Chapter 10

EECE 471 Power Flow Analysis 1

POWER FLOW ANALYSIS 10.0 INTRODUCTION The power flow analysis problem is solved to obtain the steady-state voltages and powers in a transmission system. The calculation is analogous to solving for the steady-state voltages and currents in a circuit. Consider the one-line diagram shown in Fig. 10.1. The load powers (SDi) represent the power loads supplied to large scale industrial or residential consumers. The powers (SGi) are injected at the generator nodes to supply the loads and losses incurred in the transmission system. A figure similar to Figure 10.1 might represent a sub-transmission network supplying various distribution substations and fed from the transmission substations. The power system must deliver the load powers in a reliable and economic manner with the voltage and frequency being within acceptable limits. In this chapter we are concerned mainly with the voltages being kept within limits. Thus the magnitudes and phases of the nodal voltages Vi are the variables that will be calculated in the power flow analysis problem. Once the voltages are calculated, the power flows or currents in the lines may be calculated and compared to their thermal or stability limits. Modifications in the operation such as changing SGi or in the design such as adding new lines may then be proposed to rectify the detected anomalies in the system condition.

V1

G

V2 SG2

G

SD1

V4

SD4

V5

SD5

G

V3 SG3

SD3

SG1

Figure 10.1 System one-line diagram

10.1 POWER FLOW EQUATIONS The per unit complex power injected at node i is defined as:

DiGii SSS −=∆

Si can be calculated as the sum of complex powers sent to particular nodes in the network. For example bus 3 in Figure 10.1, S3 the net complex power sent to nodes 2 and 5 as visualized in Fig. 10.2.


G

V3 SG3

SD3 S35

S32

G V3 SG3

SD3 S35

S32

V3

S3

Figure 10.2 Significance of S3

In general, using conservation of complex power for the ith bus,

∑=

=n

kiki SS

1 i=1, 2 … n (10.1)

We also define the injected node or bus current at bus i in per unit as

∑=

=−=n

kikDiGii IIII

1 i=1, 2 … n

Using I= Ybus V we get for the ith component:

∑=

=n

kkiki VYI

1 i=1, 2 … n (10.2)

Next we calculate the ith bus power as: Si = ViIi*

∑==

n

kkikii VYVS

1

** (10.3)

Suppose we let

1ij V ji i i

ik i k

ik ik ik

V V e V e

Y G jB

θ

θ θ θ

∆∠

∆

∆

= =

= −

= +

So using a polar representation of the complex voltages and a Cartesian one for the admittances then (10.3) becomes as follows:

∑=

−=n

kikik

jkii jBGeVVS ik

1)(θ

∑=

−+=n

kikikikikki jBGjVV

1))(sin(cos θθ i=1, 2 … n (10.4)


Resolving (10.4) into real and imaginary parts:

∑=

+=n

kikikikikkii BGVVP

1)sincos( θθ

∑=

−=n

kikikikikkii BGVVQ

1)cossin( θθ (10.5)

10.2 THE POWER FLOW PROBLEM The power flow problem is based on mathematical as well as operational consideration. At a generator bus the active power PGi and the voltage magnitude |Vi| are specified. At all buses the complex power demands SDi are specified. Buses with generators are known as generator or PV buses and buses with no generators are load or PQ buses. For example in Figure 10.1 buses 1, 2, and 3 are generator buses and buses 4 and 5 are load buses. However, the power at all generator buses cannot be specified because the losses are not known prior to solving the power flow equations. Hence the power at one of the generator buses, selected arbitrarily, is left open to take the slack and balance the active power. This bus, known as the slack or swing bus, is selected as Bus 1 to simplify the presentation and, instead of specifying the active power, the phase θ1 at this bus is specified in addition to the voltage magnitude |V1|. In summary there are three types of sources at the different buses:

1. A voltage source, assumed at Bus 1. 2. P, |V| sources, at the other generator buses. 3. P, Q sources, at the load buses.

In developing the methodology for solving the power flow problem we present two cases. Case I is a simpler problem to solve; it has one slack bus and the remaining buses are PQ buses: Case I: Given V1, S2, S3, …, Sn find S1, V2, V3,…, Vn

Case II is more typical with PV and PQ buses. The PV buses are numbered from 2 to m, and the PQ buses are from m+1 to n. Case II: Given V1, (P2, |V2|), …, (Pm, |Vm|), Sm+1, ..., Sn

find S1, (Q2, θ2), …, (Qm, θm), Vm+1, ..., Vn

Before we discuss the mathematical formulation note the following points:


• Case I corresponds to the one-generator case. Case II is the more typical case.

• In both cases we assume that two out of four variables at each bus are given (complex Vi specifies |Vi| and θi, while complex Si specifies Pi and Qi) and are asked to find the remaining two variables.

• Once the stated problem has been solved, we know all the Vi and can then solve for power flows or currents on individual transmission lines.

• In Case II we may specify Qimin ≤ Qi ≤ Qi

max, i = 2, 3, … , m.

10.3 SOLUTION BY GAUSS ITERATION Consider Case I of the power flow problem. The equations to be solved are a form of (10.3) repeated for convenience as:

∑=

=n

kkkVYVS

1

**111 (10.3a)

∑=

=n

kkikii VYVS

1

** i= 2 … n (10.3b)

Note that when the voltages are known we can solve for S1 from (10.3a). The key issue now is to solve the n-1 equations given in (10.3b). Taking complex conjugates of (10.3b) and dividing by Vi* and separating the Yii term we obtain:

∑≠=

+=n

ikk

kikiiii

i VYVYVS

1*

* i= 2 … n (10.6)

By rearranging

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−= ∑

≠=

n

ikk

kiki

i

iii VY

VS

YV

1*

*1 i=2, 3, …, n (10.8)

Thus we get n – 1 implicit nonlinear algebraic equations of a general form:

),,,(~

),,,(~),,,(~

32

3233

3222

nnn

n

n

VVVhV

VVVhV

VVVhV

L

M

L

L

=

=

=

(10.9)

where the functions ih~ are given by (10.8). In vector notation (10.9) takes the form

x = h(x) (10.10)


Equation (10.10) is solved by iteration. In the simplest case we use the formula: xv+1 = h(x v) v=0, 1, … (10.11)

where the superscript v indicates the iteration number. Thus starting at x0 the iterative process generates a sequence x0, x1, x2, … If the series converges (i.e., xv → x*), then

x* = h(x*) In practice we stop the iterations when the changes in x v become very small. Thus defining vvv xxx −=∆ +1 , we stop when the norm of is smaller than arbitrarily small positive number ε:

ε≤∆ vx

The norm may be

• a sup norm: )max ii

v x∆=∆x

• or an Euclidian norm: 2/1

1

2)( ⎥⎦

⎤⎢⎣

⎡∆=∆ ∑

=

N

ii

v xx

Using a scalar version of (10.11) the Gaussian iteration with v= 0, 1, 2 … may be written as:

),,,(

),,,(

),,,(

211

2121

2

2111

1

vN

vvN

vN

vN

vvv

vN

vvv

xxxhx

xxxhx

xxxhx

L

M

L

L

=

=

=

+

+

+

(10.12)

An improved version is the Gauss-Seidel iteration given by:

),,,,(

),,,(

),,,(

),,,(

11

12

11

1

12

112

13

21

121

2

2111

1

vN

vN

vvN

vN

vN

vvv

vN

vvv

vN

vvv

xxxxhx

xxxhx

xxxhx

xxxhx

+−

+++

+++

++

+

=

=

=

=

L

M

L

L

L

(10.12)

Example 10.2 Use Gauss iteration to solve the system shown in Figure E10.2, i.e. find S1 and V2.


V1=1∠0° V2

SD2= 0.5+j1.0

G

SD1 S1

ZL=j0.5 SG1

S2

jQG2=j1.0

Figure E10.2 Solution We iterate on V2 using (10.8). With n= 2 we have only equation:

⎥⎦

⎤⎢⎣

⎡−=+

121*2

*2

22

12 )(

1 VYVS

YV v

v

We next calculate the matrix Ybus:

⎥⎦

⎤⎢⎣

⎡−

−=⎥

⎦

⎤⎢⎣

⎡=

2222

2221

1211

jjjj

YYYY

Ybus

Thus putting the values for S2, V1, Y11 and Y21 we get:

0.1)(

25.0*

2

12 +−=+

vv

VjV

Starting with a guess 0102 ∠=V we get convergence in about 6 steps. The results

are listed in Table E10.2.

Table E10.2

Iteration V2 0 1.0∠0 1 1.030776∠-14.036243 2 0.970143∠-14.036249 3 0.970261∠-14.931409 4 0.966235∠-14.931416 5 0.966236∠-14.995078 6 0.965948∠-14.995072


A simple modification of the procedures developed to handle Case I works for Case II. For buses i= 2 to m the reactive power v

iQ is not specified and thus to evaluate (10.8) the reactive power at node i, at iteration v is estimated by:

⎥⎦

⎤⎢⎣

⎡= ∑

=

n

k

vkik

vi

vi VYVQ

1

** )(Im~ i=2, 3, …, n (10.14)

Using this estimate the voltages at buses i= 2 to m are estimated by:

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−

−= ∑

≠=

+n

ikk

vkikv

i

vii

ii

vi VY

VQjP

YV

1*

1

)(

~1~ i=2, 3, …, n (10.15)

Since |Vi| is specified for the generator buses, we then replace 1~ +viV by speciV and

obtain Viv+1.

Example 10.3 Now consider the simple example shown in Figure E10.3 to illustrate Gaussian iteration for Case II.

V1=1∠0°

SD2= 1.0+j0.5

G

SD1

S1

ZL=j0.5

SG1

S2

G SG2=0.25+jQG2

V2=1∠θ2

Figure E10.3 Solution The iteration formula (10.15) with Yii= Y22= -j2, and Yik= Y21= j2 gives

*2

*2

12 )(2

~75.012)(

~75.02

1~v

vi

v

viv

VjQjj

VQj

jV +

+=⎥⎥⎦

⎤

⎢⎢⎣

⎡−

−−−=+

where viQ~ is the estimate of the reactive power at the bus at iteration v given by

(10.14) as

[ ]{ }

)Re1(2

22Im

)(Im~

2

222

*2

*22

*1

*2122

v

vv

vvv

V

VjVj

VYVYVQ

−=

⎟⎠⎞

⎜⎝⎛ +−=

+=

Starting with a guess 0102 ∠=V we obtain the iterations in Table E10.3.


Table E10.2

Iteration V2v vQ2

~ 12

~ +vV

0 1.0∠0 0 1.0680∠-20.5560°

1 1.0∠-20.5560° 0.1273 1.0003∠-21.9229°

2 1.0∠-21.9229° 0.1446 1.0000∠-22.0169°

3 1.0∠-22.0169° 0.1459 1.0000∠-22.0238°

4 1.0∠-22.0238° 0.1459

To complete the solution we can solve for S1 using (10.3a).

[ ]

1459.075.0)0238.221)(2(2

*2

*12

*1

*1111

jjj

VYVYVS

+=∠−=

+=o

10.4 NEWTON-RAPHSON ITERATION The Newton-Raphson iteration method to solve a problem of the f(x) = 0. It is based on the Taylor series expansions for f(x + ∆x) given by

f(x + ∆x) = f(x) + f’(x) ∆x + h.o.t. (10.21) where “h.o.t.” stands for “higher-order terms”. When n scalar equations are to be solved they are denoted by the following:

f(x) = 0 (10.22) In this case, the Taylor series of the n scalar equations are given by:

h.o.t.)(...)()()(

h.o.t.)(...)()()(

h.o.t.)(...)()()(

11

21

1

222

11

1

111

+∆∂

∂++∆

∂∂

+=∆+

+∆∂

∂++∆

∂∂

+=∆+

+∆∂∂

++∆∂

∂+=∆+

nn

nnnn

nn

nn

xx

fxx

fff

xx

fxx

fff

xx

fxx

fff

xxxxx

xxxxx

xxxxx

M

(10.23)

Using matrix notation we get f (x + ∆x) = f(x) + J(x) ∆x + h.o.t. (10.24)

where


⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

∂∂

∂∂

∂∂

∂∂

=∆

n

nn

n

xf

xf

xf

xf

)()(

)()(

)(

1

1

1

1

xx

xx

xJL

MOM

L

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

∆

∆∆

=∆∆

nx

xx

M2

1

x (10.25)

J(x) is called the Jacobian matrix of f evaluated at x. By neglecting the h.o.t. and if the inverse of J(x) exists we get the general N-R iteration formula:

[ ] )()(11 vvvv xfxJxx−+ −= (10.26)

In some places we will use the simpler notation Jv rather than J(xv). The iterative procedure defined by (10.26) will generate a series of values x0, x1 … xv that will make f(xv) progressively smaller as v increases. The iterative process would stop when the norm )(xf is smaller than some small number ε.

Note that in (10.26) we need to update matrix J at every iteration and invert it. In large systems calculating inverses is computationally expensive. Instead we define

vvv xxx −=∆ +∆

1 (10.27) And we write (10.26) as

)()( vvv xfxxJ −=∆ (10.28)

We then for ∆xv by putting J in its upper and lower factors and doing forward and backward substitution as discussed in Section 9.2.

Example 10.5 Given the direct-current (dc) system shown in figure E10.5, use the Newton-Raphson method to find the (dc) bus voltages V1 and V2 and find PG1.

Figure E 10.5

R = 0.01 V1 = 1.0 PD1 = 0.5 PD2 = 1.0 PD3 = 0.5


Solution Note that the considerations and techniques developed for ac systems hold as well for dc systems. For simplicity we are using the same notation, although all the variables are real quantities. Forming Ybus we have

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−−−−

=211121112

100busY

In what corresponds to (10.3), we write the power flow equations

P1 = 200V12 –100V1V2 – 100V1V3

P2 = -100V2V1 +200V22 – 100V2V3

P3 = -100V3V1 – 100V3V2 + 200V32

Bus 1 is the slack bus, with V1 known and P1 unknown. We (temporarily) strip away the first equation and solve the remaining two for the unknown V2 and V3. In using the Newton-Raphson the next-step is to put the equations in the form:

f(x) = 0. The simplest way is to subtract the left sides from the right sides, or vise-versa. Subtracting left from right and putting in the known values of P2 = -PD2 = -1.0, P3 = -PD3 = -0.5, V1 = 1.0, we get

f1(x) = 1.0 – 100V2 + 200V22 – 100V2V3 = 0

f2(x) = 0.5 – 100V3 – 100V2 V3 + 200V32 = 0

The first component of x is V2, the second is V3. Next we find the Jacobian matrix,

⎥⎦

⎤⎢⎣

⎡+−−−

−−+−=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

∂∂

∂∂

∂∂

∂∂

=∆

322

232

2

2

1

2

2

1

1

1

4141

100)(VVV

VVV

xf

xf

xf

xf

xJ

In the 2×2 case it is easy to find the inverse and we will solve using (10.26). In higher-dimensional cases we would almost certainly use (10.28). Starting with a “flat-profile” (i.e. V2 = 1, V3 = 1), we get

⎥⎦

⎤⎢⎣

⎡−

−=

2112

1000J

( ) ⎥⎦

⎤⎢⎣

⎡=

−

2112

300110J

⎥⎦

⎤⎢⎣

⎡=

5.00.1

)( 0xf

Then using (10.26) yields


⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡−⎥

⎦

⎤⎢⎣

⎡=

993333.0991667.0

5.00.1

2112

3001

111x

Continuing to the next iteration gives us

⎥⎦

⎤⎢⎣

⎡−

−=

981667.199333.0991667.0973333.1

1001J ( ) ⎥⎦

⎤⎢⎣

⎡=

−

973333.199333.0991667.0981667.1

54.292111J

⎥⎦

⎤⎢⎣

⎡=

00323.000843.0

)( 1xf

Note that x1 is really a very good estimate. We want f(x) = 0 and already f(x1) is almost zero! Note the great improvement from f(x0) to f(x1) in just one step. Continuing the iteration we have

⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡−⎥

⎦

⎤⎢⎣

⎡=

993283.0991599.0

00323.000843.0

973333.1993333.0991667.0981667.1

54.2921

993333.0991667.02x

Noting that the change between x1 and x2 is less than 0.0005, we stop the iteration. Alternatively, we can compute f(x2). We get

⎥⎦

⎤⎢⎣

⎡=

000040.0000053.0

)( 2xf

which is small compared to the power levels given for this problem. Finally we find P1 by using the first equation in (10.29). We find P1 = 1.511800. Note that since PD1 + PD2 = 1.5, the I2R loss in the transmission system is 0.011800.

10.5 APPLICATION TO POWER FLOW EQUATIONS The power flow equations that are solved in the Newton-Raphson (N-R) iterations are obtained from (10.5) and repeated here as

∑=

+=n

kikikikikkii BGVVP

1)sincos( θθ

∑=

−=n

kikikikikkii BGVVQ

1)cossin( θθ i=1, 2 … n (10.5)

The first equations corresponding to the slack bus are removed since at this bus P1 and Q1 are not specified. The left hand sides of (10.5) are specified values given by DiGii PPP −= and DiGii QQQ −= . The right hand sides represent the calculated power flowing to the network from the bus. They are functions of |Vi| and θi, i= 2, 3 … n. It is convenient to define the vectors θ, |V| and their composite vector x as follows:


⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

nθ

θM2

θ ⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

nV

VM2

V ⎥⎦

⎤⎢⎣

⎡=

Vθ

x (10.31)

At the solution point the calculated active and reactive powers are equal to the powers specified at the bus. We then represent the RHS of (10.5) with a simpler notation, remove the equations of the slack bus, and rearrange to get 0)( =− ii PP x

0)( =− ii QQ x i= 2, 3 … n (10.34)

Equations (10.34) have 2n-2 components and correspond to the vector function f(x) in (10.22). In expanded matrix notation (10.34) take the following form

0

x

x

x

x

xf =

⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−

−−−−−−−−−

−

−

=

nn

nn

QQ

QQ

PP

PP

)(

)(

)(

)(

)(

22

22

M

M

(10.35)

Next we need to identify the Jacobian matrix J(x) of f(x) given in a partitioned format as follows:

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

=⎥⎦

⎤⎢⎣

⎡=

||)(

||)()()(

||)(

||)()()(

||)(

||)()()(

||)(

||)()()(

)(

22

2

2

22

2

2

22

2

2

22

2

2

2221

1211

n

nn

n

nn

nn

n

nn

n

nn

nn

VQ

VQQQ

VQ

VQQQ

VP

VPPP

VP

VPPP

xxxx

xxxx

xxxx

xxxx

JJJJ

xJ

LL

MOMMOM

LL

LL

MOMMOM

LL

θθ

θθ

θθ

θθ

(10.36)

Each partition of matrix J(x) has dimensions (n-1)×(n-1). The terms of The Jacobian matrix J(x) can be evaluated explicitly using the RHS of (10.5). Finally to determine the form of N-R iterations we define the so called mismatch vectors:


⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−

−=∆

nn PP

PP

)(

)()(

22

x

xxP M

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−

−=∆

nn QQ

QQ

)(

)()(

22

x

xxQ M (10.37)

The f(x) can be expressed in terms of the mismatch vectors as follows:

⎥⎦

⎤⎢⎣

⎡∆∆

=)()(

)(xQxP

xf (10.38)

So finally by using (10.31), (10.36), and (10.38) in (10.28) we get:

⎥⎦

⎤⎢⎣

⎡

∆∆

−=⎥⎦

⎤⎢⎣

⎡

∆∆

⎥⎦

⎤⎢⎣

⎡

)()(

||2221

1211v

v

v

v

vv

vv

xQxP

Vθ

JJJJ (10.39)

Starting from an initial guess x0= [θ0; |V|0] at every step (v) of the N-R iteration we evaluate the mismatch vectors ∆P(xv) and ∆Q(xv) and the Jacobian matrix J(xv) and solve (10.39) to obtain ∆xv = [∆θv; ∆|V|v] and then use xv+1= xv + ∆ xv. As the iterations proceed the mismatch vectors are expected to go to zero. Next we calculate the elements of the Jacobian matrices as follows:

)sincos()(

)sincos()(

)cossin()(

)cossin()(

12

1

12

11

1

11

ikikikikik

iik

iii

n

kikikikikk

i

iii

ikikikikkik

iik

n

ikk

ikikikikkii

iii

BGVV

PJ

GVBGVV

PJ

BGVVPJ

BGVVPJ

θθ

θθ

θθθ

θθθ

+=∂∂

=

++=∂∂

=

−=∂∂

=

+−=∂

∂=

∑

∑

=

≠=

x

x

x

x

(10.40a)

)cossin()(

)cossin()(

)sincos()(

)sincos()(

22

1

22

21

1

21

ikikikikik

iik

iii

n

kikikikikk

i

iii

ikikikikkik

iik

n

ikk

ikikikikkii

iii

BGVV

QJ

BVBGVV

PJ

BGVVQJ

BGVVQJ

θθ

θθ

θθθ

θθθ

−=∂∂

=

−−=∂∂

=

+−=∂

∂=

+=∂

∂=

∑

∑

=

≠=

x

x

x

x

(10.40b)


From these equations we can deduce some properties of the Jacobian matrix J. The J matrix is real but has a dimension of (2n-2)×(2n-2). The off-diagonal elements in each submatrix involve only elements directly connecting nodes in the network. If there is no direct connection between two buses, the corresponding entry in the submatrix is zero. Thus each partition of J has a structure similar to that of Ybus and as such is called a sparse matrix. For large networks the number of zero elements could be higher than 90%! In a Case II problem, at the generator or PV buses we do not solve for |Vi| because these are specified and hence we have less variables to solve for in this case. Hence for every generator bus we remove the reactive power equation involving Qi since it is not specified and hence cannot be equated to the calculated power Qi (x). Given that n and m are the number of buses and the number of generators, the number rows and columns of the Jacobian matrix J is n+m-2. This is illustrated by the structure of J(x) matrix for the system shown in Figure 10.1 as follows:

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

=

5

5

4

5

5

5

4

5

3

5

2

5

5

4

4

4

5

4

4

4

3

4

2

4

5

5

4

5

5

5

4

5

3

5

2

5

5

4

4

4

5

4

4

4

3

4

2

4

5

3

4

3

5

3

4

3

3

3

2

3

5

2

4

2

5

2

4

2

3

2

2

2

)()()()()()(

)()()()()()(

)()()()()()(

)()()()()()(

)()()()()()(

)()()()()()(

)(

VQ

VQQQQQ

VQ

VQQQQQ

VP

VPPPPP

VP

VPPPPP

VP

VPPPPP

VP

VPPPPP

xxxxxx

xxxxxx

xxxxxx

xxxxxx

xxxxxx

xxxxxx

xJ

θθθθ

θθθθ

θθθθ

θθθθ

θθθθ

θθθθ

The vector of variables x is given by:

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

=

5

4

5

4

3

2

VVθθθθ

x

It could happen that at the end of the N-R power flow analysis one of the generators is providing reactive power outside its capability limits discussed in Chapter 6. In this case, the reactive power generated is set at the limit that was


violated and the bus is treated as a PQ bus. The voltage magnitude at the bus is reintroduced as a variable and the Jacobian matrix and the mismatch vector are augmented to reflect the change in the bus type, then the solution process is repeated.

Example 10.6

Use the Newton-Raphson method to find θ2, θ3, |V3|, then calculate SG1 and QG12 for the system shown in Figure E10.6. The shunt elements of transmission lines are equal to j0.02 and their series impedances are equal to j0.1.

V1=1∠0°

G

|V2|=1.05

PG2=0.6661 G

V3

SD3=2.8653+j1.2244

SG1

Figure E10.6 Solution The Ybus matrix of the system is given by:

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−

−=

98.1910101098.1910101098.19

bus

jjjjjjjjj

Y

Bus 1 is the slack bus, Bus 2 is a PV bus, and Bus 3 is a PQ bus. We have 3 variables to determine during the N-R iterations and the Jacobian is a 3×3 matrix. We then write the expression for P2(x), P3(x), and Q3(x) using the RHS of (10.5):

)sin(5.10sin5.10

)sin()sin()(

3232

3223321221212

θθθ

θθθθ

−+=

−+−=

V

BVVBVVP x (10.42a)

)sin(5.10sin10

)sin()sin()(

23323

2332231331133

θθθ

θθθθ

−+=

−+−=

VV

BVVBVVP x (10.42b)

]98.19)cos(5.10cos10[

])cos()cos([)(2

323333

332

32332231331133

VVV

BVBVVBVVQ

−−+−=

+−+−−=

θθθ

θθθθx

(10.42c)


The vector of unknown x and the Jacobian J(x) matrix are given by:

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

3

3

2

Vθθ

x

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

=

3

3

3

3

2

3

3

3

3

3

2

3

3

2

3

2

2

2

)(

VQQQVPPPVPPP

θθ

θθ

θθ

xJ (10.43)

We can now calculate the elements of the Jacobian matrix:

)cos(5.10cos5.10

)cos()cos(

3232

3223321221212

2

θθθ

θθθθθ

−+=

−+−=∂∂

V

BVVBVVP

)cos(5.10

)cos(

323

3223323

2

θθ

θθθ

−−=

−−=∂∂

V

BVVP

)sin(5.10

)sin(

32

322323

2

θθ

θθ

−=

−=∂∂ BVVP

)cos(5.10 2332

3 θθθ

−−=∂∂ VP

)cos(5.10cos10 233333

3 θθθθ

−+−=∂∂ VVP

)sin(5.10sin10 2333

3 θθθ −+=∂∂VP

)sin(5.10 2332

3 θθθ

−−=∂∂ VQ

)sin(5.10sin10 233333

3 θθθθ

−+=∂∂ VVQ

32333

3 96.39)cos(5.10cos10[ VVQ

−−+−=∂∂

θθθ


Let us note the scheduled powers at the buses: P2= PG2= 0.6661 P3= -PD3= -2.8653 Q3= -QD3= -1.2244 We start the N-R iterations by making an initial guess for the unknowns:

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

1000

3

3

20

Vθθ

x

We then calculate the mismatch vector:

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

+−+−

=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−

=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

∆∆∆

7044.08653.26661.0

2244.152.08653.206661.00

)()()(

30

3

30

3

20

20

3

3

2

QQPPPP

QPP

xxx

As expected, at the initial guess the mismatch is large. Iteration 1: We start the iteration by calculating the Jacobian matrix J0 using (10.43):

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−

−=

46.190005.205.1005.1021

)( 0xJ

The inverse of the Jacobian is given by:

[ ]⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=⎥

⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡= −

−−−

0514.00000656.00328.000328.00640.0

)( 122

111

1

22

1110

J00J

J00J

xJ

The correction on the initial guess is obtained:

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−

=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−=∆

0362.001660.00513.0

7044.08653.26661.0

0514.00000656.00328.000328.00640.0

0x

Next we update the values of the variables:

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−−

=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−

+⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=∆+=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

=9638.001660.00513.0

0362.001660.00513.0

100

00

13

13

12

1 xxxVθθ


We calculate the mismatch vector using the new values of variables:

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−

=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

∆∆∆

2251.01145.00463.0

)()()(

31

3

31

3

21

21

3

3

2

QQPPPP

QPP

xxx

The elements of the mismatch vector has have been reduced by a factor of 10. But we judge that they are not small enough and we continue iterating. Iteration 2: We then calculate the Jacobian matrix J1 using (10.43):

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−

−=

2199.187508.21582.18541.25589.190534.10

2017.10534.105396.20)( 1xJ

The inverse of the Jacobian is given by:

[ ]⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

−

0561.00084.00009.00087.00696.00336.00010.00336.00651.0

)(11xJ

The new values of the variables are given by:

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−−

=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−−

=9502.01744.00524.0

2251.01145.00463.0

0561.00084.00009.00087.00696.00336.00010.00336.00651.0

9638.01660.00513.0

2x

We then calculate the mismatch vector:

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

∆∆∆

0031.00023.00019.02

3

3

2

QPP

The mismatch is judged small enough and on this basis we may stop iterating. We finally calculate SG1 and QG12 using (10.5):

1987.2

)1744.0sin(502.9)0524.0sin(5.10)sin()sin( 31133121122111

=+=

−+−== θθθθ BVVBVVPPG


1365.0

]98.19)1744.0cos(502.9)0524.0cos(5.10[])cos()cos([ 11

2131133121122111

=−+−=

+−+−−== BVBVVBVVQQG θθθθ

6395.1

]028.22)122.0cos(977.9)0524.0cos(5.10[])cos()cos([ 22

2232233212211222

−=−+−−=

+−+−−== BVBVVBVVQQG θθθθ

10.6 DECOUPLED POWER FLOW In Example 10.6 we notice that the elements of the off-diagonal submatrices J12 and J21 are much smaller than elements of matrices J11 and J22. The reason may seen by looking at some terms. For example consider a typical term in J12:

)sincos()(12ikikikiki

k

iik BGV

VP

J θθ +=∂∂

=x

Transmission lines are mostly reactive and Gik are small, and under normal operating conditions the angle θik= θi - θk is relatively small making the term involving Gik also quite small. The same reasoning applies to elements of J21:

)sincos()(21ikikikikki

k

iik BGVV

QJ θθ

θ+−=

∂∂

=x

However typical terms in the diagonal submatrices J11 and J22 are not small as can concluded by examining typical terms of these matrices in (10.40). Note that in these terms the dominant Bik is being multiplied by cos θik. The implication is that changes in active power ∆Pi depends mostly on θi (and not so much on |Vi|), and changes in reactive power depends mostly on |Vi| (and not so much on θi). Based on this discussion the Jacobian matrix can be approximated as follows:

⎥⎦

⎤⎢⎣

⎡≅ v

vv

22

11

J00JJ

The N-R iterative equations (10.39) take the following form:

)(11vvv xPθJ ∆−=∆ (10.51a)

)(21vvv xQVJ ∆−=∆ (10.51b)

Again, another observation in Example 10.6 is that matrices J0 and J1 did not significantly change and as such further simplification can be made to equations 10.51. This is due to the fact that in transmission systems under normal operating conditions the phase angle differences across lines are small (i.e. θik ≈0) and the


voltage magnitudes are around 1 per unit (i.e. |Vi|≈|Vk|≈1). Under these conditions matrices J11 and J22 are approximated by matrices H and L, respectively. The elements of these matrices are given by:

∑≠

=ik

ikii BH and ikik BH −= (10.52a)

iiik

ikii BBL 2−−= ∑≠

and ikik BL −= (10.52b)

Therefore equations (10.51a) and (10.51b) are replaced by the following:

)( vv xPθH ∆−=∆ (10.53a)

)( vv xQVL ∆−=∆ (10.53b)

It is important to observe that matrices H and L are constant matrices that are independent of the iteration count, which may be approximated by the negative of the B matrix, the imaginary part of Ybus. Thus matrices H and L are formed and inverted (or factorized) only once at the onset of the iterative process. These equations are known as the fast-decoupled power flow equations. In programming the fast-decoupled power flow usually one full iteration of the active power equation (10.53a) is carried out followed by one full iteration of the reactive power equation (10.53b) using the updated values of phase angle as follows:

vvv

vvv

θθθ

VθPHθ

∆+=

∆−=∆+

−

1

1 ),( (10.54a)

vvv

vvv

VVV

VθQLV

∆+=

∆−=∆+

+−

1

11 ),( (10.54b)

The above equations (10.54a and 10.54b) imply a kind of block successive displacement which allows for a faster convergence and is illustrated in the following example.

Example 10.7

Use the fast-decoupled Newton-Raphson method to find θ2, θ3, |V3|, of the system of Example 10.6 then calculate SG1 and QG12. The system diagram and data are repeated in Figure 10.7 for convenience. The shunt elements of transmission lines are equal to j0.02 and their series impedances are equal to j0.1.


V1=1∠0°

G

|V2|=1.05

PG2=0.6661 G

V3

SD3=2.8653+j1.2244

SG1

Figure E10.7 Solution The Ybus matrix of the system was found to be:

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−

−=

98.1910101098.1910101098.19

bus

jjjjjjjjj

Y

Thus (10.54a) for this problem is:

⎥⎥⎦

⎤

⎢⎢⎣

⎡

∆

∆−=⎥

⎦

⎤⎢⎣

⎡∆∆

⎥⎦

⎤⎢⎣

⎡−

−

),(),(

98.19101098.19

3

2

3

2vv

vvv

PP

VθVθ

θθ

In this simple example we will use matrix inversion to determine the variable updates as follows:

⎥⎥⎦

⎤

⎢⎢⎣

⎡

∆

∆⎥⎦

⎤⎢⎣

⎡−=⎥

⎦

⎤⎢⎣

⎡∆∆

),(),(

0668.00334.00334.00668.0

3

2

3

2vv

vvv

PP

VθVθ

θθ

The updated phase angles are then updated as:

vvv

v⎥⎦

⎤⎢⎣

⎡∆∆

+⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡=

++

3

2

3

21

3

21

θθ

θθ

θθ

θ

And (10.54b) for this problem is:

),(98.19 133

vvv QV Vθ +∆−=∆

)(0501.0 33vv QV x∆−=∆

Note that the calculation of ∆P2(θv, |V|v), ∆P3(θv, |V|v), and ∆Q3(θv, |V|v) are based on equations (10.42a), (10.42b), and (10.42c), respectively. The results of the iterations are given in Table E10.7.


Table E10.7

Iteration v

θ2 θ3 |V3| ∆P2 ∆P3 ∆Q3

0 0 0 1 -0.6661 2.8653 0.9198 1 -0.0513 -0.1691 0.9540 -0.0272 0.0829 0.0683 2 -0.0522 -0.1737 0.9505 -0.0050 0.0131 0.0094 3 -0.0523 -0.1744 0.9501 -0.0007 0.0019 0.0014

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