Power System Analysis Lab Manual

POWER SYSTEM ANALYSIS LAB MANUAL

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) UNIVERSITY OF ENGINEERING & TECHNOLOGY LAHORE, PAKISTAN ASAD NAEEM 2006‐RCET‐EE‐22


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

06

07 ASAD NAEEM 2006‐RCET‐EE‐22

Bus elimination of a 4-BUS system using MATLAB

To study the Concept of Modifications of an Existing BusImpedance Matrix & Implementing in MATLAB

Application of Gauss-Siedal and Newton-Raphson method for load flow studies on a three bus system using MATLAB


08

09

10

11

12

13

14

15

ASAD NAEEM 2006‐RCET‐EE‐22

Harmonic Load Modeling using built-in and user defined models of ETAP Impact of personal computer load on power distribution network of RCET Flow of triplen harmonics (zero-sequence harmonics) during 5 different schemes of connection for a 3-phase transformer with presence of large non-linear load using ETAP Three phase short circuit analysis (3-phase faults-device duty) for a given power system using ETAP Three phase short circuit analysis (3-phase faults-30 cycle network) for a given power system using ETAP Three phase short circuit analysis (LG, LL, LLG, & 3-Phase Faults - ½ Cycle) for a given power system using ETAP Three phase short circuit analysis (LG, LL, LLG, & 3-Phase Faults - 1.5 to 4 Cycle) for a given power system using ETAP Three phase short circuit analysis (LG, LL, LLG, & 3-Phase Faults - 30 Cycle) for a given power system using ETAP


EXPERIMENT#01 To plot the daily load curve for the given data using MATLAB Given data: Interval from 12 A.M 2 6 9 12 P.M 2 4 6 8 10 11

To 2 A.M 6 9 12 2 P.M 4 6 8 10 11 12 A.M

Requirements: 1. 2. 3. 4.

Find average value of load Find peak value of load Find the load factor Plot the load curve


Load MW 6 5 10 15 12 14 16 18 16 12 6


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



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;



23 24 6]; 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 ASAD NAEEM 2006‐RCET‐EE‐22


18

16

14

P,MW

12

10

8

6

4 0

5

10

15

20

25

TIME,Hr

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.



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 raceway

Paste elements from a Dumpster Cell to the one‐line diagram or U/G 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

Check Continuity

Check the system continuity for non‐energized elements



Power Calculator

Help

Activate PowerStation Calculator that relates MW, MVAR, MVA, kV, Amp, and PF together with either kVA or MVA units

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

















Example implementation:



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:



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 I1= (y10+y14+y13) V1 + 0V2 + (-y13) V3+ (-y14) V4



• Applying KCL at node-2: I2= (V2-0) y20 + (V2-V3) y23+ (V2-V4) y24 I2= 0V1+ (y20+y23+y24) V2 + (-y23) V3+ (-y24) V4 • Applying KCL at node-3: I3= (V3-0) y30 + (V3-V1) y31+ (V3-V4) y34 + (V3-V2) y32 I3= (-y31) V1+ (-y32) V2+ (y30+y31+y34) V3 + (-y34) V4 • Applying KCL at node-4: 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:

Where:



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



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.



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 ASAD NAEEM 2006‐RCET‐EE‐22


allows you to select from several different methods in order 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 ¾ ¾ ¾

Modeling of the main network Modeling of composite networks Running of load flow analysis Complete report from ETAP load flow analyzer



MODELING OF BASIC RCET NETWORK



MODELING OF COMPOSITE NETWORKS STAFF COLONY:



OLD BUILDING:

NEW BUILDING:



HOSTEL-A,B:



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 Study the load flow analysis for that network



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 • The resultant diagram now can be shown as:



Part‐1: Elimination of Bus‐3&4

MATRIX FORM

Where:



MATLAB CODE >>YBUS= [0-9.80i 0

0+4.00i

0+5.00i;

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];

0

>>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


POWER SYSTEM ANALYSIS LAB MANUAL 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]; ASAD NAEEM 2006‐RCET‐EE‐22


>>R=[-18.0556i]; >>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 ASAD NAEEM 2006‐RCET‐EE‐22


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 Adding Adding Adding


Zb Zb Zb Zb

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


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) Zorg=[1 2 3 4;2 5 6 7;3 6 8 9;4 7 9 10] ASAD NAEEM 2006‐RCET‐EE‐22


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



CASE‐2: ADDING Zb FROM A NEW BUS‐P TO AN EXISTING BUS‐K This condition is explained in the following diagram:

Clearly, Vp-Vk,new=Ip*Zb Vp=Vk,new+Ip*Zb Where, Vk,new=Vk,org+Ip*Zkk Vp= Vk,org+Ip(Zkk+Zb) Hence the modified matrix will take the form as:



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 ASAD NAEEM 2006‐RCET‐EE‐22


MATLAB RESULTS

CASE‐3: ADDING Zb FROM AN EXISTING BUS‐K TO REFERENCE BUS This condition is explained in the following diagram:



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 ASAD NAEEM 2006‐RCET‐EE‐22


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 for a=1:l for b=1:l K(a,b)=Znew(a,b); end ASAD NAEEM 2006‐RCET‐EE‐22


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



CASE‐4: ADDING Zb BETWEEN TWO EXISTING BUSES This condition is explained in the following diagram:

In this case, Zbb=Zb+Zjj+Zkk-2Zjk Hence the modified matrix will take the form as:



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); ASAD NAEEM 2006‐RCET‐EE‐22


elseif i==l+1 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 ASAD NAEEM 2006‐RCET‐EE‐22


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



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 bus • Addition of new bus • Addition of new reference bus • Addition of new

impedance from a new bus to reference impedance from a new bus to existing impedance from an existing bus to impedance between two existing buses

In this experiment we learn how to modify the bus impedance matrix for all four cases using MATLAB. ASAD NAEEM 2006‐RCET‐EE‐22


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 ASAD NAEEM 2006‐RCET‐EE‐22


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 V3(0)=1.0+j0.0



GAUSS-SIEDEL FARMULA

SLACK1=conj(V1)*[V1*(y12+y13)-( y12*V2+y13*V3)] Sij=Vi*conj(Iij) Iij=yij*(Vi-Vj)

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



%CODE FOR PART-B 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 ASAD NAEEM 2006‐RCET‐EE‐22


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



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.

Where,impedances are replaced by admittances as:



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); ASAD NAEEM 2006‐RCET‐EE‐22


V3=Vr3+j*Vi3 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



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 ASAD NAEEM 2006‐RCET‐EE‐22


V3 = 1.039961668920058 - 0.008929007616053i 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



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



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 ASAD NAEEM 2006‐RCET‐EE‐22

-10+j20

-10+j30


-10+j20

26-j52

-10+j30

-16+j32

-16+j32 26-j62];

FARMULA’S P1=V1^2*Y11*cos(Ѳ11)+V1*V2*Y12*cos(Ѳ12-d1+d2)+... V1*V3*Y13*cos(Ѳ13-d1+d3) Q1=-V1^2*Y11*sin(Ѳ11)-V1*V2*Y12*sin(Ѳ12-d1+d2)-... V1*V3*Y13*sin(Ѳ13-d1+d3) Q3=-V3*V1*Y31*sin(Ѳ31)-d3+d1)-V3*V2*Y32*... sin(Ѳ32-d3+d2)-V3^2*Y33*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))]; ASAD NAEEM 2006‐RCET‐EE‐22


Q=-V(2)*V(1)*Y(2,1)*sin(t(2,1)-d(2)+d(1))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))



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

DX = -0.0453 -0.0077 -0.0265 V= 1.0500 0.9735 1.0400 d=

0

-0.0453 -0.0077 ASAD NAEEM 2006‐RCET‐EE‐22

49.7200


2ND ITERATION DC =

-0.0992 0.0217 -0.0509

J= 51.7247 -31.7656

21.3026

-32.9816

65.6564 -15.3791

-28.5386

17.4028

DX = -0.0018 -0.0010 -0.0018 V= 1.0500 0.9717 1.0400 d= 0 -0.0471 -0.0087


48.1036


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

DX = 1.0e-005 * -0.3856 -0.2386 -0.4412 V= 1.0500 0.9717 1.0400 d=

0

-0.0471 -0.0087 ASAD NAEEM 2006‐RCET‐EE‐22

47.9549


FINAL RESULTS: 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 nonlinear equations, and can use finite-difference estimates of the derivatives to evaluate the gradients. ASAD NAEEM 2006‐RCET‐EE‐22


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



• 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



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 ASAD NAEEM 2006‐RCET‐EE‐22


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



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) Total harmonic distortion (THD)



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



BUILT-IN MODELS OF ETAP

ROCKWELL (12-Pulse-VFD)





TOSHIBA (PWM-ASD)

TYPICAL (LCI) ASAD NAEEM 2006‐RCET‐EE‐22


TYPICAL-IEEE (12-Pulse1)




TYPICAL-IEEE (18-Pulse-CT)

TYPICAL-IEEE (18-Pulse-VT)







TYPICAL-IEEE (Fluorescent)

TYPICAL-IEEE (Large-ASD)



TYPICAL-IEEE (SPC)

TYPICAL-IEEE (XFMR-Magnet)



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 Enter the spectrum components in the table



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). ASAD NAEEM 2006‐RCET‐EE‐22


PC LOAD THD’S Harmonic No. rd

%THD

3

91.63

5th

86.61

7th

69.87

th

9

44.76

11th

54.81

13th

46.44

15th

46.44

th

17

33.05

19th

24.70

23rd

11.74

th

25

7.900

29th

5.120

%THD

178.97

ONE LINE DIAGRAM



LOAD FLOW DIAGRAM ASAD NAEEM 2006‐RCET‐EE‐22




LOAD FLOW ANALYSIS REPORT

HARMONIC ANALYSIS RESULTS ASAD NAEEM 2006‐RCET‐EE‐22


VOLTAGE WAVEFORMS & SPECTRAS

BUS-5:

BUS-11: ASAD NAEEM 2006‐RCET‐EE‐22






CURRENT WAVEFORMS & SPECTRAS ASAD NAEEM 2006‐RCET‐EE‐22


CABLE-2:

CABLE-7: ASAD NAEEM 2006‐RCET‐EE‐22








HARMONIC ANALYSIS REPORT

COMMENTS ASAD NAEEM 2006‐RCET‐EE‐22


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 • How to get harmonic spectra at required locations



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. 3rd


%THD 91.63


5th

86.61

7th

69.87

9th

44.76

11th

54.81

th

13

46.44

15th

46.44

17th

33.05

th

19

24.70

23rd

11.74

25th

7.900

th

5.120

29

%THD

178.97

5 DIFFERENT SCHEMES OF TRANSFORMER WINDING PRIMARY SIDE

SECONDARY SIDE

Y-Grounded Y-Grounded Y-Grounded Y-Ungrounded Delta

Y-Ungrounded Y-Grounded Delta Delta Delta

1 2 3 4 5

HARMONIC ANALYSIS RESULTS ASAD NAEEM 2006‐RCET‐EE‐22


CASE-1: Transformer primary side Y-Grounded and secondary Y-Ungrounded VOLTAGE SPECTRA ON LT-SIDE

VOLTAGE SPECTRA ON HT-SIDE

VOLTAGE WAVEFORM ON LT-SIDE



VOLTAGE WAVEFORM ON HT-SIDE

CURRENT SPECTRA ON LT-SIDE



CURRENT SPECTRA ON HT-SIDE

CURRENT WAVEFORM ON LT-SIDE



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 Yconnection. The magnitude of remaining harmonic components is reduced on the HT side of transformer.

CASE-2: ASAD NAEEM 2006‐RCET‐EE‐22


Transformer primary side Y-Grounded and secondary Y-Grounded VOLTAGE SPECTRA ON LT-SIDE














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



Transformer primary side Y-Grounded and secondary Delta VOLTAGE SPECTRA ON LT-SIDE


VOLTAGE WAVEFORMS ON LT-SIDE












OBSERVATIONS: Harmonic source is connected on the LT side of transformer, so the triplen harmonics are blocked due to deltaconnection. The triplen harmonics are also blocked on the HT side of transformer as there are no triplen harmonics on secondary side of transformer.



Transformer primary side Y-Ungrounded and secondary Delta VOLTAGE SPECTRA ON LT-SIDE














OBSERVATIONS: Harmonic source is connected on the LT side of transformer, so the triplen harmonics are blocked due to deltaconnection. The triplen harmonics are also blocked on the HT side of transformer as there are no triplen harmonics on secondary side of transformer.



Transformer primary side Delta and secondary Delta VOLTAGE SPECTRA ON LT-SIDE














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



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.



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. ASAD NAEEM 2006‐RCET‐EE‐22


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

ONE LINE DIAGRAM

FAULTY POINT • BUS-15 There is a short circuit fault on bus-15. ASAD NAEEM 2006‐RCET‐EE‐22


LOAD FLOW DIAGRAM



SHORT CIRCUIT ANALYSIS DIAGRAM



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: Current Power flow

Before fault 568A 345KW

After fault 4.8KA 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.



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

FAULTY POINT ASAD NAEEM 2006‐RCET‐EE‐22


• BUS-15 There is a short circuit fault on bus-15.

LOAD FLOW DIAGRAM




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: Current Power flow ASAD NAEEM 2006‐RCET‐EE‐22







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 sequence networks.



ONE LINE DIAGRAM

FAULTY POINT • BUS-15 There is a short circuit fault on bus-15.



LOAD FLOW DIAGRAM




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: Current Power flow






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 shortcircuit 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 sequential networks.



ONE LINE DIAGRAM




LOAD FLOW DIAGRAM









After fault 5.7KA 5KW




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 shortcircuit 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.



ONE LINE DIAGRAM




LOAD FLOW DIAGRAM













COMPARISON OF SHORT CIRCUIT ANALYSIS

3‐phase faults‐ device duty 3‐phase faults‐30 cycle network LG, LL, LLG, & 3‐ Phase Faults ‐ ½ Cycle LG, LL, LLG, & 3‐ Phase Faults ‐ 1.5 to 4 Cycle LG, LL, LLG, & 3‐ Phase Faults ‐ 30 Cycle


CURRENT

POWER

4.8

3.6 3.6

3.6 6.2

5.1

5.7

5

4.8

4.8

Power System Analysis Lab Manual

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