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10EEL78 POWER SYSTEM SIMULATION LABORATORY MANUAL
Department of Electrical and Electronics Engineering
Canara Engineering College
Benjanapadavu-574219
Bantwal Tq
August 2013- December 2013
2
10EEL78 POWER SYSTEM SIMULATION LABORATORY
Prerequisite:
10CCP13 Computer concepts and C programming
10CPL16 Computer programming lab
10ES34 Network Analysis
10EE53 Transmission and Distribution
10EE61 Power System Analysis and Stability
10EE665 Object Oriented Programming using C++
10EE71 Computer Techniques in Power System Analysis
Objectives:
At the end of the semester/program the students must be able to
I. Study the behavior of power system by developing its model and simulating it using the
software packages MATLAB/C++
and MiPower
II. Develop the model for any given transmission line and analyse its behavior.
III. Simulate and study various short circuit faults on the transmission system.
IV. Appreciate the usage of software packages in performing load flow analysis for a given
power system.
V. Analyse the behavior of the machines connected to the power system.
VI. Appreciate modeling and simulation of any system using these software packages.
VII. Explore the potential applications of these softwares to design and solve various
Engineering problems.
Number of practical hours/week: 06
Total number of practical hours: 42
Evaluation Procedure:
Internal assessment marks: 25{Records-10 marks, Preparation and performance-5marks, IA
exam-10marks}
Final Exam marks:50
3
List of Experiments:
1. Y Bus formation for power system with and without mutual coupling, by singular
transformation.
2. Y Bus formation for power system with and without mutual coupling, by inspection
method
3. Analysis of Short Transmission Line using ABCD constants
4. Analysis of Medium Transmission Line using ABCD constants using T circuits
5. Analysis of Medium Transmission Line using ABCD constants using π circuits
6. Power Angle Diagrams for salient and non salient pole synchronous machines
7. Solution of Swing Equation
8. Economic load dispatch using B-coefficient.
9. Load flow analysis using Gauss Seidel method.
10. Load flow analysis using Newton Raphson Method.
11. Short Circuit Fault analysis
4
Experiment No:1
Y Bus formation for power system with and without mutual coupling, by singular
transformation.
Aim: 1.To formulate the bus admittance matrix, Y Bus by singular transformation for the
power system given.
2. To write a program in MATLAB/C++
to verify the same.
Network:
Single line diagram of a power system
Equivalent graph representation
Oriented Connected Graph Tree and cotree of the oriented graph
Impedance parameters:
Element Bus code Self impedance Bus code Mutual impedance
1 0-1 0.2
2 1-2 0.3 0-1 0.05
3 2-3 0.4
4 3-0 0.5
Steps:
1. Form [A] matrix using Tree and cotree of the oriented graph.
2. Form [y] admittance matrix using Tree and cotree of the oriented graph.
3. Find transpose matrix of [A] i.e [A]│
4. Find Ybus=[A] [y] [A]│
load
1 3 2
0 1
2 3
4
1
2
3
0 1
2 3
4
1
2
3
5
Experiment No:2
Y Bus formation for power system with and without mutual coupling, by inspection
method
Aim: 1.To formulate the bus admittance matrix, Y Bus by inspection method for the network
given.
2. To develop a program in MATLAB/C++
to verify the same.
Network:
0.025+j0.125
0.02+j0.06
0.03+j0.1
0.015+j0.09
0.018+j0.072
j0.02
j0.02
j0.02
j0.02
j0.02
j0.02j0.02
j0.02
11 2
34
j0.02
j0.02
Impedance parameters:
Sl.No Element No Transmission line Self
impedance
Half line charging
admittance From
Bus
To
Bus
1 1 1 2 0.02+j0.06 j0.02
2 2 2 3 0.025+j0.125 j0.02
3 3 3 4 0.03+j0.1 j0.02
4 4 4 1 0.015+j0.09 j0.02
5 5 4 2 0.018+j0.072 j0.02
6
Experiment 3
Analysis of Short Transmission Line using ABCD constants
Aim:1.Toanalyse the given short transmission line using ABCD constants and hence find
the sending end voltage, current and regulation.
2.To verify the same by writing program in MATLAB/C++
.
Theory: Classification of transmission lines
Transmission lines are classified as short, medium and long. When the length of the
line is less than about 80Km the effect of shunt capacitance and conductance is neglected
and the line is designated as a short transmission line. For these lines the operating voltage
is less than 20KV. For medium transmission lines the length of the line is in between 80km
- 240kmand the operating line voltage will be in between 21KV-100KV.In this case the
shunt capacitance can be assumed to be lumped at the middle of the line or half of the shunt
capacitance may be considered to be lumped each end of the line. The two representations of
medium length lines are termed as nominal-T and nominal-π respectively.
Lines more than 240Km long and line voltage above 100KV require calculations in terms of
distributed parameters. Such lines are known as long transmission lines. This classification
on the basis of length is more or less arbitrary and the real criterion is the degree of accuracy
required.
Problem statement:
A 50 Hz, 3Φ, transmission line has total series impedance of 10+j50 Ω and total admittance
of j10-3
mho. Receiving end power is 100MW at 132kV with 0.8 pf lagging. Find the ABCD
constants, sending end current, power and voltage, efficiency and regulation. Use short
transmission line approximation by writing the circuit diagram.
R+jX
VrVs
IrIs
Short transmission line representation in Two Port Network form
Formulae to be used:
7
s
r
s
s
sssss
r
rs
rsrs
rs
r
rl
rl
rr
P
P
factorpower end sendingφ cos
power end sending P
voltageendsendingVwhere)(cosIV3P
x100V
VVregulation %
asevoltage/phendreceivingVwherejX)(RIVV
II
factorpower end receiving cos
power end receiving P
voltagelineendreceivingV wherecosV3
PI
r
r
Find ABCD constants
r
r
s
s
I
V
DC
BA
I
V Vs=AVr+ BIr ; I s= CVr+ DIr
Vs=(R+jX) Ir+Vr
A=1 B=R+jX
Is=Ir
C=0 D=1
Vs
ɸs
Is
Ir ɸr IrR IrX
Vr
8
Experiment 4
Analysis of Medium Transmission Line using ABCD constants using nominal T circuits
Aim: 1.To analyse the given medium transmission line using nominal T circuits and hence
find the sending end voltage, current and regulation.
2. To verify the same by writing program in MATLAB/C++
.
Problem statement:
A 200 Km long, 3Φ overhead line has a resistance of 48.7 Ω/phase, inductive reactance
of 80.2 Ω/phase and capacitance (line to neutral) 8.42 nF/Km. It supplies a load of 13.5
Mw at a voltage of 88 kV at 0.9 pf lagging. Using nominal T circuit, find the sending end
voltage , current and regulation.
R/2+jX/2
VrVs
IrIs
R/2+jX/2
C
Ic
Vc
V1
Transmission line representation in Two Port Network form (T circuit)
Formulae to be used:
mhoCjωY
Cjω
1X
factorpower end receiving cos
power end receiving P
voltagelineendreceivingV wherecosV3
PI
c
c
r
rlrl
rr
r
r
crrc
'
c
rrr
'
YjX/2)(R/2IVYVI
asevoltage/phendreceivingVwherejX/2)(R/2IVV
)1(jXRZwhere2
Y1IYV
YjX/2)(R/21IYV
YjX/2)(R/2IYVIIII
c
rcr
crcr
crcrrcrs
Z
9
)2(4
ZY1ZI
2
ZY1V
2Y
21IYV
2IV
voltageendsendingtheisVwhere2
I2
IVjX/2)(R/2IVV
c
r
c
r
crcrrr
ssrrs
'
s
ZZZ
ZZ
2
YZ1D
YC
4
ZY1ZB
2
YZ1A
I
V
D C
BA
I
V
constantsABCDoftermsinequationstandardwithcomparingon
I
V
2
YZ1Y
4
ZY1Z
2
YZ1
I
V
asformmatrixthein(2)and(1)equationsPlacing
c
c
c
c
r
r
s
s
r
r
c
c
cc
s
s
s
r
s
sssss
r
rs
P
P
factorpower end sendingφ cos
power end sending P)(cosIV3P
x100V
VVregulation %
Vs
ɸs Is
ɸr IrR/2 IrX/2
V│ Ic
IsR/2
IsX/2
Vr
10
Experiment 5
Analysis of Medium Transmission Line using ABCD constants using π circuits
Aim: 1.To analyse the given medium transmission line using nominal π circuits and hence
find the sending end voltage, current and regulation.
2. To verify the same by writing program in MATLAB/C++
.
Problem statement:
A 200 Km long, 3Φ overhead line has a resistance of 48.7 Ω/phase, inductive reactance
of 80.2 Ω/phase and capacitance (line to neutral) 8.42 nF/Km. It supplies a load of 13.5
Mw at a voltage of 88 kV at 0.9 pf lagging. Using nominal π circuit, find the sending end
voltage, current and regulation.
R+jX
VrVs
IrI s
C1 C2
Ic1Ic2I'
Transmission line representation in Two Port Network form (π circuit)
Formulae to be used:
mhoCjωY
C jω
1X
C/2CC
factorpower end receiving cos
power end receiving P
voltagelineendreceivingV wherecosV3
PI
c
c
21
r
rl
rl
r
r
r
r
)1(2
YVIIII c
rrc2r
'
)2(jXRZwhereZI2
ZY1V
Z2
YVIVZIVZIVV
r
c
r
c
rrr
'
r
'
rs
11
)3(2
ZY1I
4
ZY1V
2
YZI
2
ZY1V
2
YVI
2
YVIIII
c
r
c
r
c
r
c
r
c
rr
c
s
'
c1
'
s
cY
2
YZ1D
4
ZY1YC
ZB
2
YZ1A
I
V
D C
BA
I
V
constantsABCDoftermsinequationstandardwithcomparingon
I
V
2
YZ1
4
ZY1Y
2
YZ1
I
V
asformmatrixthein(3)and(2)equationsPlacing
c
c
c
c
r
r
s
s
r
r
cc
c
c
s
s
Z
s
r
s
sssss
r
rs
P
P
factorpower end sendingφ cos
power end sending P)(cosIV3P
x100V
VVregulation %
Ir
12
Experiment 6
Power Angle Diagrams for salient and non salient pole synchronous machines
Aim: 1.To determine the excitation e.m.f, reluctance power for a given synchronous
generator.
2. To plot the power angle diagram for the same generator
3. To verify the same by writing program in MATLAB.
Problem statement:
1. A 95.2 kV, 165 MVA, synchronous generator has armature resistance 0.2 Ω/ph,
synchronous reactance xd=2Ώ/ph is operating at 0.8 power factor lagging.
Determine the excitation e.m.f, regulationand also plot power angle diagram (δ vs
Power).
Calculation Details:
Ra
jXa
Ef
Vt
Ia
Per phase equivalent circuit of a non salient pole synchronous generator
δsin
X
VE*3PPowerElectrical
d
tf
e
WhereEf=generator e.m.f
Vt=terminal voltage
δ=machine angle=p0wer angle
Xd=Machine Direct axis reactance
3
VV
V3
MVAI
t
a
13
Φ=power factor angle =cos-1
(power factor)
V= line to line voltage in kV.
t
tf
aaatf
V
VERegn
)jX(RIVE
2. A 95.2 kV, 165 MVA, synchronous generator has armature resistance 0.2 Ω/ph, direct
axis reactance xd=2Ώ/ph quadrature axis reactance xq=2Ω/ph is operating at 0.8 power
factor lagging. Determine the excitation e.m.f, reluctance power and also plot power
angle diagram.
) (2δsin
X2X
)X(XVδsin
X
VE*3PPowerElectrical
qd
qd2t
d
tfe = pdf+pdh
Where Ef=generator e.m.f
Vt=terminal voltage
δ=machine angle
Xd=Machine Direct axis reactance
Xq=Machine quadrature axis reactance
3
VV
V3
MVAI
t
a
Φ=power factor angle =cos-1
(power factor)
V= line to line voltage in kV.
2. A 34.64 kV, 60 MVA synchronous generator has a direct axis reactance of 13.5 ohms
and quadrature axis reactance of 9.333 ohms operating at 0.8 pf lag. Determine the
excitation emf, regulation, reluctance power and also plot the power angle diagram.
Neglect the armature reactance
14
Phasordiagram of a salient synchronous generator supplying inductive load
x100V
VERegulation%
IXXEEemfExcitation
δsinII
neglected)being(RδEX I jVE
t
tf
dqdqf
ad
aqqatq
) (2δsinX2X
)X(XVpowerReluctance
qd
qd2t
Note: Show one sample calculation for δ=10º
15
Experiment 7
Solution of Swing Equation
Aim: write a program in Matlab / C++
1.Toplot the Swing curve for a single machine connected to infinite bus when a 3Φ to ground
fault occurs at bus.
2. Also to find its transient stability conditions.
Problem statement:
For the system given below:
Generator is generating 120MW and 80 VAR. The voltage of infinite bus is 01 . The
line reactance is 0.08pu on 100MVA base. The machine transient reactance is 0.2pu,
inertia constant H is 4 pu. Plot swing curve and determine the transient stability when a
3Φ to ground fault occurs at bus for duration of (a) 0.1 sec (b) 0.3 sec. Take
Pm=Pe1=1.2 (pre fault), Pe2=0 (during fault), Pe3=1.9(post fault). Take time increment as
0.02 sec and simulation time 0.5 sec. Use Runge-Kutta 4 (RK-4) method.
Calculation Details:
Consider the situation in which the synchronous machine is operating in steady state
delivering a power Peequal to Pmwhen there is a fault occurs in the system. Opening up of the
circuit breakers in the faulted section subsequently clears the fault. The circuit breakers take
about 5/6 cycles to open and the subsequent post-fault transient last for another few cycles.
The input power, on the other hand, is supplied by a prime mover that is usually driven by a
steam turbine. The time constant of the turbine mass system is of the order of few seconds,
while the electrical system time constant is in milliseconds. Therefore, for all practical
purpose, the mechanical power is remains constant during this period when the electrical
transients occur. The transient stability study therefore concentrates on the ability of the
power system to recover from the fault and deliver the constant power Pm with a possible new
load angle δ.
16
The angle δ is the angle of the internal emf of the generator and it dictates the amount of
power that can be transferred. This angle is therefore called the load angle.
From equal area criterion of transient stability, A1=A2
)δ(δP)cosδ(cosδP)dδPsinδ(PA
)δ(δPdδPA
cmmmcmax
δ
δ
mmax2
0cm
δ
δ
m1
m
c
c
0
Equating A1 and A2
000
1
c
tlei
0maxm
0m
m0m
max
mc
δcosδsin)2δ(πcosδ
xxxwherex
PPmaxwhereδsinPP
δπδwhere
δcosδδP
Pδcos
In general swing equation is given by )P(P2H
ω
dt
δdem
s
2
2
Consider the period during which fault occurs i.e. Pe=0.
H
sinδPP
H
PP
2H
ω
dt
dω
P2H
ω
dt
δd
maxea
m
s
m
s
2
2
whereH is the normalized inertia constant given by
ratingMVAGenerator
MJinspeedssynchronouatenergykineticstoredH
Integrating the equation twice gives
17
02
ms δtP
4H
ωδ
Replacing δ by δc and t by tc ,
ms
0cc
02c
msc
Pω
)δ4H(δt
δt4H
Pωδ
Runge-Kutta4 method:
Runge-Kutta 4th order method is a numerical technique to solve ordinary differential
equation of the form
0yy(0)y),f(x,dx
dy
So only first order ordinary differential equations can be solved by using Runge-Kutta 4th
order method.
)2
hky,
2
hf(xk
)y,f(xkwhere
k)k2(k(k6
1yy
1ii2
ii1
4321i1i
h)kyh,f(xk
)2
hky,
2
hf(xk
3ii4
2ii3
To find load angle δ0
Pmax sin (δ0) =Pm
max
m10
P
Psinδ
18
Δt)ksin(δP(PH
fπ
Δt)2
ksin(δP(P
H
fπ
Δt)2
ksin(δP(P
H
fπ
Δt)sinδP(PH
fπ
where
Δt)(ωh)kyh,f(xk
Δt)2
(ω)2
hky,
2
hf(xk
Δt)2
(ω)2
hky,
2
hf(xk
Δt)(ω)y,f(xkwhere
k)k2(k(k6
1
30maxe2
20maxe3
10maxe2
0maxe1
303ii4
20
2ii3
10
1ii2
0ii1
4321i1i
where δ is the load angle i.e angular position of the rotor
ω is the angular rotor speed
Note: Show one calculation of δ using R-K4 method
))2((6
1ωω 4321i1i
19
Experiment 8
Economic load dispatch using B-coefficient.
Aim: 1. To perform optimal generator scheduling for thermal power plants to meet the total
load demand using software MiPower.
Theoretical background:
Economic load dispatch problem is allocating loads to plants for minimum cost while
meeting the constraints. It is formulated as an optimizing problem of minimizing the total
fuel cost of all committed plants while meeting the load demand and losses. The Economic
dispatch solution has two different problems to be solved:
1. Unit commitment or pre-dispatch problem to find the specified margin of operating
reserve over a specified period of time.
2. On-line economic dispatch wherein it is required to distribute the load among the
generating units actually paralleled with the system in such a manner to minimize the
total cost of supplying the minute-minute requirements of the system.
The basic economic dispatch problem can be described mathematically as a minimization
problem of minimizing the total fuel cost of all committed plants subject to the constraints
n
1i
ii )(PFMinimize
Fi(Pi) is the fuel cost equation of ith
plant. It is the variation of fuel cost in Rs with generated
power in MW. Normally they are of quadratic form equation
iii2iiii cPbPa)(PF
Optimize the above equation subjected to the constraints. There are two types of constraints.
They are
(i) Equality constraints: They are basic power flow equations.
The total generation=total demand + loss
(ii) Inequality constraints:
(a) Generator constraints: MVA loading should be within the acceptable limits.
(MW)2+(MVAR)
2 ≤ (MVArate)
2
Pgmin ≤Pgen ≤ Pgmax and Qgmin ≤ Qgen ≤ Qgmax
(b) Voltage constraints: Voltage magnitudes and phase angles at various nodes
should vary within certain limits.
|Vpmin| ≤ |Vp|≤ |Vpmax|
(c) Transformer tap setting constraints:
20
tap_min ≤ tap tap_max
(d) Line Loading constraints:
(e) L_loading ≤ Lmax_loading
Economic dispatch problem can be defined as to minimize the total generating costs
ng
1n
nFMinimize TF
Subject to
Total demand=sum of all generator generations
ng
1n
nD PP whereng is the number of generators.
Making use of Lagrangian multiplier,
ng
1n
nDT PPλFF where λ is the Lagrangian multiplier.
Problem statement:
Cost equation and loss co-efficient of different units in a plant are given below. Determine
economic generation for total load demand of 240 MW.
Unit No Cost of fuel input in Rs/hr
1 C1=0.05 P12+20 P1+800 0 ≤ P1 ≤ 100
2 C2=0.06 P22+15 P2+1000 0 ≤ P2 ≤ 100
3 C1=0.07 P32+18 P3+900 0 ≤ P3 ≤ 100
Loss coefficients:
B11=0.0005 B12=0.00005 B13=0.0002
B22=0.0004 B23=0.00018 B33=0.0005
B21=B12 B23=B32 B13=B31
Algorithm:
1. Assume a suitable value of λ. This value should me more than the largest intercept of
the incremental production cost of various generators.
2. Calculate the generations based on equal incremental production cost.
3. Calculate the generation at all the buses using the equation
21
nnnn
nm
nommnn
n
2Bλ
F
BPB2λ
f1
P
4. Check if the difference in power at all generator buses between two consecutive
iterations is less than the prescribed value. If not, go back to step 3.
5. Calculate the losses using the relation
oommommn
m n
nL BPBPBPP
6. Also calculate DLG PPPΔP
7. If |ΔP| is less than ε, stop calculation and calculate the cost of generation with these
values of powers.
8. If |ΔP| is more than ε , update the value of λ and go back to step 3.
Procedure :
1. To solve Economic Dispatch by using MiPower Package, invoke “MiPower Tools in
the MiPower main screen and select “ Economic Dispatch by B-coefficient ”
22
2. Select new to create new file
3. Select location to save the file and give the file name
23
4. Enter the values of total demand as 240 MW and No of generators as 3. Select
Generator number as 1 in generator details and enter corresponding values of Pmin,
Pmax, Pscheduled and enter corresponding values of Pmin, Pmax, Pscheduled and corresponding
C0, C1, C2 values.
Similarly enter the values of Pmin, Pmax and Psch and C0, C1 and C2 values for other two
generators
24
5. Enter initial value of Lambda as 5. Enter the values of B11 as 0.0005 and save the
value.
Enter B10=B01=B20=B02=B03=B30=B00=0 and λ= value between 5-25.
Save the values
6. Click and save button to save all values. Now click on execute to run economic
dispatch study.
25
Experiment 9
Load flow analysis using Gauss Seidel method.
Aim: 1. To obtain the load flow analysis solution for a given power system by Gauss Seidel
method using software MiPower.
Theoretical background:
Load Flow analysis, is the most frequently performed system study by electric utilities. This
analysis is performed on a symmetrical steady-state operating condition of a power system
under “normal” mode of operation and aims at obtaining bus voltages and line / transformer
flows for a given load condition. The network consists of a number of buses (nodes)
representing either generating stations or bulk power substations, switching stations
interconnected by means of transmission lines or power transformers. Load Flow analysis is
essentially concerned with the determination of complex bus voltages at all buses, given the
network configuration and the bus demands.
To determine: The complex voltages at all the buses. The steady state of the system is given
by the state vector X defined as
X = (δ1 δ2…… δV1 V2 …….VN)T = (δ
TV
T)T
Classification of Buses:
Slack bus: While specifying a generation 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. This leaves us with the only alternative of
specifying two variables δs and |Vs| pertaining to a generator bus (usually a large capacity
generation bus is chosen and this is called as slack bus) and solving for the remaining (N-1)
complex bus voltages from the respective (N-1) complex load flow equations. Incidentally
the specification of |Vs| helps us to fix the voltage level of the system and the specification of
δs as zero, makes Vs as reference phasor. Thus for the slack bus, both δ and |V| are specified
and PG and QG are to be computed only after the iterative solution of bus voltages is
completed.
P-V buses: In order to maintain a good voltage profile over the system, it is customary to
maintain the bus voltage magnitude of each of the generator buses at a desired level. This
can be achieved in practice by proper Automatic Voltage Regulator (AVR) settings. These
generator buses and other Voltage-controlled buses with controllable reactive power source
such as SVC buses are classified as P-V buses since PG and |V| are specified at these buses.
Only one state variable, δ is to be computed at this bus. The reactive power generation QG at
26
this bus which is a dependent variable is also to be computed to check whether it lies within
its operating limits.
P-Q buses: All other buses where both PI and QI are specified are termed as P-Q buses and
at these buses both δ and |V| are to be computed.
The Gauss Seidel method is an iterative algorithm for solving a set of non-linear load flow
equations. The non-linear load flow equations are given by
n
1pq
qpq
1p
1q
1kqpq1-i*
p
pp
pp
1kp VYVY
V
jQP
Y
1V k
where p=1,2,3….n
Note: 1.The above equation is applicable for load buses since in load bus changes in both
magnitude and phase of voltages are allowed.
2. In generator bus, the magnitude of the voltage remains constant. The above
equation is used to calculate the phase of the voltage.
3. In case of slack bus, the voltage will not change. Hence above equation is not used.
The variables in the equations for p=1,2,3….n are the node voltages V1, V2,V3…..Vn.
1.In Gauss Seidel method, initial voltages are assumed and they are denoted as V10, V2
0,
V30…..Vn
0.
2.Substituting these initial values , V11 is computed.
3.The revised value of bus voltage V11 is replaced for initial value V1
0 and the revised bus-2
voltage V21 is computed.
4.Replace V11for V1
0 and V2
1 for V2
0 and compute V3
1.
5.The iteration process is repeated till the bus voltage converges with prescribed accuracy.
6. For a generator bus, the reactive power is not specified. Estimate the reactive power at
(k+1)th
iteration, using the equation
n
pq
kqpq
1p
1q
1kqpq
*kp
1kp VYVYVIm*)1(Q
7. For generator buses a lower and upper limit for reactive powers will be specified. If the
estimated reactive power of the bus is violates the specified limits, then the reactive power of
the bus is equated to the limit violated and it is treated as load bus.
8. If it does not violate then continue it as generator bus.
9. Compute slack bus power after computing the bus voltages upto the accuracy using the
equation
27
n
1q
qpq*ppp VYVjQP
10 . Line flows are the power fed to the buses into various lines and they are calculated as
shown below. Consider a line connecting bus-p and bus-q. Y
pq/2 Ipq1Ipq2
Ipq
Ypq
Ypq/2
p q
2
YV)YV(VIII
pqppqqppq2pq1pq
Complex power injected by bus-p in line pq
2
YV)YV(VVjQPS
pqppqqp
*ppqpqpq
Similarly complex power injected by bus-q in line –pq
2
YV)YV(VVjQPS
pqqpqpq
*qqpqpqp
Power loss in transmission line –pq is
Spqloss=Spq+Sqp
Acceleration Factor:
Experience has shown that the number of iterations required for convergence can be
considerably reduced if the correction in bus voltage computed at each iteration is multiplied
by a factor greater than unity (termed as acceleration factor) to bring the voltage closer to the
value to which it is converging.
Problem statement:
A single line diagram of a 3 bus power system with generators at bus 1 and 2 are shown
in the Figure given below. The system parameters are given in table(A), load and
generation data are in Table(B). Line impedances are marked in pu on a 100 MVA
base. Line charging suceptances are neglected. The voltage at bus 3 is maintained at
1.04 pu. Considering bus 1 as a slack bus, obtain the load flow solution using Gauss
Seidel method.
28
12
3
Load
Table (A)
Bus code Impedance
1-2 0.02+j0.04
1-3 0.01+j0.03
2-3 0.0125+j0.025
Table (B)
Bus No Bus Voltage Generation Load
MW MVAR MW MVAR
1 1.05+j0 - - 0 0
2 - 0 0 400 250
3 1.04+j0 200 - 0 0
Procedure:
1. Open Power System Network Editor. Select menu option Database → Configure.
Configure Database dialog is popped up as shown below. Click Browse button.
29
2. Open dialog box is popped up as shown below, where you are going to browse the
desired directory and specify the name of the database to be associated with the single
line diagram. Click open button after entering the desired database name. Configure
Database dialog will appear with path chosen.
3. Click OK button on the Configure database dialog. The dialog as shown appears.
Uncheck the Power System Libraries and Standard Relay Libraries. For the problem
chosen, the standard libraries are not needed because all the data is given on pu for
power system libraries ( like transformer, line/cable, generator), and relay libraries are
required only for relay co-ordination studies. If libraries are selected, standard libraries
will be loaded along with the database. Click Electrical Information tab. Since the
impedances are given on 100MVA base, check the pu status. Enter the Base MVA ad
Base frequency as shown below. Click on Breaker Ratings button to give breaker ratings.
Click OK button to create the database to return to Network Editor.
30
Bus Base Voltage Configuration
In the network editor, configure the base voltages for the single line diagram, Select menu
option Configure → Base voltage. The dialog shown below appears. If necessary change the
Base-voltages, color, Bus width and click OK.
Procedure to Draw First Element – Bus
Click on Bus icon provided on power system tool bar. Draw a bus and a dialog appears
prompting to give the Bus ID and Bus Name. Click OK. Database manager with
corresponding Bus Data form will appear. Modify the Area number, Zone number and
contingency Weightage data if it is other than the default values. If this data is not furnished,
keep the default values. Usually the minimum and maximum voltage ratings are ±5% of the
rated voltage. If these ratings are other than this, modify these fields. Otherwise keep the
default values. Bus description field can be effectively used if the bus name is more than 8
characters. If bus name is more than 8 characters, then a short name is given in the name field
and the bus description field can be given as Slack and the bus description field can be Slack.
31
After entering data click save which invokes Network Editor.Follow the same procedure for
remaining buses. Following table gives the data for other buses.
Note: Since the voltages are mentioned in pu, any kV can be assumed. So the base voltage is
chosen as 11kV.
Procedure to Draw Transmission Line
Click on Transmission Line icon provided on power system tool bar. To draw the line, click
in between two buses and to connect to the from bus clicking LMB (Left Mouse Button) on
the From Bus and join it to another bus by double clicking the mouse button on the To Bus.
Element ID dialog will appear.
Enter Element ID number and click OK. Database manager with corresponding Line/cable
Data form will be open. Enter the details of that line as shown below.
32
After entering data Save and close. Line/Cable Data form will appear. Click Save which
invokes Network Editor update next element. Data for remaining elements given in the
following table.
33
Transmission Line Element Data
Transmission Line Library Data
Procedure to Draw Generator
Click on Generator icon provided on power system tool bar. Connect it to bus 1 by clicking
the LMB on Bus 1. The Element ID dialog will appear. Enter ID number and click OK.
Database with corresponding Generator Data form will appear. Enter details as shown
below.
Since generator at bus 1 is mention as slack bus, only specified voltage will have importance.
Note: At Slack bus, only voltage and angle are mentioned. Scheduled power, real power
minimum and maximum constraints do not have much importance.
If the bus is a PV bus (like bus 3), then scheduled power, specified voltage, minimum and
maximum real and reactive power data is must.
34
Enter Manufacturer Ref. No as 1 and click on Generator Library button. Generator library
form will appear.
Procedure To Enter Load Data
Click on Load icon provided on power system tool bar. Connect load 1 at BUS2 by clicking
LMB on Bus2 Element ID dialog will appear. Give ID No as 1 and say OK. Load Data form
will appear. Enter load details as shown below. Then click Save button, which invokes
Network Editor
35
Similarly enter second load data as 60+j25
Solve Load Flow Analysis
Select Menu option Solve → Load Flow Analysis. Following dialog will appear.
When study Info button is clicked, following dialog will open. Select Gauss-Seidel Method
and enter acceleration factor as 1.4 and P-Tolerance and Q-Tolerance as 0.0001. Click OK.
36
Experiment 10
Load flow analysis using Newton Raphson Method.
Aim: 1. To obtain the load flow analysis solution for a given power system by Newton
Rapson Method using software MiPower.
Theoretical background:
The Newton-Raphson method is a powerful method of solving nonlinear algebraic equations.
It works faster and converges faster compared to GS method. The number of iterations
required to obtain a solution is independent of the system size. For a typical bus of the power
system, the current entering bus iis given by
jiji VYI
Expressing the above equation in polar form,
jijjiji δθVYI
The complex power at bus i is
jijjijii
i*iii
δθVYδV
IVjQP
Separating the real and imaginary parts,
)δδsin(θYVVQ
)δδcos(θYVVP
jiijijj
n
1j
ii
jiijijj
n
1j
ii
These equations are nonlinear. Expanding in Taylor’s series about the initial estimate and
neglecting all higher order terms results in following set of linear equations.
(k)n
(k)2
(k)n
(k)2
n
(k)n
2
(k)n
2
(k)n
2
(k)n
43
n
(k)2
2
(k)2
n
(k)2
2
(k)2
n
(k)n
2
(k)n
2
(k)n
2
(k)n
1
n
(k)2
2
(k)2
n
(k)2
2
(k)2
(k)n
(k)2
(k)n
(k)2
VΔ
VΔ
Δδ
Δδ
V
Q
V
Q
δ
Q
δ
Q
JJ
V
Q
V
Q
δ
Q
δ
Q
V
P
V
P
δ
P
δ
P
J2J
V
P
V
P
δ
P
δ
P
ΔQP
ΔQ
ΔP
ΔP
37
Bus 1 is assumed to be slack bus. The Jacobian matrix gives the linearised relationship
between small changes in voltage angle Δδi(k)
and voltage magnitudeΔ׀Vi(k)
with small׀
changes in real and reactive power ΔPi(k)
and ΔQi(k)
. The above matrix can be represented in
its simplified form as
VΔ
Δδ
JJ
JJ
ΔQ
ΔP
43
21
The diagonal and off-diagonal elements of J1 are
ijfor )δδsin(θYVVδ
P
)δδsin(θYVVδ
P
jiijijjij
i
jiijijj
ij
ii
i
The diagonal and off-diagonal elements of J2 are
ijfor )δδ(θ cos YVP
)δδcos(θYVθcosYV2V
P
jiijijii
ij
jiijijjiiijii
i
jV
The diagonal and off-diagonal elements of J3 are
ijfor )δδcos(θ YVVδ
Q
)δδcos(θYVVδ
Q
jiijijjij
i
jiijijj
ij
ii
i
The diagonal and off-diagonal elements of J4 are
ijfor )δδ(θsin YVV
Q
)δδsin(θYVθsinYV2V
Q
jiijijij
i
ij
jiijijjiiijii
i
The terms ΔPi(k)
and ΔQi(k)
are the difference between the scheduled and calculated values
known as power residuals given by
ΔPi(k)
= Pisch
-Pik
ΔQi(k)
= Qisch
-Qik
Calculation of change in bus voltage and angle by
38
1
43
21
k
k
ki
k
JJ
JJ
ΔQ
ΔP
VΔ
Δδ
The new estimate for bus voltages are
(k)i
(k)i
1)(ki
(k)i
(k)i
1)(ki
VΔVV
Δδδδ
The process is continued until the residuals ΔPi(k)
and ΔQi(k)
are less than specified accuracy.
εΔQ
εΔP
(k)i
(k)i
Problem statement:
A single line diagram of a 3 bus power system with generators at bus 1 and 2 are shown
in the Figure given below. The system parameters are given in table(A), load and
generation data are in Table(B). Line impedances are marked in pu on a 100 MVA
base. Line charging suceptances are neglected. The voltage at bus 3 is maintained at
1.04 pu. Considering bus 1 as a slack bus, obtain the load flow solution using Newton-
Raphson methodmethod.
12
3
Load
Table (A)
Bus code Impedance
1-2 0.02+j0.04
1-3 0.01+j0.03
2-3 0.0125+j0.025
39
Table (B)
Bus No Bus Voltage Generation Load
MW MVAR MW MVAR
1 1.05+j0 - - 0 0
2 - 0 0 400 250
3 1.04+j0 200 - 0 0
Algorithm:
Step-1: Choose the initial values of the voltage magnitudes |V| (0)
of all npload buses and n −
1 angles δ (0)
of the voltages of all the buses except the slack bus.
Step-2: Use the estimated |V|(0)
and δ (0)
to calculate a total n − 1 number of injected real
power Pcalc(0)
and equal number of real power mismatch ΔP (0)
.
Step-3: Use the estimated |V| (0)
and δ (0)
to calculate a total np number of injected reactive
power Qcalc(0)
and equal number of reactive power mismatch ΔQ (0)
.
Step-3: Use the estimated |V| (0)
and δ (0)
to formulate the Jacobian matrix J (0)
.
Step-4:Solve forδ (0)
and Δ |V| (0)
÷ |V| (0)
.
Step-5 : Obtain the updates
Step-6: Check if all the mismatches are below a small number. Terminate the process if yes.
Otherwise go back to step-1 to start the next iteration with the updates given by equations.
Procedure to enter the data in MiPower:
1. Open Power System Network Editor. Select menu option Database → Configure.
Configure Database dialog is popped up as shown below. Click Browse button.
40
2. Open dialog box is popped up as shown below, where you are going to browse the
desired directory and specify the name of the database to be associated with the single
line diagram. Click open button after entering the desired database name. Configure
Database dialog will appear with path chosen.
3. Click OK button on the Configure database dialog. The dialog as shown appears.
Uncheck the Power System Libraries and Standard Relay Libraries. For the problem
chosen, the standard libraries are not needed because all the data is given on pu for
power system libraries ( like transformer, line/cable, generator), and relay libraries are
41
required only for relay co-ordination studies. If libraries are selected, standard libraries
will be loaded along with the database. Click Electrical Information tab. Since the
impedances are given on 100MVA base, check the pu status. Enter the Base MVA ad
Base frequency as shown below. Click on Breaker Ratings button to give breaker ratings.
Click OK button to create the database to return to Network Editor.
Bus Base Voltage Configuration
In the network editor, configure the base voltages for the single line diagram, Select menu
option Configure → Base voltage. The dialog shown below appears. If necessary change the
Base-voltages, color, Bus width and click OK.
42
Procedure to Draw First Element – Bus
Click on Bus icon provided on power system tool bar. Draw a bus and a dialog appears
prompting to give the Bus ID and Bus Name. Click OK. Database manager with
corresponding Bus Data form will appear. Modify the Area number, Zone number and
contingency Weightage data if it is other than the default values. If this data is not furnished,
keep the default values. Usually the minimum and maximum voltage ratings are ±5% of the
rated voltage. If these ratings are other than this, modify these fields. Otherwise keep the
default values. Bus description field can be effectively used if the bus name is more than 8
characters. If bus name is more than 8 characters, then a short name is given in the name field
and the bus description field can be given as Slack and the bus description field can be Slack.
43
After entering data click save which invokes Network Editor.Follow the same procedure for
remaining buses. Following table gives the data for other buses.
Note: Since the voltages are mentioned in pu, any kV can be assumed. So the base voltage is
chosen as 11kV.
Procedure to Draw Transmission Line
Click on Transmission Line icon provided on power system tool bar. To draw the line, click
in between two buses and to connect to the from bus clicking LMB (Left Mouse Button) on
the From Bus and join it to another bus by double clicking the mouse button on the To Bus.
Element ID dialog will appear.
Enter Element ID number and click OK. Database manager with corresponding Line/cable
Data form will be open. Enter the details of that line as shown below.
44
After entering data Save and close. Line/Cable Data form will appear. Click Save which
invokes Network Editor update next element. Data for remaining elements given in the
following table.
45
Transmission Line Element Data
Transmission Line Library Data
Procedure to Draw Generator
Click on Generator icon provided on power system tool bar. Connect it to bus 1 by clicking
the LMB on Bus 1. The Element ID dialog will appear. Enter ID number and click OK.
Database with corresponding Generator Data form will appear. Enter details as shown
below.
Since generator at bus 1 is mention as slack bus, only specified voltage will have importance.
Note: At Slack bus, only voltage and angle are mentioned. Scheduled power, real power
minimum and maximum constraints do not have much importance.
If the bus is a PV bus (like bus 3), then scheduled power, specified voltage, minimum and
maximum real and reactive power data is must.
46
Enter Manufacturer Ref. No as 1 and click on Generator Library button. Generator library
form will appear.
Procedure To Enter Load Data
Click on Load icon provided on power system tool bar. Connect load 1 at BUS2 by clicking
LMB on Bus2 Element ID dialog will appear. Give ID No as 1 and say OK. Load Data form
will appear. Enter load details as shown below. Then click Save button, which invokes
Network Editor
47
Sim
ilarl
y enter second load data as 60+j25
Solve Load Flow Analysis
Select Menu option Solve → Load Flow Analysis. Following dialog will appear.
When study Info button is clicked, following dialog will open. Select Newton-Raphson
Method and enter acceleration factor as 1.4 and P-Tolerance and Q-Tolerance as 0.0001.
48
Experiment 11
Short Circuit Fault analysis
Aim: 1. To determine the fault currents and voltages in a single transmission line system with
star-delta transformers at a specified location for
(a) Single line to ground fault
(b) Line to line fault
(c) Double line to ground fault
using software MiPower.
Theoretical background:
Single line to ground fault:
The single line to ground fault on a power system can be represented by connecting three
stubs on the three transmission lines as shown
Ia
Ib
Ic
a
b
c
F
The following relations exist at the fault
Ib=0 Ic=0 Va=0
Therefore for a single line to ground fault,
Ia1=Ia2=Ia0=Ia/3=Vf / (Z1+Z2+Z0)
The above equations indicates that the three sequence networks should be connected in series
through the fault point in order to simulate a single line to ground fault as shown below.
+-
Vpf Z1 Z2 Z0
Va1 Va2 Va0
- +
Ia1
Ia1
Ia2=Ia1
+ - + -
Line to Line fault:
The line to linefault on a power system can be represented by connecting three stubs on the
three transmission lines as shown
49
Ia
Ib
Ic
a
b
c
F
The following relations exist at the fault
Ib=-IcIa=0
Therefore for a line to line fault
Va1=Va2
)Z(Z
VI
21
pfa1
The above equations indicates that the sequence networks should be connected in
parallelthrough the fault point in order to simulate a line to linefault as shown below.
Vp
f
Z1
Z2
Va
1
Va
2
-
+
Ia1
Ia2
Ia1=-Ia2
+
-
+
-
Double Line to ground fault:
The line to double line to ground fault on a power system can be represented by connecting
three stubs on the three transmission lines as shown
50
Ia
Ib
Ic
a
b
c
F
If
The following relations exist at the fault
Vb=Vc =0 Ia=0
Therefore for a line to line fault
Va1=Va2=Va0
)Z(Z
VI
21
pfa1
The above equations indicate that the sequence networks should be connected in parallel
through the fault point in order to simulate a double line to ground fault as shown below.
Vp
f
Z1
Z2
Va1
Va2
-
+
Ia1
Ia2
+
-
+
-
Va0
Ia0
+
-
Z0
Problem statement:
For the given single transmission line system, find the fault currents and voltages for the
following types of fault at bus 3.
(a) Single line to ground fault
(b) Line to line fault
(c) Double line to ground fault
51
For the transmission line, assume X1=X2 and X0=2.5 XL. Take MVA rating as 100, base
voltage=220, Ea=1p.u
G
Bus 1 Bus 2 Bus 3
11 kV 220 kV 220 kVxd
'=0.1 xt=0.1xL=0.4
The sequence diagram is as shown below
0.1 0.1 0.1 0.1 0.10.4 0.4
Positive sequence Negative sequence Zero sequenceF+ F-
Fz
0.1 1
Calculations:
Single line to ground fault:
aa0a2a1
3I
1III
.impedancefaulttheisZwhereZ
E
3Z)ZZ(Z
EI f
eq
a
f021
aa1
Z1+Z2+Z0=Zeq (Zf=0)
eq
aa1
Z
EI
Ia=3*Ia1inp.u
KVBase3
KVABasecurrentBase
Actual current Ia=Ia in p.u*Base current
Fault MVA= Base voltage *fault current in p.u
The sequence voltages at the fault point are
Va1=1-Z1 Ia1=
Va2=-Z2Ia2
Va0=-Z0Ia0
The lines to neutral voltages at the fault point are
52
a2
a1
a0
2
2
c
b
a
V
V
V
aa1
aa1
111
V
V
V
all in p.u
Thus actual values of line to neutral voltages at the fault point are
p.u)in(V3
BaseKVVp.u)in(V
3
BaseKVV0V ccbba
Line-Line fault:
Ic=-IbIa=0
p.uincurrentfault)Z(Z
E3
)ZZ(Z
E3II
21
a
f21
acb
KVBase3
KVABasecurrentBase
Actual fault current = Base current*fault current in p.u
Fault MVA= Base voltage *fault current in p.u
)Z(Z
EI
21
aa1
Ia2= -Ia1
Va0=0
Va1=Va2
Thus line to neutral voltages at the fault point are
a2
a1
a0
2
2
c
b
a
V
V
V
aa1
aa1
111
V
V
V
Thus actual values of line to neutral voltages at the fault point are
p.u)in(V3
BaseKVVp.u)in(V
3
BaseKVV0V ccbba
Double Line to ground fault:
)3ZZ(Z
)3Z(ZZZ
EI
f02
f021
aa1
2
a11
2
aa2
Z
IZ
Z
EI
53
0
a11
0
aa0
Z
IZ
Z
EI
Ia=Ia1+Ia2+Ia0
a0
a2
a1
2
2
c
b
a
I
I
I
aa1
aa1
111
I
I
I
Fault MVA in phase B= Base voltage *fault current Ib in p.u
Fault MVA in phase C= Base voltage *fault current Ic in p.u
KVBase3
KVABasecurrentBase
Actual sequence currents = Sequence Currents in p.u*Base currents
Actual currents = Currents in p.u*Base currents
The sequence voltages are
Va1=Va2=Va0=1-Ia1Z1
Thus line to neutral voltages at the fault point are
a2
a1
a0
2
2
c
b
a
V
V
V
aa1
aa1
111
V
V
V
Thus actual values of line to neutral voltages at the fault point are
p.u)in(V3
BaseKVVp.u)in(V
3
BaseKVV0V ccbba
Procedure to enter the data in MiPower:
Open Power System Network Editor. Select menu option Database → Configure. Configure
Database dialog popped up. Click Browse button.
MiPower Database Configuration:
54
Open dialog box is popped up as shown below where you are going to browse the desired
directory and specify the name of the database to be associated with the single line diagram.
Click Open button after entering the desired name. Configure Database dialog will appear
with path chosen.
Click OK button in the Configure database dialog, the following dialog appears
55
Uncheck the Power System Libraries and Standard Relay Libraries. For this example these
standard libraries are not needed because all the data is given on pu for power system
libraries (like transformer, line/cable, generator) and relay libraries are required only for relay
co-ordination studies. If libraries are selected, standard libraries will be loaded along with the
database. Click Electrical Information tab. Since the impedances are on 100 MVA base,
check the pu status. Enter the Base MVA and Base frequency as shown below. Click
Breaker Ratings tab. If the data is furnished, modify the breaker ratings for required voltage
levels. Otherwise accept the default values. Click OK button to create the database to return
to Network Editor.
Bus Base Voltage Configuration
In the Network Editor, configure the base voltages for the single line diagram. Select menu
option Configure → Base Voltage. Dialog shown below appears. If necessary change the
Base-voltages, color, Bus width and click OK.
56
Procedure to Draw First Element – Bus
Click on Bus icon provided on power system tool bar. Draw a bus and a dialog appears
prompting to give Bus ID and Bus Name. Click OK. Database manager with corresponding
Bus Data form will appear. Modify the Area number, Zone number and Contingecy
Weightage data if it is other than the default values. If this data is not furnished, keep the
default values. Usually the minimum and maximum voltage ratings are ±5% of the rated
voltage. If these ratings are other than this, modify these fields. Otherwise keep the default
values.
Bus description field can be effectively used if the bus name is more than 8 characters. If bus
name is more than 2 characters, then a short name is given in the bus name field and bus
description field can be used to abbreviate the bus name. For example let us say the bus name
is Northeast then bus name can be given as NE and bus description field can be North East.
57
After entering data click Savewhich invokes Network Editor. Follow the same procedurefor
remaining buses. Following table gives the data for other buses.
Procedure to Draw Transmission line
Click on Transmission Line icon provided on power system tool bar. To draw the line, click
in between the buses and to connect to the from bus, double click LMB(Left Mouse Button)
on the From Bus and join it to another bus by double clicking the mouse button on the To
Bus Element ID dialog will appear.
58
Enter Element ID number and click OK. Database manager with corresponding Line/cable
Data form will open. Enter the details of that line as shown below.
Enter Structure Ref No as 1 and click on Transmission Line Library>> button. Line &
Cable Library form will appear. Enter transmission line library data.
59
After entering data, Save and close. Line/ Cable Data form will appear. Click Save which
invokes Network Editor.
Procedure to Draw Transformer
Click on Two Winding Transformer icon provided on power system tool bar. To draw the
transformer click in between two buses and to connect to the from bus, double click
LMB(Left Mouse Button) on the From Bus and join it to another bus by double clicking the
mouse button on the To Bus. Element ID dialog will appear. Click OK
60
Transformer library form will be open. Enter the data as shown below. Save and close library
screen. Transformer element data form will appear. Click Save button, which invokes
Network Editor.
Procedure to Draw Generator
Click on Generator icon provided on power system tool bar. Draw the generator by clicking
LMB(Left Mouse Button) on the Bus 1. Element ID dialog will appear. Click OK.
61
Generator Data form will be opened. Enter the Manufacture Ref. Number as 1. Enter
Generator data in the form as shown below.
62
Click on Generator Library >>button. Enter generator library details as shown below
Save and Close the library screen. Generator data screen will be reopened. Click Save button
and invokes Network Editor.
Note: To neglect the transformer resistance, in the multiplication factor table give the X to R
ratio as 9999.
To solve short circuit studies choose menu option Solve → Short Circuit Analysis or click
on SCS button on the toolbar on the right side of the screen. Short circuit analysis screen
appears.
63
Study Information:
In Short Output Options select the following
64
Afterwards click Execute. Short circuit study will be executed. Click on Report to view the
report file.