Power System Analysis Lab Manual
May 3, 2017 | Author: chethan | Category: N/A
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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
POWER SYSTEM ANALYSIS LAB MANUAL
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
POWER SYSTEM ANALYSIS LAB MANUAL
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
POWER SYSTEM ANALYSIS LAB MANUAL
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
ASAD NAEEM 2006‐RCET‐EE‐22
Load MW 6 5 10 15 12 14 16 18 16 12 6
POWER SYSTEM ANALYSIS LAB MANUAL
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
ASAD NAEEM 2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
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;
ASAD NAEEM 2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
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
POWER SYSTEM ANALYSIS LAB MANUAL
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.
ASAD NAEEM 2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
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
ASAD NAEEM 2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
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
ASAD NAEEM 2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
ASAD NAEEM 2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
ASAD NAEEM 2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
ASAD NAEEM 2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
ASAD NAEEM 2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
ASAD NAEEM 2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
ASAD NAEEM 2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
ASAD NAEEM 2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
Example implementation:
ASAD NAEEM 2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
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:
ASAD NAEEM 2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
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
ASAD NAEEM 2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
• 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:
ASAD NAEEM 2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
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
ASAD NAEEM 2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
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.
ASAD NAEEM 2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
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
POWER SYSTEM ANALYSIS LAB MANUAL
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
ASAD NAEEM 2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
MODELING OF BASIC RCET NETWORK
ASAD NAEEM 2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
MODELING OF COMPOSITE NETWORKS STAFF COLONY:
ASAD NAEEM 2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
OLD BUILDING:
NEW BUILDING:
ASAD NAEEM 2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
HOSTEL-A,B:
ASAD NAEEM 2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
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
ASAD NAEEM 2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
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:
ASAD NAEEM 2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
Part‐1: Elimination of Bus‐3&4
MATRIX FORM
Where:
ASAD NAEEM 2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
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
ASAD NAEEM 2006‐RCET‐EE‐22
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
POWER SYSTEM ANALYSIS LAB MANUAL
>>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
POWER SYSTEM ANALYSIS LAB MANUAL
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
ASAD NAEEM 2006‐RCET‐EE‐22
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
POWER SYSTEM ANALYSIS LAB MANUAL
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
POWER SYSTEM ANALYSIS LAB MANUAL
Zb=17; l=length(Zorg); for i=1:l+1 for j=1:l+1 if i
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