Power System Analysis Lab Manual

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POWER SYSTEM ANALYSIS

SUBMITTED TO:

ENGR.M.JUNAID

SUBMITTED BY:

ASAD NAEEM

2006-RCET-EE-22

DEPARTMENT OF ELECTRICAL ENGINEERING

(A CONSTITUENT COLLEGE: RACHNA COLLEGE OF ENGINEERING & TECHNOLOGY GUJRANWALA)

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01

To plot the daily load curve for the given data using MATLAB

02 Introduction to basics of Electrical Transients Analyzer _{Program (ETAP)}

03 Evaluate the value of voltages for a 4-BUS system using node _{equations in MATLAB}

04 Modeling and Load flow analysis of RCET power distribution

network using ETAP

05

Bus elimination of a 4-BUS system using MATLAB

06

To study the Concept of Modifications of an Existing Bus-Impedance Matrix & Implementing in MATLAB

07 Application of Gauss-Siedal and Newton-Raphson method for

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08

Harmonic Load Modeling using built-in and user defined models of ETAP

09

Impact of personal computer load on power distribution network of RCET

10 _{different schemes of connection for a 3-phase transformer with}Flow of triplen harmonics (zero-sequence harmonics) during 5

presence of large non-linear load using ETAP

11 _{Three phase short circuit analysis (3-phase faults-device duty)}

for a given power system using ETAP

12 _{Three phase short circuit analysis (3-phase faults-30 cycle}

network) for a given power system using ETAP

13

Three phase short circuit analysis (LG, LL, LLG, & 3-Phase Faults - ½ Cycle) for a given power system using ETAP

14

Three phase short circuit analysis (LG, LL, LLG, & 3-Phase Faults - 1.5 to 4 Cycle) for a given power system using ETAP

15

Three phase short circuit analysis (LG, LL, LLG, & 3-Phase Faults - 30 Cycle) for a given power system using ETAP

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EXPERIMENT#01

To plot the daily load curve for the given data using

MATLAB

Given data:

Interval from

To

Load MW

12 A.M

2 A.M

6

2

6

5

6

9

10

9

12

15 12 P.M

2 P.M

12

2

4

14

4

6

16

6

8

18

8

10

16

10

11

12

11 12 A.M

6 Requirements:

1. Find average value of load

2. Find peak value of load

3. Find the load factor

4. Plot the load curve

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Theory

Loads:

Loads of power systems are divided into three main categories that are given below.

1. Industrial Loads 2. Commercial Loads 3. Residential Loads

Very large industrial loads are served through the transmission lines. Large industrial loads are served directly from the sub-transmission level. And small industrial loads are served directly from the primary distribution network. The industrial loads are composite loads and induction motors from a high proportion of these loads. These composite loads are functions of voltage and frequency and form a major part of the system load. Commercial and residential load consist largely of lighting, heating and cooling. These loads are independent of frequency and consume negligibly small reactive power.

The real power of loads is expressed in terms of kilowatts or megawatts. The magnitude of load varies throughout the day and power must be available to the consumer on demand.

The daily load curve of a utility is a composite of demands made by various classes of users. The greatest value of load during a twenty four hours is called the peak or maximum demand. Smaller peaking generators may be commissioned to meet the peak load that occurs for only a few hours. In order to asses the usefulness of the generating plant the load factor is defined.

The load factor is the ratio of average load over a designated period of time to the peak load occurring in that period. Load factor may be given for a day, a month or an year. Yearly or annual load factor is the most useful since a year represents a full cycle of time. The daily load factor is

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Daily load factor = average load / peak load

Multiplying the numerator and denominator by a time period of 24 hr we have

Daily load factor= average load*24 hr / (peak load*24 hrs) = energy consumed during 24 hr/ (peak load*24 hr) The annual load factor is

Annual load factor = total annual energy / (peak load*8760 hr) Today’s typical system load factors are in range of 55-70%. In Pakistan WAPDA standard for urban areas load factor is 60% and that of rural areas is 65%.

Matlab code:

data=[0 2 6; 2 6 5; 6 9 10; 9 12 15; 12 14 12; 14 16 14; 16 18 16; 18 20 18; 20 22 16; 22 23 12;

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p=data(:,3); Dt=data(:,2)-data(:,1); w=p'*Dt; pavg=w/sum(Dt) peak=max(p) LF=pavg/peak*100 L=length(data); tt = [data(:,1) data(:,2)]; t = sort(reshape(tt, 1, 2*L)); for n = 1:L pp(2*n-1)=p(n); pp(2*n)=p(n); end plot(t,pp)

xlabel('TIME,Hr'),ylabel('P,MW')

Matlab results:

pavg =11.5417

peak =18

LF =64.1204

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0 5 10 15 20 25 4 6 8 10 12 14 16 18 TIME,Hr P,M W

COMMENTS:

In this experiment we learn how to find the daily load curve for any power system using MATLAB. Load curve is very important as we can achieve very important information from it like:

• Peak load • Average load • Load factor

These quantities are very helpful for understanding any power system.

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EXPERIMENT#02

Introduction to basics of Electrical Transients Analyzer

Program (ETAP)

What is ETAP?

ETAP is the most comprehensive analysis platform for the design, simulation, operation, control, optimization, and automation of

generation, transmission, distribution, and industrial power systems.

Project Toolbar

The Project Toolbar contains icons that allow you to perform shortcuts of many commonly used functions in PowerStation. Create Create a new project file Open Open an existing project file Save Save the project file Print Print the one‐line diagram or U/G raceway system Cut Cut the selected elements from the one‐line diagram or U/G raceway system to the Dumpster Copy Copy the selected elements from the one‐line diagram or U/G raceway system to the Dumpster Paste Paste elements from a Dumpster Cell to the one‐line diagram or U/G raceway system Zoom In Magnify the one‐line diagram or U/G raceway system Zoom Out Reduce the one‐line diagram or U/G raceway system Zoom to Fit Page Re‐size the one‐line diagram to fit the window

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Power Calculator Activate PowerStation Calculator that relates MW, MVAR, MVA, kV, Amp, and PF together with either kVA or MVA units Help Point to a specific area to learn more about PowerStation

Mode Toolbar

ETAP offers a suite of fully integrated software solutions including arc flash, load flow, short circuit, transient stability, relay coordination, cable ampacity, optimal power flow, and more. Its modular functionality can be customized to fit the needs of any company, from small to large power systems.

Edit Mode

Edit mode enables you to build your one‐line diagram, change system connections, edit engineering properties, save your project, and generate schedule reports in Crystal Reports formats. The Edit Toolbars for both AC and DC elements will be displayed to the right of the screen when this mode is active. This mode provides a wide variety of tasks including: ∙ Drag & Drop Elements ∙ Connect Elements ∙ Change IDs ∙ Cut, Copy, & Paste Elements ∙ Move from Dumpster ∙ Insert OLE Objects ∙ Cut, Copy & OLE Objects ∙ Merge PowerStation Project ∙ Hide/Show Groups of Protective Devices ∙ Rotate Elements ∙ Size Elements ∙ Change Symbols ∙ Edit Properties ∙ Run Schedule Report Manager

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Example implementation:

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EXPERIMENT#03

Evaluate the value of voltages for a 4BUS system using

node equations in MATLAB

GIVEN ONE LINE DIAGRAM

REACTANCE DIAGRAM

In the first step, we draw the reactance diagram of the given one-line diagram as shown below:

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SOURCE TRANSFORM

• After making the reactance diagram, we apply source transformation on the given network by replacing the voltage sources with current sources

• Replace all the reactance by admittances using the relation:

• Y=1/X

• The resultant diagram now can be shown as:

NODE EQUATIONS

Now, using the above figure write the node equations of the system:

• Applying KCL at node-1:

I1= (V1-0) y10 + (V1-V4) y14+ (V1-V3) y13

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I2= (V2-0) y20 + (V2-V3) y23+ (V2-V4) y24

I2= 0V1+ (y20+y23+y24) V2 + (-y23) V3+ (-y24) V4

I3= (V3-0) y30 + (V3-V1) y31+ (V3-V4) y34 + (V3-V2) y32

I3= (-y31) V1+ (-y32) V2+ (y30+y31+y34) V3 + (-y34) V4

0= (V4-V1) y14+ (V4-V3) y43 + (V4-V2) y42

0= (-y14) V1+ (-y42) V2 + (-y34) V3+ (y14+y43+y42) V4

Matrix form of the node equations is:

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CALCULATIONS

MATLAB CODE

YBUS= [0-9.80i 0 0+4.00i 0+5.00i; 0 0-8.30i 0+2.50i 0+5.00i; 0+4.00i 0+2.50i 0-15.30i 0+8.00i; 0+5.00i 0+5.00i 0+8.00i 0-18.00i]; I= [0-1.20i; 0-0.7200-0.9600i; 0-1.2000i; 0];

ZBUS=inv (YBUS); V=ZBUS*I

MATLAB RESULTS

V = 1.4111 - 0.2668i 1.3831 - 0.3508i 1.4059 - 0.2824i 1.4010 - 0.2971i

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COMMENTS:

In this experiment we learn that using the bus impedance or admittance matrix we can find the voltages and currents for all buses of a given power system.

Moreover, we use MATLAB for the calculation of these quantities by just entering the bus impedance matrix and one given quantity (current or voltage) and MATLAB gives the results of very complex networks within no time.

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EXPERIMENT#04

Modeling and Load flow analysis of RCET power

distribution network using ETAP

INTRODUCTION:

LOAD FLOW STUDIES

In power engineering, the power flow study (also known as

load-flow study) is an important tool involving numerical

analysis applied to a power system. Unlike traditional circuit

analysis, a power flow study usually uses simplified notation such as a one-line diagram and per-unit system, and focuses on various forms of AC power (i.e: reactive, real, and

apparent) rather than voltage and current. It analyses the power systems in normal steady-state operation. There exist a number of software implementations of power flow

studies.

The great importance of power flow or load-flow studies is in the planning the future expansion of power systems as well as in determining the best operation of existing systems. The principal information obtained from the power flow study is the magnitude and phase angle of the voltage at each bus and the real and reactive power flowing in each line.

LOAD FLOW STUDIES IN ETAP

ETAP load flow analysis software calculates bus voltages, branch power factors, currents, and power flows throughout the electrical system. ETAP allows for swing, voltage

regulated, and unregulated power sources with multiple power grids and generator connections. It is capable of performing analysis on both radial and loop systems. ETAP

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to achieve the best calculation efficiency and accuracy.

Run Load Flow Studies Update Cable Load Currents

Load Flow display Option Alert View

Report Manager

Halt current calculations

Net on line data

STEPS

¾ Modeling of the main network ¾ Modeling of composite networks ¾ Running of load flow analysis

¾ Complete report from ETAP load flow analyzer

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MODELING OF BASIC RCET NETWORK

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MODELING OF COMPOSITE NETWORKS

STAFF COLONY:

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OLD BUILDING:

NEW BUILDING:

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HOSTEL-A,B:

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HOSTEL-E:

Complete ETAP load flow analysis report of the given

network is attached with this experiment.

COMMENTS:

In this experiment we learn how to:

• Model a power system in ETAP

• Model composite networks in a basic network • Assign properties of components added

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EXPERIMENT#05

Bus elimination of a 4BUS system using MATLAB

REACTANCE DIAGRAM

It is given that the transformer and generator at bus-3 are disconnected, so the reactance diagram now becomes:

SOURCE TRANSFORM

• After making the reactance diagram, we apply source transformation on the given network by replacing the voltage sources with current sources

• Replace all the reactance by admittances using the relation:

• Y=1/X

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Part‐1: Elimination of Bus‐3&4

MATRIX FORM

Where:

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MATLAB CODE

>>YBUS= [0-9.80i 0 0+4.00i 0+5.00i; 0 0-8.30i 0+2.50i 0+5.00i; 0+4.00i 0+2.50i 0-14.5i 0+8.00i; 0+5.00i 0+5.00i 0+8.00i 0-18.00i]; >>K= [0-9.80i 0; 0 0-8.30i];

>>L= [0+4.00i 0+5.00i; 0+2.50i 0+5.00i]; >>M= [0-14.5i 0+8.00i; 0+8.00i 0-18.00i]; >>LT= [0+4.00i 0+2.50i; 0+5.00i 0+5.00i]; >>N=inv (M); >>P=L*N*LT; >>Ybus=K-P

MATLAB RESULTS

Ybus = 0 - 4.8736i 0 + 4.0736i 0 + 4.0736i 0 - 4.8736i

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Part-2: Elimination Bus-4

MATLAB CODE:

>>Ybus=[-9.8i 0 4.0i 5i; 0 -8.3i 2.5i 5i; 4i 2.5i -14.5i 8i; 5i 5i 8i -18i]; >>K=[-9.8i 0 4i;0 -8.3i 2.5i;4i 2.5i -14.5i]; >>L=[5i;5i;8i]; >>M=[-18i]; >>P=L'; >>T=inv(M); >>A=K-L*T*P

MATLAB RESULTS

A=

0 -11.1889i 0 - 1.3889i 0 + 1.7778i 0 - 1.3889i 0 - 9.6889i 0 + 0.2778i 0 + 1.7778i 0 + 0.2778i 0 -18.0556i

Part-3: Elimination Bus-3

MATLAB CODE:

>>P=[-11.1889i -1.3889i;-1.3889i -9.6889i]; >>Q=[1.7778i;0.2778i];

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>>S=Q'; >>T=inv(R); >>B=P-Q*T*S

MATLAB RESULTS

B = 0 -11.3639i 0 - 1.4163i 0 - 1.4163i 0 - 9.6932i

COMMENTS:

Bus impedance matrix is a very important tool for the calculation of voltages and currents at all the buses of a given network. Suppose that any fault occurs in the power system then we can get a task to modify the bus impedance matrix by eliminating the faulty node which will reduce the order of matrix by eliminating the faulty node.

In this experiment we learn how to: • Eliminate last two nodes together • Eliminate only one last node

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EXPERIMENT#06

To study the Concept of Modifications of an Existing Bus

Impedance Matrix & Implementing in MATLAB

IMPEDANCE MATRIX

Impedance matrix is a very important tool in power system analysis. Using this matrix we can find:

• Voltages at all buses when currents are given • Currents at all buses when voltages are given So it is very important that how to modify the bus

impedance matrix when any new impedance is add into the original system.

Suppose a power system with n-buses having the impedances matrix of order n*n:

There are four cases that can take place while adding a new impedance Zb in the system:

• Adding Zb from a new bus-P to reference bus • Adding Zb from a new bus-P to an existing bus-K • Adding Zb from an existing bus-K to reference bus • Adding Zb between two existing buses

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MODIFICATION CASES

CASE‐1: ADDING Zb FROM A NEW BUS TO REFERENCE BUS

This condition is explained in the following diagram:

Clearly,

Vp-0=Ib*Zb Vp=Ib*Zb

Hence the modified matrix will take the form as:

MATLAB CODE

function [Z]=Case1(Zorg,Zb)

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Zb=17; l=length(Zorg); for i=1:l+1 for j=1:l+1 if i<=l && j<=l Znew(i,j)=Zorg(i,j); elseif i==l+1 && j==l+1 Znew(i,j)=Zb; else Znew(i,j)=0; end end end Znew

MATLAB RESULTS

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CASE‐2: ADDING Zb FROM A NEW BUS‐P TO AN EXISTING BUS‐K

Clearly, Vp-Vk,new=Ip*Zb Vp=Vk,new+Ip*Zb Where, Vk,new=Vk,org+Ip*Zkk Vp= Vk,org+Ip(Zkk+Zb)

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MATLAB CODE:

function [Z]=CASE2(Zorg,Zb) Zorg=[1 2 3 4;2 5 6 7;3 6 8 9;4 7 9 10] Zb=5; l=length(Zorg); row =Zorg(l,:); column =Zorg(:,l); for i=1:l+1 for j=1:l+1 if i<=l && j<=l Znew(i,j)=Zorg(i,j); elseif i==l+1 for p=1:l Znew(i,p)=row(p); end elseif j==l+1 for q=1:l Znew(q,j)=column(q); end end if i==l+1 && j==l+1 Znew(i,j)=Zb+Zorg(l,l); end end end Znew

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MATLAB RESULTS

CASE‐3: ADDING Zb FROM AN EXISTING BUS‐K TO REFERENCE

BUS

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Here we can apply the same case as in case-2 and then put Vp=0. This task can be achieved by eliminating the last row and column of the Znew matrix.

Now this matrix is of the order (n+1)*(n+1), we have to achieve a matrix of order n*n using formula:

Zkj(new)=Zkj(org)-(Zk(n+1)Z(n+1)j/Zkk+Zb) In this case, K=n

MATLAB CODE:

function [Z]=CASE3(Zorg,Zb) Zorg=[1 2 3 4;2 5 6 7;3 6 8 9;4 7 9 10] Zb=5; l=length(Zorg); row =Zorg(l,:); column =Zorg(:,l); for i=1:l+1 for j=1:l+1

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Znew(i,j)=Zorg(i,j); elseif i==l+1 for p=1:l Znew(i,p)=row(p); end elseif j==l+1 for q=1:l Znew(q,j)=column(q); end end if i==l+1 && j==l+1 Znew(i,j)=Zb+Zorg(l,l); end end end Znew for a=1:l for b=1:l K(a,b)=Znew(a,b); end

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end for a=1:l L(a,1)=Znew(a,5); end M=Znew(l+1,l+1); P=L'; T=inv(M); Zwithnewbusrefferenced=K-L*T*P

MATLAB RESULTS

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CASE‐4: ADDING Zb BETWEEN TWO EXISTING BUSES

In this case,

Zbb=Zb+Zjj+Zkk-2Zjk

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Here again we have to eliminate the last row and column to achieve the final matrix.

MATLAB CODE:

function [Z]=CASE4(Zorg,Zb) Zorg=[1 2 3 4;2 5 6 7;3 6 8 9;4 7 9 10] Zb=5; l=length(Zorg); R1 =Zorg(l,:); C1 =Zorg(:,l); R2 =Zorg(l-1,:); C2 =Zorg(:,l-1); for i=1:l+1 for j=1:l+1 if i<=l && j<=l Znew(i,j)=Zorg(i,j);

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for p=1:l Znew(i,p)=R1(p)-R2(p); End elseif j==l+1 for q=1:l Znew(q,j)=C1(q)-C2(q); End End if i==l+1 && j==l+1 Znew(i,j)=Zb+Zorg(l,l)+Zorg(l-1,l-1)-(2*Zorg(l,l-1)); end end end Znew for a=1:l for b=1:l K(a,b)=Znew(a,b); end end for a=1:l

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L(a,1)=Znew(a,5); end M=Znew(l+1,l+1); P=L'; T=inv(M); Zfinal=K-L*T*P

MATLAB RESULTS

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COMMENTS:

Bus impedance matrix is a very important tool for the calculation of voltages and currents at all the buses of a given network. Suppose that any improvement occurs in the power system then we can get a task to modify the bus

impedance matrix by adding the new impedance in the system. The new impedance can be added in four different conditions:

• Addition of new impedance from a new bus to reference bus

• Addition of new impedance from a new bus to existing bus

• Addition of new impedance from an existing bus to reference bus

• Addition of new impedance between two existing buses In this experiment we learn how to modify the bus

impedance matrix for all four cases using MATLAB.

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EXPERIMENT#07

Application of GaussSiedal and NewtonRaphson method for load flow studies on a three bus system using MATLAB(Implimentation of example#6.7,6.8 & 6.10 from POWER SYSTEM ANALYSIS by Hadi Saadat)

EXAMPLE 6.7

Given figure shows the one line diagram of a simple three bus system with generation at bus-1. The magnitude of voltage at bus-1 is adjusted to 1.05 per unit. The scheduled loads at buses-2 and 3 are as marked on the diagram. Line impedances are marked in per unit on a 100-MVA base and the line charging susceptances are neglected.

(A) Using the Gauss-Siedal 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 flows and line losses. Construct a power flow diagram showing the direction of line flow

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SOLUTION

Line impedances are converted to admittances:

At the P-Q buses, the complex loads expressed in per unit are:

S2sch=-(256.6+j110.2)/100= -2.566-j1.102 pu

S3sch=-(138.6+j45.2)/100= -1.386-j0.452 pu

Starting from an initial estimate of V2(0)=1.0+j0.0 and

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GAUSS-SIEDEL FARMULA

SLACK

1

=conj(V

1

)*[V

1

*(y

12

+y

13

)-( y

12

*V

2

+y

13

*V

3

)]

S

_ij

=V

_i

*conj(I

_ij

)

I

ij

=y

ij

*(V

i

-V

j

)

MATLAB CODE

y12=10-j*20; y13=10-j*30; y23=16-j*32; V1=1.05+j*0;

%CODE FOR PART-A

iter=0; s2=-2.566-j*1.102; s3=-1.386-j*0.452; V2=1+j*0.0; V3=1+j*0.0; for I=1:10; iter=iter+1; V2=(conj(s2)/conj(V2)+y12*V1+y23*V3)/(y12+y23); V3=(conj(s3)/conj(V3)+y13*V1+y23*V2)/(y13+y23); end V2 V3

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Pslack=conj(V1)*[V1*(y12+y13)-(y12*V2+y13*V3)]

%CODE FOR PART-C

I12=y12*(V1-V2) I21=-I12 I13=y13*(V1-V3) I31=-I13 I23=y23*(V2-V3) I32=-I23 s12=V1*conj(I12) s21=V2*conj(I21) s13=V1*conj(I13) s31=V3*conj(I31) s23=V2*conj(I23) s32=V3*conj(I32) SL12=s12+s21 SL13=s13+s31 SL23=s23+s32

MATLAB RESULTS

PART-A RESULTS V2 = 0.9800 - 0.0600i V3 = 1.0000 - 0.0500i PART-B RESULTS SLACK-BUS POWER Pslack = 4.0949 - 1.8900i

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PART-C RESULTS I12 = 1.9000 - 0.8000i I21 = -1.9000 + 0.8000i I13 = 2.0000 - 1.0000i I31 = -2.0000 + 1.0000i I23 = -0.6400 + 0.4800i I32 = 0.6400 - 0.4800i LINE FLOWS s12 = 1.9950 + 0.8400i s21 = -1.9100 - 0.6700i s13 = 2.1000 + 1.0500i s31 = -2.0500 - 0.9000i s23 = -0.6560 - 0.4320i s32 = 0.6640 + 0.4480i LINE LOSSES SL12 = 0.0850 + 0.1700i SL13 = 0.0500 + 0.1500i SL23 = 0.0080 + 0.0160i

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EXAMPLE 6.8

Given figure shows the one line diagram of a simple three bus system with generators at buses-1 and 3. The

magnitude of voltage at bus-1 is adjusted to 1.05pu. voltage magnitude at bus-3 is fixed at 1.04 pu with a real power generation of 200MW. A load consisting of 400MW and

250MVAR is taken from bus-2. Line impedances are marked in per unit on a 100MVA base, and the line charging

susceptances are neglected. Obtain the power flow solution by the Gauss-Siedal method including line flows and line losses.

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FARMULA’S

S3=conj(V3)*(y33*V3-y13*V1-y23*V2) Q3=-imag(conj(V3)*(y33*V3-y13*V1-y23*V2))

MATLAB CODE

y12=10-j*20; y13=10-j*30; y23=16-j*32; y33=y13+y23; V1=1.05+j*0; format long iter=0; s2=-4.0-j*2.5; p3=2; V2=1+j*0.0; Vm3=1.04; V3=1.04+j*0; for I=1:10; iter=iter+1; E2=V2; E3=V3; V2=(conj(s2)/conj(V2)+y12*V1+y23*V3)/(y12+y23) DV2=V2-E2; Q3=-imag(conj(V3)*(y33*V3-y13*V1-y23*V2)) s3=p3+j*Q3; Vc3=(conj(s3)/conj(V3)+y13*V1+y23*V2)/(y13+y23); Vi3=imag(Vc3); Vr3=sqrt(Vm3^2-Vi3^2);

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DV3=V3-E3; end V2 V3 Q3 format short I12=y12*(V1-V2); I21=-I12; I13=y13*(V1-V3); I31=-I13; I23=y23*(V2-V3); I32=-I23; s12=V1*conj(I12); s21=V2*conj(I21); s13=V1*conj(I13); s31=V3*conj(I31); s23=V2*conj(I23); s32=V3*conj(I32); I1221=[I12,I21]; I1331=[I13,I31]; I2332=[I23,I32]; SL12=s12+s21 SL13=s13+s31 SL23=s23+s32 S1=(s12+s13) S2=(s23+s21) S3=(s31+s32) S12=s12 S21=s21 S13=s13 S31=s31 S23=s23 S32=s32

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MATLAB RESULTS

ITERATION RESULTS:

1ST_ITERATION V2 = 0.974615384615385 - 0.042307692307692i Q3 = 1.160000000000002 V3 = 1.039987148574197 - 0.005170183798502i 2ND_ITERATION V2 = 0.971057059512953 - 0.043431876337850i Q3 = 1.387957731052817 V3 = 1.039974378708180 - 0.007300111679686i 3RD_ITERATION V2 = 0.970733708554698 - 0.044791724463619i Q3 = 1.429040300785471 V3 = 1.039966679445820 - 0.008325001047174i 4TH_ITERATION V2 = 0.970652437281433 - 0.045329920732880i Q3 = 1.448333275594840 V3 = 1.039963173621928 - 0.008752000354604i 5TH_ITERATION V2 = 0.970623655331095 - 0.045554240372625i Q3 = 1.456209166612119

(59)

6TH_ITERATION V2 = 0.970612037114234 - 0.045646940090561i Q3 = 1.459469889628077 V3 = 1.039961037734205 - 0.009002221658867i 7TH_ITERATION V2 = 0.970607253520093 - 0.045685276728252i Q3 = 1.460818201396914 V3 = 1.039960775170297 - 0.009032502820155i 8TH_ITERATION V2 = 0.970605276281561 - 0.045701131870879i Q3 = 1.461375872168914 V3 = 1.039960666313617 - 0.009045027392915i 9TH_ITERATION V2 = 0.970604458527297 - 0.045707689707255i Q3 = 1.461606535170454 V3 = 1.039960621244008 - 0.009050207830587i 10TH_ITERATION V2 = 0.970604120282796 - 0.045710402176455i Q3 = 1.461701943643423 V3 = 1.039960602594413 - 0.009052350604469i

(60)

FINAL RESULTS:

V2 = 0.970604120282796 - 0.045710402176455i V3 = 1.039960602594413 - 0.009052350604469i Q3 = 1.461701943643423 SL12 = 0.0839 + 0.1679i SL13 = 0.0018 + 0.0055i SL23 = 0.0985 + 0.1969i S1 = 2.1841 + 1.4085i S2 = -3.9999 - 2.5000i S3 = 2.0000 + 1.4618i S12 = 1.7936 + 1.1874i S21 = -1.7096 - 1.0195i S13 = 0.3906 + 0.2212i S31 = -0.3887 - 0.2157i S23 = -2.2903 - 1.4805i S32 = 2.3888 + 1.6775i

(61)

EXAMPLE 6.10

Given figure shows the one line diagram of a simple three bus system with generators at buses-1 and 3. The

magnitude of voltage at bus-1 is adjusted to 1.05pu. voltage magnitude at bus-3 is fixed at 1.04 pu with a real power generation of 200MW. A load consisting of 400MW and

250MVAR is taken from bus-2. Line impedances are marked in per unit on a 100MVA base, and the line charging

susceptances are neglected. Obtain the power flow solution by the Newton-Raphson method including line flows and line losses.

Where,impedances are replaced by admittances as:

The bus impedance matrix can be constructed as: YBUS=[20-j50 -10+j20 -10+j30

(62)

-10+j20 26-j52 -16+j32 -10+j30 -16+j32 26-j62];

FARMULA’S

P

1

=V

1

^2*Y

11*cos(Ѳ11

)+V

1

*V

2

*Y

12*cos(Ѳ12

-d

1

+d

2

)+...

V

1

*V

3

*Y

13*cos(Ѳ13

-d

1

+d

3

)

Q

1

=-V

1

^2*Y

11*sin(Ѳ11

)-V

1

*V

2

*Y

12*sin(Ѳ12

-d

1

+d

2

)-...

V

1

*V

3

*Y

13*sin(Ѳ13

-d

1

+d

3

)

Q

3

=-V

3

*V

1

*Y

31*sin(Ѳ31

)-d

3

+d

1

)-V

3

*V

2

*Y

32

*...

sin(Ѳ32

-d

3

+d

2

)-V

3

^2*Y

33*sinѲ33

MATLAB CODE

V=[1.05;1.0;1.04]; d=[0;0;0]; Ps=[-4;2.0]; Qs=-2.5; YB=[20-j*50 -10+j*20 -10+j*30 -10+j*20 26-j*52 -16+j*32 -10+j*30 -16+j*32 26-j*62]; Y=abs(YB); t=angle(YB); iter=0;

pwracur=0.00025; %power accuracy

DC=10; %set the maximun power residue to a high value while max(abs(DC))>pwracur iter=iter+1 P=[V(2)*V(1)*Y(2,1)*cos(t(2,1)-d(2)+d(1))+V(2)^2*Y(2,2)*cos(t(2,2))+... V(2)*V(3)*Y(2,3)*cos(t(2,3)-d(2)+d(3)); V(3)*V(1)*Y(3,1)*cos(t(3,1)-d(3)+d(1))+V(3)^2*Y(3,3)*cos(t(3,3))+... V(3)*V(2)*Y(3,2)*cos(t(3,2)-d(3)+d(2))];

(63)

V(2)^2*Y(2,2)*sin(t(2,2))-... V(2)*V(3)*Y(2,3)*sin(t(2,3)-d(2)+d(3)); J(1,1)=V(2)*V(1)*Y(2,1)*sin(t(2,1)-d(2)+d(1))+... V(2)*V(3)*Y(2,3)*sin(t(2,3)-d(2)+d(3)); J(1,2)=-V(2)*V(3)*Y(2,3)*sin(t(2,3)-d(2)+d(3)); J(1,3)=V(1)*Y(2,1)*cos(t(2,1)-d(2)+d(1))+2*V(2)*Y(2,2)*cos(t(2,2))+... V(3)*Y(2,3)*cos(t(2,3)-d(2)+d(3)); J(2,1)=-V(3)*V(2)*Y(3,2)*sin(t(3,2)-d(3)+d(2)); J(2,2)=V(3)*V(1)*Y(3,1)*sin(t(3,1)-d(3)+d(1))+... V(3)*V(2)*Y(3,2)*sin(t(3,2)-d(3)+d(2)); J(2,3)=V(3)*Y(2,3)*cos(t(3,2)-d(3)+d(2)); J(3,1)=V(2)*V(1)*Y(2,1)*cos(t(2,1)-d(2)+d(1))+... V(2)*V(3)*Y(2,3)*cos(t(2,3)-d(2)+d(3)); J(3,2)=-V(2)*V(3)*Y(2,3)*cos(t(2,3)-d(2)+d(3)); J(3,3)=-V(1)*Y(2,1)*sin(t(2,1)-d(2)+d(1))-2*V(2)*Y(2,2) *sin(t(2,2)) ... V(3)*Y(2,3)*sin(t(2,3)-d(2)+d(3)); DP=Ps-P; DQ=Qs-Q; DC=[DP;DQ] J DX=J\DC d(2)=d(2)+DX(1); d(3)=d(3)+DX(2); V(2)=V(2)+DX(3); V, d, delta=180/pi*d; end P1=V(1)^2*Y(1,1)*cos(t(1,1))+V(1)*V(2)*Y(1,2)*cos(t(1,2)- d(1)+d(2))+... V(1)*V(3)*Y(1,3)*cos(t(1,3)-d(1)+d(3)) Q1=-V(1)^2*Y(1,1)*sin(t(1,1))-V(1)*V(2)*Y(1,2)*sin(t(1,2)-d(1)+d(2))-... V(1)*V(3)*Y(1,3)*sin(t(1,3)-d(1)+d(3)) Q3=-V(3)*V(1)*Y(3,1)*sin(t(3,1)-d(3)+d(1))-V(3)*V(2)*Y(3,2)*... sin(t(3,2)-d(3)+d(2))-V(3)^2*Y(3,3)*sin(t(3,3))

(64)

MATLAB RESULTS

1ST_ITERATION DC = -2.8600 1.4384 -0.2200 J = 54.2800 -33.2800 24.8600 -33.2800 66.0400 -16.6400 -27.1400 16.6400 49.7200 DX = -0.0453 -0.0077 -0.0265 V = 1.0500 0.9735 1.0400 d = 0 -0.0453 -0.0077

(65)

DC = -0.0992 0.0217 -0.0509 J = 51.7247 -31.7656 21.3026 -32.9816 65.6564 -15.3791 -28.5386 17.4028 48.1036 DX = -0.0018 -0.0010 -0.0018 V = 1.0500 0.9717 1.0400 d = 0 -0.0471 -0.0087

(66)

3RD_ITERATION DC = 1.0e-003 * -0.2166 0.0382 -0.1430 J = 51.5967 -31.6939 21.1474 -32.9339 65.5976 -15.3516 -28.5482 17.3969 47.9549 DX = 1.0e-005 * -0.3856 -0.2386 -0.4412 V = 1.0500 0.9717 1.0400 d = 0 -0.0471 -0.0087

(67)

P1 = 2.1842 Q1 = 1.4085 Q3 = 1.4618

COMMENTS:

Power system calculations are mostly very complex for large power systems. To analyze such power systems, there are two very important iterative methods:

• Gauss Siedel Method • Newton Raphson method

In this experiment we learn how to apply these two methods using MATLAB.

Gauss-Seidel iteration has two advantages:

• Errors do not accumulate during the calculation. If the procedure converges, it approaches the correct answer without rounding errors such as can occur during

inversion of large matrices.

• The method can be used for nonlinear sets of equations.

While Newton Raphson method is readily applied to non-linear equations, and can use finite-difference estimates

(68)

EXPERIMENT#08

Harmonic Load Modeling using builtin and user defined

models of ETAP

HARMONIC ANALYSIS

Because of the wide and ever increasing applications of power

electronic devices, such as variable speed drives, uninterruptible

power supplies (UPS), static power converters, etc., power system

voltage and current quality has been severely affected in some

areas. In these areas components other than that of fundamental

frequency can be found to exist in the distorted voltage and

current waveforms. These components usually are the integer

multipliers of the fundamental frequency, called harmonics. In

addition to electronic devices, some other non‐linear loads, or

devices including saturated transformers, arc furnaces,

fluorescent lights, and cycloconverters are also responsible for the

deterioration in power system quality.

HARMONIC SOURCES

The following components can be modeled as a harmonic voltage

source in PowerStation:

• Power Grid

• Synchronous Generator

• Inverter

• Charger/Converter

(69)

• Static Load

IMPORTANT DEFINITIONS

Transients

The term transient has long been used in the analysis of power

system variations to denote an event that is undesirable and

momentary in nature. Transient is “that part of the change in a

variable that disappears during transition from one steady state

operating condition to another.”

Impulsive transient

An impulsive transient is a sudden; non–power frequency change

in the steady‐state condition of voltage, current, or both that is

unidirectional in polarity (primarily either positive or negative).

Oscillatory transient

An oscillatory transient is a sudden, non–power frequency change

in the steady‐state condition of voltage, current, or both, that

includes both positive and negative polarity values.

Long-Duration Voltage Variations

Long‐duration variations encompass root‐mean‐square (rms)

deviations at power frequencies for longer than 1 minute.

Overvoltage

An overvoltage is an increase in the rms ac voltage greater than

110 percent at the power frequency for duration longer than 1

(70)

min. Over voltages are usually the results of load switching (e.g.,

switching off a large load or energizing a capacitor bank).

Under voltage

An under voltage is a decrease in the rms ac voltage to less than

90 percent at the power frequency for a duration longer than 1

min. Under voltages are the results of switching events that are

the opposite of the events that cause over voltages.

Short-Duration Voltage Variations

This category encompasses the IEC category of voltage dips and

short interruptions. Each type of variation can be designated as

instantaneous, momentary, or temporary, depending on its

duration.

Short‐duration voltage variations are caused by fault conditions,

the energization of large loads which require high starting

currents, or intermittent loose connections in power wiring.

Interruption

An interruption occurs when the supply voltage or load current

decreases to less than 0.1 pu for a period of time not exceeding 1

min.

Sags (dips)

Sag is a decrease to between 0.1 and 0.9 pu in rms voltage or

current at the power frequency for durations from 0.5 cycle to 1

min.

Swells

(71)

A swell is defined as an increase to between 1.1 and 1.8 pu in rms

voltage or current at the power frequency for durations from 0.5

cycle to 1 min.

Voltage Imbalance

Voltage imbalance (also called voltage unbalance) is sometimes

defined as the maximum deviation from the average of the three‐

phase voltages or currents, divided by the average of the three‐

phase voltages or currents, expressed in percent.

Waveform Distortion

Waveform distortion is defined as a steady‐state deviation from

an ideal sine wave of power frequency principally characterized

by the spectral content of the deviation.

Harmonics

Harmonics are sinusoidal voltages or currents having frequencies

that are integer multiples of the frequency at which the supply

system is designed to operate (termed the fundamental

frequency usually 50 or 60 Hz).

Interharmonics

Voltages or currents having frequency components that are not

integer multiples of the frequency at which the supply system is

designed to operate (e.g., 50 or 60 Hz) are called Interharmonics.

Odd harmonics

(72)

Voltages or currents having frequency components that are odd

integer multiples of the frequency at which the supply system is

designed to operate (e.g., 50 or 60 Hz) are called odd harmonics.

Even harmonics

Voltages or currents having frequency components that are even

integer multiples of the frequency at which the supply system is

designed to operate (e.g., 50 or 60 Hz) are called even harmonics.

Voltage Fluctuation

Voltage fluctuations are systematic variations of the voltage

envelope or a series of random voltage changes, the magnitude of

which does not normally exceed the voltage ranges specified by

ANSI C84.1 of 0.9 to 1.1 pu.

Power Frequency Variations

Power frequency variations are defined as the deviation of the

power system fundamental frequency from it specified nominal

value (e.g., 50 or 60 Hz).

Power factor, displacement

The power factor of the fundamental frequency components of

the voltage and current waveforms

Power factor (true)

The ratio of active power (watts) to apparent power (volt

amperes)

(73)

The ratio of the root mean square of the harmonic content to the

rms value of the fundamental quantity, expressed as a percent of

the fundamental.

Triplen harmonics

A term frequently used to refer to the odd multiples of the third

harmonic, which deserve special attention because of their

natural tendency to be zero sequence.

ONE LINE DIAGRAM

MODELING OF HARMONIC LOAD

• Double click on the charger

• Select the harmonics section

• Select the type of harmonics from the given library of harmonics

(74)

BUILT-IN MODELS OF ETAP

ROCKWELL (12-Pulse-VFD)

(75)

ROCKWELL (6-Pulse-VFD)

TOSHIBA (PWM-ASD)

(76)

TYPICAL-IEEE (12-Pulse1)

(77)

TYPICAL-IEEE (18-Pulse-CT)

(78)

TYPICAL-IEEE (6-Pulse1)

(79)

TYPICAL-IEEE (Fluorescent)

(80)

TYPICAL-IEEE (SPC)

(81)

USER DEFINED MODELING

• Go to the library tab given on the main window of ETAP • Select the harmonic section

• Click on add tab

• Enter the name of new harmonic model • Click on edit tab

(82)

COMMENTS

In this experiment, we learnt:

• How to model a harmonic load using built-in models • How to built a user-defined harmonic model

• How to model a harmonic load using user-defined models

So, ETAP is a very powerful tool for harmonic analysis of any power distribution network.

EXPERIMENT#09

Impact of personal computer load on power distribution

network of RCET

Harmonics

Harmonics are sinusoidal voltages or currents having frequencies

that are integer multiples of the frequency at which the supply

system is designed to operate (termed the fundamental

frequency usually 50 or 60 Hz).

(83)

PC LOAD THD’S

Harmonic No. %THD

3rd _91.63 5th _86.61 7th _69.87 9th _44.76 11th _54.81 13th _46.44 15th _46.44 17th _33.05 19th _24.70 23rd _11.74 25th _7.900 29th _5.120 %THD 178.97

(84)

(85)

(86)

LOAD FLOW ANALYSIS REPORT

(87)

VOLTAGE WAVEFORMS & SPECTRAS

BUS-5:

(88)

(89)

(90)

(91)

CABLE-2:

(92)

(93)

(94)

(95)

(96)

In this experiment, we learnt:

• How to apply harmonic analysis to a power distribution network

• How to perform load flow analysis on a power distribution network

• How to perform harmonic analysis on a power distribution network

(97)

EXPERIMENT#10

Flow of triplen harmonics (zerosequence harmonics)

during 5 different schemes of connection for a 3phase

transformer with presence of large nonlinear load using

ETAP

ONE LINE DIAGRAM

HARMONIC MODEL USED

PC load is used as a source of harmonics in this experiment that has the following range of THD’S.

Harmonic No. %THD

(98)

5th _86.61 7th _69.87 9th _44.76 11th _54.81 13th _46.44 15th _46.44 17th _33.05 19th _24.70 23rd _11.74 25th _7.900 29th _5.120 %THD 178.97

5 DIFFERENT SCHEMES OF TRANSFORMER

WINDING

PRIMARY

SIDE SECONDARY

SIDE

1 Y-Grounded Y-Ungrounded 2 Y-Grounded Y-Grounded 3 Y-Grounded Delta 4 Y-Ungrounded Delta 5 Delta Delta

(99)

CASE-1:

Transformer primary side Y-Grounded and

secondary Y-Ungrounded

VOLTAGE SPECTRA ON LT-SIDE

VOLTAGE SPECTRA ON HT-SIDE

(100)

VOLTAGE WAVEFORM ON HT-SIDE

(101)

CURRENT SPECTRA ON HT-SIDE

(102)

CURRENT WAVEFORM ON HT-SIDE

OBSERVATIONS:

Harmonic source is connected on the LT side of transformer, so the triplen harmonics are blocked due to ungrounded Y-connection. The magnitude of remaining harmonic

components is reduced on the HT side of transformer.

CASE-2:

(103)

Transformer primary side Y-Grounded and

secondary Y-Grounded

(104)

(105)

(106)

OBSERVATIONS:

Harmonic source is connected on the LT side of transformer, so the triplen harmonics are not blocked due to grounded Y-connection. The triplen harmonics are also present on the HT side of transformer as that is also Y-grounded.

(107)

Transformer primary side Y-Grounded and

secondary Delta

(108)

(109)

(110)

OBSERVATIONS:

Harmonic source is connected on the LT side of transformer, so the triplen harmonics are blocked due to

delta-connection. The triplen harmonics are also blocked on the HT side of transformer as there are no triplen harmonics on secondary side of transformer.

(111)

Transformer primary side Y-Ungrounded and

secondary Delta

(112)

(113)

(114)

OBSERVATIONS:

delta-connection. The triplen harmonics are also blocked on the HT side of transformer as there are no triplen harmonics on secondary side of transformer.

(115)

Transformer primary side Delta and secondary

Delta

(116)

(117)

(118)

OBSERVATIONS:

delta-connection. The triplen harmonics are also blocked on the HT side of transformer as there are no triplen harmonics on secondary side of transformer. Moreover, primary side is also delta-connected.

(119)

EXPERIMENT#11

Three phase short circuit analysis (3phase faultsdevice

duty) for a given power system using ETAP

SHORT CIRCUIT ANALYSIS

The power station short circuit analysis program analyze the effect of three phase, line to ground, line to line, and line to line to ground faults on the electrical distribution networks. The program calculates the total short circuit currents as well as the contributions of individual motors, generators, and utility ties in the system. Fault duties are in compliance with the latest editions of the ANSI/IEEE standards and IEC standards.

The ANSI/IEEE Short-Circuit Toolbar and IEC Short-Circuit Toolbar sections explain how you can launch a short-circuit calculation, open and view an output report, or select display options. The Short-Circuit Study Case Editor section

explains how you can create a new study case, what

parameters are required to specify a study case, and how to set them. The Display Options section explains what options are available for displaying some key system parameters and the output results on the one-line diagram, and how to set them.

Short-Circuit Toolbar

This toolbar is active when you are in Short-Circuit mode and the standard is set to ANSI in the Short-Circuit Study Case Editor.

(120)

3-Phase Faults - Device Duty

Click on this button to perform a three-phase fault study per ANSI C37 Standard. This study calculates momentary

symmetrical and asymmetrical rms, momentary

asymmetrical crest, interrupting symmetrical rms, and

interrupting adjusted symmetrical rms short-circuit currents at faulted buses. The program checks the protective device rated close and latching, and adjusted interrupting capacities against the fault currents, and flags inadequate devices.

(121)

sequence sub-transient reactance.

ONE LINE DIAGRAM

FAULTY POINT

• BUS-15

(122)

LOAD FLOW DIAGRAM

(123)

SHORT CIRCUIT ANALYSIS DIAGRAM

(124)

COMMENTS:

In this experiment, we use three phase fault-device duty analysis to analyze the effect of fault on the system. Following results are obtained in this experiment:

At bus-15:

Before fault After fault

Current 568A 4.8KA

Power flow 345KW 3.6KW

We observe that the current flowing through bus-15 is increased up to many times as compared to the current before fault.

We observe that the power flowing through bus-15 is decreased up to many times as compared to the power before fault due to the short circuit at bus-15 as the load connected to that bus is now shorted and no power is flowing into that load.

(125)

EXPERIMENT#12

Three phase short circuit analysis (3phase faults30 cycle

network) for a given power system using ETAP

3-Phase Faults – 30-Cycle Network

Click on this button to perform a three-phase fault study per ANSI standards. This study calculates short-circuit currents in their rms values after 30 cycles at faulted buses.

Generators are modeled by their positive sequence transient reactance’s, and short-circuit current contributions from motors are ignored.

ONE LINE DIAGRAM

(126)

• BUS-15

There is a short circuit fault on bus-15.

LOAD FLOW DIAGRAM

(127)

SHORT CIRCUIT ANALYSIS DIAGRAM

COMMENTS:

At bus-15:

(128)

(129)

EXPERIMENT#13

Three phase short circuit analysis (LG, LL, LLG, & 3Phase

Faults ½ Cycle) for a given power system using ETAP

LG, LL, LLG, & 3-Phase Faults - ½ Cycle

Click on this button to perform line-to-ground, line-to-line, line-to-line-to-ground, and three-phase fault studies per ANSI standards. This study calculates short-circuit currents in their rms values at ½ cycles at faulted buses.

Generators and motors are modeled by their positive, negative, and zero sequence sub-transient reactance.

In all the unbalanced fault calculations (½ cycle, 1.5-4 cycle and 30 cycles), it is assumed that the negative sequence impedance of a machine is equal to its positive sequence impedance. Generator, motor, and transformer grounding types and winding connections are taken into consideration when constructing system positive, negative, and zero

(130)

ONE LINE DIAGRAM

FAULTY POINT

• BUS-15

(131)

LOAD FLOW DIAGRAM

(132)

SHORT CIRCUIT ANALYSIS DIAGRAM

COMMENTS:

(133)

(134)

EXPERIMENT#14

Three phase short circuit analysis (LG, LL, LLG, & 3Phase

Faults 1.5 to 4 Cycle) for a given power system using

ETAP

LG, LL, LLG, & 3-Phase Faults - 1.5 to 4 Cycle

Click on this button to perform three-phase, line-to-ground, line-to-line, line-to-line-to-ground, and three-phase fault studies per ANSI standards. This study calculates short-circuit currents in their rms values between 1.5 to 4 cycles at faulted buses.

Generators are modeled by their positive, negative, and zero sequence sub-transient reactance, and motors are modeled by their positive, negative and zero sequence transient reactance. Generator, motor and transformer grounding types and winding connections are taken into considerations when constructing system positive, negative, and zero

(135)

ONE LINE DIAGRAM

FAULTY POINT

• BUS-15

(136)

LOAD FLOW DIAGRAM

(137)

SHORT CIRCUIT ANALYSIS DIAGRAM

COMMENTS:

(138)

At bus-15:

Power flow 345KW 5KW

(139)

EXPERIMENT#15

Three phase short circuit analysis (LG, LL, LLG, & 3Phase

Faults 30 Cycle) for a given power system using ETAP

LG, LL, LLG, & 3-Phase Faults - 30 Cycle

Click on this button to perform three-phase, line-to-ground, line-to-line, line-to-line-to-ground, and three-phase fault studies per ANSI standards. This study calculates short-circuit currents in their rms values at 30-cycles at faulted buses.

Generators are modeled by their positive, negative, and zero sequence reactance, and short-circuit current contributions from motors are ignored. Generator, motor, and

transformer grounding types and winding connections are taken into consideration when constructing system positive, negative, and zero sequence networks.

Power System Analysis Lab Manual