ee76 power system simulation lab

EINSTEIN COLLEGE OF ENGINEERING Sir.C.V.Raman Nagar, Tirunelveli-12

Department of Electrical and Electronics Engineering

Subject Code: EE76

Power System Simulation Lab

Name : ……………………………………

Reg No : ……………………………………

Branch : ……………………………………

Year & Semester : ……………………………………

EE76 Power System Simulation Lab

©Einstein College of Engineering

of 57

TABLE OF CONTENTS

S.No Date Name of the Experiment Page No

Marks Remarks Staff Initial

1 Computation of Line Parameters

2 Formation of Bus Admittance and Impedance Matrices and Solution of Networks

3 Solution of Load Flow and Related Problems Using Gauss-Seidel Method.

4 Solution of Load Flow and Related Problems Using Newton-Raphson Method

5 Solution of Load Flow and Related Problems Using Fast Decoupled Methods

6 Symmetrical Fault Analysis

7 Unsymmetrical Fault Analysis

8 Economic Dispatch in Power Systems

9 Load Frequency Dynamics of Single and Two Area Power System

10 Electromagnetic Transients in Power Systems

11 Transient and Small Signal Stability Analysis of Single Machine Infinite Bus System

12 Transient Stability Analysis of Multi- Machine System


of 57 ©Einstein College of Engineering

COMPUTATION OF LINE PARAMETERS

AIM: To determine the positive sequence line parameters L and C per phase per kilometer of a three phase single and double circuit transmission lines for different conductor arrangements . Objectives: i) To understand the modeling and performance of short, medium and long Transmission lines. ii) To write a MATLAB program to determine the Line parameters Software required:

MATLAB 6.1

Formula: 1. Single Phase - Two Wire System

GMD = D GMR = re-1/4 = r’

r = radius of conductor 2. Three Phase - Symmetrical Spacing

GMD = D

GMR = re-1/4 = r’

r = radius of conductor

3. Three Phase - Asymmetrical Transposed GMD = (DAB DBC DCA)1/3

GMR = re-1/4 = r’

r = radius of conductors 4. Composite Conductor Lines The inductance of composite conductor - x., is given by

0.2 lnxx

GMDLGMR

Where

GMD = [(Daa’ Dab’ … Dam’ ) …… (Dna’ Dnb’….. Dnm’ )]mn GMRx = [(Daa Dab … D an) …… (Dna Dnb….. D nn)] n2 r’a = ra.e-1/4

5. Bundle Conductors: GMR for two sub conductor Dsb = [Ds x d]1/2

GMR for three sub conductor Dsb = (Ds x d2)1/3

GMR for four sub conductor Dsb = 1.09 (Ds x d3)1/4

Where



Ds is the GMR of each sub conductor

d is the bundle spacing

6. Capacitance:

GMR= rb for bundled conductors

rb=[r*d2]1/3 for 3 conductor bundle

7. Three-phase – Asymmetrical - transposed : GMD = [DAB DBC DCA] 1/3

GMR = r ; for solid conductor

GMR = Ds for stranded conductor

= rb for bundled conductor where

rb = [r*d]1/2 for 2 conductor bundle

rb = [r*d2]1/3 for 3 conductor bundle (1.20)

rb = 1.09 [r*d3]1/4 for 4 conductor bundle

Where

r = radius of each sub conductor

d = bundle spacing

7. Three-phase – double circuit - transposed :

GMRc = [rA rB rC]1/3

Where rA rBand rC are GMR of each phase group obtained as rA =[rb Da1az]1/2

rB =[rb Db1bz]1/2

rC =[rb Dc1cz]1/2

Where

rb =GMR of for bundled conductor

EXERCISES: 1) A single phase line has two parallel conductor 2meters apart. the diameter of each

conductor is 1.2 cm. calculate the loop inductance per km of the line.

2) A single phase transmission line has two parallel conductor 3m apart, the radius of each conductor being 1cm.Calculate the loop inductance per km length of the line if the material of the conductor is(i)Copper(ii)Steel with relative permeability of 100.



3) Find the inductance per km of a 3phase transmission line using 1.24cm diameter conductors when there are placed at the corners spacing of an equilateral triangle of each side 2m.

4) The 3 conductors of a 3phase line are arranged at the corners of a triangle of sides 2m,2.5m&4.5m.Calculate the inductance per km of the line when the conductors are regularly transposed. The diameter of each conductor is 1.24cm

5) Two conductors of a single phase line ,each of 1cm diameter, are arranged in a vertical plane with one conductor mounted 1m above the other. A second identical line is mounted at the same height as the first and spaced horizontally 2.5m apart from it. The two upper &the two lower conductors are connected in parallel. Determine the inductance per km of the resulting double circuit line.

6) The spacing of a double circuit 3phase overhead line. The phase sequence is ABC & the line is completely transposed, the conductor radius in 1.3 cm .Find the inductance/phase/km.

7) A single phase line has two parallel conductors 3m apart, radius of each conductor being 1cm. Calculate the capacitance of the line /km. given that £0 =8.854*10-12 F/M.

RESULT:



FORMATION OF BUS ADMITTANCE AND IMPEDANCE

MATRICES AND SOLUTION OF NETWORKS. Aim: To understand the formation of bus admittance matrix(Ybus) of a given power system network, to effect certain required changes on this matrices & to obtain network solution using this matrices. Objectives:

1. To write a program in MATLAB to determine the bus admittance matrix(Ybus) 2. To obtain the modified Ybus to effect specified modifications in the configuration of

the network. 3. To determine the bus impedance matrices Zbus

Software Required: MATLAB

Formulae: Y.V = I Z.I = V

TWO RULE METHOD FOR YBUS FORMATION: Rule 1:

Matrix form: Ybus=

Y11 Y12 Y13 Y21 Y22 Y23 Y31 Y32 Y33

Where, Y11 = y11 + y12 +y13

Y22 = y11 + y12 +y13 Y33 = y11 + y12 +y13 Y12 = -y12 Y13 = -y13 Y23= Y32

Ykk = y/a2 ; Ymm = y ; Ykm = Ymk = -y/a

Algorithm: Step 1: Initialize Y with all elements set to zero. Step 2: Read the line list, one line at-a-time and update Y by adding the respective



Contribution. Step 3: Read the transformer list, one transformer at-a-time and update Y by adding the

respective contribution. Step 4: Read the shunt element list, one element at-a-time and update Y by adding the

respective contribution. EXERCISES:

1. Find the Y bus for the given power system. The impedances are Z12=j0.4 p.u; Z13=j0.3 p.u and Z23=j0.2p.u.

2.to find the bus admittance matrix given the admittance value.

Line Admittance

1-2

1-3

2-3

2-4

3-4

2-j8

1-j4

0.666-j2.664

1-j4

2-j8

3.To find bus admittance value given the impedance value.

Line R(p.u) X(p.u)

1

3

2



1-2

1-3

1-4

2-4

3-4

0.05

0.10

0.20

0.10

0.05

0.15

0.30

0.40

0.30

0.15

Result:



SOLUTION OF LOAD FLOW AND RELATED PROBLEMS USING GAUSS-SEIDEL METHOD.

Aim: To understand, in particular, the mathematical formulation of load flow model in complex form& a simple method of solving flow problems of small sized system using gauss seidal iterative algorithm. Objectives:

1. To build mathematical model of load flow problem using Gauss-Seidel method. 2. To write MATLAB program to solve the load flow problem by using Gauss-Seidel

method.

Software Required : MATLAB 6.1 Formulae:

n

ij

kij

i

j

kijk

kii

ii

kjj

i

iVYVY

VjQP

YV

1

1

1

1*

11 ..1

n

ij

kij

i

j

kij

ki

kjji

VYVYValP ...Re1

1

1*1

n

ij

kij

i

j

kij

ki

kjji

VYVYVaQ ...Im1

1

1*1

Algorithm for GSLF: Step1: Read the input data Step2: Find the admittance matrix Step3: Choose the flat voltage profile 1+j*0 Step4: set the iteration count p=0 and bus count i=1 Step5: check the slack bus, if it is the generator bus then go to the next step otherwise go to step7. Step6: Before the check for the slack bus if it is slack bus then go to step11 otherwise go to next step. Step7: Check the reactive power of the generator bus within the given limit. Step8: If the reactive power violates a limit then treat the bus as load bus. Step9: Calculate the phase of the bus voltage on load bus. Step10: Calculate the change in bus voltage of the repeat step mentioned above until all the bus voltages are calculated. Step11: Stop the program and print the results. Exercise: 4-BUS, 5-LINES POWER SYSTEM The systemdata for a load flow solution are given in table. Determine the voltages at the end of first iteration by gauss seidal method. Given α=1.6.



Line Admittance

1-2

1-3

2-3

2-4

3-4

2-j8

1-j4

0.666-j2.664

1-j4

2-j8

Bus specification:

Bus code P Q V Remarks

1

2

3

4

-

0.5

0.4

0.3

-

0.2

0.3

0.1

1.06<0”

-

-

-

Slack

PQ

PQ

PQ

Result:



SOLUTION OF LOAD FLOW AND RELATED PROBLEMS USING NEWTON-RAPHSON METHOD

Aim: To develop a software program to obtain real and reactive power flows,bus voltage magnitude and angles by using N-R method. Objectives:

1. To procure the knowledge of steady state analysis of power flow in a power system. 2. To build mathematical model of load flow problem using Newton Raphson(NR) method

and Fast Decoupled method. 3. To become proficient in the usage of software for practical problem solving in the areas

of power system planning and operation. 4. To become proficient in the usage of the software in solving problems using Newton-

Raphson and Fast Decoupled load flow methods.

Software Required: MATLAB Formulae:

Real Power, )cos(.||.|||| ijijijjii YVVP

Reactive Power, )sin(.||.|||| ijijijjii YVVQ

Algorithm: Step1: Form the admittance(Ybus) matrix Step2: Assume initial values of bus voltages|Vp|0 and phase angles p for load buses and phase

angles for PV buses.Normally we set the assumed bus voltage magnitude and its phase

angle equal slack bus quantities|V1|0= 1.0, 1 =00

Step3: Compute Pp and Qp for each load bus using formulae’s Step4: Compute the scheduled errors ∆Pp and ∆Qp for each load bus from the following

relations.

∆Ppk=Ppsp-Pkpcal p=2,3…..n

∆Qpk= Qpsp-Qkpcal p=2,3…..n For PV buses , the excat value of Qp is not specified, but its limits are known. If the

calculated value ofQp is within limits, only ∆Pp is calculated. If the calculated value of

Qp is beyond the limits, then an appropriate limit is imposed and ∆Qp is also calculated

by subtracting the calculated value of Qp from the appropriate limit. The bus under

consideration is now treated as a load on(PQ) bus.

Step5: Compute the elements of the jacobian matrix using the estimated |Vp| and p from step 2. Step6:Obtain ∆ and ∆|Vp| from eqns calculated in step 6,modify the voltage magnitude and



phase angle at all loads by th

Step7: using the values of ∆ p and ∆|Vp| in step 6,modify the voltage magnitude and phase

angle at all loads using jacobian matrix. Start the next itration cycle at step2 with thse

modified p and |Vp|

Step8: Continue until scheduled errors ∆Ppk and ∆Qpk for all load buses are within a specied

tolerance, ie, ∆Ppk <€, ∆Qpk<€. Where € denotes the tolerance level for load buses

Step9: Calculate line flows and power at the slack bus exactly in the same manner as in the gauss

seidal method.

Exercise: The load factor data for the sample power system are given below. The voltage magnitude at bus 2 is maintained at 1.04 p.u. the maximum and minimum reactive power limits of the generator at bus 2 are 0.35 and 0 p.u respectively. Determine the set of load flow equations of the end of first iteration by using NR method. V2=1.04 p.u; 0≤Q2≤0.35 Bus code Impedance 1-2 1-3 2-3

0.08+j0.24 0.02+j0.06 0.06+j0.18

Bus code Assumed voltage Generation Load MW MVAR MW MVAR

1 2 3

1.06+j0 1+j0 1+j0

0 0.2 0

0 0 0

0 0 0.65

0 0 0.25

RESULT:



SOLUTION OF LOAD FLOW AND RELATED PROBLEMS USING FAST DECOUPLED METHODS

Aim: To develop a software program to obtain real and reactive power flows,bus voltage magnitude and angles by using N-R method. Objectives:

1. To procure the knowledge of steady state analysis of power flow in a power system. 2. To build mathematical model of load flow problem using Newton Raphson(NR) method and Fast Decoupled method. 3. To become proficient in the usage of software for practical problem solving in the areas of power system planning and operation. 4. To become proficient in the usage of the software in solving problems using Newton- Raphson and Fast Decoupled load flow methods.

Software Required: MATLAB 6.1 Formulae:

Real Power, )cos(.||.|||| ijijijjii YVVP

Reactive Power, )sin(.||.|||| ijijijjii YVVQ

Algorithm: Step1: Read the slack bus voltages, real bus powers and reactive bus powers,bus voltage magniyude and reactive power limits; Step2: Form the admittance(Ybus) matrix without line charging admittance and shunt admittance. Step3:Form B matrix , from Ybus matrix obtained in step 2. Step4:Form Ybus matrix with double line charging admittance. Step5: Form B’’ matrix , from Ybus matrix obtained in step 4. Step6: Calculate the inverse of B’& B’’ matrices. Step7:Initialize the bus voltages. Step8:Calculate [∆P/|V|],[∆Q,|V|] Step9:If ∆P/|V| & ∆Q/|V| are less than or equal to tolerance limit, solution has convergence and go to step 12 otherwise increase iteration count and go to step 10. Step 10: Calculate[∆ δ]=[B’]-1[∆P/|V|] [∆|V|]=[B’’]-1[∆Q/|V|] Step11:Update [ δ]& [|V|] for all buses except slack bus. [ δ]new= [ δ]old+ [∆ δ] [|V|]new=[|V|]old+[∆|V|] Exercise: The load factor data for the sample power system are given below. Taking bus 1 as slack bus . Determine the Voltage at various buses.FDLF at the end of first iteration. V2=1.04 p.u; 0≤Q2≤0.35



Bus code Impedance Half line charging admittance 1-2 1-3 2-3

0.06+j0.18 0.02+j0.06 0.04+j0.12

j0.05 j0.06 j0.05

Bus code

Assumed voltage Generation Load MW MVAR MW MVAR

1 2 3

1.06+j0 1+j0 1+j0

0 0.2 0

0 0 0

0 0 0

0 0 0.25

RESULT:



SYMMETRICAL FAULT ANALYSIS Aim: To develop a software program to carry out simulation study of a symmetrical three phase short circuit on a given power system. Objectives: 1. To become familiar with modelling and analysis of power systems under faulted

condition and to compute the fault level, post-fault voltages and currents for different types of faults, both symmetric and unsymmetrical.

2. To calculate the fault current, post fault voltage and fault current through the branches for a three phase to ground fault in a small power system and also study the effect of neighboring system. Check the results using available software Software Required: MATLAB6.1 Formulae: Fault current,If=V/(Zf+Zpp) Fault voltage Vf =V(1-(Zbus/(Zf+Zpp)) Where Zf=Fault impedance Zpp= Line impedance Algorithm: Step1: Read line data,machine data, transformer data, fault impedance etc. Step2: Form the admittance[Ybus] matrix and calculate [Ybus]modi. Step3:Form [Zbus] by inverting the[Ybus] modified. Step4:Initialize count I=0.. Step5: Find the bus at which fault occurs I=I+1. Step6: Compute fault current at faulted bus and bus voltage at all buses.. Step7: Compute all line and generator currents. Step8:Check if I<number of buses, if yes go to step 5 else go to step 9. Step9:Print the results and stop the program. Exercise: 1)Consider a 4 bus system fault occurs on bus4. Where 1&2 are generator buses ,3&4 are load buses. Find fault current and fault voltage & line currents

Bus code Impedance Half line charging admittance

1-2 1-3 1-4 2-3 2-4

j5 j0.15 j10 j0.1 j0.15

j0.1(bus1) j0.1(bus2)

2) A generator is connected through a transformer to a synchronous motor. The sub transient reactance’s of generator and motor are 0.15 & 0.35 respectively. The leakage reactance of the



transformer is 0.1 p.u. All the reactance are calculated on a common base. A 3phase fault occurs at the terminals of the motor when the terminal voltage of the generator is is 0.9 p.u. The output current of generator is 1p.u. &0.8 pf leading. Find the sub transient current in p.u. in the fault, generator & motor. Use the terminal voltage of generator as reference vector. RESULT:



UNSYMMETRICAL FAULT ANALYSIS

Aim: To calculate the sub transient current in the faulted phase when unsymmetrical fault taken place. Objectives:

1. To become familiar with modeling and analysis of power systems under faulted condition and to compute the fault level, post-fault voltages and currents for different types of faults, both symmetric and unsymmetrical.

2. To calculate the fault current, post fault voltage and fault current through the branches for a three phase to ground fault in a small power system and also study the effect of neighboring system. Check the results using available software. Software Required: MATLAB Formulae:

i) single line to ground fault:

i)321

1 ZZZEI a

a ;

ii)b

bb MVA

KVI2

iii) fbactual IIIf *

iv) 3/1

b

b

KVKVAIb

ii)Line to line fault

i) 21

1 ZZEI a

a

ii) 210 IaIaIaIa iii) 1211110 ;;0 VaVaZIaEaVaVa iv) VbVcaVaVaaVaVbVaVaVaVa ;; 21

20210

v) ;;; VaVcVcaVcVbVbcVbVaVab phase value or bas voltage=(KVb)1/3

iii)Double line to ground fault: i) 011 IaIaIa

ii)

02

021

1

ZZZZ

Z

EaIa

iii) 1101 ZIaEaVaVa



iv)1

22 Z

VaIa

Algorithm: Step1: Read line data,machine data, transformer data, fault impedance etc. Step2: Find the line at which fault occurs. Step3: Compute fault current at faulted lines. Step4:Print the results and stop the program Exercises:

1) A salient pole generator without damber is rated 20MVA, 13.8KV and has a reactance of 0.25 p.u the subtransient. The negative zero sequence reactance are 0.35&0.10 p.u. The neutral of the generator is solidly grounded. Determine the sub transient current in the generator and the line to line voltages for the sub transient conditions when a single line to ground fault occurs at the generator terminals with generator operating at rated voltage. Neglect resistance.

2) Find the sub transient currents &line to line voltages at the fault under sub transient conditions when a line to line fault between phases b&c occurs at the terminals of the generator. Assume that the generator is unloaded and operating at rated terminal when the fault occurs. Neglect resistances.

3) Find the sub transient currents &line to line voltages at the fault under sub transient conditions when a double line to line fault between phases b&c occurs at the terminals of the generator. Assume that the generator is unloaded and operating at rated terminal when the fault occurs. Neglect resistances.

RESULT:



ECONOMIC DISPATCH IN POWER SYSTEMS Aim:

(i). To understand the basics of the problem of Economic Dispatch (ED) of optimally adjusting the generation schedules of thermal generating units to meet the system load which are required for unit commitment and economic operation of power systems.

(ii). To understand the development of coordination equations (the mathematical model for ED) without and with losses and operating constraints and solution of these equations using direct and iterative methods

Objectives: 1. Determine the economic generation schedule of each unit and incremental cost of

received power for a sample power system, for a given load cycle.

2. Determine transmission loss for a sample system, for the given load levels.

Software required: MATLAB 6.1

Theory: The constant is called the incremental cost in Rs/MWhr.

The characteristic equation is C=a+bP+cP2

1

1

dPdC =

2

2

dPdC =……………

n

n

dPdC =

Optimum generation as a function of and cost coefficient

Let C1 = a1+b1P1+c1P12

C2 = a2+b2P2+c2P22

….. …………

Ck = ak+bkPk+ckPk2

1

1

dPdC = = b1+2P1c1

2

2

dPdC = = b2+2P2c2 ………….. (5)

…………..…………..

k

k

dPdC = = bk+2Pkck …………..(6)



Rearranging = 1

1

C +

2

1

C + ……. +

Ck

1 = 1

1

cb +

2

2

C

b +…..+ k

k

cb +

2(P1+P2+….Pk)……… (7)

=

k

n nc1

1 =

k

n n

n

cb

1

+ 2PT …………..(8)

=

k

n n

T

k

n

c

P

1

1 n

n

1

2cb

Pn = n

n

cb

2

Iterative Procedure: The incremental transmission loss depends on the power output from all the units.

The co-ordination equation is k

k

dPdC +

K

L

PP

= ……..(1)

The co-ordination equation cannot be solved directly.An iterative procedure including a method

of successive approximation for the nth unit is given by n

n

dPdC +

n

L

PP ……(2)

C1 = a1+b1P1+c1P12 ; Ck = an+bnPn+cnPn

2

n

n

dPdC = = bn+2Pncn ……………….(3)

Substituting (3) in (1), bn + 2Pncn + K

L

PP

But K

L

PP =

KP

k

m

k

nnmnm PBP

1 1

m n

Pn(2cn + 2 Bnn) = -bn- nm

mnmBP2



Divide by => Pn = nnn

mnm

Bc

BP

22

2-bn-

Pn = nn

n

nmmnm

Bc

BP

22

2-bn-1

………(4)

For a system with 2 units of power output P1 and P2,

P1 = 11

1

1221

22

2-b-1

Bc

BP

………(5)

P2 = 22

2

2112

22

2-b-1

Bc

BP

………(6)

Steps: 1. Assume P2=0

2. Find P1 from eqn(5)

3. Find P2 from eqn(6) using the value of P1 from prevo\ious steps

4. Substitute P2 in eqn(5)

5. Repeat steps(3) and (4) one after the other

Algorithm for iterative process:- Step 1: Get the value of Ng, cost coefficients, B-Coefficients ,total demand, tolerance limit, initial iteration value

Step 2: Find the value assuming no losses using formula =

Ng

n i

D

Ng

n

c

P

1

1 i

i

1

2cb

Step 3: Calculate the power using Pi(k) = iii

i

Bkbbk))((2

)(

Step 4: Find losses using PL(k) =

Ng

iiii kPB

1

)(2



Step 5: Find power mismatch using .P(k) = PD + PL -

Ng

ii kP

1

)(

Step 6: Check the power mismatch P(k).Find if it is less than or equal to tolerance limit. If yes

end the process else find the value of change in using .k

iP

=

2))((2 iii

iiii

BkcBbc

)(k = kNg

i

iPkP

1

)(

Step 7: Find the value of new

(k+1) = (k) + (k)

Step 8: Substitute equal to new value and repeat from step 2.

Formulae used: Neglecting losses:

1. =

k

n n

T

k

n

c

P

1

1 n

n

1

2cb

2. Pn = n

n

cb

2

Including Losses:

1. Pi(k ) = iii

i

Bkbbk))((2

)(

2. PL (k) =

Ng

iiii kPB

1

)(2

3. .P(k) = PD + PL -

Ng

ii kP

1

)(

4. k

iP

=

2))((2 iii

iiii

BkcBbc

5. )(k = kNg

i

iPkP

1

)(



6. (k+1) = (k) + (k)

Exercises (1) The power plant has the following characteristics

F1=0.00889P12+10.333P1+200 Rs/hr

F2=0.00741P22+10.833P2+240 Rs/hr

Determine the economic schedule to meet the demand of 150MW and transmission loss PL=0.001P1

2+0.0002P22-2x0.0002P1P2 using iterative techniques.

(2) The power plant has the following characteristics

F1=0.05P12+21.5P1+800 Rs/hr

F2=0.1P22+27P2+500 Rs/hr

F3=0.07P32+16P3+900 Rs/hr

Determine the economic schedule to meet the demand of 150MW.

(3)

Bmn =

Three plants A,B,C supply powers of 50 MW,100MW & 200 MW respectively. Calculate the transmission loss in the network in p.u. value and incremental transmission loss of the three plants. Assume base value=200MW.

Result:

0.01 -0.0003 -0.0002

-0.0003 0.0025 -0.0005

-0.0002 -0.0005 0.0031



LOAD FREQUENCY DYNAMICS OF

SINGLE AND TWO AREA POWER SYSTEM

AIM:

To obtain the frequency response of single and two area power system using MATLAB

SOFTWARE REQUIRED:

MATLAB 6.1

FORMULA USED:

1)GG = KG

1+ STG

2)GT = KT

1+ STT

3)GP = KP

1+ STP

4)Kp=1/D

Where

D = damping coefficient Where GG – gain of generator

GT - gain of turbine

GP - gain of power

KP – power system constant

KT- turbine constant

KG- generator constant

TP – power system time constant

TG- generator time constant



SIMULINK RESULTS:

Single Area Power System

BLOCK DIAGRAM

OUTPUT RESPONSE:

:



SINGLE AREA POWER SYSTEM BLOCK DIAGRAM:

OUTPUT RESPONSE:



TWO AREA POWER SYSTEM:

BLOCK DIAGRAM:

OUTPUT RESPONSE (1):



OUTPUT RESPONSE (2):

Result:



ELECTROMAGNETIC TRANSIENTS IN POWER SYSTEMS

AIM:

To plot the electro magnetic transients in the power system using MATLAB

SOFTWARE REQUIRED:

MATLAB 6.1

FORMULA USED:

Case(i):

REACTIVE TERMINATION:LINE TERMINATED BY INDICATOR:

it(s ) = 2E f /L

s(s+Zc/L)

et(s) = 2Ef

(s+Zc/L)

it(s) = 2E (1-e –Zct/L)

Zc

ir = E (1-2e-Zct/L)

Zc

Case (ii):LINE TERMINATED BY CAPACITANCE: it(s) = 2E (1/S+I/ZcC )



Zc

Case (iii):LINE TERMINATED BY A RESISTANCE EQUAL TO SURGE IMPEDANCE: it(s) = 2E

ZcS

ef(s) = E

Case (iv):OPEN CIRCUITED LINE: et = 2ef

er = ef

ir = - if

Case (v):SHORT CIRCUITED LINE:

it = 2if

er = -ef

ir = if

Where,

ef , if - forward voltage and current

et , it - transmitted voltage and current

er , ir - reflected voltage and current

PROGRAM: %line terminated by inductor for voltage Ef=10000;

L=0.004;

Zc=400;

n1=[2*Ef0];

d1=[1 Zc/L];

t=0:0.00001:0.0001;

Et=step(n1,d1,t); plot(t,Et,'r');

Er=Et-Ef;

hold on;

plot(t,Er,'b');



OUTPUT RESPONSE:

%line terminated by capacitor for voltage

Ef=10000;

C=0.000000009;

Zc=400;

n1=[2*Ef/(Zc*C)];

d1=[1 1/(Zc*C)];

t=0:0.00001:0.0001;

Et=step(n1,d1,t);

plot(t,Et,'r');

Er=Et-Ef;

hold on; plot(t,Er,'b');



OUTPUT RESPONSE:

%line terminated by capacitor for current Ef=10000;

C=0.000000009;

Zc=400;

n1=[2*Ef/Zc 0)]; d1=[1 1/(Zc*C)];

t=0:0.00001:0.0001;

It=step(n1,d1,t);

plot(t,It,'r');

hold on;

If=Ef/Zc;

Ir=It-If;

plot(t,Ir,'b');



OUTPUT RESPONSE:

%line terminated by inductor for current

Ef=10000;

L=0.004;

Zc=400;

n1=[2*Ef0];

d1=[1 Zc/L];

tf(n1,d1)

t=0:0.00001:0.0001; n2=[2*Ef/L];

d2=[1 Zc/L;

It=step(n2,d2,t);

plot(t,It,'r');

If=Ef/Zc;

Ir=It-If;

hold on;



plot(t,Ir,'b');

OUTPUT RESPONSE:

RESULT:



TRANSIENT AND SMALL SIGNAL STABILITY ANALYSIS OF SINGLE-MACHINE INFINITE BUS SYSTEM

AIM: To become familiar with various aspects of the transient and small signal stability analysis of

Single-Machine Infinite Bus (SMIB) system. SOFTWARE REQUIRED:

A.U Power lab or equivalent Exercises:

For a typical power system comprising a generating, step-up transformer, double-circuit transmission line connected to infinite bus: Transient Stability Analysis

1. Hand calculation of the initial conditions necessary for the classical model of the synchronous machine.

2. Hand computation of critical clearing angle and time for the fault using equal area criterion.

3. Simulation of typical disturbance sequence: fault application, fault clearance by opening of one circuit using the software available and checking stability by plotting the swing curve.

4. Determination of critical clearing angle and time for the above fault sequence through trial and error method using the software and checking with the hand computed value.

5. Repetition of the above for different fault locations and assessing the fault severity with respect to the location of fault.

6. Determination of the steady-state and transient stability margins. Small-signal Stability Analysis:

7. Familiarity with linearised swing equation and characteristic equation and its roots, damped frequency of oscillation in Hz, damping ratio and undamped natural frequency.

8. Force-free time response for an initial condition using the available software.

9. Effect of positive, negative and zero damping.

THEORETICAL BACK GROUND Stability: Preamble [1]

The stability of systems in general, is intimately connected with maintenance of equilibrium in the presence of opposing forces. Once equilibrium is established, stability will not be a major concern if the system is free from disturbance or if the system is in steady-state. It is only when the steady-state is disturbed, then the stability of the system is inquired, i.e whether the system will return to the pre-disturbance steady-state. The pre-disturbance steady-state plays an important part in determining whether the system will return to this state after a disturbance. Power system stability - Definition [2]



It is the property of the system that enables it to remain in a state of operating equilibrium under normal operating conditions and to regain an acceptable state of equilibrium after being subjected to a disturbance. Manifestation of power system instability

Power system instability manifests as loss of synchronism between rotating inertias connected to the system and /or unacceptably low voltage. Both situations, if countermeasures are not active, can lead to total system black out or total voltage collapse. The former involves dynamics of generator rotor angles under input-output power balance. The collapse of voltage can occur without accompanying loss of synchronism. Causes, nature and effects of disturbances

The types of disturbances outlined earlier can further be subdivided as: Natural causes such as a tornado that can cause a flashover across insulators Inadvertent causes such as maloperation of protection Intended actions such as opening/closing of circuit breakers by the operator

All disturbances involve one or more phase conductors and/or ground and always result in imbalance between mechanical power of the turbine and output electrical power of the generator. The power imbalance triggers the dynamics which results in deviation in generator rotor speeds, bus voltages and possibly other variables from their nominal values. If the disturbance is not removed, the protection system may act to isolate that portion of the system where the deviations are excessive. Basic assumptions made in stability studies

1. Only synchronous frequency currents and voltages are considered in the machine and the network. Direct current offsets and harmonics are neglected.

2. Symmetrical components are used in the representation of unbalanced faults. 3. Generator voltages are unaffected by machine speed variations.

Modeling For Transient Stability Consider a single machine connected to an infinite bus shown in Fig. 1. An infinite bus is a source of constant frequency and voltage.

Fig 1. Single machine connected to infinite bus system

The equivalent circuit with the generator represented by classical model and all resistances neglected is shown in fig. 2. A simplified equivalent circuit is shown in Fig.3.



Fig 2. Equivalent circuit

Fig.3. Simplified Equivalent Circuit

Following expressions are valid for figures 2 and 3:

__________________(1)



Computation of Initial Conditions:

Assume that the generator output power (Pe,Qe) and the terminal voltage magnitude Et are specified. The equivalent circuit shown in Fig.2 with parallel combination of X1 and X2 replaced by X3 is shown in Fig. 4.

Fig.4. Equivalent circuit with parallel combination of X1 and X2 replaced by X3

Assume Et as reference, i.e, Et = Et <0o, the computation of initial conditions consists of

following steps: Swing Equation

During any disturbance in the system, the rotor will accelerate or decelerate with respect to synchronously rotating axis and the relative motion begins. The equation describing the

(2)

(3)

(4)

(4.a)

(5)

(5.a)



relative motion is called as swing equation. The following assumptions are made in the derivation of swing equation:

(i) Machine represented by classical model (ii) Controllers are not considered (iii) Loads are constants (iv) Voltage and currents are sinusoids

The fundamental equation of motion of the rotor of the synchronous machine is given by

Equation (6) can be rewritten as two first order equations in state variable form:

(6)

(7)

(8)



Numerical Integration Techniques

The differential equations (8) are to be solved using numerical techniques. There are several techniques available and two of them, modified Euler and fourth order Runge- Kutta methods, are illustrated taking a simple example of a first order equation in a single variable:

(9)



Determination of Critical Clearing Time

Critical clearing time is the maximum allowable time between the occurrence of a fault and clearing of the fault for which the system will be stable. For a given load condition and specified fault, the critical clearing time for a system is found out by trial and error method as explained. Start with a fault clearing time which is stable. Increase the clearing time in steps till instability results. The trial value of clearing time just before instability was detected is the critical clearing time. This will give you the coarse value of the critical clearing time. By varying the clearing time around this point in small steps till you find the clearing time which is just critical. The clearing time margin for a fault may be defined as Clearing time margin = critical clearing time – clearing time specified

Stability Margin in MW

Assume that the machine connected to infinite bus delivers Po MW (Fig.1) and a fault is specified at the end of line no. 1 with a clearing time tc = 0.07 seconds. Suppose the MW output of the machine and the load power are increased in steps and stability is checked for each step of load with the same fault clearing time. Let the system remain stable up to a maximum loading of say PLmax and a small increase in load beyond PLmax causes instability. Then the MW stability margin is defined as Ps = PLmax – Po Critical Clearing Time and Clearing Angle from Equal Area Criterion

This method is suitable for hand computations and is applicable only to single machine connected to infinite bus system. Nonetheless, it gives a clear physical picture of the dynamics of rotor motion when subjected to a disturbance. Consider the system shown in Fig.3. and its model in Fig 5. The power-angle relationship is given by equation (1) and the power angle curves for various operating condition is given in Fig. 5.



?

Fig. 5. Power angle curve ` The steady state operating condition is given by point a and the corresponding rotor angle is o. Consider a three phase fault at location F on line 2 as shown in fig 1.The fault is cleared by opening the circuit breakers at both ends of the line. The P- ? plots for three network conditions are shown in fig 5.

When the fault occurs, the operating point changes from a to b. Since Pm > Pe, the rotor accelerates until the operating point reaches c where the fault is cleared at 1. The operation shifts to e. Now Pe>Pm, the rotor decelerates, but continues to increase until the kinetic energy gained during the period of acceleration (Area A1) is transferred to the system. The operating point moves from e to f such that area A2 is equal to A1. The rotor angle will oscillate back and forth around the point of intersection of the straight line representing Pm and curve B at its natural frequency such that area A1 = area A2. This is known as equal area

criterion.

Fig. 6. Determination of critical clearing angle using equal area criterion



Critical Clearing Angle and Time

With delayed fault clearing as shown in Fig.6, the area A2 just equals to A1 at clearing angle equal to cr. Any further delay in clearing causes area A2 above Pm to be less than A1 resulting in loss of synchronism. The angle cr for which A1 = A2 is called critical clearing angle. An expression for critical clearing angle from Fig.6 can be derived as follows:

Applying equal area criterion,

Modelling For Small Signal Stability The electrical power output of the generator in p.u. is

If speed is expressed in p.u. the air-gap torque is equal to air-gap power, Hence

Linearising (12) about an initial operating condition at = o

(11)

(12)

(13)

(13.a)



is called the synchronising torque co-efficient.

The state equations (8) are rewritten as

Linearising equation (14) and using equation (13.a) and (15) we get

The block diagram representation of Eq.(17) is showb in fig(9)

7 Block Diagram for Equation (17)

Taking Laplace transform of the above equation

(14)

(15)

(16)

(17)



(18)

(19)

(20)

(21)



EXERCISES 1. A power system comprising a thermal generating plant with four 555 MVA, 24kV, and 60HZ units’ supplies power to an infinite bus through a transformer and two transmission lines (refer Fig 8)

(22)

(23)



Fig.8 Single Machine Infinite Bus System

The data for the system in per unit on a base of 2220 MVA, 24 kV is given below: Xd’=0.3p.u, H=3.5MW/MVA, transformer X=0.15p.u, line1 X=0.5p.u, line 2 X=0.93p.u. Plant operating conditions P=0.9p.u, p.f=0.9lag, Et=1p.u. Case (1) It is proposed to examine the transient stability of the system for a 3 phase to ground fault at the end of the line 2 near H.T bus occurring at t=0s.The fault is cleared in 0.07s by simultaneous opening of two CBs at the end of the both lines 2. Case (2) Find the initial condition necessary for a classical model of the m/c for the above pre-fault operating conditions. Find the critical clearing angle and time using equal angle criterion. Result:



TRANSIENT STABILITY ANALYSIS OF MULTIMACHINE POWER SYSTEMS Aim :

To become familiar with modelling aspects of synchronous machines and network, state-of-the-art algorithm for simplified transient stability simulation, system behaviour when subjected to large disturbances in the presence of synchronous machine controllers and to become proficient in the usage of the software to tackle real life problems encountered in the areas of power system planning and operation.

Exercises: For typical multi-machine power system:

7.1 Simulation of typical disturbance sequence: fault application, fault clearance by opening of a line using the software available and assessing stability with and without controllers.

7.2 Determination of critical clearing angle and time for the above fault sequence through trial and error method using the software.

7.3 Determination of transient stability margins.

7.4 Simulation of full load rejection with and without governor.

7.5 Simulation of loss of generation with and without governor.

Software Required: Math lab or AU Power lab

THECHNICAL BACKGROUND:

1. INTRODUCTION

Multi-machine equations can be written similar to the one-machine system connected to the infinite bus. In order to reduce the complexity of the transient stability analysis, similar simplifying assumptions are made as follows.

-Each synchronous machine is represented by a constant voltage source behind the direct axis transient reactance. This representation neglects the effect of saliency and assumes constant flux linkages. -The governor’s action are neglected and the input powers are assumed to remain constant during the entire period of simulation.

-Using the pre-fault bus voltages, all loads are converted to equivalent admittances to ground and are assumed to remain constant.

-Damping or asynchronous powers are ignored.



-The mechanical rotor angle of each machine coincides with the angle of the voltage behind the machine reactance.

-Machines belonging to the same station swing together and are said to be coherent. A group of coherent machines is represented by one equivalent machine.

2. MATHEMATICAL MODEL OF MULTIMACHINE TRANSIENT STABILITY ANALYSIS The first step in the transient stability analysis is to solve the initial load flow and to

determine the initial bus voltage magnitudes and phase angles. The machine currents prior to disturbance are calculated from,

Where

m= is the number of generators

Vi- is the terminal voltage of the ith generator

Pi and Qi are the generator real and reactive powers. All unknown values are determined from the initial power flow solution. The generator

armature resistances are usually neglected and the voltages behind the transient reactance are then obtained,

Next, all load are converted to equivalent admittances by using the relation

To include voltages behind transient reactance, m buses are added to the n bus power

system network. The equivalent network with all load converted to admittances is shown in Fig.1,



Nodes n+1, n+2, . . ., n+m are the internal machine buses, i.e., the buses behind the

transient reactances. The node voltage equation with node 0 as reference for this network, is

Or

Where

Ibus is the vector of the injected bus currents Vbus is the vector of bus voltages measured from the reference node.

The diagonal elements of the bus admittance matrix are the sum of admittances connected to it, and the off-diagonal elements are equal to the negative of the admittance between the nodes. The reference is that additional nodes are added to include the machine voltages behind transient reactances. Also, diagonal elements are modified to include the load admittances.



To simplify the analysis, all nodes other than the generator internal nodes are eliminated using Kron reduction formula . To eliminate the load buses, the bus admittance matrix in (4) is partitioned such that the n buses to be removed are represented in the upper n rows. Since no current enters or leaves the load buses, currents in the n rows are zero. The generator currents are denoted by the vector Im and the generator and load voltages are represented by the vector E’ m and Vn, respectively. Then, Equation (4), in terms of sub matrices, becomes

The voltage vector Vn may be eliminated by substitution as follows.

The reduced bus admittance matrix has the dimensions (m x m), where m is the number

of generators. The electrical power output of each machine can now be expressed in terms of the machine’s internal voltages



Expressing voltages and admittances in polar form, i.e., I

and substituting for Ii in (12), result in

The above equation is the same as the power flow equation. Prior to disturbance, there is

equilibrium between the mechanical power input and the electrical power output, and we have

The classical transient stability study is based on the application of a three-phase fault. A solid three-phase fault at bus k in the network results in Vk = 0. This is simulated by removing the kth row and column from the prefault bus admittance matrix. The new bus admittance matrix is reduced by eliminating all nodes except the internal generator nodes. The generator excitation voltages during the fault and postfault modes are assumed to remain constant. The electrical power of the ith generator in terms of the new reduced bus admittance matrices are obtained from (14). The swing equation with damping neglected, for machine i becomes

Where

Yij are the elements of the faulted reduced bus admittance matrix Hi is the inertia constant of machine I expressed on the common MVA base SB. If HGi is the inertia constant of machine I expressed on the machine rated MVA SGi, then Hi is given by



Showing the electrical power of the ith generator by Pef and transforming (16) into state variable mode yields

In transient stability analysis problem, we have two state equations for each generator.

When the fault is cleared, which may involve the removal of the faulty line, the bus admittance matrix is recomputed to reflect the change in the networks. Next the post-fault reduced bus admittance matrix is evaluated and the post-fault electrical power of the ith generator shown by Ppf i is readily determined from (14). Using the post-fault power Ppf i, the simulation is continued to determine the system stability, until the plots reveal a definite trend as to stability or instability. Usually the slack generator is selected as the reference machines are plotted. Usually, the solution is carried out for two swings to show that the second swing is not greater than the first one. If the angle differences do not increase, the system is stable. If any of the angle differences increase indefinitely, the system is unstable.



Result:



Viva questions 1. What are the components of power system?

2. Define per unit value.

3. What is the need for base values?

4. What are the advantages of per unit computations?

5. What is impedance & reactance?

6. What is a bus?

7. Name the diagonal & off diagonal elements of bus admittance matrix?

8. What is bus admittance matrix?

9. What is bus impedance matrix?

10. What are the methods available for forming bus admittance matrix?

11. Write the symmetrical components of 3 phase system?

12. What is meant by positive negative & zero sequence impedance? 13. What is power flow study or load flow study?

14. What is the need for load flow study?

15. What are the different types of buses in a power system?

16. What are the information that are obtained from a load flow study?

17. What is swing bus?

18. What is the need for slag bus?

19. What do you mean by a flat voltage start?

20. When the generator buses treated as load bus?

21. What is jacobian matrix?

22. What is infinite bus? 23. How the reactive power of generator is controlled?

24. How the faults are classified?

25. List the symmetrical & unsymmetrical faults?

26. What is the need for short circuit studies?

27. Define transient & sub transient reactance.

28. Name the various unsymmetrical faults in a power system.

29. Define stability.

30. Define steady state & transient stability.



31. Explain steady state stability limit

32. Explain transient stability limit.

33. How stability studies are classified. What are they?

34. Define swing curve.

35. Define power angle.

36. Define equal area criterion.

37. What is AGC?

38. Explain unit commitment.

39. What is economic dispatch?

40. Explain participation factor.

ee76 power system simulation lab

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