CHAPTER

PoWER FLOW ANALYSIS

5.1. INTRODUCTION

Power flo analy sis is one of the basic tools used in power system studies. It is concemed
with the steady state analysis of the system when it is working under a normal balanced
operating condition. Load flow or power flow analysis is determination of the voltage,
current. real power and reactive power at various points in electrical network.
Significance in Load Fiow or Power Flow Analysis
The infomation of load flow is essential for analyzing the effective alternative plans for
system expansion to meet increase load demand.
The load flow studies are very important for planning. economic scheduling, control and
operations of existing systems as well as planning its future expansion depends upon
know ing the effect of interconnections, new loads, new generating stations, or new
transmission lines, etc., before they are installed.
With the help of load flow studies, we can determine the best size as well as the most
favourable locations for the power system capacitors both for the improvement of the power
factor and raising the bus voltages of the electrical network. It helps us to determine the
capacity of the proposed generating stations, substations, or new lines. Problems such as
optimization of system losses, stability check and general security of the system can be
studied with the heip of load flow studies.
The information obtained from the loadflow studies are the magnitude and phase angle of
voltages at each bus, active and reactive power flow in each line, and power loss in the line. It
also gives the initial conditions of the systerm when the transient behaviour of the system is to
be studied.
The mathematical formulation of the load flow problem results in a system of non-linear
équations. These equations can be written in terms of either the bus admittance matrix or bus
impedance matrix. Using bus admittance matrix, is amenable to digital computer analysis.
because it could be formed and modified for network changes in subsequent cases, and
requires less computation time and memory.
The load flow analysis can be carried out for small and medium size power systems. It
Suits for radial distribution systems with high R/X ratio. The load flow analysis helps to
identify the overloaded'underloaded lines and transformers as well as overvoltage
undervoltage buses in the system.
| 5.2 Power System Analys
It is used to study the optimum location of capacity is and their size to
unacceptable voltage profile. nprove
5.2. NEED FOR LOAD FLOW ANALYSIS (OR) IMPORTANCE OF POWER FLOW
ANALYSIS IN PLANNING AND OPERATION OF POWER SYSTEMS
Load flow analysis is performed on a symmetrical steady-state operating condition of a
power system under normal mode of operation. The solution of load flow gives bus vollages
and line/transformer power flows for a given loadcondition. This intormation is essential for
long term planning and operational planning.
Long Term Planning
Load flow analysis helps in investigating the effectiveness of alternative plans and
choosing the best plan for system expansion to meet the projected operating state.
Operational Planning
It helps in choosing the best unit commitment plan and generation schedules to run the
system efficiently for the next day's load condition without violating the bus voltages and
line flow operating limits.
Steps for Load Flow Study
The following work has to be performed for a load flow study.
> Representation of the system by single line diagram.
> Determining the impedance diagram using the information in single
line diagram.
> Formulationof network equations.
> Solution of network equations.
5.3. CLASSIFICATIONOF BUSES
In the network of power system, buses
each bus.
become nodes and a voltage can be specified tor
The power flow equation is
N
P, +j Q, = V, X Y, V, i= 1,2,
....., N ...(5.1)
Complex bus voltage V, = |V,|8, ...(5.2)
From equation (5.1) and (5.2), we know that
power system is associated with tou!
quantities and they are real power (P), Reactive power
phase angle of voltage (). In load flow problem, two (Q), Voltage magnitude V an0
and the remaining two quantities are obtained by quantities are specified for each bus
are classified based on the variables solving the load flow equations. The buse
specified. There are three types of buses.
1. Slack bus or swing bus or reference bus.
2. Generator bus or voltage controlled bus or P-V bus
or regulated bus.
3. Load bus or P-Q bus.
 Powr Flow Anahsis 5.3

The following table gives the quantities specified and the quantities to be specified for
cach bus.
S.No. Bus Quantities specified Quantities to be specified
Slack bus V.S P,Q
P.Vbus (or)
Generator bus P,| V Q. 8
P-Q bus (or)
Load bus P, Q |V|,8
where P= Pa-PL
Q = QG -
Pa = Real power generated by generator connected to the bus.
QG = Reactive power generated by generator connected to thebus.
PL = Real power drawn by the load.
Q = Reactive power drawn by the load.
Slack Bus
In slack bus, voltage magnitude and phase angle of voltages are specified pertaining to a
generator bus usually a large capacity generation bus is chosen. We assume voltage (V) as
reference phasor.
ile., j = 0
where 8 = Phase angle of voltage.
This bus makes up the difference between the scheduled loads and generated power that
are caused by the losses in the network.
Obtain (N - I)complex bus voltages from the (N - 1) load flow equations. Incidentally
the specification of | Vdack l helps us to fix the voltage level of the system. In power flow
study, at all buses net complex flow into the network is not known in advance.
While specifying ageneration schedule for a given system demand, one can fix up the
generation setting of all the generation buses except one bus because of the limitation of not
knowing the transmission loss in advance. Therefore, it is necessary to have one bus called
slack bus.
Power Balance equation is
N

P = FP, = Pa .(5.3)

Real power loss Tots! generation Total lcd

P depends on PR loss in the transmission line and transformers of the nerwork The
individual currents in the various lines of the network cannot be calculated until after the
voltage magnitude and angle are known at every bus of the system. Therefore P, is initialy
unknown. Real and reactive power are not specified for slack bus.
 5.4 Power System Analyi,
bus at which
In powCr Mow problem, we assume one generator bus as slack
generation is not prespecified. After the power flow problem has been solved,
the power
between the total specified real power going into the system at all the other buses and s dif erence
total power consumed by loads plus 1²R losses are assigned to the slack bus. Therefo
generator bus must be selected as the slack bus. The slack bus is needed to account
transmission line losses.
Generator Bus or P-VBus or Voltage Controiled Bus or Regulated Bus
Atthese buses, the real power and voltage magnitudes are specified. The phase angles of
of th.
the voltages and the reactive power are to be determined. The limits on the value
reactive power are also specified.
In order to maintaina good voltage profile over the system, Automatic Voltage Regulator
(AVR) is used.
Static VAR compensator buses are called as P-V buses because real power and voltag:
magnitudes are specified at these buses.
Load Bus or P-Q Bus
At these buses, the active and reactive powers are specified. The magnitude and phas:
angle of the voltage are unknown. These are called as load bus.

5.4. DESCRIPTION OF LOAD FLOW PROBLEM

Generator Generato
In load flow analysis, the system is
considered to be operating under steady
state balanced condition and per phase Tr. Line
analysis is used. A sample power
system may be represented by apower
network as shown in Fig.5.1.
The network consists of a number of
buses (Generator, load buses] repre Transformer
senting either generating station or bulk
Load Load
power substation, switching stations
interconnected by means of transmis Fig. 5.1. Sample powersystem network
sion lines or power transformers. Each transmission line is characterized by its t equia
circuit. The transformer with off-nominal tap ratio is characterized by their I equ
circuit. Shunt compensating capacitors or reactors are represented as shunt susceptance.
Load flow analysis is essentially concerned with the determination of comples
The us
voltages at all buses, given the network configuration and the bus demands. Out ofch
generation and demand are characterized by complex powers flowing into and
buses respectively.
Power Flow Analsis 5.5

Ageneration schedule is nothing but a combination of MW generation of the various

generators should match the given system demand plus the transmission losses. It should be
noted that there are many generation schedules available to match the given system demand
and one such schedule is chosen for load flow analysis.
5.4.1. IDEAL LOAD FLOW PROBLEM (OR) STATEMENT OF IDEAL POWER FLOW PROBLEM
Load flow problem is stated as follows:
Given : The netvork confgration(line impedance and half line charging admittance)
and all the bus power injections.
P, = PG-Pp
where P, Bus power injection.
PG Bus generation.
Pp = Bus demand.
To determine the complex voltages at all the buses :
The state vector X is defined as
X = [V V2: ... (5.4)
Once the voltages at all the buses are known, then we can compute slack bus power,
power flows in the transmission lines and power loss in the transmission lines.
5.4.2. PRACTICAL LOAD FLOW PROBLEM OR
STATEMENT OF PRACTICAL POWER FLOW PROBLEM
Practical load flow problem can be stated as follows :
Given: The nerwork configuration, complex power demands for all buses, real power
generation schedles and voltage magnitudes of allthe P-V buses and voltage magnitude of
the slack bus.
To determine .:
> Bus admittance matrix.
Bus voltage phase angles of all buses except the slack bus and bus voltage
magnitudes of all the P-Q buses.
State vector X =[V, V ......, VE o, ð,
5.5. POWERFLOW EQUATION (PFE)/DEVELOPMENT OF POWER FLOW MODEL IN
COMPLEX VARIABLE FORM AND POLAR VARIABLE FORM
The power flow or load flow model in complex form is obtained by writing one complex
power matching equation at each bus for the Fig.5.2.
PGij QGi ( Ppi j QDi P;+jQ, =(PGi-Poi)+j (QGi -Qoi)

V;

Fig. 5.2. Complex pOwer balancing at a bus

5.6 Power System Analy
Net power injected into the bus i.
S, = SG Spi
PG,tjQG,-(Pp, tj Qp)
= PG- Pp,tj (Qi-Qp)
P, +jQ,
We know, P, + Q, = ...(5.5)
Consider two bussystem as shown in Fig.5.3.
Let I, be the net or bus current entering into bus I.
Let I, be the net current entering into bus 2.
[1] = [Y}[VI

|Y21 Z12
Yy = y10t y12 Y10

Y2= Y20 y2i

Y1 = Y21-Y21 Fig. 5.3. Two bus system
In general Y, = |Y,|Z0,= |Y,|cos ti|Y,|sin 0, , ... (5.6)
1 = Y| V,+Yn V,
I, = Y1V,+ Y2 V,
In general, the net current entering into h bus
N
I, = Y,,V, +Y, V,t. +Y,N VN = E Y, V, (5.7)
Substituting I, from equation (5.7) in equation (5.l), we get
S, = P, +jQ, = V, 1
S, = P,-jQ, =v*I,
N
P,-jQ, V where i =1,2, N
There are Ncomplex variable
V,. V, ......,. V can be determined.equations from which the N unknown complex
variables

Substituting Y,, fromequation (5.6), we get,

P,-jQ, = v EIY,| Z0, V,
Power Flow Analysis 5.7

where V, = |V,|2 8,, v*=| V,|2-8,

V, = |V,|26,; 8, =Phase angle of voltage
N
.. P,-j Q, = £ |V,||Y,l|V,| (0,, +8,- 8)
Equating real and reactive parts, we obtain
P, - E |V,||Y,||V,|cos (0,, +å, -8,)
N

Q, = - £ |V,||Y,V,|sin (®,, +8, - 8,)

We can write the above equations as
N

P, = |V,PiY,,|cos 8,, +E |V,|Y,IV,|cos (0,, +8,- 8,) ... (5.8)

Q. = -|V, P|Y|sin 6,,- E |V,||Y, |V,| sin (0,, +8, -6,) .(5.9)

The P, and Q, equations are called as polar form of the power flow equations (or) static
load flow equations.
For an N' bus system, there will be 2N power flow equations. Each bus is characterized
by four variables P, Q, V,and ×, resulting in a total of 4N variables. The power flow
equations can be solved for 2N variables if the remaining 2N variables are specified.

5.6. SOLUTION TO LOAD FLOW PROBLEM

Anumber of methods are available for solving load flow problem. In allthese methods,
voltage solution is initially assumed and then improved upon using some iterative process
until convergence is reached.
The load flow methods are given by,
(i) Gauss-Seidel Load Flow Method (GSLF).
(ii) Newton-Raphson Load Flow Method (NRLF).
(iii) Fast-decoupled Load Flow Method (FDLF),

5.7. ITERATIVE SOLUTION USING GAUSS-SEIDEL METHOD TO 1

LOAD FLOW PROBLEM-INCLUDINGQ-LIMIT CHECK FOR 2

VOLTAGE CONTROLLED BUSES

It is also known as the method of successive displacements.

Consider N-bus system as shown in Fig.5.4.
N
Bus 1to M are machine or generator bus
Bus M + | to N are load buses. Fig. 5.4. N-bus system
 5.8 Power System Analysis
Flat Votage Siart
Phase angle. 8,° = 0. for i= 1.2, ........, N (for all buses except slack)
Voltage. |V°| = 1.0 for i = M +1. ......N (for all P-V buses)
|V,=|V,spec for all P-V buses and slack bus.
Bus I is a generator bus and take it as a reference bus or slack bus. Here the voltage is
specified.
In load buses, assume initial value of voltage as l 0° and findthe new value of voltages.
We start our calculation from bus 2 onwards. In the generator buses, first check for
generator limit and find the voltages.
Injected bus power is given by,
S, = P,-jQ, = V; I,
N
=

j=1
N

P,-jQ, = v* Y, V, + v E Y, V,
j=1

V Y,V, = P,-jQ,-V; j=1

2 Y, V,
1 P-j9
V, = ... (5.10)
j =1

i= 1,2, 3, ........, Nexcept slack bus

Let yola yold , vold be the initial bus
voltages. On
substituting initial values in the
above equation, we can find Vhe. ynew
2 3
replace the old values by the new values.
V.After calculating each voltages.
Therefore equation (5.10) becomes,
= P,-jQ, i-1
V
,* old
Y, ve ... (5.11)
For load bus,
equation (5.11) is applicable to find|V | and 8 values.
For slack bus,
the voltage is specified, so itwill not change in each iteration.
Power Flow Analysis 5.9

For PV or generator bus,

ovaBue is not specificd for PV bus. Adjustng the complex voltage V, e, +jf, to
correct the voltage magnitude to the spccified value | V, Ispee:
v"e= |V, snee Z$cal

where 8 calculated = tan

(ii) Compute the reactive power generation using the Ve as

i-1
vd .. (5.12)
j=1

QGi =
Qgal + QD
If Qamin) S QGi < QGi (max) > set Q, =Qa-QD
then compute V.
If Qa < QGi (min) set QGi = QGr(min) then compute VN using equation (5.1).
It QG > QGi (max) set QG = QGi(ma then compute V using equation (5.|1).
Acceleration Factor (a)
In Gauss Seidel method, the number of iterations required for convergence can be reduced
if the correction in bus voltage computed at each iteration is multiplied by a factor greater
which it is
than unity, called as acceleration factor to bring the voltage closer to the value to
converging. The range of 1.3 to 1.7is found to be satisfactory for typical systems.
vold + a ynew - vlid ... (5.13)

where vd = Voltage value obtained in the previous iteration.

= Acceleration factor.
Vnew = New value of voltage obtained in the current iteration.
Convergence Check
For the power mismatch is small and acceptable, a very tight tolerance must be specified
on both real and imaginary components of voltage.
Iteration process continues until the magnitudes AP and AQ<0.001 p.u. (specified value).
Voltage accuracy is 0.00001.
AVmor max { Ae +jf, k=1, 2, ......., Nslack)
5.7.1. COMPUTATION OF SLACK BUS POWER
Since all the voltages are known, we can compute slack bus power (P, and Q) and thus
PGi and QG which are unknowns at the slack bus.
N

Slack bus power P-i = V Y, V, (5.14)

 5.10 Pawer System Anaysis
5.7.2. COMPUTATION OF LINE FLOWS

+Ipi
Ypi

Fig. 5.5. Line connecting two buses

Consider the line connecting two buses i and í as shown in Fig.5.5. The line can be
represented by the series admittance Ys and the two shunt admittances (half line chargina
admittance) Yp, and Yp
Line current (forward) I,, = l+lp,
= (V,-V) Y, +V, x Yp
Lire current (reverse) 1,, = -Is t Ip,
(V,-V) Y, + V, Yp,
Line power (forward) S,, = P tjQ, = V, I;
V,[(V,-) Y, +V, Yp, J*
V, {(V,*-V,) Y + | V, 2Y,}
Line power (reverse) S,, = V, 1
... (5.15)
...(5.16)
5.7.3. COMPUTATION OF TRANSMISSION LOSS
Power loss in the transmission line ij.
Sy(Los) = S, +S,
= P,tjQ, +P,, +j Q, ...(5.17)
Real power loss = P,+PH
Reactive power loss =
Qy +Q,
5.74. COMPUTATIONOF TRANSFORMER AND LINE FLOW EQUATION
After finding complex bus voltages, the active
and
lines/transformers are to be computed. A common equivalent reactive flows in all tne
circuit for transmission line
and transformer is shown in Fig.5.6.
For line,
For transformer, b.
Power Flow Analysis 5.11

V;
(U:1
Pi, *jQi I, Yijs Pji+jQji
I,

jb: jbc

Fig. 5.6. Equivalent circuit of atransmission line/transformer

Power flow from jhbus to the th bus, measured at the th bus is given by,
Pi, tjQ, = V, I; =V, I ... (5.18)
We know = a V, = V, ... (5.19)

I, =(V,- V) Yus + V, (j b) ... (5.20)

Substituting equation (5.19)and (5.20) in equation (5.18),

P, tjQ, (b)* ... (5.21)

Similarlypower flow from jh bus to ¡th bus,

=
(5.22)
For transformer :
Substituting b, = 0inequation (5.21) and (5.22), we get

P,tj Q, ... (5.23)

P,, +j Q = v[v-(], ..
(5.24)

Real power loss P oss= Pi; tP

Reactive power loss Loss +Qji
5.7.5. ALGORITHM FOR ITERATION METHOD
Step I: Form Y-bus matrix.
Step 2: Assume V = V7(spec) Z0° at allgenerator buses.
Step 3: Assume V,=120°=|+j0 at all load buses.
5.12 Power System Ana
Step 4: Set iterationcount = 1(iter = 1).
Step 5: Let bus number i= .
Step 6: If refers to generator bus go tostep no. 7, otherwise go to sten
Step 7(a): Ifi refers to the slack bus go to step 9. Otherwise go to step 7(b).
Step 7(b) : Compute Q, using,
N

Q
cal
- -Im j= |
v,Y, V,
QGi Q +QL
Check for Q limit violation.
cal
" min)< QGi < Q, (max) > then i (spec) Q
If Q, (min) < QGi then i(spec) Q, (min) -Li
If Q; (max) <QGi then Q, (spec) = Q; (max) QLi
If Qimit is violated, then treat this bus as P-Q bus tillconvergence is obtained.
Step 8: Compute V, using the equation,
v new =
Pi(spec) - i(spec) Y,, Vncw Vold
i vold * i=j+|

Step 9: Ifiis less than number of buses, increment i by l and go to step 6.

Step 10: Compare two successive iteration values for V,.
If Vne
i
-vod < tolerance, go to step l2.
Step 1l : Update the new voltage as
ynew = yold+ a (Vnew - yold)
yold Vnew
iter = iter + 1 : go to step 5
Step 12 : Compute relevant quantities.
N

Slack bus power, S, = P,-jQ, = V*I = v* E Y, V,

j=1
Line flows S, = P, +jQy
j series t | V,|2 Y
Poss + Pi
QLoss Qj
Step 13: Stop the execution.
 S.13

FLOWCHHART FOR GAUSS-SEIDEL METHOD INCLUDING PV BUS ADJUSTMENT

sZ6.

Start

Bus Data

Qfor=V-1

intaze bs votaes

Yes

Yes

No

=2tN
csses reae
Cacuz:ane Sons totane
ter ter 1
5.14 Power System Analysis
SOLVED EXAMPLES

Example S.1 Perform power flow of one iteration for the system as shown in
using Gauss-Seidal method. Deternine slack bus power, line flows and line Fig,
base MVAas 100. (a= 1.I).
Take losses.

1.05 20° MHE 2

0.0839 + j0.5183
90 + j20 j0.005
j0.0636 j0.0636
30 + j10 MVA

Solution :
Step l: Formulate Ypus
When the switch is open, there is no connection of capacitor at bus 2.
Take the bus 2 as load bus.
0.3044 -jl.816 -0.3044 +jl.88
Y bus
0.3044 +jl.88 0.3044 -jl.816 J
Step 2: Initialize bus voltages.
vold 1.05 Z0° p.u
yold 1.0 Z0° p.u
2

Step 3: Calculate ynew

2

-30
P, = -30 MW = 100 P.u = -0.3 p.u

Q = - 10MVAR = - 100
10
P:lu = -0.lp.u
1
Y
fP2-j2
vold * Ya1 Vhew
2

- 0.3 +j0.1
0.3044 -jl.816 1.0 -0°0.3044 +jl.88) 1.05
1.0054-j0.1577 = 1.018 2-8.9150
Step 4: Calculate V Using acceleration factor.
vhew
2acc 2

1.0+ 1.1 [1.0054 -j0.1577-1]|

1.0059 -j0.173 = 1.0207 Z-9,78°
AHerFlowAnalysis 5.15

Slack bus power.

Siep5.
S, = P: V[Y V +Yn Vl
1.05 Z-0° (0.3044 -j l.816) 1.0S + (-0.3044 + i1.88) (1.0207 2-9.78))
- 0.3556 +j0.0388 p.u = 35.56+j3.88
MVA
p. = 35.56 MW, Q=3.88 MVAR
Real power generation Pa1 = P + P
=35.56 + 90 = 125.56 MW
Reactive power generation QGI = Q, +Q1
= -3.88 + 20 = 16.12 MVAR

Step 6: Line flows,

*
Bus
S, =PytjQ, =V, [V; - v÷|Y series +|V, 2Y:
Pi
From To
2 S12 = series +|V, |* Y*10
= 1.05 [1.05 -0°- (1.0059 +j0.173)] x
(0.3044 +jl.88) + 1.0S2 x (-j0.0636)
= 0.3556 -j0.0383 p.u
P12 = 0.3556 p.u = 35.56 MW
Q12 -0.0383 p.u = -3.83 MVAR
2 1
S21 = V, [V; -V; ]Y2 series +|V,P Y,20
(1.0059 -j0.173) [1.0059 +j0.173 - 1.05] x
(0.3044 +jl.88) + 1.02072 x (-j0.0636)
= -0.3459 -j0.038p.u
= -0.3459 p.u = -34.59 MW
P21 MVAR
Q21 = -0.038 p.u = -3.8
`tep 7: Iransmission line loss (Sij Loss - S, +S,)
= 0.97 MW
P2 Loss P + P1 = 35.56 - 34.59
MVAR
Q12 Loss Q2tQ =-3.83 +(-3.8) = -7.63
5.16 Power System Analysis
Example 5.2| Using Gauss-Seidal method, determine bus voltages, slack bus na.
line flows and line losses for the Fig. shown.
60 MW
- 20 < Q<100 MVAR

1.05 Z0° 2 1.02 p.uvolt

0.0839 +j0.5183
90 + j20
j0.0636 j0.0636

Solution:
Step I : Formulate Yhus
0.3044 -jl.816 0.3044 +jl.88 1
Y bus -

-0.3044 +jl.88 0.3044 -j1.816J

Step 2 : Initialize bus voltages.
V= 1.05 Z0° p.u
vold = 1,02 Z0° p.u
Step 3 : Calculate Q,.

= -Im V} (V, Y1+ V, Y2]

= -Im 1.02 -0° [1.05 Z0° x (0.3044 +j 1.88) +
1.02 Z0 (0.3044 -jl.816))
= -0.124 p.u = -12.4 MVAR
Check for Oimnit Violation :
al
Q2(min) < Q<Q2 (max)
-20 < -12.4 <100
This bus acts as generator bus.
Q, = - 12.4 MVAR = -0.124 p.u
Step 4:Calculate ynew
2

22 L - Y vi
0.6 +j0.124
0.3044 -jl.816 L.02 /-0 (-0.3044 +j1l.88) * 1.05
0.9078-jl.8524
0.3044 -iL816 1.073+j0.32 = 1.12 Z16.6]0

=V2spec) = 1.02 ZI6.6|°

Power Flow Analsis 5.17

Step 5: Using acceleration factor.

ew
vold
1.02 + 1.1|L.02 Z16.61° - 1.02]
= L.02+ 1.I[0.977 + j0.292 - 1.02]
= 0.973+j0.32= 1.02 218.24°
Bus voltages are V, = 1.05 Z0°
V, = 1.02 Z18.24°
Step 6: Slack bus power
S, = P-jQ; =VY, V, + Y1; V]
= 1.05 Z-0° [(0.3044 -j1.816) x 1.05 + (0.3044 +j1.88) (1.02 18.24°)]
= -0.607-j0. 18 p.u
= -60.7-jl8 p.u
.. P, = -60.7 MW, Q, = 18 MVAR
Real power generation Pa = P +PI
= -60.7 + 90 = 29.3 MW
Reactive power generation QGI = Q, +QI
= 18+ 20 = 38 MVAR
Step 7:Line flows.
Bus Sj =P, +j Q, =V,IV} -V|Y series

From To where Yp= Yio

2
S12
=
series +| V Y
= 1.05 [1.05 -0°-(0.973 -j0.32)] x
(0.3044 +jl.88) + 1.052 x (-j0.0636)
-0.607 +j0.18 p.u
P2 = 0.607 p.u = -60.7 MW, Q2=0.18 p.u = 18MVAR
2
S, = V, [V-VJY 21 series +|V,P Y20
(0.973 +j0.32) [0.973 -j0.32 - 1.0S] x
(0.3044 +jl .88) + 1.022 x (-j0.0636)
0.64 -j0.1167 p.u
P1 0.64 p.u = 64 MW
Q -0.1167 p.u = - 11.67 MVAR
 Power System Analy
5.18

Step &: Transmission loss (S,Loss = S,, +S,;)

=
P2 Loss = Pp t P1 = -60.7 + 64 :3.3 MW
Q12 Loss
=
Q2tQ = 18 - ll.67 =
6.33 MVA
Using Gauss-Seidal method, determine bus voltages, for the F
Example 5.3
shown. TakeBase MVA-100, a= 1.1.
60 MW
10 < Q2< 100 MVAR

V= 1.05 Z0* 2 1.02 volt

0.0839 +j0.5183
90 +j20 jo 0636
j0.0636

9 Solution :
Step 1: Formulate Ybus
0.3044-jl.816 -0.3044 +jl.88 1
Ybus =
0.3044 +jl.88 0.3044 -j 1.816 J
Step 2: Initialize bus voltages.
= 1.05 Z0° p.u
= 1.02 20° p.u
Step 3 :CalculateQ value for generator bus.
= -Im {V; [V, Y2+V, Y2 ]}
= -Im {1.02 [1.05 Z0° x (-0.3044
+jl.88)+
1.02 Z0° (0.3044 -jl.310)
= -0.124 p.u
Check for Oimi violation :
2(min) = 10 10
MVAR, Q3 (min) = 100 = 0.1 p.u
< Q2min)
i.e., -0.124 < 0.1
.. Bus 2 will act as load bus.
Q Q2imini =0.1 p.u
vd = L0 Z0° p.u
P = 60 MW = 0.6
p.u
PowerFlowAnalsis 5.19
yhew
Step 4 : Calculate 2

P2 - Y2 V
0.6 -j0.! -(-0.3044 +j1.88) 1.05
0.3044 -jl.816 1.0 Z0°

1.842 -80.49o [0.9196 -j2.074]

1.842 -80 490 2.2687-66.087°]

= 1.2316 Z14.403° = 1.193 +j0.306
Step 5: Using acceleration factor.
Vhew
2 acc vod + a[ V
= 1+1.1 [1.193 +j0.306 - 1]
1.2098 +j0.33
= 1.254 Z15.26°
Bus voltages are V, = 1.0S Z0°
V, = 1.254 Z15.26°

Example 5.4 Using Gauss-Seidal nethod, determine bus voltages and reactive
power generation for the Fig. shown. Take Base MVA = 100.
V=1.05 Z0°
0.0839 +j0.5183
2
H
30 +j20
j0.0636 j0.0636 j0.005

Solution :
Step 1: Formulate Ybus
Y1o+ Yi2 -Y12
Yhus -Y12 Yo t Y2ct Y12
Admittance of shunt element
where Y2c =

0.3044 -jl.816 -0.3044 +j1.88

=

-0.3044 +jl.88 0.3044-jl.8|1 J

SBep 2: Initialize bus voltages.
$.20
Power System Analysi
vo 1.05 Z0° p.u (slack bus)
jold
V = 1.0 Z0° p.u (P-V bus because capacitor generates
Step 3 :Calculate Q,. power reactive
ocal -Im {V, [V, Y tV, Y2]}
- Im {1.0 Z 0° [1.05 x (-0.3044 +j1.88) + 1.0 (0.3044-jl.811)
= -0.l63 p.u
Step 4 : Calculate vhew
Y2
30
P = PG - PL2 100 - 0.3 p.u; Q,=-0.163 p.u
new P,-j2
vold
-Y yhew
1
2

-0.3 +j0.163
0.3044 -jl.811 1.0 - 0° -(-0.3044 +j1.88) x 1.05
=
0.973 -j0.1611 = 0.986 Z-9.4°
= 1.0-9.4o
Reactive power generation QG2 QytQL2
= -0.163 + 0.2 = 0.037 p.u = 3.7 MVAR
Reactive bus power Q2 -0.163 p.u MVAR = - 16.3 MVAR
Example 5.5 For the system shown in Fig., determine the voltages at the end of the
first iteration by Gauss-Seidel nethod and also find the slack bus power, line
flows,
transmission loss. Assume base MVA as 100.

j0.4

j0.3 j0.2

Solution :
Bus Generator Load Omin
No.
Voltage P
MVAR
Q P MVAR
1.05 Z 0° p.u
100
2 1.02 p.u 0.3 p.u -10
3. 0.4 p.u 0.2 p.u
PowerFlowAnalvsis 5.21
Y bus.
Step1: From
1
-1
j0.4j0.3 j0.4
-1 1 -1 -1 -1
Y21
Y22 Y23 j0.4 jO.4 i0.2 j0.2
-1 - 1 1
Y31 Y32 Y33 j0.3 j0.2 j0.3j0.2
f-j5.8333 j2.5 j3.3333
j2.5 -j7.5 j5
j3.3333 j5 -j8.3333
Step 2: Initialize bus voltages.
yold 1.05 Z 0° p.u [Bus I is a slack bus i.e., V and8 isspecified)
= 1.02Z0° p.u [Bus 2 is a PV bus i.e., P and V is specified]
vold = 1.0 Z0° p.u [Bus 3 is a load bus i.e., P and Q is specified]
Note For slack bus, the specified voltage willnot change in any iteration.
For generation bus, calculate vie using the formula and write
v =

Vspecified ZOcalculated value

Step 3 :Calculate Qvalue for all generator buses.
i-1 N
Qsu =-Im vold* + S Y, vold
j =I

Im yold
2 Y V +Y, Vd+ Y,Vold
= -Im 1.02 Z 0° j2.5 x 1.05 Z 0° +(-j7.5 x 1.02 Z 0°) +j5 x 1Z0°1

= -Im L.02 0° [j2.625 -j7.65 +j5]

cal
Q = 0.025 p.u

NOw Q(min) S Q* sQ2 (max)

is within the specified limit.
Step 4: Calculate Vhe.
Vnew = 1.05 Z0° p.u
 Power System Analysis
N

YyV - Y V
P 0.3 pu (Given): Q,0.025 p.u
0.3-0.025 -j2.5 x 1.05 0° -j5 x I 0°
-j7.$ 1.020°
L0199 +0.0392
1.0207
S =1.02 22.2° = 1.0192 +j0.0392
P; Ps-PË: =0-0.4 = -0.4 p.u
Q:= Q - =0-0.2 =-0.2 p.u
P;-jQs
Y V-Y;, Vhcw
0.4 +j0.2
-j8.3333 1009-j3.3333 x 1.05Z0°-j5 x 1.02 Z 2.2°

-j8.3333l-0.4+j0.2 -j3,4999 -j5.096 + 0.196]

= 1.0075 -j0.0244 = 1.0078 -1.39°
= l0075 -j0.0244
= 1.0078 - 39°
Step 5: Slack Bus Power

S, =P, -jQ, = V} 2 Y, V,
S, = V}YH VË+Y; V,+Y3 V;]
= L05 -j5.8333 x 1.05 0°
+j2.5 x (1.0192 +j0.0392)
= -0.0175-j0.2295 p.u +j3.3333 (1.0075 -j0.0244)J
P, = - 0.0175 p.u = -1.75 MW
Q, = 0.2295 p.u = 22.95 MVAR
Step 6: Line Flo
S,, =P, +jQ, = V, [N} -v*IY;y series V, 2 Y.
P
Line îlow from bus l to2.
S = P: iQ: = VË[V}-V} Y series
PowerFlowAnalysig
5.23
= 1.05 [(1.05 Z- 0°)-1.0192 +
-0.1029 +j0.0808 p.u j0.0392] j2.5
S = P2 tjQ21 =
V,[ V} -v] Y 21 series
= 1.0192 +j0.0392 [1.0192
= 0.1029 -j0.0746 -j0.0392 1.05]j2.5
p.u
S3 P23 +j Q23
Sy3 = V,[V; -V] Y Series

1.0192 +j0.0392 |1.0192 j0.0392 -1.0075

= 0.3218 +j0.072 p.u -j0.0244}jS
S32 = P32 tj Q32
S32 =

V,[ VË-V}]Y, 32 series

= 1.0075 -j0.0244 [1.0075 +j0.0244 - 1.0192
+j0.0392]j5
=

-0.3218-j0.05 12 p.u
S13 = P3tjQi3
Si3 = V, [V-v] Y13 series

= 1.05[1.05 -0°-1.0075 -j0.0244 ] j3.3333

= 0.085 +j0.148 p.u
S31 = P3 +j31
S31 = V [ VË-vY31 series
= 1.0075 -j0.0244 x [1.0075 +j0.0244 1.0S] x j3.3333
-0.085-j0.1407 p.u
TransmissionLoss
Sij Loss S,+Sj
For line 1-2,
S12 Loss = P12 Loss +j Q12 Loss S12 + S1
S12 Loss =-0.1029 +j0.0808 + 0.1029 -j0.0746
= 0+j0.0061
Q12nss = 0.0061 p.u = 0.61 MVAR
Pi2 Loss # 0,
For line 2-3,
P23 Loss t j Q23 Loss S3+ S32
S3 Loss
0+j0.021
0.3218 +j0.072 +(-0.3218-j0.0512) =
Loss = 0, Q23 Loss = 0.021I p.u =
2.1 MVAR