Gokaraju Rangaraju Institute of Engineering and Technology PSS Lab... · 2019. 12. 28. · Author: GRIET-JSD Created Date: 5/9/2017 1:29:36 PM

POWER SYSTEMS SIMULATION LAB

___YEAR __ SEM

EEE

By

Dr. J. Sridevi

Gokaraju Rangaraju Institute of Engineering &

Technology

Bachupally

GOKARAJU RANGARAJU INSTITUTE OF

ENGINEERING AND TECHNOLOGY

(Autonomous)

Bachupally , Hyderabad-500 072

CERTIFICATE

This is to certify that it is a record of practical work done in the Power

Systems Simulation Laboratory in _____ sem of ________________

year during the year ________________________.

Name:

Roll No:

Branch: EEE

Signature of staff member

INDEX

List of Experiments

S.No Date Name of the Experiment Page no Signature

1 Sinusoidal Voltages and currents 1

2 Computation of line parameters 6

3 Modelling of transmission lines 11

4 Formation of bus admittance matrix 18

5 Load Flow solution using gauss seidel

method 23

6 Load flow solution using Newton

Rapshon method in Polar coordinates 30

7 Load flow solution using Newton

Rapshon method in Rectangular coordinates 34

8 Transient stability analysis of single-

machine infinite bus 38

system

9 Power flow solution of 3 – bus system 43

10

a)Optimal dispatch neglecting losses b) Optimal dispatch including losses

48 54

11

Three phase short circuit analysis in a

synchronous machine(symmetrical fault

analysis) 58

12 Unsymmetrical fault analysis-LG,LL,LLG

fault 64

13 Z- Bus building algorithm 73

14

a)Obtain symmetrical components of a set of unbalanced currents b)Obtain the original unbalanced phase voltages from symmetrical components

78 82

15 Short circuit analysis of a power system

with IEEE 9 bus system 85

16 Power flow analysis of a slack bus

connected to different loads 89

17 Load flow analysis of 3 motor systems

connected to slack bus 94

GRIET/EEE Power Systems Simulation Lab

1

Date: Experiment-1

SINUSOIDAL VOLTAGES AND CURRENTS

Aim: To determine sinusoidal voltages and currents

Apparatus: MATLAB

Theory : The RMS Voltage of an AC Waveform

The RMS value is the square root of the mean (average) value of the squared function of the

instantaneous values. The symbols used for defining an RMS value are VRMS or IRMS.

The term RMS, refers to time-varying sinusoidal voltages, currents or complex waveforms were

the magnitude of the waveform changes over time and is not used in DC circuit analysis or

calculations were the magnitude is always constant. When used to compare the equivalent RMS

voltage value of an alternating sinusoidal waveform that supplies the same electrical power to a

given load as an equivalent DC circuit, the RMS value is called the “effective value” and is

presented as: Veffor Ieff.

In other words, the effective value is an equivalent DC value which tells you how many volts or

amps of DC that a time-varying sinusoidal waveform is equal to in terms of its ability to produce

the same power. For example, the domestic mains supply in the United Kingdom is 240Vac. This

value is assumed to indicate an effective value of “240 Volts RMS”. This means then that the

sinusoidal RMS voltage from the wall sockets of a UK home is capable of producing the same

average positive power as 240 volts of steady DC voltage as shown below.

RMS Voltage Equivalent


2

Circuit diagram:

Fig: Simulink model for voltage and current measurement

Procedure:

1. Open Matlab-->Simulink--> File ---> New---> Model

2. Open Simulink Library and browse the components

3. Connect the components as per circuit diagram

4. Set the desired voltage and required frequency

5. Simulate the circuit using MATLAB

6. Plot the waveforms.


3

Graph:

Calculations:


4


5

Result:


6

Signature of the faculty

Exp.No:2 Date:

COMPUTATION OF LINE PARAMETERS AIM

To determine the positive sequence line parameters L and C per phase per kilometre of a three

phase single and double circuit transmission lines for different conductor arrangements and to

understand modeling and performance of medium lines. SOFTWARE REQUIRED: MAT LAB THEORY

Transmission line has four parameters namely resistance, inductance, capacitance and

conductance. The inductance and capacitance are due to the effect of magnetic and electric fields

around the conductor. The resistance of the conductor is best determined from the manufactures

data, the inductances and capacitances can be evaluated using the formula. Inductance The general formula

L = 0.2 ln (Dm / Ds)

Where, Dm = geometric mean distance (GMD)

Ds = geometric mean radius (GMR) I. Single phase 2 wire system GMD = D GMR = re-1/4 = r′ Where, r = radius of conductor II. Three phase – symmetrical spacing GMD = D GMR = re-1/4 = r′ Where, r = radius of conductor III. Three phase – Asymmetrical Transposed GMD = geometric mean of the three distance of the symmetrically placed conductors = (DABDBCDCA)1/3 GMR = re-1/4 = r′ Where, r = radius of conductors Composite conductor lines The inductance of composite conductor X, is given by Lx = 0.2 ln (GMD/GMR)


7

where, GMD = (Daa Dab )…….(Dna …….Dnm ) GMR = n2

(Daa Dab…….Dan )…….(DnaDnb…….Dnn) where, r’a

= ra e(-1/ 4) Bundle Conductors The GMR of bundled conductor is normally calculated

GMR for two sub conductor, Dsb = (Ds×d)1/2

GMR for three sub conductor, Dsb = (Ds×d2)

1/3 GMR

for four sub conductor, Dsb = 1.09 × (Ds×d2)1/4

where, Ds is the GMR of each subconductor d is bundle spacing Three phase – Double circuit transposed The inductance per phase in mH per km

is L = 0.2×ln(GMD / GMRL) mH/km

where, GMRL is equivalent geometric mean radius and is given by

GMRL = (DSADSBDSC)1/3

where, DSADSB and DSC are GMR of each phase group and given by

1/2 DSA = 4 sb Da1a2)2 = [Dsb Da1a2]

1/2 DSB = 4 sb Db1b2)2 = [Dsb Db1b2]

DSC = 4 sb Dc1c2)2 = [Dsb Dc1c2]1/2

where, Dsb =GMR of bundle conductor if conductor a1, a2….. are bundled conductor. Dsb = ra1’= rb1= ra’2 = rb’2 = rc’2 if a1, a2……. are bundled conductor GMD is the equivalent GMD per phase” & is given by GMD = [DAB * DBC *

DCA]1/3

where, DAB, DBC&DCA are GMD between each phase group A-B, B-C, C-A which are given by

DAB = [Da1b1 * Da1b2 * Da2b1 * Da2b2]1/4

DBC = [Db1c1 * Db1c2 * Db2c1 * Db2c2]1/4

DCA = [Dc1a1 * Dc2a1 * Dc2a1 * Dc2a2]1/4

Capacitance A general formula for evaluating capacitance per phase in micro farad per km of a transmission line is given by C = 0.0556/ ln (GMD/GMR) F/km Where, GMD is the “Geometric mean distance” which is same as that defined for inductance under various cases.


8

PROCEDURE

Enter the command window of the MAT LAB. Create a new M – file by selecting File - New – M – File. Type and save the program in the editor window. Execute the program by pressing Tools – Run. View the results.

EXERCISE 1.A 500kv 3 φ transposed line is composed of one ACSR 1,272,000-cmil, 45/7 bittern conductor

per phase with horizontal conductor configuration as show in fig.1. The conductors have a

diameter of 1.345in and a GMR of 0.5328in. Find the inductance and capacitance per phase per

kilometer of the line and justify the result using MAT LAB.

a b c

D12 =35’ D23 =35’

D13=70’

Fig.1

2.The transmission line is replaced by two ACSR 636,000-cmil, 24/7 Rook conductors which have

the same total cross-sectional area of aluminum as one bittern conductor. The line spacing as

measured from the centre of the bundle is the same as before and is shown in fig.2. The conductors

have a diameter of 0.977in and a GMR of 0.3924in.Bundle spacing is 18in. Find the inductance

and capacitance per phase per kilometer of the line and justify the result using MAT LAB.

a b c

D12 =35’ D23 =35’

D13=70’


9

3.A 345- KV double –circuit three- phase transposed line is composed of two ACSR, 1,431,000-

cmil, 45/7 Bobolink conductors per phase with vertical conductor configuration as shown in fig.3.

The conductors have a diameter of 1.427in and a GMR of 0.564 in .the bundle spacing in 18in.

find the inductance and capacitance per phase per kilometer of the line and justify the result using

MAT LAB.

a a’ S11 =11m

H12=7m

b S22=16.5m b’

H23=6.5m

S33=16.5m

c c’

PROGRAM

[GMD, GMRL, GMRC] = gmd;

L = 0.2*log(GMD/GMRL) C = 0.0556/log(GMD/GMRC)

OUTPUT:


10

RESULT:

Thus the positive sequence line parameters L and C per phase per kilometre of a three phase single and double circuit transmission lines for different conductor arrangements were determined and verified with MAT LAB software. The value of L and C obtained from MAT LAB program are:

Case1: L= C=

Case2: L= C=

Case3: L= C=



11

Exp.No:3 Date:

MODELLING OF TRANSMISSION LINES AIM:

To understand modeling and performance of Short, Medium and Long transmission lines.

SOFTWARE REQUIRED: MAT LAB

THEORY:

The important considerations in the design and operation of a transmission line are the

determination of voltage drop, line losses and efficiency of transmission. These values are greatly

influenced by the line constants R, L and C of the transmission line. For instance, the voltage drop

in the line depends upon the values of above three line constants. Similarly, the resistance of

transmission line conductors is the most important cause of power loss in the line and determines

the transmission efficiency.

A transmission line has three constants R, L and C distributed uniformly along the whole

length of the line. The resistance and inductance form the series impedance. The capacitance

existing between conductors for 1-phase line or from a conductor to neutral for a 3-phase line

forms a shunt path throughout the length of the line. Short Transmission Line:

When the length of an overhead transmission line is upto about 50km and the line voltage is comparatively low (< 20 kV), it is usually considered as a short transmission line. Medium Transmission Lines:

When the length of an overhead transmission line is about 50-150 km and the line voltage is moderatly high (>20 kV < 100 kV), it is considered as a medium transmission line. Long Transmission Lines:

When the length of an overhead transmission line is more than 150km and line voltage is very high (> 100 kV), it is considered as a long transmission line. Voltage Regulation:

The difference in voltage at the receiving end of a transmission line between conditions of

no load and full load is called voltage regulation and is expressed as a percentage of the receiving

end voltage.

Performance of Single Phase Short Transmission Lines

As stated earlier, the effects of line capacitance are neglected for a short transmission line. Therefore, while studying the performance of such a line, only resistance and inductance of

6


12

the line are taken into account. The equivalent circuit of a single phase short transmission line is

shown in Fig. (i).Here, the total line resistance and inductance are shown as concentrated or lumped

instead of being distributed. The circuit is a simple a.c. series circuit. Let I = load current R = loop resistance i.e., resistance of both conductors XL = loop reactance VR = receiving end voltage cos ϕR = receiving end power factor (lagging) VS = sending end voltage cos ϕS = sending end power factor The phasor diagram of the line for lagging load power factor is shown in Fig. (ii). From the right angled traingle ODC, we get,

(OC)2 = (OD)2+ (DC)2

VS2 = (OE + ED)2 + (DB + BC)2

= (VR cos ϕR + IR)2 + (VR sinϕR + IXL)2


13

An approximate expression for the sending end voltage VS can be obtained as follows. Draw

perpendicular from B and C on OA produced as shown in Fig. 2. Then OC is nearly equal to OF

i.e.,

OC = OF = OA + AF = OA + AG + GF = OA + AG + BH

VS = VR + I R cos ϕR + I XL sin ϕR Medium Transmission

Line Nominal T Method

In this method, the whole line capacitance is assumed to be concentrated at the middle

point of the line and half the line resistance and reactance are lumped on its either side as shown

in Fig.1, Therefore in this arrangement, full charging current flows over half the line. In Fig.1, one

phase of 3-phase transmission line is shown as it is advantageous to work in phase instead of line-

to-line values.

Let IR = load current per phase R = resistance per phase


14

XL = inductive reactance per phase C = capacitance per phase cos ϕR = receiving end power factor (lagging) VS = sending end voltage/phase V1 = voltage across capacitor C The phasor diagram for the circuit is shown in Fig.2. Taking the receiving end voltage VR as the reference phasor, we have, Receiving end voltage, VR = VR + j 0 Load current, IR = IR (cos ϕR - j sin ϕR)

Nominal π Method In this method, capacitance of each conductor (i.e., line to neutral) is divided into two halves;

one half being lumped at the sending end and the other half at the receiving end as shown in

Fig.3. It is obvious that capacitance at the sending end has no effect on the line drop. However,

it’s charging current must be added to line current in order to obtain the total sending end

current.


15

IR = load current per phase R = resistance per phase XL = inductive reactance per phase C = capacitance per phase cos R = receiving end power factor (lagging) VS = sending end voltage per phase The phasor diagram for the circuit is shown in Fig.4. Taking the receiving end voltage as the reference phasor, we have, VR = VR + j 0 Load current, IR = IR (cos R - j sin R)


16

EXERCISE:

1. A 220- KV, 3φ transmission line is 40 km long. The resistance per phase is 0.15 Ω per

km and the inductance per phase is 1.3623 mH per km. The shunt capacitance is negligible.

Use the short line model to find the voltage and power at the sending end and the voltage

regulation and efficiency when the line supplying a three phase load of a) 381 MVA at 0.8

power factor lagging at 220 KV. b) 381 MVA at 0.8 power factor leading at 220 KV.

PROGRAM

VRLL=220; VR=VRLL/sqrt(3);

Z=[0.15+j*2*pi*60*1.3263e-

3]*40; disp=('(a)') SR=304.8+j*228.6; IR=conj(SR)/(3*conj(VR)); IS=IR; VS=VR+Z*IR; VSLL=sqrt(3)*abs(VS) SS=3*VS*conj(IS)

REG=((VSLL-

VRLL)/(VRLL))*100 E

disp=('(b)') SR=304.8-

j*228.6;

IR=conj(SR)/(3*conj(VR));

IS=IR; VS=VR+Z*IR; VSLL=sqrt(3)*abs(VS) SS=3*VS*conj(IS) REG=((VSLL-

VRLL)/(VRLL))*100

EFF=(real(SR)/real(SS))*100

FF=(real(SR)/real(SS))*100


17

OUTPUT: RESULT: Thus the program for modeling of transmission line was executed by using MAT LAB

and the output was verified with theoretical calculation. The value of the voltage and power at the

sending end, voltage regulation and efficiency obtained from the MAT LAB program are Voltage VS =

Power SS = Voltage regulation REG = Efficiency EFF =



18

Expt. No.:4 Date:

FORMATION OF BUS ADMITTANCE MATRIX

AIM To determine the bus admittance matrix for the given power system Network

SOFTWARE REQUIRED: MAT LAB

THEORY

FORMATION OF Y BUS MATRIX

Bus admittance matrix is often used in power system studies.In most of power system

studies it is necessary to form Y-bus matrix of the system by considering certain power system

parameters depending upon the type of analysis. For example in load flow analysis it is necessary

to form Y-bus matrix without taking into account the generator impedance and load impedance.

In short circuit analysis the generator transient reactance and transformer impedance taken in

account, in addition to line data. Y-bus may be computed by inspection method only if there is no

natural coupling between the lines. Shunt admittance are added to the diagonal elements

corresponding to the buses at which these are connected. The off diagonal elements are unaffected.

The equivalent circuit of tap changing transformer may be considered in forming[y-bus] matrix. FORMULA USED Yij=ΣYij for i=1 to n Yij= -Yij= -1/Zij Yij=Yji Where Yii=yjj= Sum of admittance connected to bus Yij= Negative admittance between buses


19

START

Read the no. Of buses , no of lines and line data

Initialize the Y- BUS Matrix

Consider line l = 1

i = sb(1); I= eb(1)

Y(i,i) =Y(i,i)+Yseries(l) +0.5Yseries(l)

Y(j,j) =Y(j,j)+Yseries(l) +0.5Yseries(l)

Y(i,j) = -Yseries(l)

Y(j,i) =Y(i,j) NO Is l =NL? YES

l = l+1 Print Y -Bus

Stop


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Exercise:

Line number

Starting Bus

Ending Bus Series Line Line Changing

Impedance

Admittance

1 1 2 0.1+0.4j 0.15j

2 2 3 0.15+0.6j 0.02j

3 2 4 0.18+0.55j 0.018j

4 3 4 0.1+0.35j 0.012j

5 4 1 0.25+0.7j 0.03j

PROGRAM function[Ybus] = ybus(zdata)

nl=zdata(:,1); nr=zdata(:,2); R=zdata(:,3); X=zdata(:,4); nbr=length(zdata(:,1)); nbus =

max(max(nl), max(nr)); Z = R + j*X; %branch impedance

y= ones(nbr,1)./Z; %branch admittance Ybus=zeros(nbus,nbus); % initialize Ybus to zero

for k = 1:nbr; % formation of the off diagonal elements if nl(k) > 0 & nr(k) > 0

Ybus(nl(k),nr(k)) = Ybus(nl(k),nr(k)) - y(k); Ybus(nr(k),nl(k)) = Ybus(nl(k),nr(k)); end

end for n = 1:nbus % formation of the diagonal elements for k = 1:nbr

if nl(k) == n | nr(k) == n Ybus(n,n) = Ybus(n,n) + y(k); else, end

end

end


21


22

OUTPUT:

RESULT: Thus the bus admittance matrix of the given power system using inspection method

was found and verified by theoretical calculation.

Y Bus:



23

Expt. No.:5 Date:

LOAD FLOW SOLUTION USING GAUSS SEIDEL METHOD

AIM: To carry out load flow analysis of the given power system network by Gauss Seidal

method

SOFTWARE REQUIRED: POWERWORLD

THEORY

Load flow analysis is the study conducted to determine the steady state operating condition of the

given system under given conditions. A large number of numerical algorithms have been developed and

Gauss Seidel method is one of such algorithm.

PROBLEM FORMULATION

The performance equation of the power system may be written of

[I bus] = [Y bus][V bus] (1) Selecting one of the buses as the reference bus, we get (n-1) simultaneous equations. The bus

loading equations can be written as Ii = Pi-jQi / Vi* (i=1,2,3,…………..n) (2)

Where,

n

Pi=Re [ Σ Vi*Yik Vk] . (3) k=1

n

Qi= -Im [ Σ Vi*Yik Vk]. (4) k=1

The bus voltage can be written in form

of

n

Vi=(1.0/Yii)[Ii- Σ Yij Vj] (5) j=1

j≠i(i=1,2,…………n)& i≠slack bus

Substituting Ii in the expression for Vi, we get

n

Vi new=(1.0/Yii)[Pi-JQi / Vio* - Σ Yij Vio] (6) J=1

The latest available voltages are used in the above expression, we get n n

Vi new=(1.0/Yii)[Pi-JQi / Vo

i* - Σ YijVjn- Σ Yij Vio] (7) J=1 j=i+1

The above equation is the required formula .this equation can be solved for voltages in interactive manner.


24

During each iteration, we compute all the bus voltage and check for convergence is carried out by comparison with the voltages obtained at the end of previous iteration. After the solutions is obtained. The stack bus real and reactive powers, the reactive power generation at other generator buses and line flows can be calculated.

ALGORITHM Step1: Read the data such as line data, specified power, specified voltages, Q limits at the generator buses

and tolerance for convergences Step2: Compute Y-bus matrix. Step3:

Initialize all the bus voltages. Step4: Iter=1 Step5: Consider i=2, where i’ is the bus number. Step6: check whether this is PV bus or PQ bus. If it is PQ bus goto step 8 otherwise go to next step. Step7: Compute Qi check for q limit violation. QGi=Qi+QLi.

7).a).If QGi>Qi max ,equate QGi = Qimax. Then convert it into PQ bus. 7).b).If

QGi


25

START

Read

1. Primitive Y matrix

2. Bus incidence matrix A

3. Slack bus voltages

4. Real and reactive bus powers Pi& Qi

5. Voltage magnitudes and their limits

Form Ybus

Make initial assumptions

Compute the parameters Ai for i=m+1,…,n and Bik for i=1,2,…,n;

k=1,2,…,n

Set iteration count r=0

Set bus count i=2 and ΔVmax=0

Test for

type of bus

Qi(r+1) >Qi,

max

Qi(r+1) <

Qi,min

Compute Qi(r+1)

Qi(r+1) = Qi,max Qi

(r+1) = Qi,min Compute Ai(r+1)

Compute Ai

Compute Vi(r+1)

Compute δi(r+1) and

Vi(r+1) =|Vi

s|/δi(r+1)


26

FLOW CHART FOR GAUSS SEIDEL METHOD

Exercise:

1. The Fig.1 shows the single line diagram of a simple three bus power system with generation at buses 1.

The magnitude of voltage at bus 1 is adjusted to 1.05 p.u. The scheduled loads at buses 2 & 3 are marked

on the diagram .line impedances are marked in per unit on a 100-MVA base and the line charging charging

susceptances are neglected. a). using the gauss-seidal method, determine the phasor values of the voltage at the load buses 2 and 3(P-

Q buses) accurate to four decimal places. b).find the slack bus real and reactive power. c) Determine the line flow and line losses, Using software write an algorithm.

Replace Vir by Vi(r+1) and

advance bus count i = i+1

Is i

GRIET/EEE Power Systems Simulation

Lab

27

1 0.02 + j0.04 2

256.6MW

0.01+j0.03 0.0125+j0.025 110.2Mvar

Slack bus v1 = 1.05L 3

138.6 MW 45.2 Mvar

PROCEDURE

Create a new file in edit mode by selecting File - New File.

Browse the components and build the bus sytem

Execute the program in run mode by selecting power flow by gauss seidel method

View the results in case information-Bus information.

Tabulate the results.

POWER WORLD bus diagram:


28


29

RESULTS: A one line diagram has been developed using POWER WORLD for the given power system by

Gauss Seidal method and the results are verified with model calculation.



30

Exp.No. :6 Date:

LOAD FLOW SOLUTION USING NEWTON RAPSHON METHOD IN

POLAR COORDINATES AIM

To carry out load flow analysis of the given power system by Newton Raphson method in

polar coordinates

SOFTWARE REQUIRED: POWERWORLD THEORY The Newton Raphson method of load flow analysis is an iterative method which approximates the

set of non-linear simultaneous equations to a set of linear simultaneous equations using Taylor’s series expansion and the terms are limited to first order approximation. The load flow

equations for Newton Raphson method are non-linear equations in terms of real and imaginary

part of bus voltages.

where, ep = Real part of Vp

fp = Imaginary part of Vp Gpq, Bpq = Conductance and Susceptances of admittance Ypq respectively. ALGORITHM Step1: Input the total number of buses. Input the details of series line impendence and line

charging admittance to calculate the Y-bus matrix. Step2: Assume all bus voltage as 1 per unit except slack bus. Step3: Set the iteration count as k=0 and bus count as p=1. Step4: Calculate the real and reactive power pp and qp using the formula

P=ΣvpqYpq*cos(Qpq+εp-εq) Qp=ΣVpqYpa*sin(qpq+εp-εa)


31

Evalute pp*=psp-pp* Step5: If the bus is generator (PV) bus, check the value of Qp*is within the limits.If it Violates

the limits, then equate the violated limit as reactive power and treat it as PQ bus. If limit is not

isolated then calculate, |vp|^r=|vgp|^rspe-|vp|r ; Qp*=qsp-qp* Step6: Advance bus count by 1 and check if all the buses have been accounted if not go to step5. Step7: Calculate the elements of Jacobean matrix. Step8: Calculate new bus voltage increment pk and fpk Step9: Calculate new bus voltage ep*h+ ep* Fp^k+1=fpK+fpK Step10: Advance iteration count by 1 and go to step3. Step11: Evaluate bus voltage and power flows through the line .

EXERCISE 1. For the sample system of Fig. the generators are connected at all the four buses, while loads are

at buses 2 and 3. Values of real and reactive powers are listed in Table 6.3. All buses other than

the slack are PQ type.

Assuming a flat voltage start, find the voltages and bus angles at the three buses using Newton

Raphson Method


32

PROCEDURE



Execute the program in run mode by selecting power flow by Newton Raphson method





33

RESULT A one line diagram has been developed using POWERWORLD for the given power system by Newton raphson method and the results are verified with model calculation



34

Exp.No. :7 Date:

LOAD FLOW SOLUTION USING NEWTON RAPSHON METHOD IN

RECTANGULAR COORDINATES AIM

To carry out load flow analysis of the given power system by Newton Raphson method in

Rectangular coordinates.

SOFTWARE REQUIRED: POWERWORLD THEORY The Newton Raphson method of load flow analysis is an iterative method which approximates the

set of non-linear simultaneous equations to a set of linear simultaneous equations using Taylor’s series expansion and the terms are limited to first order approximation. The load flow

equations for Newton Raphson method are non-linear equations in terms of real and imaginary

part of bus voltages.

where, ep = Real part of Vp

fp = Imaginary part of Vp Gpq, Bpq = Conductance and Susceptances of admittance Ypq respectively. ALGORITHM Step1: Input the total number of buses. Input the details of series line impendence and line

charging admittance to calculate the Y-bus matrix. Step2: Assume all bus voltage as 1 per unit except slack bus. Step3: Set the iteration count as k=0 and bus count as p=1. Step4: Calculate the real and reactive power pp and qp using the formula

P=ΣvpqYpq*cos(Qpq+εp-εq) Qp=ΣVpqYpa*sin(qpq+εp-εa)


35

Evalute pp*=psp-pp* Step5: If the bus is generator (PV) bus, check the value of Qp*is within the limits.If it Violates

the limits, then equate the violated limit as reactive power and treat it as PQ bus. If limit is not

isolated then calculate, |vp|^r=|vgp|^rspe-|vp|r ; Qp*=qsp-qp* Step6: Advance bus count by 1 and check if all the buses have been accounted if not go to step5. Step7: Calculate the elements of Jacobean matrix. Step8: Calculate new bus voltage increment pk and fpk Step9: Calculate new bus voltage ep*h+ ep* Fp^k+1=fpK+fpK Step10: Advance iteration count by 1 and go to step3. Step11: Evaluate bus voltage and power flows through the line .

EXERCISE 1. For the sample system of Fig. the generators are connected at all the four buses, while loads are at buses 2 and 3. Values of real and reactive powers are listed in Table 6.3. All buses other than the slack are PQ type. Assuming a flat voltage start, find the voltages and bus angles at the three buses using Newton Raphson Method


36

PROCEDURE



Execute the program in run mode by selecting power flow by Newton Raphson method





37

RESULT A one line diagram has been developed using POWERWORLD for the given power system by Newton raphson method and the results are verified with model calculation



38

Exp.no.:8 Date:

TRANSIENT STABILITY ANALYSIS OF SINGLE-MACHINE INFINITE BUS SYSTEM

AIM To become familiar with various aspects of the transient stability analysis of Single-Machine-Infinite

Bus (SMIB) system OBJECTIVES The objectives of this experiment are: 1. To study the stability behavior of one machine connected to a large power system subjected to a

severe disturbance (3-phase short circuit) 2. To understand the principle of equal-area criterion and apply the criterion to study the stability of

one machine connected to an infinite bus 3. To determine the critical clearing angle and critical clearing time with the help of equal-area

criterion 4. To do the stability analysis using numerical solution of the swing equation. SOFTWARE REQUIRED: MAT LAB 7.7 THEORY The tendency of a power system to develop restoring forces to compensate for the disturbing forces to

maintain the state of equilibrium is known as stability. If the forces tending to hold the machines in

synchronism with one another are sufficient to overcome the disturbing forces, the system is said to

remain stable. The stability studies which evaluate the impact of disturbances on the behavior of synchronous

machines of the power system are of two types – transient stability and steady state stability. The

transient stability studies involve the determination of whether or not synchronism is maintained after

the machine has been subjected to a severe disturbance. This may be a sudden application of large load,

a loss of generation, a loss of large load, or a fault (short circuit) on the system. In most disturbances,

oscillations are such magnitude that linearization is not permissible and nonlinear equations must be

solved to determine the stability of the system. On the other hand, the steady-state stability is concerned

with the system subjected to small disturbances wherein the stability analysis could be done using the

linearized version of nonlinear equations. In this experiment we are concerned with the transient stability of power systems. A method known as the equal-area criterion can be used for a quick prediction of stability of a one-

machine system connected to an infinite bus. This method is based on the graphical interpretation of


39

energy stored in the rotating mass as an aid to determine if the machine maintains its stability after a

disturbance. The method is applicable to a one-machine system connected to an infinite bus or a two-

machine system. Because it provides physical insight to the dynamic behavior of the machine, the

application of the method to analyze a single-machine system is considered here. Stability: Stability problem is concerned with the behavior of power system when it is subjected to

disturbance and is classified into small signal stability problem if the disturbances are small and

transient stability problem when the disturbances are large. Transient stability: When a power system is under steady state, the load plus transmission loss equals

to the generation in the system. The generating units run at synchronous speed and system frequency,

voltage, current and power flows are steady. When a large disturbance such as three phase fault, loss

of load, loss of generation etc., occurs the power balance is upset and the generating units rotors

experience either acceleration or deceleration. The system may come back to a steady state condition

maintaining synchronism or it may break into subsystems or one or more machines may pull out of

synchronism. In the former case the system is said to be stable and in the later case it is said to be

unstable.

EXERCISE: A 60Hz synchronous generator having inertia constant H = 9.94 MJ/MVA and a direct axis transient

reactance Xd’= 0.3 per unit is connected to an infinite bus through a purely resistive circuit as shown

in fig.1. Reactances are marked on the diagram on a common system base. The generator is delivering

real power of 0.6% unit, 0.8 pf lagging and the infinite bus at a voltage of 1 per unit. Assume the p.u

damping power coefficient d=0.138. Consider a small disturbance change in delta=100 . Obtain the

equation describes the motion of the rotor angle and generation frequency


40

If the generation is operating at steady state at when the input power is increased by

small amount . The generator excitation and infinite bus bar voltage are same

as before Obtain a simulink block diagram of state space mode

and simulate to obtain the response.


41


42

RESULT:



43

Exp.No: 9 Date:

POWER FLOW SOLUTION OF 3 – BUS SYSTEM

Aim: To perform power flow solution of a 3-Bus system.

Apparatus: MATLAB-PSAT

Theory:

Slack Bus: To calculate the angles θi (as discussed above), a reference angle (θi = 0) needs to

be specified so that all the other bus voltage angles are calculated with respect to this reference

angle. Moreover, physically, total power supplied by all the generation must be equal to the sum

of total load in the system and system power loss. However, as the system loss cannot be

computed before the load flow problem is solved, the real power output of all the generators in

the system cannot be pre-specified. There should be at least one generator in the system which

would supply the loss (plus its share of the loads) and thus for this generator, the real power

output can’t be pre-specified. However, because of the exciter action, Vi for this generator can

still be specified. Hence for this generator, Vi and θi(= 0) are specified and the quantities Pi and

Qi are calculated. This generator bus is designated as the slack bus. Usually, the largest generator

in the system is designated as the slack bus.

In the General Load Flow Problem the Bus with largest generating capacity is taken as the Slack

Bus or Swing Bus. The Slack Bus Voltage is taken to be 1.0 + j 0 P.U. and should be capable of

supplying total Losses in the System. But usually the generator bus are only having station

auxiliary which may be only up to 3% of total generation . If the Generation at Slack Bus is

more it can take more load connected to the slack bus.

A slack bus is usually a generator bus with a large real and reactive power output. It is assumed

that its real and reactive power outputs are big enough that they can be adjusted as required in

order to balance the power in the whole system so that the power flow can be solved. A slack

bus can have load on it because in real systems it is actually the bus of a power plant, which can

have its own load. It also takes care of the Line losses


44

Circuit Diagram:

Procedure :

Create a new file in MATLAB-PSAT

Browse the components in the library and build the bus system.

Save the file and upload the data file in PSAT main window

Execute the program and run powerflow

Get the network visualization.


45


46


47

Result:



48

Exp.No: 10(a) Date:

OPTIMAL DISPATCH NEGLECTING LOSSES

Aim : To develop a program for solving economic dispatch problem without transmission losses for a given load condition using direct method and Lambda-iteration method.

Apparatus: MATLAB

Theory:

As the losses are neglected, the system model can be understood as shown in Fig, here n number

of generating units are connected to a common bus bar, collectively meeting the total power

demand PD. It should be understood that share of power demand by the units does not involve

losses.

Since transmission losses are neglected, total demand PD is the sum of all generations of n-

number of units. For each unit, a cost functions Ci is assumed and the sum of all costs computed

from these cost functions gives the total cost of production CT.

Fig : System with n-generators

where the cost function of the ith unit, from Eq. (1.1) is:

Ci = αi + βiPi + γiPi2

Now, the ED problem is to minimize CT, subject to the satisfaction of the following equality

and inequality constraints.

Equality constraint

The total power generation by all the generating units must be equal to the power demand.

http://my.safaribooksonline.com/9788131755914/navPoint-12#C1F7http://my.safaribooksonline.com/9788131755914/navPoint-9#C1E1


49

where Pi = power generated by ith unit

PD = total power demand.

Inequality constraint

Each generator should generate power within the limits imposed.

Pimin ≤ Pi ≤ Pi

max i = 1, 2, … , n

Economic dispatch problem can be carried out by excluding or including generator power

limits, i.e., the inequality constraint.

The constrained total cost function can be converted into an unconstrained function by using

the Lagrange multiplier as:

The conditions for minimization of objective function can be found by equating partial

differentials of the unconstrained function to zero as

Since Ci = C1 + C2+…+Cn

From the above equation the coordinate equations can be written as:

The second condition can be obtained from the following equation:


50

Equations Required for the ED solution

For a known value of λ, the power generated by the ith unit from can be written as:

which can be written as:

The required value of λ is:

The value of λ can be calculated and compute the values of Pi for i = 1, 2,…, n for optimal

scheduling of generation.

Exercise:


51

𝑀𝑊 𝐿𝑖𝑚𝑖𝑡𝑠 = [10 8510 8010 70

] 𝑐𝑜𝑠𝑡 = [200 7 0.008180 6.3 0.009140 6.8 0.007

]

Find the optimal dispatch neglecting losses.

Procedure :



Execute the program in run mode by selecting tools-opf areas-select opf

Run the primal lp

View the results in case information-Generator fuel costs.


Power World diagram:


52

Results:

Calculations


53

Result



54

Exp.No: 10(b) Date:

OPTIMAL DISPATCH INCLUDING LOSSES

Aim : To develop a program for solving economic dispatch problem including transmission losses for a given load condition using direct method and Lambda-iteration method.

Apparatus: MATLAB

Theory : When the transmission losses are included in the economic dispatch problem, we

can modify (5.4) as

LOSSNT PPPPP 21

where PLOSS is the total line loss. Since PT is assumed to be constant, we have

LOSSN dPdPdPdP 210

In the above equation dPLOSS includes the power loss due to every generator, i.e.,

N

N

LOSSLOSSLOSSLOSS dP

P

PdP

P

PdP

P

PdP

2

2

1

1

Also minimum generation cost implies dfT = 0 as given Multiplying by and combing we get

N

N

LOSSLOSSLOSS dPP

PdP

P

PdP

P

P

2

2

1

1

0

Adding with we obtain

N

i

i

i

LOSS

i

T dPP

P

P

f

1

0

The above equation satisfies when

NiP

P

P

f

i

LOSS

i

T ,,1,0

Again since

NiP

df

P

f

i

T

i

T ,,1,


55

we get NN

N

i

LdP

dfL

dP

dfL

dP

df 2

2

21

1

where Li is called the penalty factor of load-i and is given by

NiPP

LiLOSS

i ,,1,1

1

Consider an area with N number of units. The power generated are defined by the vector

TNPPPP 21

Then the transmission losses are expressed in general as

BPPP TLOSS

where B is a symmetric matrix given by

NNNN

N

N

BBB

BBB

BBB

B

21

22212

11211

The elements Bij of the matrix B are called the loss coefficients. These coefficients are not

constant but vary with plant loading. However for the simplified calculation of the penalty factor

Li these coefficients are often assumed to be constant.

Exercise:


56

𝑀𝑊 𝐿𝑖𝑚𝑖𝑡𝑠 = [10 8510 8010 70

] 𝑐𝑜𝑠𝑡 = [200 7 0.008180 6.3 0.009140 6.8 0.007

]

Find the optimal dispatch including losses.

Procedure :



Execute the program in run mode by selecting tools-opf areas-select opf

Run the primal lp

View the results in case information-Generator fuel costs.


Power World Diagram:


57

Results:

Result



58

Exp.No: 11 Date:

THREE PHASE SHORT CIRCUIT ANALYSIS OF A

SYNCHRONOUS MACHINE

Aim: To Analyze symmetrical fault

Apparatus: MATLAB

Theory: The response of the armature current when a three-phase symmetrical short circuit occurs at the terminals of an unloaded synchronous generator.

It is assumed that there is no dc offset in the armature current. The magnitude of the current

decreases exponentially from a high initial value. The instantaneous expression for the fault

current is given by

where Vt is the magnitude of the terminal voltage, α is its phase angle and

is the direct axis subtransient reactance

is the direct axis transient reactance

is the direct axis synchronous reactance

with .

The time constants are


59

is the direct axis subtransient time constant

is the direct axis transient time constant

we have neglected the effect of the armature resistance hence α = π/2. Let us assume that the

fault occurs at time t = 0. From (6.9) we get the rms value of the current as

which is called the subtransient fault current. The duration of the subtransient current is

dictated by the time constant Td . As the time progresses and Td´´< t < Td´ , the first

exponential term will start decaying and will eventually vanish. However since t is still

nearly equal to zero, we have the following rms value of the current

This is called the transient fault current. Now as the time progress further and the second

exponential term also decays, we get the following rms value of the current for the sinusoidal

steady state

In addition to the ac, the fault currents will also contain the dc offset. Note that a symmetrical

fault occurs when three different phases are in three different locations in the ac cycle.

Therefore the dc offsets in the three phases are different. The maximum value of the dc offset

is given by

where TA is the armature time constant.

Procedure:






6. Plot the waveforms


60

Circuit Diagram:


61

Graph:

Calculations:


62


63

Result



64

Exp.No: 12 Date:

UNSYMMETRICAL FAULT ANALYSIS

Aim: To analyze unsymmetrical faults like LG,LL,LLG

Apparatus: MATLAB

Theory:

Single Line-to-Ground Fault

The single line-to-ground fault is usually referred as “short circuit” fault and occurs when

one conductor falls to ground or makes contact with the neutral wire. The general

representation of a single line-to-ground fault is shown in Figure 3.10 where F is the fault

point with impedances Zf. Figure 3.11 shows the sequences network diagram. Phase a is

usually assumed to be the faulted phase, this is for simplicity in the fault analysis calculations.

[1]

a

c

b

+

Vaf

-

F

Iaf Ibf = 0 Icf = 0

n

Zf

F0

Z0

N0

+

Va0

-

Ia0

F1

Z1

N1

+

Va1

-

Ia1

+

1.0

-

F2

Z2

N2

+

Va2

-

Ia2

3Zf

Iaf

General representation of a single Sequence network diagram of a

line-to-ground fault. single line-to-ground fault

Since the zero-, positive-, and negative-sequence currents are equals as it can be

observed in Figure 3.11. Therefore,

0 1 2

0 1 2

1.0 0

3a a a

f

I I IZ Z Z Z


65

With the results obtained for sequence currents, the sequence voltages can be obtained

from

0 0

2

1 1

2

22

0 1 1 1

1.0 0 1

0 1

a a

b a

ac

V I

V a a I

a a IV

By solving Equation

0 0 0

1 1 1

2 2 2

1.0

a a

a a

a a

V Z I

V Z I

V Z I

If the single line-to-ground fault occurs on phase b or c, the voltages can be found by the

relation that exists to the known phase voltage components,

0

2

1

2

2

1 1 1

1

1

af a

bf a

cf a

V V

V a a V

V a a V

as

2

0 1 2

2

0 1 2

bf a a a

cf a a a

V V a V aV

V V aV a V

Line-to-Line Fault

A line-to-line fault may take place either on an overhead and/or underground

transmission system and occurs when two conductors are short-circuited. One of the

characteristic of this type of fault is that its fault impedance magnitude could vary over a wide

range making very hard to predict its upper and lower limits. It is when the fault impedance is

zero that the highest asymmetry at the line-to-line fault occurs

The general representation of a line-to-line fault is shown in Figure 3.12 where F is the

fault point with impedances Zf. Figure 3.13 shows the sequences network diagram. Phase b and

c are usually assumed to be the faulted phases; this is for simplicity in the fault analysis

calculations [1],


66

a

c

b

F

Iaf = 0 Ibf Icf = -Ibf

Zf

F0

Z0

N0

+

Va0 = 0

-

Ia0 = 0

Zf

F1

Z1

N1

+

Va1

-

Ia1

F2

Z2

N2

+

Va2

-

Ia2

+

1.0 0o

-

Sequence network diagram of a line-to-line fault Sequence network diagram of a single

line-to-line fault.

It can be noticed that

0afI

bf cfI I

bc f bfV Z I

And the sequence currents can be obtained as

0 0aI

1 2

1 2

1.0 0a a

f

I IZ Z Z

If Zf = 0,

1 2

1 2

1.0 0a aI I

Z Z

The fault currents for phase b and c can be obtained as

13 90bf cf aI I I

The sequence voltages can be found as

0

1 1 1

2 2 2 2 1

0

1.0 -

a

a a

a a a

V

V Z I

V Z I Z I

Finally, the line-to-line voltages for a line-to-line fault can be expressed as


67

ab af bf

bc bf cf

ca cf af

V V V

V V V

V V V

Double Line-to-Ground Fault

A double line-to-ground fault represents a serious event that causes a significant

asymmetry in a three-phase symmetrical system and it may spread into a three-phase fault when

not clear in appropriate time. The major problem when analyzing this type of fault is the

assumption of the fault impedance Zf , and the value of the impedance towards the ground Zg.

The general representation of a double line-to-ground fault is shown in Figure 3.14

where F is the fault point with impedances Zf and the impedance from line to ground Zg . Figure

3.15 shows the sequences network diagram. Phase b and c are assumed to be the faulted phases,

this is for simplicity in the fault analysis calculations.

a

c

b

F

Iaf = 0 Ibf Icf

n

Zf Zf

Zg Ibf +Icf

N

Zf +3Zg

F0

Z0

N0

+

Va0

-

Ia0 Zf

F1

Z1

N1

+

Va1

-

Ia1 Zf

F2

Z2

N2

+

Va2

-

Ia2

+

1.0 0o

-

General representation of a Sequence network diagram

double line-to-ground fault. of a double line-to-ground fault

It can be observed that

0

( )

( )

af

bf f g bf g cf

cf f g cf g bf

I

V Z Z I Z I

V Z Z I Z I

The positive-sequence currents can be found as

12 0

1

2 0

1.0 0

( )( 3 )( )

( ) ( 3 )

af f g

f

f f g

IZ Z Z Z Z

Z ZZ Z Z Z Z

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z


68

20 1

2 0

( )[ ]( ) ( 3 )

fa a

f f g

Z ZI I

Z Z Z Z Z

An alternative method is,

0 1 2

0 1 2

0

( )

af a a a

a a a

I I I I

I I I

If Zf and Zg are both equal to zero, then the positive-, negative-, and zero-sequences can

be obtained from

12 0

1

2 0

1.0 0

( )( )( )

( )

aIZ Z

ZZ Z

02 1

2 0

( )[ ]( )

a aZ

I IZ Z

20 1

2 0

( )[ ]( )

a aZ

I IZ Z

The current for phase a is

0afI

2

0 1 2

2

0 1 2

bf a a a

cf a a a

I I a I aI

I I aI a I

The total fault current flowing into the neutral is

03n a bf cfI I I I

The resultant phase voltages from the relationship given in Equation 3.78 can be

expressed as

0 1 2 13

0

af a a a a

bf cf

V V V V V

V V

And the line-to-line voltages are

0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V


69

Circuit Diagram:

Procedure:






6. Plot the waveforms

Graph:


70

Calculations:


71


72

Result



73

Exp.No: 13 Date:

Z-BUS BUILDING ALGORITHM

Aim: To determine the bus impedance matrix for the given power system network.

Apparatus: MATLAB

Theory:

Formation of Z BUS matrix

Z-bus matrix is an important matrix used in different kinds of power system study such

as short circuit study, load flow study etc. In short circuit analysis the generator uses transformer

impedance must be taken into account. In quality analysis the two-short element are neglected

by forming the z-bus matrix which is used to compute the voltage distribution factor. This can

be largely obtained by reversing the y-bus formed by inspection method or by analytical method.

Taking inverse of the y-bus for large system in time consuming; Moreover modification in the

system requires whole process to be repeated to reflect the changes in the system. In such cases

is computed by z-bus building algorithm.

Algorithm

Step 1: Read the values such as number of lines, number of buses and line data, generator data

and transformer data.

Step 2: Initialize y-bus matrix y-bus[i] [j] =complex.(0.0,0.0)

Step 3: Compute y-bus matrix by considering only line data.

Step 4: Modifies the y-bus matrix by adding the transformer and the generator admittance to the

respective diagonal elements of y-bus matrix.

Step 5: Compute the z-bus matrix by inverting the modified y-bus matrix.

Step 6: Check the inversion by multiplying modified y-bus and z-bus matrices to check

whether the resulting matrix is unit matrix or not.

Step 7: Print the z-bus matrix.


74

Procedure:

Enter the command window of the MATLAB.

Create a new M – file by selecting File - New – M – File.

Type and save the program in the editor Window.

Execute the program by pressing Tools – Run.

View the results.

Read the no. Of buses , no of

lines and line data

START

Form Y bus matrix using the algorithm

Y(i,i) =Y(i,i)+Yseries(l) +0.5Yseries(l)

Y(j,j) =Y(j,j)+Yseries(l) +0.5Yseries(l)

Y(i,j) = -Yseries(l)

Y(j,i) =Y(i,j)

Modifying the y bus by adding generator

and transformer admittances to respective

diagonal elements

Compute Z bus matrix by inverting

modified Y bus

Multiply modified Y bus and Z bus and check

whether the product is a unity matrix

Print all the results

STOP


75

MATLAB Program: clc

clear all

z01 = 0.2j;

z02 = 0.4j;

z13 = 0.4j;

z23 = 0.4j;

z12 = 0.8j;

disp('STEP1: Add an element between reference node(0) and node(1)')

zBus1 = [z01]

disp('STEP2: Add an element between existing node(1) and new node(2)')

zBus2 = [zBus1(1,1) zBus1(1,1);

zBus1(1,1) zBus1(1,1)+z12]

disp('STEP3: Add an element between existing node(2) and reference node(0)')

zBus4 = [zBus2(1,1) zBus2(1,2) zBus2(1,2);

zBus2(2,1) zBus2(2,2) zBus2(2,2);

zBus2(1,2) zBus2(2,2) zBus2(2,2)+z02]

disp('Fictitious node(0) can be eliminated')

zz11 = zBus4(1,1) - ((zBus4(1,3) * zBus4(3,1)) / zBus4(3,3));


zz21 = zz12;


zBus5 = [zz11 zz12;

zz21 zz22]

disp('STEP4: Add an element between existing node(2) and new node(3)')

zBus6 = [zBus5(1,1) zBus5(1,2) zBus5(1,2);

zBus5(2,1) zBus5(2,2) zBus5(2,2);

zBus5(2,1) zBus5(2,2) zBus5(2,2)+z23]

disp('STEP5: Add an element between existing nodes (3) and (1)')

zBus7 = [zBus6(1,1) zBus6(1,2) zBus6(1,3) zBus6(1,3)-zBus6(1,1);

zBus6(2,1) zBus6(2,2) zBus6(2,3) zBus6(2,3)-zBus6(2,1);

zBus6(3,1) zBus6(3,2) zBus6(3,3) zBus6(3,3)-zBus6(3,1);

zBus6(3,1)-zBus6(1,1) zBus6(3,2)-zBus6(1,2) zBus6(3,3)-zBus6(1,3) z23+zBus6(1,1)+zBus6(3,3)-

2*zBus6(1,3)]

disp('Fictitious node(0) can be eliminated')

zzz11 = zBus7(1,1) - ((zBus7(1,4) * zBus7(4,1)) / zBus7(4,4));


76



zzz21 = zzz12;



zzz31 = zzz13;

zzz32 = zzz23;


disp('RESULT:')

zBus = [zzz11 zzz12 zzz13;

zzz21 zzz22 zzz23;

zzz31 zzz32 zzz33]

OUTPUT:

Calculations:


77

Result:



78

Exp.No: 14(a) Date:

SYMMETRICAL COMPONENTS

Aim: To obtain symmetrical components of set of unbalanced currents

Apparatus: MATLAB

Theory:

Before we discuss the symmetrical component transformation, let us first define the a-

operator.

2

3

2

10120 jea j

Note that for the above operator the following relations hold

on so and

1

2

3

2

1

22403606005

1203604804

3603

2402

000

000

0

0

aeeea

aeeea

ea

ajea

jjj

jjj

j

j

Also note that we have

02

3

2

1

2

3

2

111 2 jjaa

Using the a-operator we can write from Fig. 7.1 (b)

111

2

1 and acab aVVVaV

Similarly

2

2

222 and acab VaVaVV

Finally

000 cba VVV

The symmetrical component transformation matrix is then given by


79

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1

Defining the vectors Va012 and Vabc as

c

b

a

abc

a

a

a

a

V

V

V

V

V

V

V

V ,

2

1

0

012

Program:

V012 = [0.6 90

1.0 30

0.8 -30];

rankV012=length(V012(1,:));

if rankV012 == 2

mag= V012(:,1); ang=pi/180*V012(:,2);

V012r=mag.*(cos(ang)+j*sin(ang));

elseif rankV012 ==1

V012r=V012;

else

fprintf('\n Symmetrical components must be expressed in a one column array in

rectangular complex form \n')

fprintf(' or in a two column array in polar form, with 1st column magnitude & 2nd column

\n')

fprintf(' phase angle in degree. \n')

return, end

a=cos(2*pi/3)+j*sin(2*pi/3);

A = [1 1 1; 1 a^2 a; 1 a a^2];

Vabc= A*V012r

Vabcp= [abs(Vabc) 180/pi*angle(Vabc)];

fprintf(' \n Unbalanced phasors \n')

fprintf(' Magnitude Angle Deg.\n')

disp(Vabcp)

Vabc0=V012r(1)*[1; 1; 1];

Vabc1=V012r(2)*[1; a^2; a];

Vabc2=V012r(3)*[1; a; a^2];


80

Procedure:

1. Open Matlab--> File ---> New---> Script

2. Write the program

3. Enter F5 to run the program

4. Observe the results in MATLAB command window.

Result:

Vabc =

1.5588 + 0.7000i

-0.0000 + 0.4000i

-1.5588 + 0.7000i

Unbalanced phasors

Magnitude Angle Deg.

1.7088 24.1825

0.4000 90.0000

1.7088 155.8175


81

Result



82

Exp.No: 14(b) Date:

UNBALANCED VOLTAGES FROM SYMMETRICAL

COMPONENTS

Aim: To obtain the original unbalanced phase voltages from symmetrical components

Apparatus: MATLAB

Theory:

abca CVV 012

where C is the symmetrical component transformation matrix and is given by

aa

aaC2

2

1

1

111

3

1

The original phasor components can also be obtained from the inverse symmetrical

component transformation, i.e.,

012

1

aabc VCV

Inverting the matrix C we get

2

1

0

1

2

1

0

2

2

1

1

111

a

a

a

a

a

a

c

b

a

V

V

V

C

V

V

V

aa

aa

V

V

V

210 aaaa VVVV

21021

2

0 bbbaaab VVVaVVaVV

2102

2

10 cccaaac VVVVaaVVV

Finally, if we define a set of unbalanced current phasors as Iabc and their symmetrical

components as Ia012, we can then define

012

1

012

aabc

abca

ICI

CII


83

Program:

Iabc = [1.6 25

1.0 180

0.9 132];

rankIabc=length(Iabc(1,:));

if rankIabc == 2

mag= Iabc(:,1); ang=pi/180*Iabc(:,2);

Iabcr=mag.*(cos(ang)+j*sin(ang));

elseif rankIabc ==1

Iabcr=Iabc;

else

fprintf('\n Three phasors must be expressed in a one column array in rectangular complex

form \n')

fprintf(' or in a two column array in polar form, with 1st column magnitude & 2nd column

\n')

fprintf(' phase angle in degree. \n')

return, end

a=cos(2*pi/3)+j*sin(2*pi/3);

A = [1 1 1; 1 a^2 a; 1 a a^2];

I012=inv(A)*Iabcr;

symcomp= I012

I012p = [abs(I012) 180/pi*angle(I012)];

fprintf(' \n Symmetrical components \n')

fprintf(' Magnitude Angle Deg.\n')

disp(I012p)

Iabc0=I012(1)*[1; 1; 1];

Iabc1=I012(2)*[1; a^2; a];

Iabc2=I012(3)*[1; a; a^2];

Result:

symcomp =

-0.0507 + 0.4483i

0.9435 - 0.0009i

0.5573 + 0.2288i

Symmetrical components

Magnitude Angle Deg.

0.4512 96.4529

0.9435 -0.0550

0.6024 22.3157


84

Result



85

Exp.No: 15 Date:

SHORT CIRCUIT ANALYSIS OF IEEE 9 BUS SYSTEM

Aim: To obtain the short circuit analysis for a IEEE 9 bus system

Apparatus: POWERWORLD

Theory:

ff IZV 44

3,2,1,44

44 iV

Z

ZIZV f

ifii

We further assume that the system is unloaded before the fault occurs and that the magnitude

and phase angles of all the generator internal emfs are the same. Then there will be no current

circulating anywhere in the network and the bus voltages of all the nodes before the fault will

be same and equal to Vf. Then the new altered bus voltages due to the fault will be given from

by

4,,1,144

4

iV

Z

ZVVV f

iifi

IEEE 9 bus system :

Sbase = 100 MVA . Vbase = 220KV

Vmax = 1.06 p.u Vmin = 1 p.u

NUMBER OF LINES = 8 NUMBER OF BUSES = 9


86

Case Data

Bus Records

Bus No Area PU Volt Volt (kV) Ang(Deg)

Load

MW

Load

Mvar Gen MW Gen Mvar

1 1 1 345 0 0 0

2 1 1 345 0 163 0

3 1 1 345 0 85 0

4 1 1 345 0

5 1 1 345 0 90 30

6 1 1 345 0

7 1 1 345 0 100 35

8 1 1 345 0

9 1 1 345 0 125 50

Line Records

Generator Data

Generator Cost Data

Number

Gen

MW IOA IOB IOC IOD

Min

MW

Max

MW Cost $/Hr IC LossSens Lambda

1 0 150 5 0.11 0 10 250 150 5 0 5

From To Xfrmr R X C

Lt A

MVA

LtB

MVA

Lt C

MVA

1 4 No 0 0.0576 0 250 250 250

8 2 No 0 0.0625 0 250 250 250

3 6 No 0 0.0586 0 300 300 300

4 5 No 0.017 0.092 0.158 250 250 250

9 4 No 0.01 0.085 0.176 250 250 250

5 6 No 0.039 0.17 0.358 150 150 150

6 7 No 0.0119 0.1008 0.209 150 150 150

7 8 No 0.0085 0.072 0.149 250 250 250

8 9 No 0.032 0.161 0.306 250 250 250

Number Gen MW Gen Mvar Min MW Max MW Min Mvar

Max

Mvar

Cost

Model Part. Fact

1 0 0 10 250 -300 300 Cubic 10

2 163 0 10 300 -300 300 Cubic 10

3 85 0 10 270 -300 300 Cubic 10


87

2 163 600 1.2 0.085 0 10 300 3053.97 28.91 0 28.91

3 85 335 1 0.1225 0 10 270 1305.06 21.83 0 21.83

Procedure :



Execute the program in run mode by selecting tools-fault analysis

Select the fault on which bus and calculate


Results

Fault current =

Fault current angle =


88

Result



89

Exp.No: 16 Date:

POWER FLOW ANALYSIS OF SLACK BUS CONNECTED TO DIFFERENT LOADS

Aim: To perform power flow analysis of a slack bus connected to different loads.


Theory:

Slack Bus: To calculate the angles θi (as discussed above), a reference angle (θi = 0) needs to

be specified so that all the other bus voltage angles are calculated with respect to this reference

angle. Moreover, physically, total power supplied by all the generation must be equal to the sum

of total load in the system and system power loss. However, as the system loss cannot be

computed before the load flow problem is solved, the real power output of all the generators in

the system cannot be pre-specified. There should be at least one generator in the system which

would supply the loss (plus its share of the loads) and thus for this generator, the real power

output can’t be pre-specified. However, because of the exciter action, Vi for this generator can

still be specified. Hence for this generator, Vi and θi(= 0) are specified and the quantities Pi and

Qi are calculated. This generator bus is designated as the slack bus. Usually, the largest generator

in the system is designated as the slack bus.




auxiliary which may be only up to 3% of total generation . If the Generation at Slack Bus is

more it can take more load connected to the slack bus.



order to balance the power in the whole system so that the power flow can be solved. A slack

bus can have load on it because in real systems it is actually the bus of a power plant, which can

have its own load. It also takes care of the Line losses


90

Circuit Diagram:


91

Procedure :







92


93

Result:



94

Exp.No:17 Date:

LOAD FLOW ANALYSIS OF 3 MOTOR SYSTEMS CONNECTED TO SLACK BUS

Aim: To perform load flow analysis of 3 motor systems connected to slack bus


Theory:

Slack Bus: To calculate the angles θi (as discussed above), a reference angle (θi = 0) needs to be

specified so that all the other bus voltage angles are calculated with respect to this reference angle.

Moreover, physically, total power supplied by all the generation must be equal to the sum of total

load in the system and system power loss. However, as the system loss cannot be computed before

the load flow problem is solved, the real power output of all the generators in the system cannot

be pre-specified. There should be at least one generator in the system which would supply the

loss (plus its share of the loads) and thus for this generator, the real power output can’t be pre-

specified. However, because of the exciter action, Vi for this generator can still be specified.

Hence for this generator, Vi and θi(= 0) are specified and the quantities Pi and Qi are calculated.

This generator bus is designated as the slack bus. Usually, the largest generator in the system is

designated as the slack bus.




auxiliary which may be only up to 3% of total generation . If the Generation at Slack Bus is more

it can take more load connected to the slack bus.



order to balance the power in the whole system so that the power flow can be solved. A slack bus

can have load on it because in real systems it is actually the bus of a power plant, which can have

its own load. It also takes care of the Line losses


95

Circuit Diagram:

Procedure :







96


97

RESULT:


Documents

Gokaraju Rangaraju Institute of Engineering and Technology PSS Lab... · 2019. 12. 28. · Author: GRIET-JSD Created Date: 5/9/2017 1:29:36 PM