BMS INSTITUTE OF TECHNOLOGY & MGMT. Yelahanka, Bangalore - 64 Department of Electrical & Electronics Engineering VI I SEMESTER POWER SYSTEM SIMULATION LABORATORY 1 8 EEL7 6 LABORATORY MANUAL NAME OF THE STUDENT : BRANCH : UNIVERSITY SE AT NO. : SEMESTER & SECTION : BATCH : Vikram Chekuri HOD Laboratory In Charge P ower System Simulation L ab oratory - 18 EEL76 VII EEE BMS Institute of Technology and Management 2 Vision of the Department To emerge as one of the finest Electrical & Electronics Engineering Departments facilitating the development of compet ent professionals, contributing to the betterment of society. Mission of the Department Create a motivating environment for learning Electrical Sciences through teaching, research, effective use of state of the art facilities and outreach activities. Pr ogram Educational Objectives: Graduates of the program will, PEO1 Have successful professional careers in Electrical Sciences, and IT enabled areas and be able to pursue higher education. PEO2 Demonstrate ability to work in multidisciplinary teams and eng age in lifelong learning. PEO3 Exhibit concern for environment and sustainable development. Program Outcomes: After the successful completion of the course, the graduate will be able to, PO1 Apply knowledge of mathematics, science and engineering princi ples to the solution of engineering problems in electrical and IT enabled areas. PO2 Identify and solve complex engineering problems using first principles of mathematics and engineering sciences. PO3 Design system components and solve complex engineeri ng problems that meet specific societal and environmental needs. PO4 Conduct experiments, analyse, and interpret data to provide valid conclusion PO5 Apply appropriate modern engineering tools to complex engineering activities with an understanding of t he limitations. PO6 Demonstrate understanding of societal health, safety, legal and consequent responsibilities relevant to the professional engineering practice. PO7 Understand the impact of engineering solutions in a societal context and demonstrate th e knowledge of and need for sustainable development. PO8 Understand social issues and ethical principles of electrical engineering practice. PO9 Function effectively as an individual and as a member or leader in diverse teams to accomplish a common goal PO10 Communicate effectively with diverse audiences and be able to prepare effective reports and design documentation. PO11 Demonstrate knowledge and understanding of engineering and management principles and apply these as a member and leader in a tea m to manage projects in multi - disciplinary environments. PO12 Recognize the need to engage in independent and lifelong learning in the context of technological change. P ower System Simulation L ab oratory - 18 EEL76 VII EEE BMS Institute of Technology and Management 3 DO’S and DON’T’S in Power System Simulation Laboratory DO’S Sit down and wait f or teacher’s instruction Make sure your hands are clean and dry when you use the computer Report any problems with your computer to the teacher. Shut down the computer properly Keep the lab clean and tidy DON’T’S Do not touch, connect or disconnect a ny plug or cable without the Faculty /laboratory technician’s permission. Do not open the system unit casing or monitor casing particularly when the power is turned on. Do not misbehave in the computer laboratory. Do not plug in external devices without sc anning them for computer viruses. Always maintain an extra copy of all your important data files. Do not take food or drinks to the lab. Avoid stepping on electrical wires or any other computer cables. Do not install or download any software or modify or d elete any system files on any lab computers P ower System Simulation L ab oratory - 18 EEL76 VII EEE BMS Institute of Technology and Management 4 B.E ELECTRICAL AND ELECTRONICS ENGINEERING (EEE) CHOICE BASED CREDIT SYSTEM (CBCS) SEMESTER - VII POWER SYSTEM SIMULATION LABORATORY Subject Code 1 8 EEL76 CIE Marks 4 0 Number of Practical Hours/We ek (L:T:P) 0:2:2 SEE Marks 60 Credits 2 Exam Hours 03 Course Learning Objectives: To explain the use of standard software package: (Ex: MATLAB/C or C ++/Scilab/ Octave/Python software) · To assess the performance of medium and long transmission li nes. To obtain the power angle characteristics of salient and non - salient pole alternator. To study transient stability of radial power systems under three phase fault conditions. To develop admittance and impedance matrices of interconnected power s ystems. To explain the use of suitable standard software package. To solve power flow problem for simple power systems. To perform fault studies for simple radial power systems. To study optimal generation scheduling problems for thermal power plants. Sl. No Experiments 1 Formation for symmetric π /T configuration for Verification of 𝐴𝐷 − 𝐵𝐶 =1, Determination of Efficiency and Regulation. 2 Determination of Power Angle Diagrams, Reluctance Power, Excitation, Emf and Regulation for Salient and Non - Salient Pole Synchronous Machines. 3 To obtain Swing Curve and to Determine Critical Clearing Time, Regulation, Inertia Constant/Line Parameters /Fault Location/Clearing Time/Pre - Fault Electrical Output for a Single Machine connected to Infinite Bus thro ugh a Pair of identical Transmission Lines Under 3 - Phase Fault On One of the two Lines. 4 Y Bus Formation for Power Systems with and without Mutual Coupling, by Singular Transformation and Inspection Method. 5 Formation of Z Bus(without mutual coupling ) using Z - Bus Building Algorithm. 6 Determination of Bus Currents, Bus Power and Line Flow for a Specified System Voltage (Bus) Profile. 7 Formation of Jacobian for a System not Exceeding 4 Buses (No PV Buses) in Polar Coordinates. 8 Load Flow Analys is using Gauss Siedel Method, NR Method and Fast Decoupled Method for Both PQ and PV Buses. 9 To Determine Fault Currents and Voltages in a Single Transmission Line System with Star - Delta Transformers at a Specified Location for LG and LLG faults by simu lation. 10 Optimal Generation Scheduling for Thermal power plants by simulation. Revised Bloom’s Taxonomy Level: L1 – Remembering, L2 – Understanding, L3 – Applying, L4 – Analysing, L5 – Evaluating, L6 – Creating. P ower System Simulation L ab oratory - 18 EEL76 VII EEE BMS Institute of Technology and Management 5 Course outcomes: At the end of the course the student will be able to: CO1: Develop a program in MATLAB to assess the performance of transmission lines and formulate bus admittance matrix of an interconnected power systems. CO2: Develop a program in MATLAB to obtain the power angle ch aracteristics of salient a nd non - salient pole alternator and assess the transient stability under three phase fault conditions. CO3: Use Mi - Power package to assess the load flow problem , faults at different locations in power systems and optimal schedulin g of thermal power plants. CO – PO Mapping: PO1 PO2 PO3 PO4 PO5 PO6 PO7 PO8 PO9 PO10 PO11 PO12 CO1 3 2 3 3 3 1 1 CO2 3 2 3 3 3 1 1 CO3 1 2 3 3 1 3 - Strong ly Related 2 - Moderately Related 1 - Weakly Related P ower System Simulation L ab oratory - 18 EEL76 VII EEE BMS Institute of Technology and Management 6 CO NTENTS Exp No. Experiment Title Page No. 1 i. Ybus formation by I nspection method. ii&iii. Ybus formation by S ingular transformation method with and without Mutual coupling. 7 2 Determination of bus currents, bus power and line flows for a specified Bus system profile. 1 3 3 Calculation of ABCD parameters 1 5 4 Determination of power angle diagram for a) Salient pole synchronous machine. b) Non salient pole synchronous machine 2 1 5 Determination of swing curve 2 4 6 Jacobian Matrix Calculation 2 7 7 MATLAB Program to Solve Load Flow Equations using Gauss - Seidel Method (open end) 2 9 8 Short circuit analysis for power system. 3 4 9 Load flow analysis for a 3bus system using Newton Raphson Method/Gauss Siedel method 40 10 Optimal generator scheduling for thermal power plants 45 11 Formation of Z Bus (without mutual coupling) using Z - Bus Building Algorithm 47 A.1 Sinusoidal Voltages and Currents 49 A.2 Unsymmetrical F ault A nalysis 51 A.3 Introduction to MATLAB and its basic commands 58 P ower System Simulation L ab oratory - 18 EEL76 VII EEE BMS Institute of Technology and Management 7 Exp t.No - 1(i) Ybus formation by I nspection method. Aim: Bus admittance matrix ( Y b us ) formation for p ower systems using inspection method. Apparatus required : PC loaded with MATLAB Theory : Bus admittance matrix or Ybus is matrix which gives the inform ation about the admittances of lines connected to the node a s well as the admittance between the nodes. Principal diagonal e lements are called self admittances of node and is equal to the algebraic sum of all the admittances terminating at the node. Off diagonal elements are called mutual admittances and are equal to the admittances between the nodes. The size of ybus is n*n . W here n is the number of buses in the system and m= n+1( the total number of buses including the reference buses ) I bus = Y bus * V bus w here I bus = vector of impressed bus currents Y bus = bus admittance matrix. V bus = vector of bus voltages mea sured with respect to reference bus. Inspection method makes use of KVL at all the nodes to get the current equations. From these equations, Ybus can be directly written. It is the simplest and direct method of obtain ing all the diagonal elements as well a s off diagonal el ements in the matrix of any power system Bus admittance matrix is a sparse matrix. It is often used in solving load flow problems. Sparsity is one of its greatest advantages as it heavily reduces computer memory and time requirements. M ATLAB Program: clc clear all n= input(‘enter no. of buses:’) ; % no. of buses excluding reference nl= input(‘enter no. of lines:’) ; % no. of transmission lines sb= input(‘enter starting bus of each line:’) ; % starting bus of a line eb = input(‘enter ending bus of each line’) ; % ending bus of a line zser= input(‘enter resistance and reactance of each line:’) ; % line resistance and reactance (R, X) yshty= input(‘enter shunt admittance of the bus:’) ; % shunt admittance i=0 k=1 while i<nl zser1(i+1)=zser(k)+j*zser(k+1) % impedance of a line (R+jX) i=i+1 k=k+2 end zser2=reshape(zser1, nl ,1); yser=ones(nl,1)./zser2; ybus=zeros(n,n); for i=1:nl ybus(sb(i),sb(i))=ybus(sb(i),sb(i))+(j*yshty(i))+yser(i); ybus(eb(i),eb(i))=ybus(eb(i),eb(i))+(j*yshty(i))+yser(i); P ower System Simulation L ab oratory - 18 EEL76 VII EEE BMS Institute of Technology and Management 8 ybus(sb(i),eb(i))= - yser(i); ybus(eb(i),sb(i))= - yser(i); end Input: n= 5 nl= 7 sb=[ 1 1 2 2 2 3 4 ] eb=[ 2 3 3 4 5 4 5 ] zser=[ 0.02 0.06 0.08 0.24 0.06 0.18 0 .06 0.18 0.04 0.12 0.01 0.03 0.08 0.24 ] yshty=[ 0.03 0.025 0.02 0.02 0.015 0.01 0.025 ] Output: bus admittance matrix ybus 6.25+( - 18.70)j - 5.00+( 15.00)j - 1.25+( 3.75)j 0.00+( 0.00)j 0.00+( 0.00)j - 5.00+( 15.00)j 10.83+( - 32.42)j - 1.67+( 5.00)j - 1 .67+( 5.00)j - 2.50+( 7.50)j - 1.25+( 3.75)j - 1.67+( 5.00)j 12.92+( - 38.70)j - 10.00+( 30.00)j 0.00+( 0.00)j 0.00+( 0.00)j - 1.67+( 5.00)j - 10.00+( 30.00)j 12.92+( - 38.70)j - 1.25+( 3.75)j 0.00+( 0.00)j - 2.50+( 7.50)j 0.00+( 0.00)j - 1.25+( 3.75)j 3.75+( - 11.21)j Result: P ower System Simulation L ab oratory - 18 EEL76 VII EEE BMS Institute of Technology and Management 9 Expt.No - 1(ii) Ybus formation by singular transformation method without Mutual coupling Aim: Bus admittance matrix ( Y Bus ) formation for p ower systems without mutual coupling using singular transformatio n method. Apparatus required : PC loaded with MATLAB Theory : The Y Matrix is designated by Ybus and called the bus admittance matrix . Y matrix is a symmetric and square matrix that completely describes the configuration of power transmission lines. I n realistic systems which are quite large containing thousands of buses, the Y matrix is quite sparse. Each bus in a real power system is usually connected to only a few other buses through the transmission lines. The Y Matrix is designated by Ybus and cal led the bus admittance matrix . Y matrix is a symmetric and square matrix that completely describes the configuration of power transmission lines. In realistic systems which are quite large containing thousands of buses, the Y matrix is quite sparse. Each b us in a real power system is usually connected to only a few other buses through the transmission lines. Ybus can be alternatively assembled by use of singular transformation given by a graph theoretical approach. This alternative approach is of great theor etical and practical significance. Steps involving singular transformation: 1. Obtain the oriented graph for the given system. 2. Get the bus incidence matrix which is the one which indicates the incidence of all the elements to nodes in connected graph. The size of this matrix is e*(n - 1) where e is the number of elements in the graph and n is the number o f nodes (A) 3. Get the primitive a dmittance matrix from the graph of size e*e. If mutual coupling between the lines is neglected then the resulting primitive matrix is a diagonal matrix(off diagonal elements are zero) ( [y]) 4. Ybus can be obtained from the equation, Y bus = A t * [y] *A MATLAB Program: clc clear all n=input(‘enter no. of buses:’); % no. of buses excluding reference nl= input( ‘enter no. of lines:’); % no. of transmission lines sb= input(‘enter starting bus of each line:’); % starting bus of a line eb= input(‘enter ending bus of each line’); % ending bus of a line zser= input(‘enter resistance and reactance of eac h line:’); % line resistance and reactance (R, X) yshty= input(‘enter shunt admittance of the bus:’); % shunt admittance i=0 ; k=1 ; while i<nl zser1(i+1)=zser(k)+j*zser(k+1) ; % impedance of a line (R+jX) i=i+1 ; P ower System Simulation L ab oratory - 18 EEL76 VII EEE BMS Institute of Technology and Management 10 k=k+2 ; end zser2 =reshape(zser 1,nl ,1); yser=ones(nl,1)./zser2; ypri=zeros(nl+n,nl+n) ; ybus=zeros(n,n) ; a=zeros(nl+n,n) ; for i=1:n a(i,i)=1; end for i=1:nl a(n+i,sb(i))=1 ; a(n+i,eb(i))= - 1 ; ypri(n+i,n+i)=yser(i) ; ypri(sb(i),sb(i))=ypri(sb(i),sb(i))+yshty( i) ; ypri(eb(i),eb(i))=ypri(sb(i),eb(i))+yshty(i) ; end at=transpose(a); ybus=at*ypri*a ; zbus=inv(ybus); Input: n= 3 nl= 3 sb=[ 1 1 2 ] eb=[ 2 3 3 ] zser=[ 0.01 0.03 0.08 0.24 0.06 0.18 ] yshty=[ 0.01 0.025 0.02 ] Output: bus admittance matrix ybus : 11 .29 + ( - 33.75)j - 10.00 + ( 30.00)j - 1.25 + ( 3.75)j - 10.00 + ( 30.00)j 11.70 + ( - 35.00)j - 1.67 + ( 5.00)j - 1.25 + ( 3.75)j - 1.67 + ( 5.00)j 2.94 + ( - 8.75)j Result: P ower System Simulation L ab oratory - 18 EEL76 VII EEE BMS Institute of Technology and Management 11 Expt.No - 1(iii) Ybus formation by singular transformation method w ith Mutual coupling Aim: Bus admittance matrix ( Ybus ) formation for p ower systems with mutual coupling using singular transformation method. Apparatus required : PC loaded with MATLAB Theory: The current flows through the coil and produces the flux i n the same coil and this flux also links with the neighboring coil. The amount of the flux linking with the second coil is called mutual coupling. Then we say the two coils are mutually coupled. The amount of the energy spent in mutual coupling is measured by an impedance called mutual impedance. Hence the matlab program (Ybus formation using singular transformation without mutual coupling) should be modified by considering mutual co upling MATLAB program: clc clear all n=input(‘enter no. of buses:’); % no. of buses excluding reference nl= input(‘enter no. of lines:’); % no. of transmission lines sb= input(‘enter starting bus of each line:’); % starting bus o f a line eb= input(‘enter ending bus of each line’); % ending bus of a line zser= input(‘enter resistance and reactance of each line:’); % line resistance and reactance (R, X) nmc= input(‘enter no. of mutual couplings:’); f1= input(‘enter first l ine no. of each mutual coupling:’); s1= input(‘enter second line no. of each mutual coupling:’); mz= i nput(‘enter mutual impedance between lines:’); i=0 k=1 while i<nl zser1(i+1)=zser(k)+j*zser(k+1) % impedance of a line (R+jX) i=i+1 k=k+2 end i=0 k=1 while i<nmc mz1(i+1)=mz(k)+j*mz(k+1) i=i+1 k=k+2 end zser2=reshape(zser1, nl ,1); mz2=reshape(mz1,2,1); zpri=zeros(nl,nl); a=zeros(nl,n) P ower System Simulation L ab oratory - 18 EEL76 VII EEE BMS Institute of Technology and Management 12 for i=1:nl zpri(i,i)=zser2(i); if sb(i)~=0 a(i,sb(i))=1; end if eb(i)~=0 a(i,eb(i))= - 1 end end for i=1:nmc zpri(f1(i),s1(i))=mz2(i); zpri(s1(i),f1(i))=mz2(i); end ypri=inv(zpri) at=transpose(a) ybus=at*ypri*a zbus=inv(ybus ) Input: n= 3 nl= 5 sb=[ 0 0 2 0 1 ] e b =[ 1 2 3 1 3 ] zser=[ 0 0 .6 0 0 .5 0 0 .5 0 0 .4 0 0 .2 ] nmc=2 f1=[ 1 1 ] s1=[ 2 4 ] mz=[0 0 .1 0 0 .2 ] Output: Bus admittance matrix is 8.021 - 0.208 - 5.000 - 0.208 4.083 - 2.000 - 5.000 - 2.000 7.000 Bus impedance matrix is 0.271 0.126 0.230 0.126 0.344 0.189 0.230 0.189 0.361 Result: P ower System Simulation L ab oratory - 18 EEL76 VII EEE BMS Institute of Technology and Management 13 Expt.No - 2 Determination of bus currents, bus power and line flows for a specified Bus system profile. Aim: To determine the bus currents, bus power and line flows for any powe r system Apparatus required : PC loaded with MATLAB Theory : The last step in the load flow analysis is computing the power flows, bus currents and bus power on the various lines of the network Consider the line connecting buses i and k. The line and transformer at each end can be represented by a circuit with series admittance Y ik and two shunt admittances Y iko and Y kio as shown below. The current fed by bus I into the line can be expressed as, I ik = I ik1 +I iko = (V i – V k )Y ik + V i Y ik0 The power fed into the line from bus i is S ik = P ik +jQ ik = V i I ik * = V i (V i * - V k *) Y ik * + V i V i *Y iko * The power fed into the line from bus k is S ki = P ki +jQ ki = V K I ki * = V k (V k * - Vi*) Y ik * + V k V k *Y kio * The power loss in the (i - k ) the line is the sum of the power flows determined from the last two equations. The transmission loss ca n be computed by summing all line flows ( i e S ik + S k i for all i, k) The slack bus power can also be found by summing the flows on the lines terminating at the s lack bus. MATLAB program clc clear all vbus=[1.05+0j ; .98 - 0.06j ; 1 - 0.05j] ; % bus volatges yline = [0 0 10 - 20 10 - 30 10 - 20 0 0 16 - 32 10 - 30 16 - 32 0 0 ] ; % line G, B n= 3 % no. of buses i=0; k=1; while i<9 yline1(i+1) =yline(k)+j*yline (k+1); i=i+1 k=k+2 Bus i Bus k I ik I ki Y ik I ik1 I ik 0 Y ik0 Y ki0 I kio S ki S ki P ower System Simulation L ab oratory - 18 EEL76 VII EEE BMS Institute of Technology and Management 14 end yline2=reshape(yline 1,3,3) % line admittance Sp=zeros(n,n) for i=1:n for k=1:n if i==k continue else I(i,k)=yline 2(i,k)*(vbus(i) - vbus(k)); % lin e current I(k,i)= - I(i,k) S(i,k)=vbus(i)*conj(I(i,k)); % line power S(k,i)=vbus(k)*conj(I(k,i)); Sl(i,k)=S(i,k)+S(k,i); % line losses Sp(i,i)=Sp(i,i)+S(i,k); % bus power end end end Sbus=[Sp(1,1); Sp(2,2); Sp(3,3)] %bus curent calculation ; - Ibus=conj((Sbus./vbus)) % bus current Input: Yline = [ 0 0 10 - 20 10 - 30 10 - 20 0 0 16 - 32 10 - 30 16 - 32 0 0 ] n= 3 Output: line current matrix is 0.00+( 0.00)j 1.90+( - 0.80)j 2.00+( - 1.00)j - 1.90+( 0.80)j 0.00+( 0.00)j - 0.64+( 0.48)j - 2.00+( 1.00)j 0.64+( - 0.48)j 0.00+( 0.00)j line flow matrix is 0.00+( 0.00)j 2.00+( 0.84)j 2.10+( 1.05)j - 1.91+( - 0.67)j 0.00+( 0.00)j - 0.66+( - 0.43)j - 2.05+( - 0.90)j 0.66+ ( 0.45)j 0.00+( 0.00)j Line losses matrix is 0.00+( 0.00)j 0.09+( 0.17)j 0.05+( 0.15)j 0.09+( 0.17)j 0.00+( 0.00)j 0.01+( 0.02)j 0.05+( 0.15)j 0.01+( 0.02)j 0.00+( 0.00)j Bus power matrix is 4.10+( 1.89)j 0.00+( 0.00)j 0.00+( 0.00)j 0.00+( 0.00)j - 2.57+( - 1.10)j 0.00+( 0.00)j 0.00+( 0.00)j 0.00+( 0.00)j - 1.39+( - 0.45)j Sbus matrix is 4.10+( 1.89)j - 2.57+( - 1.10)j - 1.39+( - 0.45)j Ibus matrix is 3.90+( - 1.80)j - 2.54+( 1.28)j - 1.36+( 0.52) j Result: P ower System Simulation L ab oratory - 18 EEL76 VII EEE BMS Institute of Technology and Management 15 Expt.No - 3 Cal culation of ABCD parameters Aim: To Calculat e ABCD parameters for a given transmission line and find regulation and efficiency. Apparatus required : PC loaded with MATLAB T heory : In any four terminal passive , linear and bilateral network , the input voltage and input current can be expressed in terms of output voltage and output current. Incidentally a transmission line is a 4 terminal network; input terminal where power enters the network and output terminal where power leaves the network Therefore the input voltage (Vs) and input current (Is) of a 3 - phase transmission line can be expressed as Vs=A*Vr+B*Ir Is=C*Vr+D*Ir where, Vs= sending end voltage per phase Is= sending end current Vr= receiving end voltage per phase Ir= receiving end current A, B, C, D are known as generalized circuit constants of transmission line. The constants A and D are dimensionless as they are simply ratios whereas constants B and C are having unit’s ohm and mho respectively. If the network is symmetrical then we have A = D. If the network is reciprocal then we have AD – BC = 1. ABCD parameters for short transmission line: A=1 ; B=Z ; C=0 ; D=1 Short transmission line: If the tr ansmission line length is less than 80 km the line is treated as sort transmission line. As the line length is small the capacitive effects are small. Due to this effect of capacitance is neglected. Thus only resistance and inductance is to be taken into a ccount while analyzing short transmission lines. ABCD parameters for medium transmission line: T network: A=D=1+YZ/2 ; B=Z(1+YZ/4) ; C=Y PI network : A=D=1+YZ/2 ; B=Z ; C=Y(1+YZ/4) Medium Transmission line: If the length of transmi ssion line is lying between 80 - 240 km it is treated as medium transmission line. In the analysis of medium transmission line, the capacitance is taken into account as the line length is appreciable. For easiness in the calculations, the distributed capacit ance of the line is divided and is lumped across the line at one or more points. ABCD parameters for long transmission line: A=D=cosh(YZ) 1/2 ; B=(Z/Y) 1/2 *sinh(YZ) 1/2 ; C=(Y/Z) 1/2 *sinh(YZ) 1/2 Long Transmission line : if the length of transmission li ne is more than 240 km the line is considered as long transmission line. In the analysis of such lines, the line constants are considered uniformly distributed over the whole length of the line. The rigorous methods are applied to obtain the solution. P ower System Simulation L ab oratory - 18 EEL76 VII EEE BMS Institute of Technology and Management 16 Calc uation of efficiency and regulation if sending end parameters are given: clc clear all len=input('enter the length of the line in KM:'); r 1 =input(‘enter the resistance/phase/Km of the line:’); x 1 =input(‘enter the reactance/phase/Km of the line:’); s= input (‘enter the suspectance/phase/Km of the line:’); vs=input(‘enter the sending end voltage / phase of the line:’); is=input(‘enter the sending end current of the line:’); z=r 1 +j*x 1 ; y=j*s; Z=z*len; Y=y*len; if len<=80 x=1; elseif len<=240 x=2; else x=3; end switch x case 1 A=1; B=Z; C=0; D=A; case 2 p=input('enter 1 for PI and 2 for T'); A=1+(Y*Z)/2; D=A; if p==1 B=Z; C=Y*(1+(Z*Y)/4); else C=Y; B=Z*(1+(Z*Y)/4); end case 3 A=cosh(sqrt(Y*Z)); B=sqrt(Z/Y)*sinh(sqrt(Y*Z)); C=sqrt(Y/Z)*sinh(sqrt(Y*Z)); D=A; end ABCD=A*D - B*C; vr=D*vs - B*is; ir= - C*vs+A*is; spower=3*vs*conj(is) rpower=3*vr*conj(ir) efficiency= real(rpower)/real(spower)*100; P ower System Simulation L ab oratory - 18 EEL76 VII EEE BMS Institute of Technology and Management 17 regulation=(abs(vs)/abs(A) - abs (vr))/abs(vr)*100; Input: z=0.153+.38j; y=0.0+.000003j; vs=63508; is=105 - 50.5j; * For short transmission line enter the length in KM 50 spower = 2.0005e+007 +9.6215e+006i rpower = 1.9693e+007 +8.8477e+006i abcd constants of transmission line are A=1 .000000+j0.000000 B=7.650000+j19.000000 C=0.000000+j0.000000 D=1.000000+j0.000000 sending end vs=63508.000000+j0.000000 sending end current=105.000000+j - 50.500000 efficiency=98.442631 reg=2.819985 ad - bc=1.000000 * For medium transmission line with pi netw ork enter the length in KM200 enter 1 for PI and 2 for T 1 spower = 2.0005e+007 +9.6215e+006i rpower = 1.8549e+007 +1.2389e+007i abcd constants of transmission line are A=0.977200+j0.009180 B=30.600000+j76.000000 C= - 0.000003+j0.000593 D=0.977200+j0.009 180 sending end vs=63508.000000+j0.000000 sending end current=105.000000+j - 50.500000 efficiency=92.720923 reg=17.475816 ad - bc=1.000000 * For medium transmission line with T network enter the length in KM 200 enter 1 for PI and 2 for T2 spower = 2.0005e+ 007 +9.6215e+006i rpower = 1.8551e+007 +1.2505e+007i P ower System Simulation L ab oratory - 18 EEL76 VII EEE BMS Institute of Technology and Management 18 abcd constants of transmission line are A=0.977200+j0.009180 B=29.902320+j75.274054 C=0.000000+j0.000600 D=0.977200+j0.009180 sending end vs=63508.000000+j0.000000 sending end current=105.000000+j - 50.5 00000 efficiency=92.731488 reg=17.253285 ad - bc=1.000000 * For long transmission line enter the length in KM300 spower = 2.0005e+007 +9.6215e+006i rpower = 1.7673e+007 +1.2772e+007i abcd constants of transmission line are A=0.949067+j0.020304 B=44.34 1614+j112.371757 C= - 0.000006+j0.000885 D=0.949067+j0.020304 sending end vs=63508.000000+j0.000000 sending end current=105.000000+j - 50.500000 efficiency=88.342221 reg=32.155708 ad - bc=1.000000 Calcuation of efficiency and regulation if receiving end paramet ers are given: clc clear all len=input('enter the length in KM'); r 1 =input(‘enter the resistance/phase/Km of the line:’); x 1 =input(‘enter the reactance/phase/Km of the line:’); s= input(‘enter the suspectance/phase/Km of the line:’); vr=input(‘enter the r eceiving end voltage/ phase of the line:’); ir=input(‘enter the receiving end current of the line:’); z=r1+j*x1; y=j*s; Z=z*len; Y=y*len; if len<=80 x=1; elseif len<=240 x=2; else x=3; end P ower System Simulation L ab oratory - 18 EEL76 VII EEE BMS Institute of Technology and Management 19 switch x case 1 A=1; B=Z; C=0; D=A; ca se 2 p=input('enter 1 for PI and 2 for T'); A=1+(Y*Z)/2; D=A; if p= =1 B=Z; C=Y*(1+(Z*Y)/4); else C=Y; B=Z*(1+(Z*Y)/4); end case 3 A=cosh(sqrt(Y*Z)); B=sqrt(Z/Y)*sinh(sqrt(Y*Z)); C=sqrt(Y/ Z)*sinh(sqrt(Y*Z)); D=A; end ABCD=A*D - B*C; vs=A*vr+B*ir; is=C*vr+D*ir; spower=3*vs*conj(is) rpower=3*vr*conj(ir) efficiency= real(rpower)/real(spower)*100; regulation=(abs(vs)/abs(A) - abs(vr))/abs(vr)*100; Input: z=0.153+.38j; y=0.0+.000003j; vr=6350 8; ir=105 - 50.5j; Short transmission line (50 km) given Vr and Ir A=1.000000+j0.000000 B=7.650000+j19.000000 C=0.000000+j0.000000 D=1.000000+j0.000000 sending end vs=65270.750000+j1608.675000 sending end current=105.000000+j - 50.500000 efficiency=98.46651 3 P ower System Simulation L ab oratory - 18 EEL76 VII EEE BMS Institute of Technology and Management 20 reg=2.806845 ad - bc=1.000000 Medium transmission line (200 km, PI) given Vr and Ir A=0.977200+j0.009180 B=30.600000+j76.000000 C= - 0.000003+j0.000593 D=0.977200+j0.009180 sending end vs=69111.017600+j7017.703440 sending end current=102.894689+j - 10.71429 5 efficiency=94.775033 reg=11.929289 ad - bc=1.000000 Medium transmission line (200 km, T ) given Vr and Ir A=0.977200+j0.009180 B=29.902320+j75.274054 C=0.000000+j0.000600 D=0.977200+j0.009180 sending end vs=69001.100927+j6976.711950 sending end current= 103.069590+j - 10.279900 efficiency=94.718111 reg=11.746427 ad - bc=1.000000 Long transmission line given Vr and Ir A=0.949067+j0.020304 B=44.341614+j112.371757 C= - 0.000006+j0.000885 D=0.949067+j0.020304 sending end vs=70603.973689+j10849.218291 sending end current=100.287831+j10.388005 efficiency=92.700537 reg=18.487449 ad - bc=1.000000 Result: