-limits have notbeen considered. To accommodate the generator Q-limits, at the beginning of each iteration, reac-tive power absorbed or produced by each generator is calculated. If the calculated reactive poweris within the specified limits, the generator is retained as PV bus, otherwise the generator bus isconverted to a PQ bus, with the voltage at this bus no longer held at the specified value. Thedetailed algorithm is as follows.

Complete NRLF (polar) algorithm


2),�� n. Let the vectors of the initial voltage magnitudes and angles be denoted as V(0) and ◊(0)respectively.

Step 2: Set iteration counter k = 1.Step 3: For i = 2, 3,��m, carry out the following operations.

a) Calculate,

Q(k)i = n�j=1

V (k−1)i V (k−1)

j Yij sin �◊(k−1)i − ◊(k−1)

j − –ij�b) If, Qmin

i ≤ Q(k)i ≤ Qmaxi ; then assign �V (k)i � = V spec

i and the ith bus is retained as PV busfor kth iteration.

c) If Q(k)i > Qmaxi , then assign Qsp

i = Qmaxi or, if Q(k)i < Qmin

i , then assign Qspi = Qmin

i . In

44

both the cases, this bus is converted to PQ bus. Hence, its voltage magnitude becomes an unknownfor the present iteration (thereby introducing an extra unknown quantity) and to solve for this extraunknown quantity, an extra equation is required, which is obtained by the new value of Qsp

i (asshown above). Therefore, when the ith 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 �◊ vectors remain the same. Therefore, the size ofmatrix J

2

becomes (n − 1) × (n −m + l), that of matrix J3

becomes (n −m + l) × (n − 1) and thematrix J

4

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 boththe vectors �X and �M (in equation (2.45)) becomes (2n −m − 1 − l) × 1. Of course, if there isno generator reactive power limit violation, then l = 0.

Step 4: Compute the vectors Pcal and Qcal with the vectors ◊(k−1) and V(k−1) thereby formingthe vector �M . Let this vector be represented as �M = [�M

1

, �M2

,��M2n−m−1−l]T .

Step 5: Compute error =max (��M1

� , ��M2

� ,�� M2n−m−1−l�).

Step 6: If error ≤ ‘ (pre - specified tolerance), then the final rotation vectors are ◊(k−1) andV(k−1) and print the results. Otherwise go to step 7.

Step 7: Evaluate the Jacobian matrix with the vectors ◊(k−1) and V(k−1).Step 8: Compute the correction vector �X by solving equation (2.45).Step 9: Update the solution vectors ◊(k) = ◊(k−1) +�◊ 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 andm = 3 (as the number of generator in 3). Initially, let us assume that there is no violation of reactivepower limit at any generator. Therefore, the sizes of The Jacobian sub-matinees are as follows;J

1

→ (4 × 4), J2

→ (4 × 2), J3

→ (2 × 4) and J4

→ (2 × 2). Therefore, the size of the combinedJacobian matrix J is (6×6). As before, the NRLF algorithm starts with flat voltage profile and withthis assumed voltage profile, the di�erent quantities in the first iteration are calculated as shown inTable 2.10:

Table 2.10: Initial calculation with NRLF (polar) in the 5 bus system

Pcal = �−0.0444 −0.1776 −0.0333 −0.0444�T × 10−14; Qcal = �−0.143 −0.143�T ;

�M = �0.5 1.0 −1.15 −0.85 −0.457 −0.257�T ; error = 1.15;

As the error is more than the tolerance value (‘ = 10−12 p.u), the algorithm proceeds and inthe next few steps, the Jacobian matrix, correction vector and the updated values of the error arecalculated as shown in Table 2.11.

As the error is still more than the tolerance value, the algorithm continues and enters 2nditeration. 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 convergesafter 4 iterations. The final converged solution is shown in Table 2.13. Comparsion of Tables 2.13and 2.4 (GSLF results) shows that the final results obtained by these two methods are identicallysame. However, NRLF achieves this solution in 4 iterations as opposed at 69 iterations required byGSLF. The convergence behavior of both GSLF and NRLF are shown in Fig. 2.17. From this figureit can be observed that, NRLF has certainly much better convergence characteristics as comparedto GSLF.

Now let us consider the case where reactive power generation of generator 3 is limited to 50MVAR. The algorithm again starts with a flat start and the initial calculation are same as shownin Table 2.10 earlier. As Q

3

calculated has not crossed the limit, the program proceeds in the sameway as shown in Table 2.11 till the end of first iteration. With the voltage magnitudes and anglesobtained at the end of first iteration, the reactive power generated by all the machines are againcalculated and the value of Q

3

is found to be 53.42 MVAR (0.5342 p.u). Hence, bus 3 (generator3) is converted to a PQ bus and hence, the PQ buses in the system now are (4, 5, 3). The vectorsof calculated injected real and reactive powers, mismatch vector and final mismatch at the end of1st iteration are shown in Table 2.14. Note that without any generator reactive power violation, thesize of �M vector was (6× 1) and with generator limit violation in Q

3

, the size has now increasedto (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: J

1

→ (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 generatorQ limit violation

J1

=��

13.0858 −7.4835 0 0−7.4835 19.0911 −7.1309 −4.4768

0 −7.1309 10.8657 −3.73480 −4.4768 −3.7348 15.6951

��; J

2

=��

0 0−1.2584 −1.129822.1921 −0.9337−0.9337 3.9047

��J

3

= �0 1.2584 −2.1921 0.93370 1.1298 0.9337 −3.9047

� ; J4

= �10.5797 −3.7348−3.7348 15.4091

�;

J =

��

13.0858 −7.4835 0 0 0 0−7.4835 19.0911 −7.1309 −4.4768 −1.2584 −1.1298

0 −7.1309 10.8657 −3.7348 2.1921 −0.93370 −4.4768 −3.7348 15.6951 −0.9337 3.90470 1.2584 −2.1921 0.9337 10.5797 −3.73480 1.1298 0.9337 −3.9047 −3.7348 15.4091

��

;

�X = �0.0331 −0.0090 −0.1275 −0.0799 −0.0783 −0.0475�T ;

�◊ = �0.0331 −0.0090 −0.1275 −0.0799�T ; �V = �−0.0783 −0.0475�T◊ = �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�T ; 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 shownin Table 2.15. As the mismatch is still more than the tolerance value, the algorithm proceeds furtherand finally, the algorithm converges in 5 iterations. The final converged values are shown in Table2.16. In this table also, the NRLF (polar) results without any reactive power limit (as shown inTable 2.13) are also reproduced for comparison. Moreover, it is observed that the final convergedvalues 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 ongenerator reactive power. For this case also, the limit on the generator at bus 6 has been maintainedat 30 MVAR. Comparison of Tables 2.8 and 2.17 shows that the results obtained by GSLF and NRLFare 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 generatorQ limit violation

J1

=��

13.1998 −7.5543 0 0−7.3995 18.3929 −6.6638 −4.3296

0 −6.3896 9.6258 −3.23630 −4.1771 −3.3143 14.3457

��; J

2

=��

0 0−0.4065 −0.80980.8909 −1.0234−0.7191 2.8658

��;

J3

= �0 1.9289 −2.9037 0.97480 1.3756 0.6628 −4.3554

� ; J4

= � 9.3239 −3.3977−3.5957 14.4487

�;

J =

��

13.1998 −7.5543 0 0 0 0−7.3995 18.3929 −6.6638 −4.3296 −0.4065 −0.8098

0 −6.3896 9.6258 −3.2363 0.8909 −1.02340 −4.1771 −3.3143 14.4567 −0.7191 2.86580 1.9289 −2.9037 0.9748 9.3239 −3.39770 1.3756 0.6628 −4.3554 −3.5957 14.4487

��

;

�X = �−0.0041 −0.0068 −0.0179 −0.0154 −0.0783 −0.0084�T ;

�◊ = �0.0331 −0.0090 −0.1275 −0.0799�T ; �V = �−0.0783 −0.0475�T ;

◊ = �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�T ; error = 0.0026 ;

NRLF to reach the same solution is much less compared to that taken by GSLF for a tolerance of10e−12 p.u.

The results for 30-bus system are shown in Table 2.18. Without any generator Q-limit, the resultsobtained by GSLF and NRLF (polar) are identical, although NRLF (polar) takes only 4 iterationsagainst 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 di�culty in convergence in this case and the algorithm convergeswith 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 polarco-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 limitviolation


(p.u) (deg) (p.u) (p.u)1 1.0 0 0.56743 0.265052 1.0 1.65757 0.5 -0.185193 1.0 -0.91206 1.0 0.688754 0.90594 -8.35088 -1.15 -0.65 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 limiton Q

3

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�T ;error = 0.1087 ;

49

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

J1

=��

13.1998 −7.5543 0 0−7.3995 18.3929 −6.6638 −4.3296

0 −6.3896 9.6258 −3.23630 −4.1771 −3.3143 14.3457

��; J

2

=��

0 0 −1.5250−0.4065 −0.8098 5.15870.8909 −1.0234 −1.9289−0.7191 2.8658 −1.3756

��;

J3

=��

0 1.9289 −2.9037 0.97480 1.3756 0.6628 −4.3554

2.1541 −3.3002 0.3747 0.7713

��; J

4

=��

9.3239 −3.3977 −6.3896−3.5957 14.4487 −4.1771−7.2296 −4.5456 19.4614

��;

J =

��

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 −1.92890 −4.1771 −3.3143 14.4567 −0.7191 2.8658 −1.37560 1.9289 −2.9037 0.9748 9.3239 −3.3977 −6.38960 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

��

;

�X = �−0.0033 −0.0022 −0.0174 −0.0069 −0.0310 −0.0173 −0.0167�T ;

�◊ = �−0.0033 −0.0022 −0.0174 −0.0069�T ; �V = �−0.0310 −0.0173 −0.0167�T ;

◊ = �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�T ;

error = 0.0073;

50

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




Table 2.17: Final Results of the 14 bus system with NRLF (Polar)




51

Table 2.18: Final Results of the 30 bus system with NRLF (Polar)


(p.u) (deg) (p.u) (p.u) (p.u) (deg) (p.u) (p.u)1 1.05 0 2.38673 -0.29842 1.05 0 2.3865 -0.293862 1.0338 -4.97945 0.3586 -0.05698 1.0338 -4.98084 0.35861 -0.045623 1.03128 -7.96653 -0.024 -0.012 1.03045 -7.95523 -0.02399 -0.0124 1.02578 -9.58235 -0.076 -0.016 1.02477 -9.56929 -0.07598 -0.0165 1.0058 -13.60103 -0.6964 0.05042 1.0058 -13.60836 -0.6964 0.055326 1.02178 -11.50296 0 0 1.02084 -11.49304 0.00003 07 1.00111 -13.9994 -0.628 -0.109 1.00055 -13.99792 -0.628 -0.1098 1.023 -12.56853 -0.45 0.12343 1.023 -12.57567 -0.45 0.151119 1.04608 -13.04088 0 0 1.04006 -13.02865 0.00001 010 1.03606 -14.88589 -0.058 -0.02 1.03117 -14.8866 -0.05797 -0.0200111 1.0913 -11.16876 0.1793 0.24018 1.07807 -11.12256 0.1793 0.212 1.04859 -13.74947 -0.112 -0.075 1.04456 -13.75755 -0.11198 -0.07513 1.0883 -12.56078 0.1691 0.31043 1.08311 -12.55854 0.1691 0.314 1.03346 -14.71704 -0.062 -0.016 1.02931 -14.73168 -0.062 -0.01615 1.02825 -14.86737 -0.082 -0.025 1.02403 -14.88097 -0.08199 -0.02516 1.0359 -14.50539 -0.035 -0.018 1.03148 -14.51198 -0.035 -0.01817 1.0306 -14.98291 -0.09 -0.058 1.02583 -14.98712 -0.09 -0.05818 1.01873 -15.58107 -0.032 -0.009 1.01422 -15.59618 -0.03199 -0.00919 1.01626 -15.81066 -0.095 -0.034 1.01159 -15.8251 -0.09499 -0.03420 1.02041 -15.63819 -0.022 -0.007 1.01568 -15.64961 -0.022 -0.00721 1.02305 -15.35955 -0.175 -0.112 1.01825 -15.36524 -0.17496 -0.1120122 1.02343 -15.35222 0 0 1.01867 -15.3581 0 023 1.0165 -15.41998 -0.032 -0.016 1.01226 -15.43637 -0.032 -0.01624 1.00939 -15.81043 -0.087 -0.067 1.00517 -15.8277 -0.087 -0.06725 1.00048 -15.84004 0 0 0.99748 -15.86849 0 026 0.9825 -16.27422 -0.035 -0.023 0.97944 -16.30534 -0.035 -0.02327 1.00379 -15.59587 0 0 1.00158 -15.62827 0 028 1.02049 -12.1474 0 0 1.01959 -12.14235 0 029 0.98353 -16.87497 -0.024 -0.009 0.98126 -16.91314 -0.024 -0.00930 0.97181 -17.79427 -0.106 -0.019 0.96951 -17.83675 -0.106 -0.019


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 imaginaryparts. Hence, let Vk = Vkej◊k = ek + jfk; Ii = ai + jci and Yik = gik + jbik.

Thus, from equation (2.25), Ii = n�k=1

YikVk. Or, ai + jci = n�k=1(gik + jbik)(ek + jfk).

Or,

ai = n�k=1(gikek − bikfk)

ci = n�k=1(bikek + gikfk)

��(2.56)

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

Pi = aiei + cifi = n�k=1[ei(gikek − bikfk) + fi(bikek + gikfk)] (2.57)

andQi = fiai − eici = n�

k=1[fi(gikek − bikfk) − ei(bikek + gikfk)] (2.58)

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

Pi = gii(e2

i + f 2

i ) + n�k =1≠ i

[ei(gikek − bikfk) + fi(bikek + gikfk)] (2.59)

Qi = −bii(e2

i + f 2

i ) + n�k =1≠ i

[fi(gikek − bikfk) − ei(bikek + gikfk)] (2.60)

Now, again let us consider a ‘n’ bus system having ‘m’ generators (bus 1 being the slack bus). Foreach of the ‘(n-m)’ PQ buses, both voltage magnitude and angle are unknown. In other words, forthese buses both real and imaginary parts of the voltages are unknown. For each of the ‘(m-1)’ PVbuses, even though the voltage magnitude is known, real and imaginary parts of the voltage are notknown as there can be many combination of ei and fi to give a specified value of Vi. Hence, foreach of the PV buses also, the real and imaginary parts of the voltage are unknown. Therefore, totalnumber 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, thequantity Q

1

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 areneeded to make the NRLF (rectangular) a well posed problem. These additional specified quantitieswould be available from the specified voltage magnitudes at each of the ‘(m-1) buses. At each ofthese ‘(m-1)’ PV buses, following relation holds good between the specified quantity (Vi) and the

53

unknown quantities (ei, fi).V 2

i = e2

i + f 2

i (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 quantitiesis also (2n − 2). The relations connecting the unknown quantities to the specified quantities aregiven by equations (2.59) - (2.61), which are solved by the standard Newton-Rabhosn technique todetermine 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) aregiven below (following the basic Newton-Raphron solution method discussed previously);

��P�Q�V2

��=��J

1

J2

J3

J4

J5

J6

��e�f� (2.62)

In equation (2.62), the sizes of di�erent 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

of the vectors, the sizes of di�erent sub-matrices J1

�� J6

can be deduced as; J1

→ ˆP

ê→

(n−1)×(n−1); J2

→ ˆP

ˆf→ (n−1)×(n−1); J

3

→ ˆQ

ê→ (n−m)×(n−1); J

4

→ ˆQ

ˆf→

(n −m) × (n − 1); J5

→ ˆV 2

ê→ (m − 1) × (n − 1); J

6

→ ˆV 2

ˆf→ (m − 1) × (n − 1).

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 eachhas 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 thatthe 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 busvoltages 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

the vector �M . Let this vector be represented as �M = [�M1

, �M2

,��M2n−2

]T .

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 thispurpose, the analytical expressions of the elements of the Jacobian matrix are needed, which arederived 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). Fromthese three equations, the Jacobian matrix elements can be obtained as follows:

Matrix J1

�= ˆP

ê� (in this case, i = 2, 3,�� n, j = 2, 3,�� n)

ˆPi

êj

= 2eigii + n�k =1≠ i

(gikek − bikfk); j = i (2.64)

ˆPi

êj

= (eigij + fibij); j ≠ i (2.65)

Matrix J2

�= ˆP

ˆf� (in this case, i = 2, 3,�� n, j = 2, 3,�� n)

ˆPi

ˆfj

= 2figii + n�k =1≠ i

(bikek + gikfk); j = i (2.66)

ˆPi

ˆfj

= (figij − eibij); j ≠ i (2.67)

Matrix J3

�= ˆQ

ê� (in this case, i = (m + 1), (m + 2),�� n, j = 2, 3,�� n)

ˆQi

êj

= −2eibii + n�k =1≠ i

(−bikek − gikfk); j = i (2.68)

ˆQi

êj

= (figij − eibij); j ≠ i (2.69)

55

Matrix J4

�= ˆQ

ˆf� (in this case, i = (m + 1), (m + 2),�� n, j = 2, 3,�� n)

ˆQi

ˆfj

= −2fibii + n�k =1≠ i

(gikek − bikfk); j = i (2.70)

ˆQi

ˆfj

= −(fibij + eigij); j ≠ i (2.71)

Matrix J5

�= ˆV 2

ê� (in this case, i = 2, 3,��m, j = 2, 3,�� n)

ˆV 2

i

êj

= 2ei for j = i; andˆV 2

i

êj

= 0 for j ≠ i; (2.72)

Matrix J6

�= ˆV 2

ˆf� (in this case, i = 2, 3,��m, j = 2, 3,�� n)

ˆV 2

i

ˆfj

= 2fi for j = i; andˆV 2

i

ˆfj

= 0 for j ≠ i; (2.73)

With these expressions of Jacobian elements, the matrix J can be evaluated at each iteration asdiscussed earlier.

Now, in the above algorithm, the violations of generation reactive power limits have not beentaken into account. As in the case of NRLF (polar), in this case also, if a generator violates its Q-limits at any particular iteration, it is treated as a PQ bus for that iteration. Otherwise, it continuesto be treated as a PV bus. The detailed step-by-step procedure for taking the generation Q limitviolation into account in given below.

Complete NRLF (rectangular) algorithm

Step 1: Initialise e(0)j = V spj for j = 2, 3,�� m and e(0)j = 1.0 for j = (m + 1), (m + 2),�� n.

Also initialise f (0)j = 0.0 for j = 2, 3,�� n. Let the vectors of the initial real and imaginary partsof the voltages be denoted as e(0) and f (0) respectively.

Step 2: Set iteration counter k = 1.Step 3: For i = 2, 3,�� m, calculate Q(k)i from equation (2.60) by using e(k−1) and f (k−1). If

the calculated Q(k)i is within its respective limit, then this bus would be retained as a PV bus in thecurrent iteration. If either the lower limit or the upper limit is violated, then this bus is converted toa PQ bus. In general, if l generator buses (l ≤ (m − 1)) violate their corresponding reactive powerlimits at step 3, then the size of �Q vector increases from (n −m) to (n −m + l) and the sizeof �V 2 vector decreases from (m − 1) to (m − l − 1). As a result, the new sizes of the matricesJ

3

, 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 totalsizes 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) withthe vectors e(k−1) and f (k−1) thereby forming the vector �M . Let this vector be represented as�M = [�M

1

, �M2

,��M2n−2

]T .Step 5: Compute error =max (��M

1

� , ��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 profileand subsequently di�erent 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�T × 10−15; Qcal = �−0.143 −0.143�T ; Vcal = �1.0 1.0�T ;

�M = �−0.5 1.0 −1.15 −0.85 −0.457 −0.257 0 0�T ; error = 1.15;

As the mismatch (= 1.15) is greater than the tolerance (= 1.0e−12), the algorithm proceeds furtherand calculates the Jacobian matrix. Following the discussion presented earlier, the sizes of variousJacobian sub matrices should be as follows: J

1

→ (4 × 4); J2

→ (4 × 4); J3

→ (2 × 4); J4

→(2×4); J

5

→ (2×4); J6

→ (2×4) and J→ (8×8). The computed Jacobian matrices are shownin Table 2.20. From this set observe that the sizes of the calculated Jacobian matrices are indeed thesame as they are indicated above. After calculating the Jacobian matrix, the algorithm calculatesthe correction vector (�X) and extracts the vectors �e and �f from �X . With these obtainedvectors �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), thealgorithm repeats all these calculation and finally converges with 5 iterations. The final load flowresult is shown in Table 2.21. Observe that the result obtained with NRLF (rectangular) methodare 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 confirmyourself that the results shown in these two tables are identically same as the corresponding resultsshown earlier with NRLF (polar) and GSLF methods.

Now, let us study the behavior of the algorithm when generator Q-limit is considered. Towardsthis end, again let us first consider the 5-bus system and as before, let us consider the limit on Q

3

tobe 50 MVAR. TO begin with, the initial calculations for this case are identically same as that shownin Table 2.19. As no violation of Q

3

is detected in this initial calculation, the algorithm proceeds asusual (without any consideration of generation Q-limit violation) and repeats the same calculationsas shown in Table 2.20. As there is still no violation in Q

3

, the algorithm advances to 2nd iterationand 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 Q

3

is 0.6824 p.u. and as a result, this bus is now convertedfrom a PV bus to a PQ bus. Hence, the PQ buses are now (4, 5, 3) and only bus 2 is retained as aPV bus. The corresponding calculated vectors Qcal, Vcal and �M are shown in Table 2.24 alongwith 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

J1

=��

3.2417 −1.8412 0 0−1.8412 4.2294 −1.2584 −1.1298

0 −1.2584 2.1921 −0.93370 −1.1298 −0.9337 3.9047

��; J

2

=��

13.0858 −7.4835 0 0−7.4835 19.0911 −7.1309 −4.4768

0 −7.1309 10.8657 −3.73480 −4.4768 −3.7348 15.6951

��;

J3

= �0 −7.1309 10.5797 −3.73480 −4.4768 −3.7348 15.4091

� ; J4

= �0 1.2584 −2.1921 0.93370 1.1298 0.9337 −3.9047

�;J

5

= �2 0 0 00 2 0 0

� ; J6

= �0 0 0 00 0 0 0

�;

J =

��

3.2417 −1.8412 0 0 13.0858 −7.4835 0 0−1.8412 4.2294 −1.2548 −1.1298 −7.4835 19.0911 −7.1309 −4.4768

0 −1.2548 2.1921 −0.9337 0 −7.1309 10.8657 −3.73480 −1.1298 −0.9337 3.9047 0 −4.4768 −3.7348 15.69510 −7.1309 10.5797 −3.7348 0 1.2584 −2.1921 0.93370 −4.4768 −3.7348 15.4091 0 1.1298 0.9337 −3.9047

2.000 0 0 0 0 0 0 00 2.0000 0 0 0 0 0 0

��

;

�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 ; �f = �−0.0331 −0.009 −0.1275 −0.0799�T ;

e = �1.0 1.0 1.0 0.9217 0.9525�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�T ;

error = 0.1809;

In the third iteration the numbers of PQ buses is 3 and the number of PV buses is 1 andhence, the sizes of matrices J

3

and J4

change to (3 × 4) and those of matrices J5

and J6

become(1 × 4). The sizes of the matrices J

1

, 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 areextracted. Subsequently, the vectors e and f are updated. It is found that Q

3

still violates the limitand 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. e f V ◊ Pinj Qinj

(p.u) (p.u) (p.u) (deg) (p.u) (p.u)1 1 0 1 0 0.56743 0.265052 0.99958 0.02893 1 1.65757 0.5 -0.185193 0.99987 -0.01592 1 -0.91206 1 0.688754 0.89634 -0.13157 0.90594 -8.35088 -1.15 -0.65 0.94034 -0.08272 0.94397 -5.02735 -0.85 -0.4

Total iteration = 5



(p.u) (p.u) (p.u) (deg) (p.u) (p.u)1 1.06 0 1.06 0 2.37259 -0.33082 1.04075 -0.09419 1.045 -5.17113 0.183 -0.1663 1.0157 -0.26348 1.04932 -14.54246 -1.19 -0.087624 1.01604 -0.18634 1.03299 -10.39269 -0.4779 -0.0395 1.02801 -0.15849 1.04015 -8.76418 -0.07599 -0.015996 1.04455 -0.232 1.07 -12.52265 0.112 0.372787 0.99277 -0.23739 1.02076 -13.44781 0 08 0.99427 -0.23818 1.0224 -13.47154 0 -0.1299 0.99146 -0.24002 1.0201 -13.60908 -0.29499 -0.1659910 0.99207 -0.24176 1.0211 -13.69541 -0.09 -0.0579911 1.01383 -0.23819 1.04144 -13.22158 -0.03501 -0.01812 1.02382 -0.24445 1.0526 -13.42868 -0.06099 -0.0159913 1.01605 -0.24401 1.04494 -13.50388 -0.135 -0.0579914 0.97979 -0.25524 1.01249 -14.60128 -0.14901 -0.05001

Total iteration = 5

All the relevant calculations pertaining to 3rd iteration are shown in Table 2.25. As the mismatchis 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 computedfor the same generator reactive power limits as taken for GSLF and NRLF (polar) techniques. Thefinal solutions are shown in Tables 2.27 and 2.28 respectively. Again cross-check for yourself that thefinal 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



(p.u) (p.u) (p.u) (deg) (p.u) (p.u)1 1.05 0 1.05 0 2.38673 -0.298422 1.0299 -0.08973 1.0338 -4.97945 0.3586 -0.056983 1.02132 -0.14293 1.03128 -7.96653 -0.024 -0.0124 1.01147 -0.17076 1.02578 -9.58235 -0.076 -0.0165 0.97759 -0.23652 1.0058 -13.60103 -0.6964 0.050426 1.00126 -0.20376 1.02178 -11.50296 0 07 0.97138 -0.24218 1.00111 -13.9994 -0.628 -0.1098 0.99849 -0.22261 1.023 -12.56853 -0.45 0.123439 1.0191 -0.23604 1.04608 -13.04088 0 010 1.00129 -0.26616 1.03606 -14.88589 -0.058 -0.0211 1.07063 -0.21138 1.0913 -11.16876 0.1793 0.2401812 1.01854 -0.24923 1.04859 -13.74947 -0.112 -0.07513 1.06225 -0.23668 1.0883 -12.56078 0.1691 0.3104314 0.99956 -0.26255 1.03346 -14.71704 -0.062 -0.01615 0.99383 -0.26383 1.02825 -14.86737 -0.082 -0.02516 1.00288 -0.25946 1.0359 -14.50539 -0.035 -0.01817 0.99556 -0.26644 1.0306 -14.98291 -0.09 -0.05818 0.98129 -0.27363 1.01873 -15.58107 -0.032 -0.00919 0.97781 -0.27689 1.01626 -15.81066 -0.095 -0.03420 0.98263 -0.27506 1.02041 -15.63819 -0.022 -0.00721 0.98651 -0.27098 1.02305 -15.35955 -0.175 -0.11222 0.98691 -0.27096 1.02343 -15.35222 0 023 0.97991 -0.27028 1.0165 -15.41998 -0.032 -0.01624 0.9712 -0.27501 1.00939 -15.81043 -0.087 -0.06725 0.96249 -0.27308 1.00048 -15.84004 0 026 0.94313 -0.27533 0.9825 -16.27422 -0.035 -0.02327 0.96684 -0.26987 1.00379 -15.59587 0 028 0.99764 -0.21474 1.02049 -12.1474 0 029 0.94118 -0.2855 0.98353 -16.87497 -0.024 -0.00930 0.92532 -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 forlimit on Q

3

Pcal = �0.5000 1.0016 −1.1502 −0.8500�T ;

Qcal = �−0.5944 −0.3988 0.6824�T ; Vcal = �1.0�;�M = �0.0 −0.0016 0.0002 0.0 −0.0056 −0.0012 −0.1824 −0.0�T ; error = 0.1824;

61

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

3

J1

=��

3.3566 −1.6230 0 0−1.9584 5.5167 −1.3701 −1.2000

0 −2.0669 2.2170 −1.32890 −1.4328 −1.1870 4.0987

��; J

2

=��

13.3023 −7.5339 0 0−7.4539 18.1609 −7.1104 −4.4586

0 −6.2313 10.1633 −3.22750 −4.1175 −3.4358 14.8052

��;

J3

=��

0 −6.2313 8.4978 −3.22750 −4.1175 −3.4358 13.8060

−7.4539 19.5569 −7.1104 −4.4586

��; J

4

=��

0 2.0669 −4.5371 1.32890 1.4328 1.1870 −5.8182

1.9584 −3.5351 1.3701 1.2000

��;

J5

= �1.9992 0 0 0� ; J6

= �0.0581 0 0 0�;

J =

��

3.3566 −1.6230 0 0 13.3023 −7.5339 0 0−1.9584 5.5167 −1.3701 −1.2000 −7.4539 18.1609 −7.1104 −4.4586

0 −2.0669 2.2170 −1.3289 0 −6.2313 10.1633 −3.22750 −1.4328 −1.1870 4.0987 0 −4.1175 −3.4358 14.80520 −6.2313 8.4978 −3.2275 0 2.0669 −4.5371 1.32890 −4.1175 −3.4358 13.8060 0 1.4328 1.1870 −5.8182

−7.4539 19.5569 −7.1104 −4.4586 1.9584 −3.5351 1.3701 1.20001.992 0 0 0 0.0581 0 0 0

��

;

�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�T ;

error = 0.0031;

62

Table 2.26: Final Results of the 5 bus system with NRLF (rectangular) with limit on Q3


(p.u) (p.u) (p.u) (deg) (p.u) (p.u)1 1 0 1 0 0.56979 0.339352 0.99956 0.02961 1 1.69679 0.5 -0.047693 0.98244 -0.01097 0.9825 -0.63991 1 0.54 0.87973 -0.12927 0.88918 -8.35906 -1.15 -0.65 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


(p.u) (p.u) (p.u) (deg) (p.u) (p.u)1 1.06 0 1.06 0 2.37188 -0.312492 1.04073 -0.09432 1.045 -5.17845 0.183 -0.10663 1.01336 -0.26312 1.04697 -14.55556 -1.19 -0.087624 1.01224 -0.18505 1.02902 -10.35987 -0.4779 -0.0395 1.0242 -0.15691 1.03615 -8.71027 -0.07599 -0.015996 1.03013 -0.22759 1.05497 -12.45871 0.112 0.37 0.9847 -0.23631 1.01266 -13.49478 0 08 0.98582 -0.23701 1.01391 -13.5185 0 -0.1299 0.98318 -0.23896 1.0118 -13.66101 -0.29499 -0.1659910 0.98261 -0.2402 1.01154 -13.73679 -0.09 -0.0579911 1.00189 -0.23533 1.02915 -13.21814 -0.03501 -0.01812 1.00965 -0.24037 1.03787 -13.39145 -0.06099 -0.0159913 1.00223 -0.24028 1.03063 -13.48166 -0.135 -0.0579914 0.96883 -0.25317 1.00136 -14.64504 -0.14901 -0.05001

Total iteration = 8

63

Table 2.28: Final Results of the 30 bus system with NRLF (rectangular) with generator Q limits


(p.u) (p.u) (p.u) (deg) (p.u) (p.u)1 1.05 0 1.05 0 2.38676 -0.293892 1.0299 -0.08977 1.0338 -4.98141 0.3586 -0.045553 1.02053 -0.14263 1.03044 -7.95615 -0.024 -0.0124 1.01051 -0.17038 1.02477 -9.57038 -0.076 -0.0165 0.97756 -0.23666 1.0058 -13.6093 -0.6964 0.055336 1.00036 -0.20342 1.02084 -11.49427 0 07 0.97083 -0.24204 1.00055 -13.99903 -0.628 -0.1098 0.99845 -0.22276 1.023 -12.57695 -0.45 0.151169 1.01327 -0.2345 1.04005 -13.0305 0 010 0.99655 -0.26495 1.03117 -14.8887 -0.058 -0.0211 1.05781 -0.208 1.07807 -11.1244 0.1793 0.212 1.01458 -0.24844 1.04456 -13.75943 -0.112 -0.07513 1.05718 -0.23554 1.08311 -12.56043 0.1691 0.314 0.99547 -0.26178 1.02931 -14.73367 -0.062 -0.01615 0.98968 -0.26302 1.02403 -14.88301 -0.082 -0.02516 0.99856 -0.25851 1.03148 -14.51398 -0.035 -0.01817 0.99092 -0.26532 1.02583 -14.98919 -0.09 -0.05818 0.97686 -0.27271 1.01421 -15.59837 -0.032 -0.00919 0.97323 -0.2759 1.01159 -15.82733 -0.095 -0.03420 0.97802 -0.27402 1.01568 -15.65181 -0.022 -0.00721 0.98184 -0.26984 1.01825 -15.36745 -0.175 -0.11222 0.98228 -0.26983 1.01867 -15.36029 0 023 0.97574 -0.26947 1.01226 -15.43846 -0.032 -0.01624 0.96705 -0.27419 1.00517 -15.82979 -0.087 -0.06725 0.95946 -0.27277 0.99748 -15.87039 0 026 0.94003 -0.27501 0.97944 -16.30724 -0.035 -0.02327 0.96454 -0.26985 1.00158 -15.62998 0 028 0.99677 -0.21448 1.01959 -12.14364 0 029 0.93881 -0.2855 0.98126 -16.91488 -0.024 -0.00930 0.9229 -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 primarilygoverned 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, butnot 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 com-pared to the susceptance (Gij << Bij).

(c) Under normal steady state operation the angular di�erences among the bus voltages arequite small (◊i − ◊j ≈ 0 (within 5o − 10o)).

(d) The injected reactive power at any bus is always much less than the reactive powerconsumed by the elements connected to this bus when these elements are shorted to the ground(Qi << BiiV 2

i ).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,

ˆPi

ˆVj

= 2ViGii + n�k =1≠ i

VkYik cos(◊i − ◊k − –ik)= 2ViGii + n�

k =1≠ i

VkYik [cos(◊i − ◊k) cos –ik + sin(◊i − ◊k) sin –ik]= 2ViGii + n�

k =1≠ i

Vk [Gik cos(◊i − ◊k) +Bik sin(◊i − ◊k)] ; j = i (2.74)

ˆPi

ˆVj

= ViYij cos(◊i − ◊j − –ij)= ViYij [cos(◊i − ◊j) cos –ij + sin(◊i − ◊j) sin –ij]= Vi [Gij cos(◊i − ◊j) +Bij sin(◊i − ◊j)] ; j ≠ i (2.75)

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

ˆPi

ˆVi

≈ 0 andˆPi

ˆVj

≈ 0 �⇒ J2

≈ 0 (2.76)

Similarly, from equations (2.52) and (2.53) we get,

ˆQi

ˆ◊j

= n�k =1≠ i

ViVk [Gik cos(◊i − ◊k) +Bik sin(◊i − ◊k)] ; j = i (2.77)

ˆQi

ˆ◊j

= −ViVj [Gij cos(◊i − ◊j) +Bij sin(◊i − ◊j)] ; j ≠ i (2.78)

65

Again in light of the natures of the quantities Gii, Gij and (◊i − ◊j) as discussed above,

ˆQi

ˆ◊i

≈ 0 andˆQi

ˆ◊j

≈ 0 �⇒ J3

≈ 0 (2.79)

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

��P�Q� = �J1

00 J

4

� ��◊

�V� (2.80)

In other words, �P depends only on �◊ and �Q depends only on �V . Thus, there is adecoupling between ‘�P - �◊’ and ‘�Q - �V ’ relations. Now, from equations (2.48) and (2.49)we get,

ˆPi

ˆ◊j

= − n�k =1≠ i

ViVkYik sin(◊i − ◊k − –ik); j = i

= ViViYii sin(◊i − ◊i − –ii) − n�k=1


= −BiiV2

i −Qi ≈ −BiiV2

i ; j = i [as Qi << BiiV2

i ] (2.81)

ˆPi

ˆ◊j

= ViVjYij sin(◊i − ◊j − –ij); j ≠ i

= ViVjYij [sin(◊i − ◊j) cos –ij − cos(◊i − ◊j) sin –ij] ; j ≠ i

= ViVj [Gij sin(◊i − ◊j) −Bij cos(◊i − ◊j)] ; j ≠ i

= −ViVjBij; j ≠ i (2.82)

Similarly, from equations (2.54) and (2.55) we get,

ˆQi

ˆVj

= −2ViBii + n�k =1≠ i

VkYik sin(◊i − ◊k − –ik); j = i

or,ˆQi

ˆVj

Vi = −2V 2

i Bii + n�k =1≠ i


or,ˆQi

ˆVj

Vi = −V 2

i Bii + n�k=1

ViVkYik sin(◊i − ◊k − –ik) = Qi − V 2

i Bii; j = i

or,ˆQi

ˆVj

Vi = −V 2

i Bii; j = i [as Qi << BiiV2

i ]or,

ˆQi

ˆVj

= −ViBii; j = i (2.83)

66

ˆQi

ˆVj

= ViYij sin(◊i − ◊j − –ij); j ≠ i

= ViYij [sin(◊i − ◊j) cos –ij − cos(◊i − ◊j) sin –ij] ; j ≠ i

= Vi [Gij sin(◊i − ◊j) −Bij cos(◊i − ◊j)] ; j ≠ i

≈ −ViBij; j ≠ i (2.84)

Combining equations (2.80)-(2.82) we get, �Pi = −Vi

n�k=1

VkBik�◊k. Or,

�Pi

Vi

= − n�k=1

VkBik�◊k (2.85)

Now, as Vi ≈ 1.0 under normal steady state operating condition, equation (2.85) reduces to,

�Pi

Vi

= − n�k=1

Bik�◊k. Or,�P

V = [−B]�◊. Or,

�P

V = �B′��◊ (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 andk = 2, 3,�� n.

Again combining equations (2.80), (2.83) and (2.84) we get,

�Qi = −Vi

n�k=1

Bik�Vk. Or,�Qi

Vi

= − n�k=1

Bik�Vk. Or,

�Q

V = �B′′��V (2.87)

Again, [B′′] is also a constant matrix having a dimension of (n−m)×(n−m). Its elements are thenegative of the imaginary part of the element (i, k) of the Y

BUS

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 notnecessary to invert these matrices in each iteration. Rather, the inverse of these matrices can bestored and used in every iteration, thereby making the algorithm faster. Further simplification inthe FDLF algorithm can be made by,

a. Ignoring the series resistances is calculating the elements of [B′]. Also, by omitting theelements of [B′] that predominantly a�ect reactive power flows, i.e., shunt reactances andtransformer o� nominal in phase taps.

b. Omitting from [B′′] the angle shifting e�ect of phase shifter, which predominantly a�ects realpower 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, thealgorithm proceeds. The Y

BUS

matrix of this system is shown in Table 2.30. In this table, thenotation Y

BUS

(∶, m ∶ n) represents the elements (of the YBUS

matrix) corresponding to all rows(denoted by the notation ‘:’) and columns spanning from mth column to nth column (denoted by thenotation ‘m:n’). From this Y

BUS

matrix, the matrices [B′] and [B′′] are constructed as shown inTable 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 4and 5). Also note that the matrices [B′] and [B′′] have been formed by taking the negative of theimaginary parts of the corresponding elements of the Y

BUS

matrix. With these constant matrices,the vectors �◊ and �V are calculated and subsequently, the vectors ◊ and V have been updated.With these updated values of ◊ and V, the final mismatch (error) is again calculated as shown inTable 2.30. As the error is still more than the tolerance, the algorithm proceeds further and finallyconverges in 19 iterations for a tolerance value of 10−12 p.u. The final solution is shown in Table2.31, 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�T × 10−14; Qcal

= �−0.1430 −0.1430�T ;

�M = �0.5000 1.0000 −1.1500 −0.8500 −0.4570 −0.2570�T ;error = 1.15 ;

Now, let us impose the reactive power limit on the generation at bus 3. Starting from the flatstart, the calculation up to 1st iteration are same and hence are not shown here. The calculationpertaining to 2nd iteration are shown in Table 2.32. From this set, observe that the calculated valueof Q

3

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 in20 iterations. The final solution is shown in Table 2.31. Comparison of this result with the earlierresults (obtained with other methods) shows that because of the approximations made in FDLF, theresults obtained with FDLF are not identically the same with those obtained by the other methodsbut are very close.

The results corresponding to 14-bus and 30-bus systems are shown in Tables 2.34 and 2.35respectively. From these tables note that the number of iterations taken by FDLF is much higher thanthose 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

YBUS

(∶, 1 ∶ 3) =��

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

0 0 −1.2584 + 7.1309i

−1.8412 + 7.4835i 0 −1.1298 + 4.4768i

��;

YBUS

(∶, 4 ∶ 5) =��

0 −1.8412 + 7.4835i

0 0−1.2584 + 7.1309i −1.1298 + 4.4768i

2.1921 − 10.7227i −0.9337 + 3.7348i

−0.9337 + 3.7348i 3.9047 − 15.5521i

��;

[B′] =��

13.0138 −7.4835 0 0−7.4835 18.9271 −7.1309 −4.4768

0 −7.1309 10.7227 −3.73480 −4.4768 −3.7348 15.5521

��; [B′′] = �10.7227 −3.7348

−3.7348 15.5521�;

�◊ = �0.0306 −0.0136 −0.1492 −0.0944�T ; �V = �−0.0528 −0.0292�T ;

◊ = �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�T ;

error = 0.3069;

execution of each iteration is much faster (as the Jacobrian matrix need not be recomputed andreversed at each iteration) and hence, the total time taken by FDLF is quiet comparable to thatneeded by NRLF. From the two tables it is further noted that, in the absence of any generationreactive power limit, the results obtained by FDLF are almost identical to those obtained by othermethods. However, in the presence of generation reactive power limits, FDLF results are quiet closeto 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 willstudy 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




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

�◊ = �−0.0010 −0.0012 0.0026 0.0034�T ; �V = �−0.0399 −0.0277�T ;

◊ = �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�T ;

error = 0.1775;

70

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

[B′] =��

13.0138 −7.4835 0 0−7.4835 18.9271 −7.1309 −4.4768

0 −7.1309 10.7227 −3.73480 −4.4768 −3.7348 15.5521

��; [B′′] =

��10.7227 −3.7348 −7.1309−3.7348 15.5521 −4.4768−7.1309 −4.4768 18.9271

��;

�◊ = �−0.0006 −0.0010 0.0008 0.0034�T ; �V = �−0.0167 −0.0090 −0.0178�T ;

◊ = �0 0.0290 −0.0157 −0.1457 −0.0875�T ;

V = �1.0 1.0 0.9822 0.8906 0.9341�T ;

Pcal = �0.5268 0.9223 −1.1174 −0.8408�T ; Qcal = �−0.5939 −0.4125�T ;

�M = �−0.0268 0.0777 −0.0326 −0.0092 −0.0061 0.0125�T ;

error = 0.0777;





71



(p.u) (deg) (p.u) (p.u) (p.u) (deg) (p.u) (p.u)1 1.05 0 2.38673 -0.29842 1.05 0 2.38678 -0.292892 1.0338 -4.97945 0.3586 -0.05698 1.0338 -4.98182 0.3586 -0.04293 1.03128 -7.96653 -0.024 -0.012 1.03026 -7.95403 -0.024 -0.0124 1.02578 -9.58235 -0.076 -0.016 1.02455 -9.56795 -0.076 -0.0165 1.005 8 -13.60103 -0.6964 0.05042 1.0058 -13.6111 -0.6964 0.056546 1.02178 -11.50296 0 0 1.0206 -11.49172 0 07 1.00111 -13.9994 -0.628 -0.109 1.00041 -13.9986 -0.628 -0.1098 1.023 -12.56853 -0.45 0.12343 1.023 -12.5786 -0.45 0.157979 1.04608 -13.04088 0 0 1.03842 -13.02612 0 010 1.03606 -14.88589 -0.058 -0.02 1.02995 -14.88705 -0.058 -0.0211 1.0913 -11.16876 0.1793 0.24018 1.07426 -11.11026 0.1793 0.1880912 1.04859 -13.74947 -0.112 -0.075 1.04384 -13.76458 -0.112 -0.07513 1.0883 -12.56078 0.1691 0.31043 1.0824 -12.56395 0.1691 0.2999114 1.03346 -14.71704 -0.062 -0.016 1.02853 -14.73984 -0.062 -0.01615 1.02825 -14.86737 -0.082 -0.025 1.0232 -14.88791 -0.082 -0.02516 1.0359 -14.50539 -0.035 -0.018 1.03054 -14.51633 -0.035 -0.01817 1.0306 -14.98291 -0.09 -0.058 1.02469 -14.98908 -0.09 -0.05818 1.01873 -15.58107 -0.032 -0.009 1.01323 -15.60231 -0.032 -0.00919 1.01626 -15.81066 -0.095 -0.034 1.01052 -15.83038 -0.095 -0.03420 1.02041 -15.63819 -0.022 -0.007 1.01457 -15.65374 -0.022 -0.00721 1.02305 -15.35955 -0.175 -0.112 1.01706 -15.36723 -0.175 -0.11222 1.02343 -15.35222 0 0 1.0175 -15.36018 0 023 1.0165 -15.41998 -0.032 -0.016 1.01136 -15.44311 -0.032 -0.01624 1.00939 -15.81043 -0.087 -0.067 1.00418 -15.83356 -0.087 -0.06725 1.00048 -15.84004 0 0 0.99677 -15.87684 0 026 0.9825 -16.27422 -0.035 -0.023 0.97872 -16.31432 -0.035 -0.02327 1.00379 -15.59587 0 0 1.00105 -15.63737 0 028 1.02049 -12.1474 0 0 1.01937 -12.14228 0 029 0.98353 -16.87497 -0.024 -0.009 0.98073 -16.92364 -0.024 -0.00930 0.97181 -17.79427 -0.106 -0.019 0.96897 -17.84827 -0.106 -0.019


72

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 equa-tions pertaining to the DC system are solved together. In the sequential method, the AC and DCsystems are solved separately and the coupling between the AC and DC system in accomplishedby injecting an equivalent amount of real and reactive power at the terminal AC buses. In otherwords, for an HVDC link existing between buses ‘i’ and ‘j’ of an AC system (rectifier at bus ‘i’and inverter at bus ’j’), the e�ect of the DC link in incorporated into the AC system by injectionsP (R)DCi and Q(R)DCi at the rectifier bus ’i’ and P (I)DCj and Q(I)DCj at bus ’j’ (the super scripts ’R’ and’I’ denote the rectifier and inverter respectively). Therefore the net injected power at bus ’i’ and’j’ are: P total

i = PACi + P (R)DCi; QT otali = QACi +Q(R)DCi; P T otal

j = PACj + P (I)DCj; QT otalj = QACj +Q(I)DCj.

With these net injected powers the AC system is again solved and subsequently, the equivalent in-jected powers (P (R)DCi, Q

(R)DCi, P

(I)DCj, Q

(I)DCi) and the total injected powers (P T otal

i , QT otali , P T otal

j , QT otalj )

are updated. This process of alternately solving AC and DC system quantities is continued till thechanges in AC system and DC system quantities between two consecutive iterations become less thena threshold value. Although simultaneous technique gives the solution of the system without anyto and fro switching between the AC and DC systems, the sequential solution technique is actuallyquite 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 assumptionsare 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∠◊t ⇒ Magnitude and angle of the terminal bus bar of the convertera ⇒ Converter transformer tap ratioEs∠◊s ⇒ Magnitude and angle of the secondary side of the converter transformerIp, Is ⇒ Primary and secondary current of the converter transformer respectivelyVd, Id ⇒ 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 entireAC-DC system. With the above notations, the basic equation governing the HVDC systems are asfollow:

For rectifier

Vdr = 3√

2fi

NrarEtr cos „r = Vdor cos „r (2.88)

Vdr = Vdor cos – − 3fi

XcrNrId (2.89)

For inverter

Vdi = 3√

2fi

NiaiEti cos „i = Vdoi cos „i (2.90)

Vdi = Vdoi cos “ − 3fi

XciNiId (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 ‘�’denotes the angular di�erence 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 reactanceof the converter transformer and the angles ‘–’ and ‘“’ denote the firing angle of the rectifier andthe extinction angle of the inverter respectively.

74

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

Vdr − Vdi

Rd

= Id (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 linkis to provide complete controllability of power over a transmission corridor, both the rectifies andthe inverter stations are suitably controlled and thus, suitable control equations also need to beincorporated in the above model. We will discuss these control equations shortly. However, to solvethe 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 – „�T (2.93)

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

xc = �Vdr Vdi Id ar ai cos – cos “ „r „i�T (2.94)

In equation (2.94), Id has been taken only once as the DC current is same at both the ends. Fromequation (2.94) it is observed that there are total 9 unknown variables which need to be solved tocompletely determine the HVDC link. However, we have only 5 independent equations as shown inequations (2.88)-(2.92). Therefore, out of 9 unknown variables, any 4 variables need to be specifiedand 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 combina-tions of control specification and some of their combination are;

i) –, Pdr, “, Vdi; ii) –, Pdr, ai, Vdi;iii) ar, Pdr, ai, Vdi; iv) ar, Pdr, “, Vdi;v) ar, Pdr, “, ai; vi) ar, Pdr, –, “;

vii) –, Id, “, Vdi; viiii) –, Vdr, “, Pdi;

With any of these four specified control values, the remaining 5 variables can be solved fromequations (2.88)-(2.92) by using standard Newton-Raphoson technique. However, for the sequentialsolution techniques, the quantities P (R)Dci , Q(R)Dci, P (I)Dcj and Q(I)Dcj can be competed in a much easierway by algebraic manipulation of equations (2.88)-(2.92). we will show this procedure by two of theeight combinations listed above.

75

Combination 1

In this case, –, Pdr, “ and Vdi are specified. With these known quantities, the calculation procedureis as follows:

Step 1: We know, Pdr = VdrId. Or, Pdr = Vdr(Vdr − Vdi)Rd

(from equation (2.92)).

Or, V 2

dr − VdrVdi −RdPdr = 0. Or,

Vdr = Vdi ±�V 2

di + 4RdPdr

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.

Vdr = 12(Vdi +�V 2

di + 4RdPdr) (2.96)

Step 2: Id is calculated as,Id = Pdr

Vdr

(2.97)

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

Vdor = Vdr + 3fi

XcrNrId

cos –(2.98)

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

cos �r = Vdr

Vdor

(2.99)

ar = Vdorfi

3√

2NrEtr

(2.100)

In equation (2.100) Etr is known as in the sequential solution method, the terminal voltages areknown from the immediate past solution of the AC system equations.

Step 5: The quantities P (R)DCi and Q(R)DCi are calculated as;

P (R)DCi = Pdr and Q(R)DCi = Pdr tan „r (2.101)

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

Vdoi = Vdi + 3fi

XciNiId

cos “(2.102)

76

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

cos „i = Vdi

Vdoi

(2.103)

ai = fiVdoi

3√

2NiEti

(2.104)

Step 8: The quantities P (I)DCj and Q(I)DCj are calculated as,

P (I)DCj = VdiId and Q(I)DCj = P (I)DCj tan „i (2.105)

With these values of P (R)DCi, Q(R)DCi, P (I)DCj and Q(I)DCj, the AC system equations are again solvedto obtain the updated values of Etr and Eti and subsequently, steps (1)-(8) are repeated again toupdate the values of P (R)DCi, Q(R)DCi, P (I)DCj and Q(I)DCj. This alternate process of solving AC and DCsystem equations are repeated till convergence in obtained.

Combination 8

In this case, –, “, Pdi and Vdr are known. With these known quantities, the calculation procedureis as follows:

Step 1: We know Pdi = VdiId = Vdi

Vdr − Vdi

Rd

= VdiVdr − V 2

di

Rd

.

Or, V 2

di +RdPdi − VdiVdr = 0. Or,

Vdi = Vdr ±�V 2

dr − 4RdPdi

2 (2.106)

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

Vdi = 12(Vdr +�V 2

dr − 4RdPdi) (2.107)

Step 2: Id is calculated as,Id = Pdi

Vdi

(2.108)

With these calculated values of Vdi and Id, steps (3)-(8) of combination-1 are followed to calculatethe Equivalent power injection values, where P (R)DCi = VdrId. With these injected power values, theAC and DC systems are continued to be solved alternately till convergence in achieved. It is to benoted that at the rectifier end, P R(DCi) = −Pdr and Q(R)DCi = −Qdr as the rectifier draws both real andreactive power from the grid. On the other hand, at the inverter end, P I(DCj) = Pdi and Q(I)DCj = −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 thissystem, it is now assumed that one bipolar HVDC link is connected between bus 4 and 5 (rectifierat bus 4 and inverter at bus 5). Other relevant data for this link are as follows; Rd = 10.0 �;Nr = Ni = 2;

3fi

Xcr = 3fi

Xci = 6.0 �. Further, let us also assume that the specified values have beentaken according to combination 1 and the values are as follows:

Combination-1

– = 5o, Pdr = 100 MW; “ = 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, �V

4

� = �V5

� = 1.0 p.u. Let us alsoassume 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. Forthis purpose, equations (2.95) - (2.105) are used to calculate the values of di�erent DC variables asfollows: Vdr = 253.938 kV; id = 393.8 Amp.; Qdr = 16.276 MVAR; Pdi = 98.45 MW; Qdi = 35.024MVAR.

Now, let us look at equations (2.95) - (2.105) more closely. Equations (2.96) - (2.99) show thatthe quantities Vdr, id, Vdor and cos �r depend only on the DC system data. As the DC system dataare constant, the calculatd values of these four quantities would also be constant (i.e. their valueswould not change from iteration to iteration). Similarly, from equations (2.102) and (2.103) it canbe seen that the quantities Vdoi and cos �i are also constant. As cos �r and cos �i are constant,from equations (2.101) and (2.105), Q(R)DCi and Q(I)DCj are also constant. Thus, the equivalent realand reactive power injections at buses 4 and 5 are constant (they need not be updated at everyiteration) and hence these values can be pre-calculated and suitably adjusted into the injected realand reactive powers at buses 4 and 5 before solving the AC system equations. Thus, for the exampleat hand, the net injected real and reactive powers at bus 4 and bus 5 can be calculated as follows;P

4

= −1.15 − 1.0 = −2.15 p.u., Q4

= −0.6 − 0.16276 = −0.76276 p.u., P5

= −0.85 + 0.9845 = 0.1345p.u. and Q

5

= −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 limithas 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 andai = 0.8149. Please note that in these two equations, the quantities Etr and Eti should be taken inactual values (i.e. in kV), not in per unit.

Let us now turn our attention to combination 8. Following the same reasonong as describedabove, from equations (2.107) - (2.108) it can be observed that the quantities Vdi and Id are con-stant. Moreover, as steps (3)-(8) of combination 8 are same as in combination 1, it immediatelyfollows 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


(p.u) (deg) (p.u) (p.u)1 1.0 0 0.68984 0.463012 1.0 -0.63995 0.5 -0.172353 1.0 -4.91128 1.0 1.541344 0.82813 -17.48682 -2.15 -0.762775 0.91332 -3.89028 0.13449 -0.75025

Total iteration = 5

the DC system) are constant. Therefore, by pre-calculating these equivalent power injections andsubsequently incorporating these calculated values into net bus power injections, standard AC loadflow 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 realand reactive power injections (representing the DC system) are always constant for any combinationof the specified control variables. However, this is not true. Depending on the specified controlvariables, the equivalent real and reactive power injections may vary from iteration to iteration andtherefore, they need to be calculated in every iteration. As an example, let us consider combination3. 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 – and cos �r using equations (2.98) and (2.99) respectively.Step 5: From the knowledge of Pdr and cos �r, calculate Q(R)DCi using equation (2.101).Step 6: Calculate Vdoi from equation (2.104).Step 7: Calculate cos “ and cos �i using equations (2.102) and (2.103) respectively.Step 8: Calculate P (I)DCj and Q(I)DCj from equation (2.105).

Please note that in steps 3-5, the quantities Vdor, cos �r and Q(R)DCi are all dependent on therectifier side AC bus voltage, Etr. Similarly, in steps 6-8, the quantities Vdoi, cos �i and Q(I)DCj are alldependent on the inverter side AC bus voltage, Eti. The quantity P (I)DCj however, depends only onthe DC system quantities and hence remain constant. Thus, the equivalent reactive power injectionsat both rectifier and inverter side depend on the AC bus voltage magnitudes (although the equivalentreal power injections at both the sides are independent of AC bus voltage magnitudes). Hence, theequivalent reactive power injections need to be updated at each iteration and with these updatedpower injection values, another iteration of AC load flow is carried out. This process is continued tillconvergence is achieved. To illustrate this procedure further, let us assume that the specified valuescorresponding 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 = 98.45MW and Id = 393.8 Amp. The calculated values for di�erent other DC quantities corresponding tofirst 3 iterations are shown in Table 2.37. In this table, the symbols ‘In’ and ‘MM’ denote ‘iterationnumber’ and ‘mismatch’ respectively. Proceeding in this fashion, the algorithm finally converges in70 iterations with a convergence threshold value of 1.0e−12. The final converged values of di�erentDC quantities are as follows: Vdr = 253.938 kV; – = 19.82o; “ = 34.84o; Q(R)DCi = 41.52 MVAR;Q(I)DCj = 72.44 MVAR. The final converged values of the AC system quantities are shown in Table2.38. Note that, as in the case of Table 2.36, in this case also, no generator reactive power violationhas been assumed.

Table 2.37: Calculated DC quantities for first three iterations in 5 bus system

In V4

V5

Vdor „r Q(R)DCi – Vdoi „i Q(I)DCj “ MM(p.u.) (p.u.) (kV) (rad.) (MVAR) (deg.) (kV) (rad.) (MVAR) (deg.)0 1.0 1.0 356.52 0.778 98.54 43.48 356.52 0.7937 100.09 44.4 2.151 0.7843 0.8731 279.63 0.432 46.11 22.33 311.29 0.638 73.04 35.08 0.55782 0.7682 0.8703 273.91 0.3842 40.43 19.20 310.29 0.634 72.37 34.82 0.038

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


(p.u) (deg) (p.u) (p.u)1 1.0 0 0.75382 0.784202 1.0 -0.99374 0.5 -0.168933 1.0 -5.52481 1.0 2.173384 0.82813 -18.69153 -2.15 -1.015225 0.91332 -3.65137 0.13449 -1.12440


For further illustration, let us now consider the 30-bus system. In this system, it is now assumedthat one bipolar HVDC link is connected between bus 9 and 28 (rectifier at bus 9 and inverter at bus28). Other relevant data for this link are as follows; Rd = 10.0 �; Nr = Ni = 2;

3fi

Xcr = 3fi

Xci = 6.0 �.The load flow has been solved for combination-1, combination-3 and combination-8 (of specified quan-tities). The specified values which have been considered are as follows;

Combination-1

80

– = 5o; Pdr = 100 MW; “ = 18o; Vdi = 250 kV

Combination-3

ar = 0.75; Pdr = 100 MW; ai = 0.75; Vdi = 250 kV

Combination-8

– = 5o; Pdi = 100 MW; “ = 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 of10−12 p.u. Furthermore, the results corresponding to combination 3 are shown in Table 2.40. It is tobe noted that for these results, no reactive power limit on the generators have been considered. Thefinal solutions of corresponding DC system quantities are also shown in these tables for these threecases. Comparison of Tables 2.18, 2.39 and 2.40 shows that because of the reactive power absorptionat both bus 9 and 28, the overall voltage profile of the system is lower in the presence of HVDClink. Moreover, when the equivalent injected real and reactive powers are constant (i.e. do not varyfrom iteration to iteration), the number of iterations taken by the algorithm is quiet comparablewith that taken by the normal NRLF (polar) method (without any HVDC link). However, whenthese equivalent injected powers vary from iteration to iteration, the number of iterations taken bythe 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 beenshown. For the remaining combinations, the DC quantities can be calculated following the procedureof 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 simultane-ous technique, the AC and DC system equations are solved together. Now, in a N-bus, M-generatorpower system having a HVDC link between lens ‘k’ and ‘l’ (bus ‘k’ being the rectifier and bus ‘l’ beingthe inverter), the total number of unknown are (N −1)+(N −M)+5 = 2N −M +4. To solve theseunknowns, we also have (N −1)+ (N −M)+5 = 2N −M +4 equations. Therefore the size of Jaco-brian 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 evaluatedin each iteration. Also, the Jacobian matrix also needs to be inverted in each iteration, therebyincreasing the computation burden appreciably. Apart from that, depending upon the combinationof specified quantities, the DC equations [(2.88)-(2.92)] need to be recasted appropriately beforestarting the solution procedure. Therefore, the simultaneous solution technique does not give anycomputational advantage vis-à-vis the sequential method and thus, this method is not further dis-cussed here.

81

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

Bus no.With combination 1 With combination 8�V � ◊ Pinj Qinj �V � ◊ Pinj Qinj

(p.u) (deg) (p.u) (p.u) (p.u) (deg) (p.u) (p.u)1 1.05 0 2.42039 -0.26099 1.05 0 2.42229 -0.260232 1.0338 -5.0387 0.3586 0.06523 1.0338 -5.04273 0.3586 0.068573 1.02345 -8.08233 -0.024 -0.012 1.02325 -8.08809 -0.024 -0.0124 1.01637 -9.72937 -0.076 -0.016 1.01613 -9.73654 -0.076 -0.0165 1.0058 -13.69708 -0.6964 0.10883 1.0058 -13.70427 -0.6964 0.110366 1.01055 -11.35395 0 0 1.01026 -11.35703 0 07 0.99437 -13.96765 -0.628 -0.109 0.99419 -13.97288 -0.628 -0.1098 1.023 -12.18127 -0.45 0.47454 1.023 -12.18265 -0.45 0.483939 1.01796 -20.00269 -1 -0.16277 1.01711 -20.13217 -1.01653 -0.1683210 1.00713 -19.52942 -0.058 -0.02 1.00633 -19.6175 -0.058 -0.0211 1.0913 -18.07884 0.1793 0.38778 1.0913 -18.20671 0.1793 0.3922412 1.04132 -16.16702 -0.112 -0.075 1.04109 -16.21744 -0.112 -0.07513 1.0883 -14.97002 0.1691 0.36696 1.0883 -15.02017 0.1691 0.3687914 1.02408 -17.34016 -0.062 -0.016 1.02379 -17.39439 -0.062 -0.01615 1.01562 -17.64022 -0.082 -0.025 1.01524 -17.69657 -0.082 -0.02516 1.01841 -17.83191 -0.035 -0.018 1.01792 -17.89732 -0.035 -0.01817 1.00471 -19.23196 -0.09 -0.058 1.00399 -19.31323 -0.09 -0.05818 0.99979 -19.02914 -0.032 -0.009 0.99925 -19.09708 -0.032 -0.00919 0.9938 -19.66267 -0.095 -0.034 0.99317 -19.73758 -0.095 -0.03420 0.99623 -19.69028 -0.022 -0.007 0.99555 -19.76851 -0.022 -0.00721 0.99497 -19.72658 -0.175 -0.112 0.99419 -19.81012 -0.175 -0.11222 0.99578 -19.62149 0 0 0.99501 -19.7033 0 023 1.00064 -18.30125 -0.032 -0.016 1.00016 -18.3595 -0.032 -0.01624 0.98942 -18.8339 -0.087 -0.067 0.98881 -18.89458 -0.087 -0.06725 0.98869 -16.70952 0 0 0.98823 -16.73336 0 026 0.97048 -17.15432 -0.035 -0.023 0.97001 -17.17859 -0.035 -0.02327 0.99711 -15.12179 0 0 0.99676 -15.12284 0 028 1.00723 -9.39718 0.98449 -0.35025 1.00683 -9.35718 1 -0.3572329 0.97669 -16.41847 -0.024 -0.009 0.97633 -16.42047 -0.024 -0.00930 0.96489 -17.35084 -0.106 -0.019 0.96452 -17.35353 -0.106 -0.019


DC system solutions DC system solutionsVdr = 253.938 kV; idr = 393.8 Amp.; Vdi = 245.93 kV; idr = 406.613 Amp.;ar = 0.7088; Qdr = 16.276 MVAR; ar = 0.6988; Pdr = 101.65 MW;

ai = 0.7389; Pdi = 98.45 MW; ai = 0.7275; Qdr = 16.831 MVAR;Qdi = 35.024 MVAR; Qdi = 35.723 MVAR;

We are now at the end of our theoretical study of various load flow techniques. However, inproduction grade implementatin of these techniques, the sparsity of the linear equations (connectingthe mismatch and solution vectors) is exploited to reduce the computation time as well as mem-ory 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.With combination 3�V � ◊ Pinj Qinj

(p.u) (deg) (p.u) (p.u)1 1.05000 0.00000 2.42241 -0.252332 1.03380 -5.04581 0.35860 0.091753 1.02180 -8.07176 -0.02400 -0.012004 1.01437 -9.71785 -0.07600 -0.016005 1.00580 -13.71987 -0.69640 0.121916 1.00804 -11.32909 0.00000 0.000007 0.99286 -13.96633 -0.62800 -0.109008 1.02300 -12.20102 -0.45000 0.555549 1.00248 -20.07763 -1.00000 -0.3380810 0.99624 -19.56171 -0.05800 -0.0200011 1.09130 -18.12405 0.17930 0.4690612 1.03787 -16.26936 -0.11200 -0.0750013 1.08830 -15.06838 0.16910 0.3937714 1.01972 -17.44774 -0.06200 -0.0160015 1.01042 -17.72802 -0.08200 -0.0250016 1.01181 -17.89806 -0.03500 -0.0180017 0.99507 -19.27534 -0.09000 -0.0580018 0.99250 -19.10546 -0.03200 -0.0090019 0.98530 -19.73134 -0.09500 -0.0340020 0.98713 -19.74984 -0.02200 -0.0070021 0.98445 -19.76760 -0.17500 -0.1120022 0.98541 -19.66188 0.00000 -0.0000023 0.99404 -18.37832 -0.03200 -0.0160024 0.98099 -18.89414 -0.08700 -0.0670025 0.98192 -16.75594 0.00000 0.0000026 0.96358 -17.20702 -0.03500 -0.0230027 0.99144 -15.15766 -0.00000 -0.0000028 1.00345 -9.34449 0.98449 -0.3836929 0.97089 -16.46954 -0.02400 -0.0090030 0.95901 -17.41321 -0.10600 -0.01900


DC system solutionsVdr = 253.938 kV; idr = 393.8 Amp.;

– = 15.21o; Qdr = 33.808 MVAR;“ = 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 YBUS matrix. As we have already seen, the o�-diagonal elements ofYBUS matrix signifies the connectivity between the nodes. To be more precise, the element (i,j) ofYBUS matrix is non-zero if there is a direct connection between node ‘i’ and node ‘j’, while it is zero ifthere is no direct connectivity between these two nodes. Now, in many power systems, generally anybus is connected to mostly 3-4 buses directly. Therefore, in a 100 buses system (say), there wouldbe at best 4-5 non-zero terms (including the diagonal) in any row of the YBUS matrix, rest of theelements being zero. Therefore, out of (100 × 100) = 10, 000 elements, only about 500 terms wouldbe non-zero and the other terms (elements) would be zero. Thus, in this case, the YBUS matrix isalmost 95 percent sparse. For any larger system, the percentage of sparsity of the associated YBUS

matrix would be even more.Because of the sparsity of the YBUS 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 seenthat all the elements of the Jacobian matrix depend on the element Yij. Therefore, if this elementYij is zero, the corresponding elements of the Jacobian matrix would also be zero. As most of theelements (Yij) of the YBUS matrix are zero, it immediately follows that most of the elements of theJacobian 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), thecorrection vector (�X) is computed by inverting the Jacobian matrix and thereafter multiplyingthe 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, computationof the direct inverse of the sparse matrix involves a lot of computational burden. Therefore, it wouldbe much less intensive if equation (2.45) can be solved exploiting the sparse nature of the Jacobianmatrix. Apart from this, storing all the elements of a highly sparse matrix also consumes the memoryunnecessarily. Therefore, if only the non-zero elements are stored in appropriate fashion, a lot ofmemory 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 expectedto handle any large size power system, enhancement in the complexity of programming is often asmall 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 schemesfor storing a sparse matrix. We will start with the Gaussian Elimination method for solving a set oflinear 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-e�cient matrix. The most obviousmethod for solving equation (3.1) is to invert matrix A, that is x = A−1b. However, equation(3.1) can also be solved indirectly by converting the matrix A into an upper triangular form withappropriate changes reflected in the vector b and then by back substitution. To illustrate the basicprocedure, 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 a

11

.

x1

+ a12

a11

x2

+ a13

a11

x3

+ a14

a11

x4

= b1

a11

(3.6)

b) Multiply equation (3.6) by a21

, a31

, a41

(one by one) and subtract the resulting expressionfrom equations (3.3), (3.4) and (3.5) respectively to yield:

�a22

− a12

a21

a11

�x2

+ �a23

− a13

a21

a11

�x3

+ �a24

− a14

a21

a11

�x4

= b2

− b1

a21

a11

(3.7)

�a32

− a12

a31

a11

�x2

+ �a33

− a13

a31

a11

�x3

+ �a34

− a14

a31

a11

�x4

= b3

− b1

a31

a11

(3.8)

85

�a42

− a12

a41

a11

�x2

+ �a43

− a13

a41

a11

�x3

+ �a44

− a14

a41

a11

�x4

= b4

− b1

a41

a11

(3.9)

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

x1

+ a12

a11

x2

+ a13

a11

x3

+ a14

a11

x4

= b1

a11

(3.10)

a(1)22

x2

+ a(1)23

x3

+ a(1)24

x4

= b(1)2

(3.11)

a(1)32

x2

+ a(1)33

x3

+ a(1)34

x4

= b(1)3

(3.12)

a(1)42

x2

+ a(1)43

x3

+ a(1)44

x4

= b(1)4

(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).a) Equation (3.11) is divided throughout by a(1)

22

.

x2

+ a(1)23

a(1)22

x3

+ a(1)24

a(1)22

x4

= b(1)2

a(1)22

(3.15)

b) Multiplying equation (3.15) by a(1)32

and a(1)42

(one by one) and subtracting the resultingexpressions from equations (3.12) and (3.13) respectively one can obtain;

�a(1)33

− a(1)23

a(1)32

a(1)22

�x3

+ �a(1)34

− a(1)24

a(1)32

a(1)22

�x4

= �b(1)3

− b(1)2

a(1)22

a(1)32

� (3.16)

�a(1)43

− a(1)23

a(1)42

a(1)22

�x3

+ �a(1)44

− a(1)24

a(1)42

a(1)22

�x4

= �b(1)4

− b(1)2

a(1)22

a(1)42

� (3.17)

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

x2

+ a(1)23

a(1)22

x3

+ a(1)24

a(1)22

x4

= b(1)2

a(1)22

(3.18)

a(2)33

x3

+ a(2)34

x4

= b(2)3

(3.19)

a(2)43

x3

+ a(2)44

x4

= b(2)4

(3.20)

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

a(2)jk = a(1)jk − a(1)j2

a(1)2k

a(1)22

for j, k = 3, 4 (3.21)

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

86

a) Equation (3.19) is divided throughout by a(2)33

.

x3

+ a(2)34

a(2)33

x4

= b(2)3

a(2)33

(3.22)

b) Multiplying equation (3.22) by a(2)43

and subtracting it from equation (3.20) one can obtain,

�a(2)44

− a(2)34

a(2)43

a(2)33

�x4

= �b(2)4

− b(2)3

a(2)33

a(2)43

� (3.23)

Equation (3.23) contains only one unknown, x4

. Therefore, the value of x4

can be calculatedfrom this equation. With the value of x

4

thus calculated, x3

can be calculated from equation (3.22).Going back in this manner, x

2

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 valuesof x

2

, 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 a

11

≠ 0.

��

a11

a12

a13

a14

a21

a22

a23

a24

a31

a32

a33

a34

a41

a42

a43

a44

��

��

x1

x2

x3

x4

��=��

b1

b2

b3

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-e�cient matrixof equation (3.24)) is carried out to obtain equation (3.6) and the resulting matrix equation is shownin equation (3.25). ��

1 a12

�a11

a13

�a11

a14

�a11

a21

a22

a23

a24

a31

a32

a33

a34

a41

a42

a43

a44

��

��

x1

x2

x3

x4

��=��

b1

�a11

b2

b3

b4

��(3.25)

Step M2

On equation (3.25), the operations (R2 −R1 ∗ a21

), (R3 −R1 ∗ a31

) and (R4 −R1 ∗ a41

) arecarried out (where ‘Ri’ denotes the ith(i = 1, 2, 3, 4) row of the co-e�cient 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

this equation, it is assumed that a(1)22

≠ 0.

��

1 a12

�a11

a13

�a11

a14

�a11

0 a(1)22

a(1)23

a(1)24

0 a(1)32

a(1)33

a(1)34

0 a(1)42

a(1)43

a(1)44

��

��

x1

x2

x3

x4

��=��

b1

�a11

b(1)2

b(1)3

b(1)4

��(3.26)

Step M3

On equation (3.26), the operation ‘R2�a(1)22

’ is carried out (corresponding to equation (3.15)) toobtain the resulting matrix equation shown in equation (3.27).

��

1 a12

�a11

a13

�a11

a14

�a11

0 1 a(1)23

�a(1)22

a(1)24

�a(1)22

0 a(1)32

a(1)33

a(1)34

0 a(1)42

a(1)43

a(1)44

��

��

x1

x2

x3

x4

��=��

b1

�a11

b(1)2

�a(1)22

b(1)3

b(1)4

��(3.27)

Step M4

On equation (3.27), the operations �R3 −R2 ∗ a(1)32

� and �R4 −R2 ∗ a(1)42

� are carried out cor-responding to the equations (3.18)-(3.21) and the resulting matrix equation is shown in equation(3.28). In this equation, it is assumed that a(2)

33

≠ 0.

��

1 a12

�a11

a13

�a11

a14

�a11

0 1 a(1)23

�a(1)22

a(1)24

�a(1)22

0 0 a(2)33

a(2)34

0 0 a(2)43

a(2)44

��

��

x1

x2

x3

x4

��=��

b1

�a11

b(1)2

�a(1)22

b(2)3

b(2)4

��(3.28)

Step M5

On equation (3.28), the operation ‘R3�a(2)33

’ is carried out to obtain the matrix equation shownin equation (3.29). ��

1 a12

�a11

a13

�a11

a14

�a11

0 1 a(1)23

�a(1)22

a(1)24

�a(1)22

0 0 1 a(2)34

�a(2)33

0 0 a(2)43

a(2)44

��

��

x1

x2

x3

x4

��=��

b1

�a11

b(1)2

�a(1)22

b(2)3

�a(2)33

b(2)4

��(3.29)

Step M6

Lastly, on equation (3.29), the operation �R4 −R3 ∗ a(2)43

� is carried out to obtain the matrix

88

equation shown in equation (3.30).

��

1 a12

�a11

a13

�a11

a14

�a11

0 1 a(1)23

�a(1)22

a(1)24

�a(1)22

0 0 1 a(2)34

�a(2)33

0 0 0 a(3)44

��

��

x1

x2

x3

x4

��=��

b1

�a11

b(1)2

�a(1)22

b(2)3

�a(2)33

b(3)4

��(3.30)

In equation (3.30), a(3)44

= a(2)44

− a(2)34

a(2)43

a(2)33

and b(3)4

= b(2)4

− b(2)3

a(2)33

a(2)43

. From this equation, the

unknowns can be easily solved by back-substitution starting from the last row of the final co-e�cientmatrix in equation (3.30). Thus, Gaussian elimination enables us to solve the unknown quantitiesin a systematic manner without inverting the co-e�cient matrix. Therefore, by adopting the sameprocedure, the correction vector (�M) can be computed from equation (2.48) without having toinvert the Jacobian matrix. When a large power system in analyzed, adopting Gaussian eliminationreduces computational burden to a large extent (as compared to inversion of the Jacobian matrix).

In the above procedure, the variables a11

, a(1)22

and a(2)33

have been assumed to be non-zero. Thesevariables, by which the rows of the co-e�cient matrix are divided, are called the ‘pivot variables’.However, during the elimination process, it is not necessary that the ‘pivot variables’ would be alwaysnon-zero. If any pivot variable turns out to be zero at any intermediate step, then the correspondingrow is interchanged with the next row so that the new pivot variable is non-zero and the eliminationprocess 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 17 18 1623 27 25 2822 32 34 3612 15 41 36

��

��

x1

x2

x3

x4

��=��

10203040

��(3.31)

Towards this goal, we proceed with the di�erent steps as follows:Step M1On equation (3.31), the operation R1�A(1, 1) (where, A(1, 1) = 11) is performed to yield,

��

1 1.5455 1.6364 1.454523 27 25 2822 32 34 3612 15 41 36

��

��

x1

x2

x3

x4

��=��

0.9091203040

��(3.32)

Step M2On 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 equa-tion is given by; ��

1 1.5455 1.6364 1.45450 −8.5455 −12.6364 −5.45450 −2.0 −2.0 4.00 −3.5455 21.3636 18.5455

��

��

x1

x2

x3

x4

��=��

0.9091−0.9091

1020.0909

��(3.33)

Step M3On 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 1.45450 1 1.4787 0.63830 −2.0 −2.0 4.00 −3.5455 21.3636 18.5455

��

��

x1

x2

x3

x4

��=��

0.90910.1064

1020.0909

��(3.34)

Step M4On 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 1.45450 1 1.4787 0.63830 0 0.9574 5.27660 0 26.6065 20.8085

��

��

x1

x2

x3

x4

��=��

0.90910.106410.212829.4681

��(3.35)

90

Step M5On 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 1.45450 1 1.4787 0.63830 0 1 5.51140 0 26.6065 20.8085

��

��

x1

x2

x3

x4

��=��

0.90910.106410.667229.4681

��(3.36)

Step M6Lastly, 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 1.45450 1 1.4787 0.63830 0 1 5.51140 0 0 −125.83

��

��

x1

x2

x3

x4

��=��

0.90910.106410.6672−254.3484

��(3.37)

In equation (3.37), the co-e�cient matrix has been converted to an upper-triangular matrix.From the last row of this equation, x

4

can be calculated as x4

= 254.3484�125.83 = 2.0214. Backsubstituting this value of x

4

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 x

3

and x4

in the second rowof equation (3.37) yields the of x

2

as x2

= 0.1064 + 1.4787 × 0.4735 − 0.6383 × 2.0214 = −0.4837.Lastly, substituion of x

2

, 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 e�ective for solving a large set of sparelinear equations without having to invert the co-e�cient matrix. Moreover, at every stage, if thecalculations pertaining to Gaussian elimination is carried out only using the non-zero terms, greatsaving in the computational burden can be achieved. However, if the elimination process is carriedout in the normal sequence, at any stage of elimination, the original zero-elements may the concertedinto a non-zero element. This is normally termed as ‘fill-in’ phenomenon. On the other hand, insteadof 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 belowillustrates this point.

In equation (3.38), an initial co-e�cient matrix is shown at the left hand side (part ‘a’) and thestructure of the co-e�cient matrix after step-1 is shown at the right hand side (part ‘b’). It is to benoted 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 zeroelements 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 × × o o3 × o × o4 × o o ×

��

��

1 2 3 4

1 1 × × ×2 o × ⊗ ⊗3 o ⊗ × ⊗4 o ⊗ ⊗ ×

��(3.38)

a) Initial ‘A’ matrix b) ‘A’ matrix after step 1

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

��

4 3 2 1

4 × o o ×2 o o × ×3 o × o ×1 × × × ×

��

��

4 3 2 1

4 1 o o ×2 o o × ×3 o × o ×1 o × × ×

��(3.39)

a) Rearranged ‘A’ matrix 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 di�cult todevelop and perhaps is impossible. As an alternative, various ‘near optimal ordering’ schemes havebeen developed. Some of them are discussed below:

Scheme 1

In this scheme, before elimination, number the rows of the co-e�cient matrix ‘A’ according tothe number of non-zero, o�-diagonal terms. Thus, the rows with only one o�-diagonal, non-zeroterm are numbered first, those with two non-zero, o�- diagonal terms are numbered second andso on. However, this scheme does not take into account the changes occurring in the co-e�cientmatrix during the elimination process. Therefore, this scheme in quite easy and straight forward toimplement.

Scheme 2

In this scheme the rows of the co-e�cient matrix ‘A’ are numbered such that at each step of theelimination procedure, the row with the fewest number of non-zero o�-diagonal terms would beoperated 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 occurringin the co-e�cient matrix in advance. Thus, this method takes longer time as compared to scheme 1to 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 non-zero o�-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 tostudy its e�ects on the co-e�cient matrix in advance. Hence, this method also takes longer timethan scheme 1.

Let us now look at another technique for solving a set of linear equations without the need ofinverting the co-e�cient 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 productof two triangular matrices as A = LU, where L is a lower triangular matrix and U is an uppertriangular matrix. As an example, let the matrix A be a 4 × 4 (N = 4) matrix. Upon triangularfactorization (or ‘LU’ decomposition), the matrix A is represented as A = LU. Or,

��

a11

a12

a13

a14

a21

a22

a23

a24

a31

a32

a33

a34

a41

a42

a43

a44

��=��

–11

0 0 0–

21

–22

0 0–

31

–32

–33

0–

41

–42

–43

–44

��×��

—11

—12

—13

—14

0 —22

—23

—24

0 0 —33

—34

0 0 0 —34

��(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)

Or,Ly = b (3.42)

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

��

–11

0 0 0–

21

–22

0 0–

31

–32

–33

0–

41

–42

–43

–44

��

��

y1

y2

y3

y4

��=��

b1

b2

b3

b4

��(3.43)

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

y1

= b1

–11

yi = 1–ii

�bi − i−1�j=1

–ijyj� ; i = 2, 3, � N (3.44)

93

Again, expanding the expression y =Ux we get,

��

—11

—12

—13

—14

0 —22

—23

—24

0 0 —33

—34

0 0 0 —34

��

��

x1

x2

x3

x4

��=��

y1

y2

y3

y4

��(3.45)

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

xN = yn

—NN

xi = 1—ii

�yi − N�j=1+1

—ijxj� ; 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, jth element of equation(3.40). In general, the element aij can be represented as (from equation (3.40)),

–i1—1j + –i2—2j +�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 smallernumber. We have in fact three distinct cases:

i < j; –i1—1j + –i2—2j +� + –ii—ij = aij (3.48)

i = j; –i1—1j + –i2—2j +� + –ii—jj = aij (3.49)

i > j; –i1—1j + –i2—2j +� + –ij—jj = aij (3.50)

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

–ii = 1; i = 1, 2, � N (3.51)

With the condition of equation (3.51), an elegant procedure called ‘Crout’s algorithm’ quiteeasily solves for the N 2 unknown ‘–′ and ‘—′ by just rearranging the equations in a certain order.The algorithm is as follows:

step 1: Set –ii = 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 —ij as

—ij = aij − i−1�k=1

–ik—kj (3.52)

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 –ij as;

–ij = 1—jj

�aij − j−1�k=1

–ik—kj� (3.53)

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

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3.4.2 Example of LU decomposition

Decompose the matrix A =��3 2 72 3 13 4 1

��in ‘LU’ form through ‘Crout’s algorithm’.

Solution: Let A =��3 2 72 3 13 4 1

��=��–

11

0 0–

21

–22

0–

31

–32

–33

��

��—

11

—12

—13

0 —22

—23

0 0 —33

��Applying ‘Crout’s algorithm; –

11

= 1; –22

= 1; –33

= 1;For j = 1;

i = 1 → —11

= a11

= 3;

i = 2 → –21

= 1—

11

(a21

− 0) = 23;

i = 3 → –31

= 1—

11

(a31

− 0) = 33 = 1;

For j = 2;

i = 1 → —12

= a12

= 2;

i = 2 → —22

= a22

− –21

—12

= 53;

i = 3 → –32

= 1—

22

(a32

− –31

—12

) = 65;

For j = 3;

i = 1 → —13

= a13

= 7;

i = 2 → —23

= a23

− –21

—13

= −1113;

i = 3 → —33

= a33

− –31

—13

− –32

—23

= −85;

Hence,

��3 2 72 3 13 4 1

��=��

1 0 02�3 1 01 6�5 1

��

��3 2 70 5�3 −11�30 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 theseschemes are described here.

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Random packing:

In this method, every non-zero element of the matrix is stored in a primary array while its row andcolumn indices are stored in two secondary arrays. As each element in individually identified, theelements 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.

A =��

0 0 0 0 00 0 2.67 0 3.12−1.25 0.29 0 0 2.31

��(3.54)

Primary array (elements) X = {0.29 3.12 -1.25 2.67 2.31 0}Secondary array (row indices) i = {3 2 3 2 3 0}

Secondary array (column indices) j = {2 5 1 3 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, thenthere is no need to adopt both row and column indices for each element. Instead some alternative,more e�cient 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 3.12 -1.25 0.29 2.31}Integer array JA = {3 5 1 2 5}Integer array IST = {1 1 3 6 }

In this case, the array of row address IST has been constructed such that the number of non-zeroelements 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 ofthe 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 Ais IST(3) - IST(2) = 3 - 1 = 2 and from the first two elements of the array ‘JA’, these two elementsare located at columns 3 and 5 of the matrix whereas the elements themselves are given by the firsttwo elements of the real array (2.67 and 3.12) (which is again true from equation (3.54)). In a similarway, the elements and the locations of the elements in the third row of the matrix A can easily beidentified from the above three arrays. Moreover, please note that, as the number of rows in thematrix A is 3, the total number of entries in array ‘IST’ is 4.

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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 variablesand the dummy variables itself could specify the row number. Moreover, the non-zero elementswould indicate the column number of the elements. Also, as before, the zero at the end of the arrayindicates the termination of the list. Hence the matrix A of equation (3.54) in this scheme looks like

Real array (elements) = {2 2.67 3.12 3 -1.25 0.29 2.31 0}Integer array JA = {0 3 5 0 1 2 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 compoundidentifier could be (n× i+ j) where n is an integer higher than the number of columns in the matrixand (i, j) denote the position of the non-zero element in the matrix. For example, the matrix A inequation (3.54) could look like

Real array (elements) = {2.67 3.12 -1.25 0.29 2.31 0}Integer array JA = {2003 2005 3001 3002 3005 9999}

In the above, ‘n’ has been chosen to be equal to 1000. Also, in the real array, the entry zeroindicates the end of the array and the corresponding entry in the integer array is 9999 (to signifiesthe end of the integer array).

However, unless compound identifiers yield necessary or highly desirable storage saving, it shouldnot 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-zerorow 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 nextlecture, we will start the discussion of short circuit analysis methods.

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Load Flow Matrices IIT Roorkee Notes NPTEL

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