canara 10EEL78_PSS_2013_Manual.pdf

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)


mhoCjωY

Cjω

1X




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


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)


mhoCjωY

C jω

1X

C/2CC




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


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.


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.


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


ijfor )δδ(θ cos YVP

)δδcos(θYVθcosYV2V

P

jiijijii

ij

jiijijjiiijii

i

jV


ijfor )δδcos(θ YVVδ

Q

)δδcos(θYVVδ

Q

jiijijjij

i

jiijijj

ij

ii

i


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.


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


42



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


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


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.


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


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.


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


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


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


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


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.


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


56



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


Procedure to Draw Transmission line


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


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.


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.

canara 10EEL78_PSS_2013_Manual.pdf

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