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Power System Simulation Laboratory Exp No:1 Date: 31.12.2009 FORMATION OF Y BUS ADMITTANCE MATRIX AND BUS IMPEDANCE MATRIX USING MATLAB AIM: To obtain the bus admittance matrix (Y bus) of the given network using MATLAB THEORY: The meeting point of various components in a power system is called bus. The bus or bus bar is a conductor made of Cu or Al having negligible resistance. Hence the bus bar will have zero voltage drop when it conducts the rated current. Therefore the buses are considered as points of constant voltage in a power system. When the power system is represented by impedance /reactance diagram it can be considered as a circuit or network. The buses can be treated as nodes and the voltages of all buses (nodes) can be solved by conventional node analysis technique. Let N be the number of major or principal nodes in the circuit or network. Since the voltages of node can be measured only with respect to reference point one of node is considered as reference node. Now the network will have (N –I) independent voltages. In nodal analysis the independent voltages are solved by writing Kirchhoff‟s current law(KCL) equations. For N-1 nodes in the circuit. of writing KCL equations, the voltage sources in the circuit should be converted to equivalent current sources. Let V1,V2…..Vn. Node voltages of nodes 1,2…..n respectively.I11,I22 ,I33 …..Inn sum of current sources connected to nodes 1,2,3…..n respectively. Y jj = Sum of admittances connected to node. Y jk = Negative of the sum of admittance connected between node j and node k. Now the N no of nodal equations for n bus system will be in the form shown below Y11V1+Y12V2+Y13V3+……..Y1nVn =I11 Department of Electrical & Electronics Engineering Power System Simulation Laboratory Y21V1+Y22V2+Y23V3+……...Y2nVn=I22 … … … … … … … … … … … … Yn1V1+Yn2V2+Yn3V3+………..YnnVn=Inn The above n‟ no of equations can be arranged in the matrix form as shown in below. I BUS =YBUS VBUS When I bus is the vector of injected bus current, the current is positive when flowing away from the bus. Vbus is the vector of bus voltages measured from the reference node. bus is known as bus admittance matrix. The diagonal elements of each node is the sum of admittance connected to it . It is called self admittance or driving point admittance n= no of independent buses in the system. YII j≠i The off diagonal element is equal to the negative of admittance between the nodes. It is known as mutual admittance or transfer admittance. Yij=Yji=-Yij When bus currents are known equation(1) can be solved for the n bus voltages. Vbus= Y -1bus Ibus. The inverse of the bus admittance matrix is known as bus impedence matrix(Zbus). The admittance matrix obtained with one of the buses as reference is non singular other wise the nodal matrix is singular. Department of Electrical & Electronics Engineering Power System Simulation Laboratory If the transformer is present in the network, get the values for the transformer line impendence and also off nominal turns ratio, and also get the line charging admittance is given in the network OFF NOMINAL TURNS RATIO When the voltage or turns ratio of the transformer is not used to decide the ratio of base KV then its voltage or turns ratio. Consider a transformer with turns ratio a:1. This can be represented as an idel autotransformer in series with an admittance. Let p-q represents the input and output buses of the transformer. The ideal autotransformer is shown between P and t buses, while series admittance is shown between t and q. Itq=Current flowing from t to q Itq=(Vt-Vq)Ypq ___________ (1 The terminal current at P Ip= (Vt-Vp) Ypq/a __________ (2) Vt= Vp/a __________ (3) The terminal current at q is similarly Iq= (Vq-Vt)Ypq __________(4) Substituting for VtIq=(Vq-Vp/a) Ypq =(aVq-Vp) Ypq/a Consider an equivalent π network mode for the transformer shown below Department of Electrical & Electronics Engineering Power System Simulation Laboratory For π network Ip= (Vp-Vq)A+VpB ______________(6) Iq=(Vq-Vp)A+VqC ______________(7) Let Vp=0 and Vq=1 and substitute in (2) and (6) Ip= - Ypq/a and Ip= -A A= Ypq/a Assuming Ep=0 and Eq =1 in equation (5) and (7) Iq=Ypq and Iq=A+C = Ypq/a +c And hence C=(1-1/a) Ypq Equating the current in equation (1) and (6) and substitute for A from equation B=1/a(1/a-1) Ypq Thus we obtain equivalent π model in term of admittance and OFF nominal turn ratio as in figure. Department of Electrical & Electronics Engineering Power System Simulation Laboratory LINE CHARGING The line having a length between 50 to 150 Km and voltage s not exceeding 110 KV. In such lines the line charging capacitance can be lumped either at both ends of the line or at the center of line to give reasonably good analysis of line . If the given value of admittance is Ypq , the line charging admittance value is taken as Ypq/2 Write a program to formulate bus admittance matrix (y bus) of the given network. Department of Electrical & Electronics Engineering Power System Simulation Laboratory Line No Between buses Impedance Line Charging Off Nominal admittance Turns Ratio 1 1-2 0.025+j0.1 J0.1 __ 2 2-3 0.02+j0.08 j0.02 __ 3 3-4 0.05+j0.2 __ __ 4 1-4 0.04+j0.16 __ 0.92 Given Values Number of buses = 4 Line impedance between bus 1-2 ,Z12 =0.025+j0.1 Ω Line impedance between bus2-3, Z23 =0.02+j0.08 Ω Line impedance between bus 3-4,Z34=0.05+j0.02 Ω Line impedance between bus1-4,Z14=0.04+j0.16 Ω Transformer impedance between buses 1 & 4 trz(1,4) =0-1.75j Department of Electrical & Electronics Engineering Power System Simulation Laboratory Off nominal turns ratio of transformer a(1,4)=0.92 Generator impedance for bus (1) g(1) =1+0.5j Line charging admittance between 1&2 lba(12) =j0.01 Line charging admittance between 2&3 lca(2,3) =j0.02 Calculation of YBus Matrix Y12 = 1/Z12 =1/(0.025+j0.1) = 2.3529-j9.4118 Y23=1/Z23 = 1/(0.02+j0.08) = 2.3529-j9.4118 =2.9412-j11.7647 Y34 = 1/Z34 = 1/(0.05+j0.2) =1.1765-j4.7059 Y14(Initial) = 1/(Z14+trZ14) = 1/(0.04+j.16-j1.175) = 0.0518+j0.6285 Y14 = Y14(Initial)/a(1,4) = 0.0158+j0.6285/.92 =0.172+j0.6832 Y11=Y11+Y14(Initial)*(a(1,4)-1)/a(1,4) = 0+(0.0158+j0.6285)*(0.92-1)/0.92 = -0.00137-j0.0547 Yg1= 1/(0,5j+1) =0.8-j0.4 Y11= Y11+Y12+Y14+Yg1+hrca12 =0.00149+j0.0594+2.3529-j9.4118+0.0172+j0.6832+(0.8-0.4j)+j.005 Department of Electrical & Electronics Engineering Power System Simulation Laboratory = 3.1716-j9.0642 Y22=Y22+Y12+Y23+hra12+hrca23 =0+(2.3529-j9.4118)+(2.9412-j11.7647)+j0.005+j0.01 =5.2941-j21.1615 Y33=Y33+Y23+Y34+hrca23 =0+(2.9412-j11.7647)+(1.1765-j4.7059)+j0.01 =4.1177-j16.4606 Y44=Y44+Y14+Y34 (0.00137-j0.0547)+(0.0172+j0,6832)+(1.765-j4.7059) =1.1923-j4.0774 Y12=Y21=-Y12 Y13+Y31=-Y13=0 Y14=Y41=-Y14= -0.0172-j0.6832 Y23=Y32= -Y23= _2.942-j11.7647 Y24=Y42= -Y24=0 Y34=Y43= -Y34= -1.1765+j4.7059 RESULTANT BUS ADMITTANCE MATRIX Department of Electrical & Electronics Engineering Power System Simulation Laboratory ALGORITHM Step 1: Get the total number of buses. Step 2: Check whether generator or transformer or line charging admittance or a combination of all these are present in the given network. Step 3: Get the values of line impedances say from bus p to q for all buses in the network. Step 4: If transformer is present in the network ,get the values for transformer line impedance and also off nominal turns ratio. Step 5: Find the diagonal and off diagonal admittance for the corresponding buses which are connected to the transformer using following formulas. Z(p,q)=Zpq+TrZ(p,q) Ypq=1/Z(p,q) Ypq=y(p,q)/a(p,q) where a(p,q) is the off nominal turns ratio. Ypp=Ypp+{y p,q)(1-a(p,q))/[a(p,q)*a(p‟q)]} Yqq=yqq+[y(p,q)*(a(p,q)-1)/a(p,q)] Step 6: If generator is present get the value for generator impedance for the corresponding bus. Y(p,p)=y(p,p)+1/g(p) Step 7: Get the line charging admittance value given in the network. Connect the given values into half line charging admittance by given value by two. Step 8: Form the diagonal & off diagonal elements which can be formatted as bus admittance matrix(Y bus) Step 9: Print the resultant Y bus matrix. Department of Electrical & Electronics Engineering Power System Simulation Laboratory FLOWCHART START CHECK WHETHER THE GENERATOR OR TRANSFORMER OR LINE CHARGING ADMITTANCE COMBINATION IS PPRESENT IN THE NETWORK GET THE VALUES OF THE LINE IMPEDENCE FROM BUS P TO Q FOR ALL BUSES IN THE NETWORK IF THE IMPEDANCE TRANSFORMER IS PRESENT IN THE NETWORK GET TRANSFORMER LINE IMPEDANCE ` FIND THE DIAGONAL AND OFF DIAGONAL ADMITTANCE 1 1 Department of Electrical & Electronics Engineering Power System Simulation Laboratory 1 1 IF GENERATOR PRESENT IN THE NETWORK GET GENERATOR IMPEDANCE GET THE LINE CHARGING ADMITTANCE OF THE NETWORK FORM THE DIAGONAL AND OFF DIAGONAL ELEMENTS WHICH CAN FORMULATE THE Y BUS PRINT Y BUS STOP Department of Electrical & Electronics Engineering Power System Simulation Laboratory Y BUS Clc; Clear all n=input(„Enter the total number of buses=‟); for j=1:n for i=1:n if(==j) Z(i,j)=0; Fprintf(„Enter the generator impedance G(%d):‟j); G(j)=input(“); G(i)=G(j); Else If (i>j) Fprintf(„Enter the value of Z(%d,%d):‟j,i); Z(j,i)=input(“); Z(i,j)=Z(j,i); fprintf(„Enter the line charging admittance,L(%d,%d):‟j,i); L(j,i)=input(“); L(i,j)=L(j,i); fprintf(„Enter the off nominal turns ratio,A(%d,%d):‟j,i); A(i,j)=input(“); A(i,j)=A(j,i); fprintf(„Enter the transformerimpedance,T(%d,%d):‟j,i); Department of Electrical & Electronics Engineering Power System Simulation Laboratory T(i,j)=input(“); T(i,j)=T(j,i); end end end end end Z for p=1:n for q=1:n if(p~=q&&Z(p,q)~=0&&A(p,q)~=0) X(p,q)=1/Z(p,q)+T(p,q); W(p,q)=X(p,q)/A(p,q); Y(p,q)= -1*W(p,q); Else if(p~=q&&Z(p,q)~=0) X(p,q)=1/Z(p,q)+T(p,q); Y(p,q)= -1*X(p,q); Else Y(p,q)=0; end end end Department of Electrical & Electronics Engineering Power System Simulation Laboratory end for i=1:n for p=1:n for q=1:n if(i==p&&p~=q&&i<q&&A(p,q)~=0&&G(p)~=0) Y(i,p)=Y(i,p)+(X(p,q)*(1-A(p,q)/A(p,q)))+(1/G(p)+0.5*L(p,q)+W(i,q); else if(i==p&&p~=q&&i>q&&A(p,q)~=0G&&(p)~=0) Y(i,p)=Y(i,p)+(X(p,q)*(A(p,q)-1)/A(p,q)))+(1/G(p)+0.5*L(p,q)+L(p,q); else if(i==p&&p~=q) Y(i,p)=Y(i,p)+(X(p,q)+0.5*L(p,q)+L(p,q); end end end end end Y Enter the total number of buses=4 Enter the generator impedence G(1):1+5i Enter the value of Z(1,2):0.025+0.1i Enter the line charging admittance,L(1,2):0.01i Enter the off-nominal turns ratio,A(1,2):0 Department of Electrical & Electronics Engineering Power System Simulation Laboratory Enter the transformer impedance,T(1,2):0 Enter the value of Z(1,3) Enter the line charging admittance ,L(1,3):0 Enter the off-nominal turns ratio,A(1,3):0 Enter the transformer impedance,T(1,3):0 Enter the value of Z(1,4):.04+.16i Enter the line charging admittance,L(1,4):0 Enter the off –nominal turns ratio,A(1,4):.92 Enter the transformer impedance,T(1,4):-1.75i Enter the generator impedance G(2):0 Enter the value of Z(2,3):.02+.08i Enter the line charging admittance,L(2,3):.02i Enter the off-nominal turns rotio,A(2,3):0 Enter the transformer impedance,T(2,3):0 Enter the value of Z(2,4):0 Enter the line charging admittance,L(2,4):,02i Enter the off-nominal turns ratio,A(2,4):0 Enter the transformer impedance,T(2,4):0 Enter the generator impedance G(3):0 Enter the value of Z(3,4):.05+.2i Enter the line charging admittance,L(3,4):0 Enter the off-nominal turns ratio,A(3,4):0 Enter the transformer impedance,T(3,4):0 Department of Electrical & Electronics Engineering Power System Simulation Laboratory Enter the generator impedance G(4):0 Z= 0 0.0250+0.1000i 0 0.0400+0.1600i 0.0250+0.1000i 0 0.0250+0.8000i 0 0 0.0200+0.0800i 0 0.0500+0.2000i 0.0400+0.1600i 0 0.0500+0.2000i 0 Y= 3.1716-9.0642i -2.3529+9.4118i 0 -0.0172-0.6832i -2.3529+9.4118i 5.2941-21.1615i -2.9412+11.7647i 0 0 -2.9412+11.7647i 4.1176-16.4606i -1.1765+4.7059i -0.0172-0.6832i 0 -1.1765+4.7059i 1.1923-4.0773i RESULT Program to formulate bus admittance matrix is executed and results are verified with that of sample calculation. Department of Electrical & Electronics Engineering Power System Simulation Laboratory Exp. No. 2 Date: 4.3.2010 SEQUENCE COMPONENTS OF POWER SYSTEM NETWORK WITH SINGLE LINE TO GROUND FAULT USING MATLAB a. DETERMINATION OF FAULT PARAMETERS AIM: To determine the fault current and fault voltage for a single line to ground fault on an unloaded generator THEORY: A Fault in a circuit is any failure, which interferes its normal flow of current. The faults are associated with abnormal change in current, voltage and frequency of the power system. The faults may cause damage to the equipments if it is allowed to persist for a long time. Faults are classified into symmetrical and unsymmetrical faults. In symmetrical faults, the fault currents are equal in all the phases and can be analyzed on per phase basis. In unsymmetrical faults, the fault currents are unbalanced and so they can be analyzed only using symmetrical components. The three-phase fault is the only symmetrical fault. All other types of faults are unsymmetrical faults. The various unsymmetrical faults are 1. line to ground fault 2. Line to line fault 3. Double line to ground fault 4. one or two open conductor faults The fault condition of the power system can be divided into sub transient, transient and steady state periods. The currents in the various parts of the system and in the fault are different in these periods. Since any unsymmetrical fault causes unbalanced currents to flow in the system, the unsymmetrical faults are analyzed using symmetrical components. Let Va, Vb, Vc = three phase unbalanced voltage vectors Va1, Vb1, Vc1 = Positive sequence components of voltages Department of Electrical & Electronics Engineering Power System Simulation Laboratory Va2, Vb2, Vc2 = Negative sequence components of voltages Va0, Vb0, Vc0 = zero sequence components of voltages Now the unbalanced voltage vectors are related to sequence component voltages by the following equations Va0 = Vb0 = Vc0 Vb1 = a2 Va1 Vc1 = aVa1 Vb2 = aVa2 Vc2 = a2 Va2 a = 1<1200 a2 = 1<2400 Let Ia, Ib, Ic be a set of three phase unbalanced current vectors. Let Ia1, Ib1, Ic1 = positive sequence components of current Ia2, Ib2, Ic2 = negative sequence components of currents Ia0, Ib0, Ic0 = zero sequence components of currents Department of Electrical & Electronics Engineering Power System Simulation Laboratory Ia0 = Ib0 = Ic0 Ib1 = a2 Ia1 Ic1 = aIa1 Ib2 = aIa2 Ic2 = a2 Ia2 a = 1<1200 a2 = 1<2400 Problem: Two 11kV, 20MVA, 3Phase, star connected generators operate in parallel as shown in figure. The positive, negative and zero sequence reactances of each being respectively, j0.18, j0.15, j0.1p.u. The star point of one of the generator is isolated and that of other is earthed through a 2- Ohm resistor. A single line to ground fault occurs at the terminals of one of the generators. Estimate 1. Fault current 2. Current in grounded resistor 3. Voltage across the grounding resistor Solution Let us choose the generator values as the base values MVAb = 20MVA kVb = 11kV Base impedance, Zb = kVb2 / MVAb = 112 / 20 = 6.05 Ohm Department of Electrical & Electronics Engineering Power System Simulation Laboratory P.U value of neutral resistance = Actual value / Base impedance = 2/6.05 = 0.3306 The single line diagram and the sequence networks of the given power systems are shown in figure. Single line diagram Department of Electrical & Electronics Engineering Power System Simulation Laboratory The Thevenin‟s equivalent of the sequence network is as shown in figure For a single line to ground fault Ia1 = Ia2 = Ia0 If = Ia = 3Ia1 Department of Electrical & Electronics Engineering Power System Simulation Laboratory Hence, the Thevenin‟s of a sequence networks are connected in series as shown in the figure. The fault current is calculated by using the prefault voltage Vpf = 1 pu From the figure we get 0 = = 0.9741 <-150 pu To find the fault current Fault current, If = Ia = 3Ia1 = 3 *0.9741<-150 Department of Electrical & Electronics Engineering Power System Simulation Laboratory = 2.9223<-150 pu Base current, Ib = = = 1049.7A Actual value of fault current = p.u value of fault current * Base current =2.9223<-150 * 1049.7 = 3067.5<-150A = 3.0675<-150kA To find the current through the neutral resistor The current through the neutral resistor is same as that of the fault current Current through the neutral resistor = 2.9823<-150 pu = 3.0675<-150 kA To find the voltage across grounding resistor From the Thevenin‟s equivalent of the zero sequence network we get The voltage across grounding resistor =3RnIa0 =3RnIa1 =3*0.3306*0.9741<-150 pu =0.9661<-150 pu Department of Electrical & Electronics Engineering Power System Simulation Laboratory Actual value of the voltage across grounding resistor = pu value of voltage * kVb/√3 Where kVb is the line value = 0.9661<-150 * 11/√3 = 6.1356<-150 kV ALGORITHM : Step 1: Get the ratings of the generator power and voltage in MVA and kV respectively. Step 2: Get the values of zero (X0) , positive (X1) and negative(X2) reactance‟s respectively Step 3: Get the values of neutral to ground resistance from the generator (R). Step 4: Z b = ( Voltage ratings of the generator)2 (Power ratings of generator) Step 5: Compute the per unit value of the neutral to ground resistance, Rn =R/ Zb Step 6 : Find the Thevenin equivalent reactance (Th) Step 7 : Assume the prefault voltage Vpf = 1 p.u Step 8: Compute the single phase to ground fault current, Ia1 = Vpf / Th Step 9: Compute the fault current, If1 = 3*Ia1 Step 10: Compute the base current, Ib= (Power rating*106)/(√3*Voltage rating*103) Step 11: Compute the actual fault current, If = If1 * Ib Department of Electrical & Electronics Engineering Power System Simulation Laboratory Step 12 : Compute voltage across the grounding resistor, Vgr = 3* Rn *Ia1 Step 13: Compute the actual voltage across the grounding resistor , Vagr = Vgr * (Voltage rating)/√3 Department of Electrical & Electronics Engineering Power System Simulation Laboratory FLOWCHART Start Get the ratings of generator Get the value of Symmetrical reactance Get the value of neutral to ground resistance Compute Base impedance Compute per unit value of neutral to ground resistance Find the Thevenin equivalent reactance Assume prefault voltage Vpf = 1p.u Compute fault currents and voltages Stop Department of Electrical & Electronics Engineering Power System Simulation Laboratory PROGRAM clc; clear all; P=input('Enter the value of the power of generator in MVA:'); V=input('Enter the value of the voltage of generator in kV:'); X0=input('Enter the value of the zero sequence reactance:i'); X1= input('Enter the value of the positive sequence reactance:i'); X2=input('Enter the value of negative sequence reactance:i'); R=input('Enter the value of nutral to ground resistance from the generator:'); Zb=V^2/P; Rn=R/Zb; TX(1)=complex(0,X1/2); TX(2)=complex(0,X2/2); TX(3)=complex(3*Rn,X0); Th=0; for i=1:3 Th=Th+TX(i); end Vpf=1; Ia1=Vpf/Th; If1=3*Ia1; Ib=(P*10^6)/(sqrt(3)*V*10^3); If=If1*Ib; fprintf('Actual fault current is'); If I1=abs(If); I2=angle(If); Magnitude=I1 Angle=I2*180/pi Vgr=3*Rn*Ia1; Vagr=Vgr*V/sqrt(3); fprintf('Actual value of voltage across grounding resistor is'); Vagr Magnitude=abs(Vagr) Angle=angle(Vagr)*180/pi Department of Electrical & Electronics Engineering Power System Simulation Laboratory OUTPUT: Enter the value of the power of generator in MVA: 20 Enter the value of the voltage of generator in kV: 11 Enter the value of the zero sequence reactance: i0.1 Enter the value of the positive sequence reactance: i0.18 Enter the value of negative sequence reactance : i0.15 Enter the value of nutral to ground resistance from the generator: 2 Actual fault current is If = 2.9638e+003 -7.9195e+002i Magnitude = 3.0678e+003 Angle = -14.9604 Actual value of voltage across grounding resistor is Vagr = 5.9276 - 1.5839i Magnitude = 6.1356 Angle = -14.9604 RESULT: Fault current and fault voltage for a single line to ground fault on an unloaded generator is determined using MATLAB. Department of Electrical & Electronics Engineering Power System Simulation Laboratory Exp No: 3 Date: 18.2.2010 MODELING OF SINGLE MACHINE POWER SYSTEM USING SIMULINK AIM: To simulate swing equation for a single machine infinite bus system. A 20 MVA, 50 Hz generator delivers 18 MW over a double circuit line to an infinite bus. The generator has kinetic energy of 2.52 MJ/MVA at rated speed the generator transient reactance is Xd` = 0.35 pu. Each transmission circuit has R=0 and reactance of 0.2 pu on a 20 MVA base. | E`| = 1.1 pu and infinite bus voltage V =1.0< 0°. A three phase short circuit occurs at the mid point of one of the transmission lines. Plot swing curve with fault cleared by simultaneous opening of breakers at both ends of the line at 2.5 cycles and 6.25 cycles after occurrence of fault. And also plot the swing curve over a period of 0.5 seconds if the fault is sustained. THEORY: Under normal operating conditions the relative position of rotor axis and the resultant magnetic field is fixed. The angle between the two is known as power angle or torque angle. During any disturbance, rotor will decelerate or accelerate with respect to the synchronously rotating air gap mmf and a relative motion begins. The equation describing this relative motion is known as swing equation. Consider a synchronous generator developing an electromagnetic torque Te and running at the synchronous speed ωsm. If Tm is the driving mechanical torque under steady state operations with losses neglected we have Tm=Te. Department of Electrical & Electronics Engineering Power System Simulation Laboratory A departure from steady state due to a disturbance results in an accelerating (T m>Te) or decelerating (Tm<Te) torque Ta on the rotor. If J is the combined moment of the prime mover and the generator neglecting frictional and damping torques. From law‟s of rotation we have J d2 Өm / dt2 = Ta = Tm-Te Where Өm is the angular displacement of the rotor with respect to the stationary reference axis on the stator. ωsm is the constant angular velocity Өm= ωsm.t+δm δm is the rotor position before disturbance at time t = 0 ωm= d Өm / dt = ωsm + d δm /dt Rotor acceleration is d2 Өm / dt2 = d2 δm / dt2 J d2 δm / dt2 = Tm-Te J ωm d2 δm / dt2 = ωm Tm - ωm Te Since angular velocity times torque is equal to the power, we right the above equation in terms of power. J ωm d2 δm / dt2 = Pm - Pe The quantity J ωm is called the inertia constant and is denoted by M. It is related to kinetic energy of the rotating masses, ωk Department of Electrical & Electronics Engineering Power System Simulation Laboratory ωk=(1/2)J ωm 2 = (1/2) M ωm M= 2 ωk /m Although M is called inertia constant, it is not really constant. When the rotor speed deviates from the synchronous speed. M = 2 ωk / ωsm The swing equation in terms of the inertia constant becomes. M d2 δm /dt2 = Pm - Pe It is more convenient to write the swing equation in terms of the electrical power angle δ. If P is the number of poles of a synchronous generator, the electrical power angle δ is related to mechanical power angle δm by δ = (p/2) δm ω = (p/2) ωm Swing equation in terms of electrical power angle is (2/p) M d2 δ / dt2 = Pm - Pe Since power system analysis is done in per unit system, the swing equation is usually expressed in per unit. Dividing by the base power SB, and substituting for M. (2/p)( 2ωk / ωsm) d2 δ / dt2 = (Pm / SB )–( Pe / SB ) H is a constant called per unit inertia constant. H = (Kinetic energy in MJ at rated speed)/ Machine rating in MVA = ωk / SB The unit of H is seconds. Department of Electrical & Electronics Engineering Power System Simulation Laboratory (2.2H/ Pωsm) d2 δ / dt2 = Pm (pu) – Pe (pu) Pm (pu) and Pe (pu) are the per unit mechanical and electrical power. The electrical angular velocity is related to the mechanical angular velocity ωsm = (2/p)ωs (2H/ ωs) d2 δ / dt2 = Pm (pu) – Pe (pu) In terms of frequency F0 and to simplify the rotation the subscript pu is omitted and the powers are understood to be in per unit. (H/ Π F0) d2 δ / dt2 = Pm - Pe Where δ is in electrical radians. If δ is expressed in electrical degrees, the swing equation becomes, (H/180) d2 δ / dt2 = Pm - Pe PROCEDURE: 1. Open Matlab - new-model file and save it 2. Click start button on Matlab- start-Simulink-library browser 3. Select the required blocks and elements from the library and drag it to the model file 4. Connect the blocks and save it. 5. Initialize the integrator 1 block with value of δ0 in rad (0.377 ≈ 21.6°) 6. Set simulation time (0.1 sec) and start simulation 7. Change the threshold values of switch from (2.5/50) to (6.5/50) 8. For sustained condition give another value higher than both 9. Click on the scope so that output is obtained Department of Electrical & Electronics Engineering Power System Simulation Laboratory Fig No : 1 Single Machine Infinite Bus System Department of Electrical & Electronics Engineering Power System Simulation Laboratory Fig No. 2 : Swing Curve for 2.5/50 Department of Electrical & Electronics Engineering Power System Simulation Laboratory Fig No. 3 : Swing Curve for 6.5/50 Department of Electrical & Electronics Engineering Power System Simulation Laboratory Fig No. 4 : Swing Curve with sustained oscillations. Department of Electrical & Electronics Engineering Power System Simulation Laboratory Fig No. 5: Scope output curve for Gain Department of Electrical & Electronics Engineering Power System Simulation Laboratory RESULT: Swing curve is generated using simulink. Department of Electrical & Electronics Engineering Power System Simulation Laboratory Exp No: 4 Date: 7.1.2010 SHORT CIRCUIT STUDIES OF POWER SYSTEM USING Mi POWER AIM: To perform the short circuit analysis on a given 5- bus system using Mi Power and get the values of short circuited parameters by using A) Gauss- Siedel Method B) Newton Raphson method THEORY: When an abnormal condition arises in a power system such as fault, an insulator flashover, or lightning stroke to the transmission tower, high current flows in the power system. These currents are sensed through the relays to isolate the faulty section of the power system. Delay in the operation of the circuit breaker might result in creating instability in the system and also cause burn out of costly equipments such as transformer, generator etc. Short circuit study is an important study which provides vital information regarding the magnitude of fault current through various components of the power system during short circuit. This helps in proper selection of the circuit breaker and relays. It also helps in relay co-ordination. In arriving at a mathematical model for short circuit studies, a number of assumptions are made which simplify the formulation of the problem. Certain valid assumptions are: 1. The load, line charge capacitances and other shunt connections to the ground are neglected. Department of Electrical & Electronics Engineering Power System Simulation Laboratory 2. The generator is represented by a voltage source in series with a reactance which is taken to be sub-transient or transient reactance. 3. All the transformers are considered to be at their nominal taps 4. If the resistance of the transmission lines are sufficiently smaller than the reactance the resistance are neglected 5. Pre-fault voltage at all the buses is assumed to be(1+j0) per unit The mathematical model of the system can be derived as follows: Fig(1).The figure shows the equivalent of the power system consisting of m generator buses Voa : Pre- fault voltage referred to phase “A” Zi‟ : Generator Transient impedance, i referes to generator bus. 1,2,… m : Generator buses n : Total number of bus o : Ground bus Department of Electrical & Electronics Engineering Power System Simulation Laboratory Consider the voltage source Voa is shorted ie.nodes o and o‟ are shorted, the whole network is reduced to a passive network. We may write the mathematical equation as [Vbus] = [Zbus][Ibus] …………………………(1) Now introduce the voltage source between o and o‟. This results in raising all the bus voltage by Voa. Hence ,we may write [Vbus] = [Zbus][Ibus] + bVoa ………………………(2) Where b = [1,1,………1]TSuppose that a three phase fault occurs at pth bus. The voltages are as shown in Figure (2). 1 2 V1(F) P n V2(F) o ZF VP(F) FF Vn(F) Fig(2) P n o Department of Electrical & Electronics Engineering Power System Simulation Laboratory Zf = Fault Impedance If = Fault Current Ip(F) = Current injected at bus „p‟ It is clear from the above figure (2) that If = -Ip(F) ……………………….(3) Vf = Vp(F) =If.Zf ………………………..(4) Ii(F) = 0; Vi(F) = unknown for i = 1,2,……….n, I # P. This follows from the fact that the venin‟s theorem is applied to n –port network shown in the above figure. Expanding equation 2 for n port description. V1(F) = Z11I1(F) + ………………..+ Z1PIp(F) +…………..+Z1nIn(F) + Voa V2(F) = Z21I1(F) + ………………+ Z2PIP(F) + …………….+Z2nIn(F) + Voa VP(F) = ZP1I1(F) +…………..+ ZPPIP(F) + …………….+ZPnIn(F) + Voa Vn(F) = Z21I1(F) +……………+Z2PIP(F) + ……………+ ZnnIn(F) + Voa Using the results 3&4 in the above equation, we get Zf IF = -ZPPIF + Voa IF = Voa ZPP + ZF Therefore VF = ZF . Voa = VP(F) ZPP + ZF The other bus voltages are obtained from the rest of the equations in equation (5). Department of Electrical & Electronics Engineering Power System Simulation Laboratory Vi(F) = Voa - ZiP . Vo a ZPP + ZF Vi(F) = Voa 1- ZiP where i= 1,2,………………..n ZPP + ZF Knowing the fault current (IF) and all the bus voltages, the line currents and generator for currents can be computed. Algorithm: Step 1: Start Step 2: Read line data, machine data, and transformer data, Fault impedance, etc. Step 3: Compute Ybus matrix and calculate modified Ybus matrix. Step 4: Form(Zbus) by inverting the modified Ybus matrix. Step 5: Initialize count 1=0. Step 6: I = I+1. This means fault occurs a bus‟I‟. Step 7:Compute fault current at faulted bus and bus voltages at all buses. Step 8: Compute all line currents and generator currents. Step 9: Check whether 1 is less than the number of buses, if so go to step-6 otherwise go to next step. Step 10: Print the results. Step 11: Stop. Department of Electrical & Electronics Engineering Power System Simulation Laboratory FLOW CHART START Read line data, bus data,machine data, Fault impedance,etc. Compute Y bus matrix & modified Y bus matrix I=0 I=I+1 Fault occurs at bus I Compute fault current at faulted bus and bus voltages at all buses. Compute all line currents and generator currents IS yes I<nb? No Print the results STOP Department of Electrical & Electronics Engineering Power System Simulation Laboratory The figure shows the single line diagram of a 6-bus system with two identical generating units, five lines and two transformers. Per unit transmission line series impedances and shunt susceptances are given on 100 MVA base, generator‟s transient impedance and transformer leakage reactances are given in the accompanying table. If a fault occurs in bus 5-4, find the fault MVA. The data is given below: Department of Electrical & Electronics Engineering Power System Simulation Laboratory If a 3 phase to ground fault occurs at bus 5 find the fault MVA. The data is given below. BUS CODE IMPEDENCE Zpq LINE CHARGING P-q Zpq Y’pq/2 3-4 0.00+j0.15 0 3-5 0.00+j0.10 0 3-6 0.00+j0.2 0 5-6 0.00+j0.15 0 4-6 0.00+j0.10 0 Generator details: G1=G2=100 MVA,11 KV with Xd‟ =10% Transformer details: T1=T2=11/110 KV, 100 MVA, leakage reactance=x= 5% All impedances are on 100 MVA base Department of Electrical & Electronics Engineering Power System Simulation Laboratory PROCEDURE: The software Mi power introduces a better interface between the human and the computer for the easier calculation of power flow problems like fault analysis, load flow analysis, contingency occurrence etc. PROCEDURE TO ENTER THE DATA: Open Power system network editor. Select menu option Data base Configure. Click Browse button. Open dialog box is popped up, where we can browse the desired directory and specify the name of the database to be associated with the single line diagram . Click the Open button after the entering the desired database name. Configure Database dialog will appear with path chosen. Click on OK button in the Configure database dialog, the following configuration information appears. Uncheck the power system libraries and Standard Relay Libraries. 1. To draw the single line diagram: To draw bus : Click on Bus icon provided on power system tool bar. Draw a bus and give Bus ID and Bus Name. Click OK Bus Data Bus No 1 2 3 4 5 6 Bus Name Bus1 Bus2 Bus3 Bus4 Bus5 Bus6 Nominal Voltage 11 11 110 110 110 110 Area Number 1 1 1 1 1 1 Zone number 1 1 1 1 1 1 Contingency weightage 1 1 1 1 1 1 Department of Electrical & Electronics Engineering Power System Simulation Laboratory 2.To Draw Transmission Line: Click on Transmission Line icon provided on power system tool bar.Enter Element ID No and click OK. Enter structure ref No. and click on Transmission Line Library button. Enter Transmission Line Library data. Transmission Line Element Data Line Number 1 2 3 4 5 Line Name Line 3-4 Line 3-5 Line 3-6 Line 4-6 Line 5-6 De-Rated 100 100 100 100 100 MVA No. of circuits 1 1 1 1 1 From Bus No. 3 3 3 4 5 To Bus No. 4 5 6 6 6 Line Length 1 1 1 1 1 From Breaker 5000 5000 5000 5000 5000 Rating To Breaker 5000 5000 5000 5000 5000 Rating Structure Ref. 1 2 3 2 1 No. Transmission Line Library Data Structure Ref. No. 1 2 3 Structure Ref. Name Line 3-4 & 5-6 Line 3-5 & 4-6 Line 3-6 Positive sequence resistance 0 0 0 Positive sequence reactance 0.15 0.1 0.2 Positive sequence susceptance 0 0 0 Thermal Rating 100 100 100 Department of Electrical & Electronics Engineering Power System Simulation Laboratory 3.To Draw Transformer Click on Two Winding Transformer icon provided on power system Tool bar. Connect the transformer in between two buses. Then Element ID dialog will appear. Click OK. Transformer Element Data form will be open. Enter the Manufacturer Ref.No. Enter Transformer Data. Click on Transformer Library button. Enter the data. Save and close Library screen. 2nd Transformer details Transformer Number 2 Transformer Name 2T2 From Bus Number 6 To Bus Number 2 Control Bus Number 2 Number of units in Parallel 1 Manufacturer ref.Number 30 De-Rated MVA 100 From Breaker Rating 5000 To Breaker Rating 350 Nominal Tap Position 5 4.To Draw Generator : Click on Generator icon provided on power system Tool Bar. Draw the generator on the bus 1. Element ID dialog will appear. Click OK. Generator Data form will be opened. Enter the Manufacturer ref.No. Enter Generator Data. Click on Generator Library button. Enter Generator Library details. Save and close Library screen. Department of Electrical & Electronics Engineering Power System Simulation Laboratory Name GEN-2 Bus Number 2 Manufacturer Ref. Number 20 Number of Generators in parallel 1 Capability Curve number 0 De-rated MVA 100 Specified voltage 11 Scheduled Power 80 Reactive Power minimum 0 Reactive Power maximum 60 Breaker rating 350 Type of modeling Infinite Department of Electrical & Electronics Engineering Power System Simulation Laboratory OUTPUT: I SHORT CIRCUIT STUDIES CASE NO : 1 CONTINGENCY : 0 SCHEDULE NO : 0 CONTINGENCY NAME : Base Case ------------------------------------------------------------------------------- LARGEST BUS NUMBER USED : 6 ACTUAL NUMBER OF BUSES : 6 NUMBER OF 2 WIND. TRANSFORMERS : 2 NUMBER OF 3 WIND. TRANSFORMERS : 0 NUMBER OF TRANSMISSION LINES : 5 NUMBER OF SERIES REACTORS : 0 NUMBER OF SERIES CAPACITORS : 0 NUMBER OF BUS COUPLERS : 0 NUMBER OF SHUNT REACTORS : 0 NUMBER OF SHUNT CAPACITORS : 0 NUMBER OF SHUNT IMPEDANCES : 0 NUMBER OF GENERATORS : 2 NUMBER OF MOTORS : 0 NUMBER OF LOADS : 0 NUMBER OF FILTERS : 0 NUMBER OF HVDC CONVERTORS : 0 ------------------------------------------------------------------------------- NUMBER OF ZONES : 1 PRINT OPTION : 3 (BOTH DATA AND RESULTS PRINT) PLOT OPTION : 0 (NO PLOT FILE GENERATION) BASE MVA : 100.000 NOMINAL SYSTEM FREQUENCY: 50.000 PREFAULT VOLTAGE OPTION: 0 (VOLTAGE OF 1.0 PU IS ASSUMED) FAULT OPTION : 1 (FAULT CONSIDERED AT SELECTED BUSES, ONE AT A TIME) FLOW OPTION : 0 (NO FAULT CONTRIBUTIONS ARE COMPUTED) Department of Electrical & Electronics Engineering Power System Simulation Laboratory FAULT TYPE : 4 (DOUBLE LINE TO GROUND FAULT FROM Y AND B PHASES) POST FAULT VOLT OPTION : 0 (NO COMPUTATION) ------------------------------------------------------------------------------- TOTAL FAULT BUSES : 2 BUSES : 2 4 ------------------------------------------------------------------------------- FAULT RESISTANCE - PHASE - 0.000000 (PU) FAULT REACTANCE - PHASE - 0.000000 (PU) FAULT RESISTANCE - GROUND - 0.000000 (PU) FAULT REACTANCE - GROUND - 0.000000 (PU) II SHORT CIRCUIT STUDIES CASE NO : 1 CONTINGENCY : 0 SCHEDULE NO : 0 CONTINGENCY NAME : Base Case ------------------------------------------------------------------------------- LARGEST BUS NUMBER USED : 6 ACTUAL NUMBER OF BUSES : 6 NUMBER OF 2 WIND. TRANSFORMERS : 2 NUMBER OF 3 WIND. TRANSFORMERS : 0 NUMBER OF TRANSMISSION LINES : 5 NUMBER OF SERIES REACTORS : 0 NUMBER OF SERIES CAPACITORS : 0 NUMBER OF BUS COUPLERS : 0 NUMBER OF SHUNT REACTORS : 0 NUMBER OF SHUNT CAPACITORS : 0 NUMBER OF SHUNT IMPEDANCES : 0 NUMBER OF GENERATORS : 2 NUMBER OF MOTORS : 0 NUMBER OF LOADS : 0 NUMBER OF FILTERS : 0 NUMBER OF HVDC CONVERTORS : 0 ------------------------------------------------------------------------------- Department of Electrical & Electronics Engineering Power System Simulation Laboratory NUMBER OF ZONES : 1 PRINT OPTION : 3 (BOTH DATA AND RESULTS PRINT) PLOT OPTION : 0 (NO PLOT FILE GENERATION) BASE MVA : 100.000 NOMINAL SYSTEM FREQUENCY: 50.000 PREFAULT VOLTAGE OPTION : 0 (VOLTAGE OF 1.0 PU IS ASSUMED) FAULT OPTION : 1 (FAULT CONSIDERED AT SELECTED BUSES, ONE AT A TIME) FLOW OPTION : 0 (NO FAULT CONTRIBUTIONS ARE COMPUTED) FAULT TYPE : 7 (3 PHASE AND SLG, ONE AFTER THE OTHER) POST FAULT VOLT OPTION : 0 (NO COMPUTATION) ------------------------------------------------------------------------------- TOTAL FAULT BUSES : 1 BUSES : 3 ------------------------------------------------------------------------------- FAULT RESISTANCE - PHASE - 0.000000 (PU) FAULT REACTANCE - PHASE - 0.000000 (PU) FAULT RESISTANCE - GROUND - 0.000000 (PU) FAULT REACTANCE - GROUND - 0.000000 (PU) ------------------------------------------------------------------------------- FAULT AT BUS NUMBER 1 : NAME Bus1 CURRENT (AMPS/DEGREE) FAULT MVA SEQUENCE (1,2,0) PHASE (A,B,C) SEQUENCE (1,2,0) PHASE (A,B,C) MAGNITUDE ANGLE MAGNITUDE ANGLE MAGNITUDE MAGNITUDE --------- ------- --------- ------- --------- --------- 3572 -89.73 3572 -89.73 1361 1361 0 -90.00 3572 150.27 0 1361 0 -90.00 3572 30.27 0 1361 R/X RATIO OF THE SHORT CIRCUIT PATH : 0.0048 PEAK ASYMMETRICAL SHORT-CIRCUIT CURRENT : 10051 AMPS Department of Electrical & Electronics Engineering Power System Simulation Laboratory PASCC = k x sqrt(2) x If , k = 1.9897 ------------------------------------------------------------------------------- FAULT AT BUS NUMBER 2 : NAME Bus2 CURRENT (AMPS/DEGREE) FAULT MVA SEQUENCE (1,2,0) PHASE (A,B,C) SEQUENCE (1,2,0) PHASE (A,B,C) MAGNITUDE ANGLE MAGNITUDE ANGLE MAGNITUDE MAGNITUDE --------- ------- --------- ------- --------- --------- 71443 -89.73 71443 -89.73 1361 1361 0 -90.00 71443 150.27 0 1361 0 -90.00 71443 30.27 0 1361 R/X RATIO OF THE SHORT CIRCUIT PATH : 0.0048 PEAK ASYMMETRICAL SHORT-CIRCUIT CURRENT : 201026 AMPS PASCC = k x sqrt(2) x If , k = 1.9897 ------------------------------------------------------------------------------- FAULT AT BUS NUMBER 3 : NAME Bus3 CURRENT (AMPS/DEGREE) FAULT MVA SEQUENCE (1,2,0) PHASE (A,B,C) SEQUENCE (1,2,0) PHASE (A,B,C) MAGNITUDE ANGLE MAGNITUDE ANGLE MAGNITUDE MAGNITUDE --------- ------- --------- ------- --------- --------- 5813 -89.17 5813 -89.17 1108 1108 0 -90.00 5813 150.83 0 1108 0 -90.00 5813 30.83 0 1108 R/X RATIO OF THE SHORT CIRCUIT PATH : 0.0144 PEAK ASYMMETRICAL SHORT-CIRCUIT CURRENT : 16187 AMPS PASCC = k x sqrt(2) x If , k = 1.9689 ------------------------------------------------------------------------------- FAULT AT BUS NUMBER 4 : NAME Bus4 CURRENT (AMPS/DEGREE) FAULT MVA SEQUENCE (1,2,0) PHASE (A,B,C) SEQUENCE (1,2,0) PHASE (A,B,C) Department of Electrical & Electronics Engineering Power System Simulation Laboratory MAGNITUDE ANGLE MAGNITUDE ANGLE MAGNITUDE MAGNITUDE --------- ------- --------- ------- --------- --------- 3871 -89.47 3871 -89.47 738 738 0 -90.00 3871 150.53 0 738 0 -90.00 3871 30.53 0 738 R/X RATIO OF THE SHORT CIRCUIT PATH : 0.0092 PEAK ASYMMETRICAL SHORT-CIRCUIT CURRENT : 10840 AMPS PASCC = k x sqrt(2) x If , k = 1.9801 ------------------------------------------------------------------------------- FAULT AT BUS NUMBER 5 : NAME Bus5 CURRENT (AMPS/DEGREE) FAULT MVA SEQUENCE (1,2,0) PHASE (A,B,C) SEQUENCE (1,2,0) PHASE (A,B,C) MAGNITUDE ANGLE MAGNITUDE ANGLE MAGNITUDE MAGNITUDE --------- ------- --------- ------- --------- --------- 3871 -89.47 3871 -89.47 738 738 0 -90.00 3871 150.53 0 738 0 -90.00 3871 30.53 0 738 R/X RATIO OF THE SHORT CIRCUIT PATH : 0.0092 PEAK ASYMMETRICAL SHORT-CIRCUIT CURRENT : 10840 AMPS PASCC = k x sqrt(2) x If , k = 1.9801 ------------------------------------------------------------------------------- FAULT AT BUS NUMBER 6 : NAME Bus6 CURRENT (AMPS/DEGREE) FAULT MVA SEQUENCE (1,2,0) PHASE (A,B,C) SEQUENCE (1,2,0) PHASE (A,B,C) MAGNITUDE ANGLE MAGNITUDE ANGLE MAGNITUDE MAGNITUDE --------- ------- --------- ------- --------- --------- 5813 -89.17 5813 -89.17 1108 1108 0 -90.00 5813 150.83 0 1108 0 -90.00 5813 30.83 0 1108 Department of Electrical & Electronics Engineering Power System Simulation Laboratory R/X RATIO OF THE SHORT CIRCUIT PATH : 0.0144 PEAK ASYMMETRICAL SHORT-CIRCUIT CURRENT : 16187 AMPS PASCC = k x sqrt(2) x If , k = 1.9689 ------------------------------------------------------------------------------- 3 phase fault level Bus No. Name BUS kV 3PH-fMVA Fault I NOMINAL kA ------- -------- -------- -------- -------- 1 Bus1 220.000 1361.2 3.572 2 Bus2 11.000 1361.2 71.445 3 Bus3 110.000 1107.6 5.814 4 Bus4 110.000 737.5 3.871 5 Bus5 110.000 737.5 3.871 6 Bus6 110.000 1107.6 5.814 ------------------------------------------------------------------------------- ------------------------------------------------------------------------------- FAULT AT BUS NUMBER 1 : NAME Bus1 CURRENT (AMPS/DEGREE) FAULT MVA SEQUENCE (1,2,0) PHASE (A,B,C) SEQUENCE (1,2,0) PHASE (A,B,C) MAGNITUDE ANGLE MAGNITUDE ANGLE MAGNITUDE MAGNITUDE --------- ------- --------- ------- --------- --------- 0 -90.00 0 -90.00 0 0 0 -90.00 0 180.00 0 0 0 -90.00 0 180.00 0 0 ------------------------------------------------------------------------------- FAULT AT BUS NUMBER 2 : NAME Bus2 CURRENT (AMPS/DEGREE) FAULT MVA SEQUENCE (1,2,0) PHASE (A,B,C) SEQUENCE (1,2,0) PHASE (A,B,C) MAGNITUDE ANGLE MAGNITUDE ANGLE MAGNITUDE MAGNITUDE Department of Electrical & Electronics Engineering Power System Simulation Laboratory --------- ------- --------- ------- --------- --------- 1 -90.00 2 -90.00 0 0 1 -90.00 0 180.00 0 0 1 -90.00 0 180.00 0 0 ------------------------------------------------------------------------------- FAULT AT BUS NUMBER 3 : NAME Bus3 CURRENT (AMPS/DEGREE) FAULT MVA SEQUENCE (1,2,0) PHASE (A,B,C) SEQUENCE (1,2,0) PHASE (A,B,C) MAGNITUDE ANGLE MAGNITUDE ANGLE MAGNITUDE MAGNITUDE --------- ------- --------- ------- --------- --------- 2953 -88.55 8860 -88.55 563 1688 2953 -88.55 0 0.00 563 0 2953 -88.55 0 0.00 563 0 ------------------------------------------------------------------------------- FAULT AT BUS NUMBER 4 : NAME Bus4 CURRENT (AMPS/DEGREE) FAULT MVA SEQUENCE (1,2,0) PHASE (A,B,C) SEQUENCE (1,2,0) PHASE (A,B,C) MAGNITUDE ANGLE MAGNITUDE ANGLE MAGNITUDE MAGNITUDE --------- ------- --------- ------- --------- --------- 1295 -89.47 3884 -89.47 247 740 1295 -89.47 0 0.00 247 0 1295 -89.47 0 0.00 247 0 ------------------------------------------------------------------------------- FAULT AT BUS NUMBER 5 : NAME Bus5 CURRENT (AMPS/DEGREE) FAULT MVA SEQUENCE (1,2,0) PHASE (A,B,C) SEQUENCE (1,2,0) PHASE (A,B,C) MAGNITUDE ANGLE MAGNITUDE ANGLE MAGNITUDE MAGNITUDE --------- ------- --------- ------- --------- --------- 1295 -89.47 3884 -89.47 247 740 Department of Electrical & Electronics Engineering Power System Simulation Laboratory 1295 -89.47 0 0.00 247 0 1295 -89.47 0 0.00 247 0 ------------------------------------------------------------------------------- FAULT AT BUS NUMBER 6 : NAME Bus6 CURRENT (AMPS/DEGREE) FAULT MVA SEQUENCE (1,2,0) PHASE (A,B,C) SEQUENCE (1,2,0) PHASE (A,B,C) MAGNITUDE ANGLE MAGNITUDE ANGLE MAGNITUDE MAGNITUDE --------- ------- --------- ------- --------- --------- 2953 -88.55 8860 -88.55 563 1688 2953 -88.55 0 0.00 563 0 2953 -88.55 0 0.00 563 0 ------------------------------------------------------------------------------- Single phase fault level Bus No. Name BUS kV 1PH-fMVA Fault I NOMINAL kA ------- -------- -------- -------- -------- 1 Bus1 220.000 0.0 0.000 2 Bus2 11.000 0.0 0.002 3 Bus3 110.000 1688.1 8.861 4 Bus4 110.000 739.9 3.884 5 Bus5 110.000 739.9 3.884 6 Bus6 110.000 1688.1 8.861 ------------------------------------------------------------------------------- ------------------------------------------------------------------------------- Positive, negative and zero sequence impedancees on 100.000 MVA base Bus No. Name BUS kV 3PH-fMVA SLG-fMVA P-Zdrive N-Zdrive Zdrive Re/Im Re/Im Re/Im ------- -------- -------- -------- -------- -------- -------- -------- 1 Bus1 220.000 1361.2 0.0 0.00035 0.00003 0.00000 Department of Electrical & Electronics Engineering Power System Simulation Laboratory 0.07347 0.01605 10099.99023 2 Bus2 11.000 1361.2 0.0 0.00035 0.00003 0.00000 0.07347 0.01605 10099.99023 3 Bus3 110.000 1107.6 1688.1 0.00130 0.00141 0.00179 0.09027 0.04597 0.04141 4 Bus4 110.000 737.5 739.9 0.00125 0.00126 0.00127 0.13558 0.09421 0.17563 5 Bus5 110.000 737.5 739.9 0.00125 0.00126 0.00127 RESULT: The contingency was detected in bus number 3 and the result summary is as shown. Department of Electrical & Electronics Engineering Power System Simulation Laboratory Exp No:5 Date: 6.1.2010 LOAD FLOW ANALYSIS USING GAUSS-SEIDEL METHOD FOR BOTH P-Q&P-V BUS USING MI-POWER AIM: To carry out load flow analysis of the given power system by Gauss-Seidel method. THEORY: Load flow analysis is the study conducted to determine the steady state operating condition of the given power system under given conditions.A large number of numerical algorithms have been developed and Gauss-Seidel method is one of such algorithms. PROBLEM FORMULATION: The performance equation of a power system may be written as [IBUS] = [YBUS] [VBUS] 1 Selecting one of the buses as the reference bus,we get (n-1) simultaneous equations .The bus loading equation can be written as 2 Ii = Pi - jQi (i=1,2,3……….n) Vi* n Where Pi = Re Σ V *Y V i ik k 3 K=1 n Qi = -Im Σ Vi*YikVk 4 K=1 The bus voltage can be written in the form of Department of Electrical & Electronics Engineering Power System Simulation Laboratory n Vi = 1.0 Ii - ΣY ij Vj 5 Yii j=1 j=i ( i=1,2,………..n ) & i≠slack bus Substituting Ii in the expression for Vi we get n Vinew = 1.0 Pi - jQi - Σ Yij Vi0 6 Yii Vi0* j=1 j≠i If the latest available voltages are used in the above expression,we get, n n Vinew = 1.0 Pi - jQi - Σ Yij Vj - n Σ Yij Vi0 7 Yii Vi0* j=1 j=i+1 The above equation is the required formula.This equation can be solved for bus voltages in an iterative manner.During each iteration, we compute all the bus voltages and check for convergence is carried out by comparison with the voltages obtained at the end of the previous iteration.After the solution is obtained , the slack bus real and reactive powers , the reactive power generation at other generator buses and line flows can be calculated This method is easier when compared with other methods , because the calculations are very simple.This is used for making a good initial start for Newton-Raphson method. Department of Electrical & Electronics Engineering Power System Simulation Laboratory ALGORITHM Step 1: Read the data such as line data,specified powers,specified voltages,Q limits at the generator buses and tolerance for convergence. Step 2: Compute Y-bus matrix. Step 3: Initialise all the bus voltages. Step 4: Iter=1. Step 5: Consider i=2,where „i‟ is the bus number. Step 6: Check whether this is P-Vbus or P-Q bus,go to step8 otherwise go to the next Step. Step 7:Compute Qi check for Q-limit violation.QGi=Qi+QLi. If QGi>Qi,max,equate to QGi to Qi max there by converting it into P-Q bus. Qi=Qimax-Qi, then go to step8. Similarly if QGi<Qi min equate QGi to Qi min there by converting it into P-Q bus. Qi=Qi min-Q Li then go to step8. If QGi is within the upper and lower limits,don‟t change QGiQi =QGi-QLi,then go to step8. Step 8: Calculate the new value of the bus voltage using Gauss-Seidel formula. i-1 n Vi = 1.0 Pi - jQi - ΣY ij Vj n - ΣY ij Vj0 Yii Vi0* j=1 j=i+1 Adjust voltage magnitude of the bus to specified magnitude if Q limits are not violated. Step 9: If all the buses are considered,go to step10.Otherwise increment the bus no i=i+1 and go to step6. Step 10: Check for convergence.If there is no convergence go to step11.Otherwise go to step12. Step 11: Update the bus voltages using the formula Vi new = Vi old + α (Vi new- Vi old ) (i=1,2,3……..n),i≠slack bus Department of Electrical & Electronics Engineering Power System Simulation Laboratory α is the acceleration factor=1.4. Increment the iteration counter Iter=Iter+1 and go to step5. Step 12: Calculate the slack bus power,Q at P-V buses,real and reactive line flows,real and reactive line losses and print all the results including all the complex bus voltages and all the bus angles. Step 13:Stop. Department of Electrical & Electronics Engineering Power System Simulation Laboratory FLOW CHART START Read line data,bus data,Vspec,Pspec,qspec,q- limits,tol,acceleration- factor Compute Y-bus matrix by inspection method Initialise all the bus voltages suitably Set Iteration count Iter=1 c Bus No i=2 b i=i+1 No Does i refer to e P-V Bus? Yes Calculate Q bus i QGi=QBus i+QLi a Department of Electrical & Electronics Engineering Power System Simulation Laboratory a e QGi>Qi max QGi<Qi min Check for Q limit violation QGi=Qimax QGi=Qimin Qi=Qi calculated Qispec=Qi max-QLi Qispec=Qi min-QLi i-1 N Vinew = 1 Pi - jQi - * Σ YijVjnew - Σ Yij Vjold Yii Vi old j=1 j=i+1 No Is this the last b bus? Yes b Department of Electrical & Electronics Engineering Power System Simulation Laboratory d Check whether Yes No |V i new - Vi |≤Tol old i=2 to nb V inew = V iold +α(V inew - V iold) i=2 to nb Calculate all the line flows,BusVoltage magnitudes,bus angles total line losses.Reactive power generated at P-V bus,slack bus power At generator buses where Q limits are not violated,adjust the magnitude of the bus voltage to the specified Print the results value. STOP V inew = (V inew *V ispec)/ |V inew | V iold = V inew i=2 to nb Iter=Iter+1 c Department of Electrical & Electronics Engineering Power System Simulation Laboratory ONE LINE DIAGRAM G 1 4 North Lake 3 Main 2 6 1 4 7 3 5 2 5 South G Elm Department of Electrical & Electronics Engineering Power System Simulation Laboratory DATA FOR SAMPLE SYSTEM IMPEDANCE AND LINE CHARGING FOR THE SAMPLE SYSTEM: Bus code Impedance Line From - (R+jX) Charging To B/2 1-2 0.02+j0.06 0.0+j 0.030 1-3 0.08+j0.24 0.0+ j0.025 2-3 0.06+j0.08 0.0+ j0.02 2-4 0.06+j0.08 0.0+ j0.02 2-5 0.04+j0.12 0.0+ j0.015 3-4 0.01+j0.03 0.0+ j0.010 4-5 0.08+j0.24 0.0+ j0.025 GENERATION ,LOADS AND BUS VOLTAGES FOR SAMPLE SYSTEM: Bus Bus Generation Generation Load Load No Voltage MW MVAR MW MVAR 1 1.06+j0.0 0 0 0 0 2 1.00+j0.0 40 30 20 10 3 1.00+j0.0 0 0 45 15 4 1.00+j0.0 0 0 40 5 5 1.00+j0.0 0 0 60 10 Department of Electrical & Electronics Engineering Power System Simulation Laboratory PROCEDURE TO ENTER THE DATA FOR PERFORMING STUDIES USING MI- POWER Mi-Power-Database-Configuration Open Power system Network Editor.Select Menu option Database Configure.Configure Database dialog is popped up.Click Browse button. Open dialog box is popped up,where you are going to browse the desired directory and specify the name of the database to be associated with the single line diagram.Click open button after entering the desired database name.Configure Database dialog will appear with path chosen. Click Ok button on the Configure Database dialog.Uncheck the Power System Libraries and Standard Relay Libraries.For this example these Standard libraries are not needed ,because all the data is given on pu for power system libraries (liketransformer ,line ,generator ),and relay libraries are required only for relay co-ordination studies.If Libraries are selected, standard libraries will be loaded along with database.Click Electrical Information tab.Since the impedances are given on 100MVA base,check the pu status.Enter the Base MVA and Base frequency.Click on Breaker Ratings button to give Breaker ratings.Click OK button to create the database to return to Network Editor. Bus Base Voltage Configuration In the network editor,Configure the base voltages for the single line diagram.Select Menu option Configure Base Voltage.If necessary change the Base Voltages ,colour,Bus width and click OK. Procedure to Draw First Element-Bus Click on Bus icon provided on power system tool bar. Draw a bus and a dialog appears prompting to give the Bus ID and Bus Name. Click OK. Database manager with corresponding Bus Data form will appear. Modify the Area number, Zone number and Contigency Weightage data if it is other than the default values. If this data is not furnished, keep the default values. Usually the minimum and maximum voltage ratings are ±5% of the rated voltage. If these ratings are other than this modify these fields. Otherwise keep the default values. Bus description field can be effectively used if the bus name gives more tha 8 characters. If Bus name is more than 8 characters ,then a short name is given in the Department of Electrical & Electronics Engineering Power System Simulation Laboratory bus name field and the Bus description field can be used to abbreviate the Bus name. For example let us say the bus name is Northeast, then bus name can be given as NE and the Bus description field can be North East. After entering data click Save which invokes Network Editor .Follow the same procedure for remaining buses. Following table gives data for other buses Bus Bus Nominal Number Name Voltage(kV) 2 South 220 3 Lake 220 4 Main 220 5 Elm 220 Procedure to Draw Transmission Line Click on Transmission Line icon provided on power system tool bar.To draw the line click in between two buses and to connect to the from bus double clicking Left Mouse Button on the From Bus and join it to another bus by double clicking the mouse button on the To Bus. Element ID dialog will appear. Enter Element ID number and click OK.Database manager with corresponding Line\Cable Data form will be open.Enter the details of that line. Enter Structure Ref No.as 1 and click on Transmission Line Library>>button. Line & cable Library form will appear.Enter Transmission line library data. After entering data Save and Close.Line\Cable Data form will appear.Click Save,which invokes Network Editor to update next element.Data for elements given in the following table. Department of Electrical & Electronics Engineering Power System Simulation Laboratory TRANSMISSION LINE ELEMENT DATA: Line From To No.Of Structure No Bus Bus circuits Ref.No 1 1 2 1 1 2 1 3 1 2 3 2 3 1 3 4 2 4 1 3 5 2 5 1 4 6 3 4 1 5 7 4 5 1 2 TRANSMISSION LINE LIBRARY DATA Structure Structure Ref Resistance Reactance Line Thermal Ref No Name charging B/2 Rating 1 Line 1-2 0.02 0.06 0.03 100 2 Line 1- 0.08 0.24 0.025 100 3&4-5 3 Line 2- 0.06 0.18 0.02 100 3&2-4 4 Line 2-5 0.04 0.12 0.015 100 5 Line 3-4 0.01 0.03 0.01 100 Department of Electrical & Electronics Engineering Power System Simulation Laboratory Procedure to Draw Generator: Click on Generator icon provided on power system tool bar.Connect it to bus 1 by clicking the Left Mouse button on Bus 1.The Element ID dialog will appear.Enter ID number and click OK.Database with corresponding Generator Data form will appear.Enter all the details Generator 1 Element Data Manufacturer Ref No 1 No.of units parallel 1 Specified Voltage 233.200 Derated MVA 100 Scheduled Power 80 Real Power Min 0 Real Power Max 80 Reactive Power Min 0 Reactive Power Max 60 Since the specified voltage is given as 1.06pu,click Compute Volt button and give 1.06 value. Voltage will be calculated and appear in the specified voltage field. Since generator at bus 1 is mentionas slack bus,only specified voltage will have importance. Enter Manufacturer Ref. No as 1 and click on Generator Library button.Generator library form will appear Generator 1 Library Data MVA Rating 100 MW Rating 80 KV Rating 220 Department of Electrical & Electronics Engineering Power System Simulation Laboratory Manufacturer Gen1 Name After entering data Save and close.In Generator Data form click Save.Network Editor screen will be invoked.Similarly connect generator 2 at bus 2.Enter its details as given in the following table. Generator 2 Element Data Manufacturer Ref No 2 No.of units parallel 1 Specified Voltage 220 Derated MVA 50 Scheduled Power 40 Real Power Min 0 Real Power Max 40 Reactive Power Min 30 Reactive Power Max 30 Generator 1 Library Data MVA Rating 100 MW Rating 80 KV Rating 220 Manufacturer Gen1 Name Procedure to Enter Load Data Click on Load icon provided on power system tool bar.Connect Load 1 at BUS 2 by clicking the Left Mouse button on Bus 2.Element ID dialog will appear.Give Id number as 1 and say OK.Load Data Department of Electrical & Electronics Engineering Power System Simulation Laboratory form will appear .Enter load details.Then click Save button,which invokes Network Editor.Enter the Load detail as given in the following table. Load Bus MW MVAR No No 1 2 20 10 2 5 60 10 3 3 45 15 4 4 40 5 Solve Load Flow Analysis Select Menu option Solve Load Flow Analysis.When Study Info button is clicked, following dialog will open.Select Gauss-Seidel method and enter acceleration factor as1.4&P-Tolerance and Q-Tolerance as 0.0001. Click OK. Execute Load Flow Analysis and click on Report in Load Flow Analysis dialog to view Report. OUTPUT: Department of Electrical & Electronics Engineering Power System Simulation Laboratory LOAD FLOW BY GAUSS-SIEDEL METHOD CASE NO : 1 CONTINGENCY : 0 SCHEDULE NO : 0 CONTINGENCY NAME : Base Case RATING CONSIDERED : NOMINAL ------------------------------------------------------------------------------- VERSION NUMBER : 6.1 LARGEST BUS NUMBER USED : 5 ACTUAL NUMBER OF BUSES : 5 NUMBER OF 2 WIND. TRANSFORMERS : 0 NUMBER OF 3 WIND. TRANSFORMERS : 0 NUMBER OF TRANSMISSION LINES : 7 NUMBER OF SERIES REACTORS : 0 NUMBER OF SERIES CAPACITORS : 0 NUMBER OF CIRCUIT BREAKERS : 0 NUMBER OF SHUNT REACTORS : 0 NUMBER OF SHUNT CAPACITORS : 0 NUMBER OF SHUNT IMPEDANCES : 0 NUMBER OF GENERATORS : 2 NUMBER OF LOADS : 4 NUMBER OF LOAD CHARACTERISTICS : 0 NUMBER OF UNDER FREQUENCY RELAY: 0 NUMBER OF GEN CAPABILITY CURVES: 0 NUMBER OF FILTERS : 0 NUMBER OF TIE LINE SCHEDULES : 0 NUMBER OF CONVERTORS : 0 NUMBER OF DC LINKS : 0 ------------------------------------------------------------------------------- LOAD FLOW WITH GAUSS-SEIDEL METHOD : 5 NUMBER OF ZONES : 1 PRINT OPTION : 3 - BOTH DATA AND RESULTS PRINT PLOT OPTION : 1 - PLOTTING WITH PU VOLTAGE NO FREQUENCY DEPENDENT LOAD FLOW, CONTROL OPTION: 0 BASE MVA : 100.000000 NOMINAL SYSTEM FREQUENCY (Hzs) : 60.000000 Department of Electrical & Electronics Engineering Power System Simulation Laboratory FREQUENCY DEVIATION (Hzs) : 0.000000 FLOWS IN MW AND MVAR, OPTION :0 SLACK BUS : 1 TRANSFORMER TAP CONTROL OPTION : 0 Q CHECKING LIMIT (ENABLED) : 0 REAL POWER TOLERANCE (PU) : 0.00010 REACTIVE POWER TOLERANCE (PU) : 0.00010 MAXIMUM NUMBER OF ITERATIONS : 15 BUS VOLTAGE BELOW WHICH LOAD MODEL IS CHANGED : 0.75000 CIRCUIT BREAKER RESISTANCE (PU) : 0.00000 CIRCUIT BREAKER REACTANCE (PU) : 0.00010 TRANSFORMER R/X RATIO : 0.05000 ------------------------------------------------------------------------------ ANNUAL PERCENTAGE INTEREST CHARGES : 15.000 .ANNUAL PERCENT OPERATION & MAINTENANCE CHARGES : 4.000 LIFE OF EQUIPMENT IN YEARS : 20.000 ENERGY UNIT CHARGE (KWHOUR) : 2.500 Rs LOSS LOAD FACTOR : 0.300 COST PER MVAR IN LAKHS : 5.000 Rs ------------------------------------------------------------------------------- ZONE WISE MULTIPLICATION FACTORS ZONE P LOAD Q LOAD P GEN Q GEN SH REACT SH CAP C LOAD ---- -------- -------- -------- -------- -------- -------- -------- 0 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1 1.000 1.000 1.000 1.000 1.000 1.000 1.000 ------------------------------------------------------------------------------- Department of Electrical & Electronics Engineering Power System Simulation Laboratory BUS DATA BUS NO. AREA ZONE BUS KV VMIN-PU VMAX-PU NAME ------- ---- ---- -------- -------- -------- -------- 1 1 1 220.000 0.950 1.050 Bus1 2 1 1 220.000 0.950 1.050 Bus2 3 1 1 220.000 0.950 1.050 Bus3 4 1 1 220.000 0.950 1.050 Bus4 5 1 1 220.000 0.950 1.050 Bus5 ------------------------------------------------------------------------------- TRANSMISSION LINE DATA STA CKT FROM FROM TO TO LINE PARAMETER RATING KMS NODE NAME* NODE NAME* R(P.U) X(P.U.) B/2(P.U.) MVA --- --- ---- -------- ---- -------- --------- --------- --------- ------ ----- 3 1 1 Bus1 2 Bus2 0.02000 0.06000 0.03000 100 1.0 3 1 1 Bus1 3 Bus3 0.08000 0.24000 0.02500 100 1.0 3 1 4 Bus4 5 Bus5 0.08000 0.24000 0.02500 100 1.0 3 1 2 Bus2 3 Bus3 0.06000 0.18000 0.02000 100 1.0 3 1 2 Bus2 4 Bus4 0.06000 0.18000 0.02000 100 1.0 3 1 2 Bus2 5 Bus5 0.04000 0.12000 0.01500 100 1.0 3 1 3 Bus3 4 Bus4 0.01000 0.03000 0.01000 100 1.0 ------------------------------------------------------------------------------- ------------------------------------------------------------------------------- TOTAL LINE CHARGING SUSCEPTANCE : 0.29000 TOTAL LINE CHARGING MVAR AT 1 PU VOLTAGE : 29.000 ------------------------------------------------------------------------------- Department of Electrical & Electronics Engineering Power System Simulation Laboratory TOTAL CAPACITIVE SUSCEPTANCE : 0.00000 pu - 0.000 MVAR TOTAL INDUCTIVE SUSCEPTANCE : 0.00000 pu - 0.000 MVAR ------------------------------------------------------------------------------- GENERATOR DATA SL.NO* FROM FROM REAL Q-MIN Q-MAX V-SPEC CAP. MVA STAT NODE NAME* POWER(MW) MVAR MVAR P.U. CURV RATING ------ ---- -------- --------- --------- --------- --------- ---- ------- ---- 1 1 Bus1 80.0000 0.0000 60.0000 1.0600 0 100.00 3 2 2 Bus2 40.0000 30.0000 30.0000 1.0000 0 50.00 3 ------------------------------------------------------------------------------- LOAD DATA SLNO FROM FROM REAL REACTIVE COMP COMPENSATING MVAR VALUE CHAR F/V * NODE NAME* MW MVAR MVAR MIN MAX STEP NO NO STAT ---- ---- -------- -------- -------- -------- ------- ------- ------- ---- ---- 1 2 Bus2 20.000 10.000 0.000 0.000 0.000 0.000 0 0 3 0 2 5 Bus5 60.000 10.000 0.000 0.000 0.000 0.000 0 0 3 0 3 3 Bus3 45.000 15.000 0.000 0.000 0.000 0.000 0 0 3 0 4 4 Bus4 40.000 5.000 0.000 0.000 0.000 0.000 0 0 3 0 ------------------------------------------------------------------------------- INCLUDING OUT OF SERVICE VALUES Department of Electrical & Electronics Engineering Power System Simulation Laboratory TOTAL SPECIFIED MW GENERATION : 120.00000 TOTAL MIN MVAR LIMIT OF GENERATOR : 30.00000 TOTAL MAX MVAR LIMIT OF GENERATOR : 90.00000 TOTAL SPECIFIED MW LOAD : 165.00000 reduced 165.00000 TOTAL SPECIFIED MVAR LOAD : 40.00000 reduced 40.00000 TOTAL SPECIFIED MVAR COMPENSATION : 0.00000 reduced 0.00000 ------------------------------------------------------------------------------- IN SERVICE VALUES TOTAL SPECIFIED MW GENERATION : 120.00000 TOTAL MIN MVAR LIMIT OF GENERATOR : 30.00000 TOTAL MAX MVAR LIMIT OF GENERATOR : 90.00000 TOTAL SPECIFIED MW LOAD : 165.00000 reduced 165.00000 TOTAL SPECIFIED MVAR LOAD : 40.00000 reduced 40.00000 TOTAL SPECIFIED MVAR COMPENSATION : 0.00000 reduced 0.00000 ------------------------------------------------------------------------------- GENERATOR DATA FOR FREQUENCY DEPENDENT LOAD FLOW SLNO* FROM FROM P-RATE P-MIN P-MAX %DROOP PARTICI BIAS NODE NAME* MW MW MW FACTOR SETTING C0 C1 C2 ------ ---- -------- -------- --------- --------- --------- --------- --------- 1 1 Bus1 80.000 0.0000 80.0000 4.0000 0.0000 0.0000 0.0000 0.0000 0.0000 2 2 Bus2 40.000 0.0000 40.0000 4.0000 0.0000 0.0000 0.0000 0.0000 0.0000 Department of Electrical & Electronics Engineering Power System Simulation Laboratory ------------------------------------------------------------------------------- Acceleration factor : 1.40 ------------------------------------------------------------------------------- Slack bus angle (degrees) : 0.00 ------------------------------------------------------------------------------- ------------------------------------------------------------------------------- Iteration count = 1 Error = 0.052537 Bus = 2 Iteration count = 2 Error = 0.015724 Bus = 5 Iteration count = 3 Error = 0.007669 Bus = 5 Iteration count = 4 Error = 0.002768 Bus = 2 Iteration count = 5 Error = 0.002594 Bus = 5 Iteration count = 6 Error = 0.001050 Bus = 4 Iteration count = 7 Error = 0.000867 Bus = 3 Iteration count = 8 Error = 0.000394 Bus = 2 Iteration count = 9 Error = 0.000217 Bus = 3 Iteration count = 10 Error = 0.000117 Bus = 3 Iteration count = 11 Error = 0.000044 Bus = 2 ------------------------------------------------------------------------------- BUS VOLTAGES AND POWERS NODE FROM V-MAG ANGLE MW MVAR MW MVAR MVAR NO. NAME P.U. DEGREE GEN GEN LOAD LOAD COMP ---- -------- ------ ------ -------- -------- -------- -------- -------- 1 Bus1 1.0600 0.00 129.534 -7.469 0.000 0.000 0.000 #< 2 Bus2 1.0475 -2.81 40.000 30.000 20.000 10.000 0.000 3 Bus3 1.0242 -5.00 0.000 0.000 45.000 15.000 0.000 4 Bus4 1.0236 -5.33 0.000 0.000 40.000 5.000 0.000 Department of Electrical & Electronics Engineering Power System Simulation Laboratory 5 Bus5 1.0180 -6.15 0.000 0.000 60.000 10.000 0.000 ------------------------------------------------------------------------------- NUMBER OF BUSES EXCEEDING MINIMUM VOLTAGE LIMIT (@ mark) : 0 NUMBER OF BUSES EXCEEDING MAXIMUM VOLTAGE LIMIT (# mark) : 1 NUMBER OF GENERATORS EXCEEDING MINIMUM Q LIMIT (< mark) : 1 NUMBER OF GENERATORS EXCEEDING MAXIMUM Q LIMIT (> mark) : 0 ------------------------------------------------------------------------------- LINE FLOWS AND LINE LOSSES SLNO CS FROM FROM TO TO FORWARD LOSS % NODE NAME NODE NAME MW MVAR MW MVAR LOADING ---- -- ---- -------- ---- -------- -------- -------- -------- -------- ------- 1 1 1 Bus1 2 Bus2 88.825 -8.610 1.4093 -2.4345 84.2# 2 1 1 Bus1 3 Bus3 40.710 1.141 1.1911 -1.8583 38.4^ 3 1 4 Bus4 5 Bus5 6.334 -2.280 0.0307 -5.1178 6.8& 4 1 2 Bus2 3 Bus3 24.690 3.535 0.3513 -3.2385 24.7& 5 1 2 Bus2 4 Bus4 27.936 2.957 0.4413 -2.9660 27.5^ 6 1 2 Bus2 5 Bus5 54.824 7.346 1.1253 0.1756 52.8$ 7 1 3 Bus3 4 Bus4 18.901 -5.166 0.0357 -1.9898 19.1& ------------------------------------------------------------------------------- ! NUMBER OF LINES LOADED BEYOND 125% : 0 @ NUMBER OF LINES LOADED BETWEEN 100% AND 125% : 0 # NUMBER OF LINES LOADED BETWEEN 75% AND 100% : 1 $ NUMBER OF LINES LOADED BETWEEN 50% AND 75% : 1 ^ NUMBER OF LINES LOADED BETWEEN 25% AND 50% : 2 & NUMBER OF LINES LOADED BETWEEN 1% AND 25% : 3 * NUMBER OF LINES LOADED BETWEEN 0% AND 1% : 0 Department of Electrical & Electronics Engineering Power System Simulation Laboratory ------------------------------------------------------------------------------- NEW SYSTEM FREQUENCY FOR ISLAND 1 : 60.000000 Hzs ------------------------------------------------------------------------------- Summary of results TOTAL REAL POWER GENERATION : 169.534 MW TOTAL REAL POWER DRAWAL -ve g : 0.000 MW TOTAL REACT. POWER GENERATION : 22.531 MVAR GENERATION pf : 0.991 TOTAL SHUNT REACTOR INJECTION : 0.000 MW TOTAL SHUNT REACTOR INJECTION : 0.000 MVAR TOTAL SHUNT CAPACIT.INJECTION : 0.000 MW TOTAL SHUNT CAPACIT.INJECTION : 0.000 MVAR TOTAL REAL POWER LOAD : 165.000 MW TOTAL REAL POWER INJECT,-ve L : 0.000 MW TOTAL REACTIVE POWER LOAD : 40.000 MVAR LOAD pf : 0.972 TOTAL COMPENSATION AT LOADS : 0.000 MVAR TOTAL HVDC REACTIVE POWER : 0.000 MVAR TOTAL REAL POWER LOSS (AC+DC) : 4.584590 MW ( 4.584590+ 0.000000) PERCENTAGE REAL LOSS (AC+DC) : 2.704 TOTAL REACTIVE POWER LOSS : -17.429301 MVAR ------------------------------------------------------------------------------- Zone wise distribution Description Zone # 1 ---------------- ---------- MW generation 169.5343 Department of Electrical & Electronics Engineering Power System Simulation Laboratory MVAR generation 22.5314 MW load 165.0000 MVAR load 40.0000 MVAR compensation 0.0000 MW loss 4.5846 MVAR loss -17.4293 MVAR - inductive 0.0000 MVAR - capacitive 0.0000 ------------------------------------------------------------------------------- Zone wise export(+ve)/import(-ve) Zone # 1 MW & MVAR ------ -------- -------- 1 ----- Area wise distribution Description Area # 1 ---------------- ---------- MW generation 169.5343 MVAR generation 22.5314 MW load 165.0000 MVAR load 40.0000 MVAR compensation 0.0000 MW loss 4.5846 MVAR loss -17.4293 MVAR - inductive 0.0000 MVAR - capacitive 0.0000 Exp No:5 b) Department of Electrical & Electronics Engineering Power System Simulation Laboratory Date: 6.1.2010 LOAD FLOW ANALYSIS USING NEWTON-RAPHSON METHOD FOR BOTH P-Q&P-V BUS USING MI-POWER AIM: To carry out load flow analysis of the given power system by Newton-Raphson method THEORY: Load flow analysis is the study conducted to determine the steady state operating condition of the given system under given conditions. A large number of numerical algorithms have been developed and Newton-Raphson method is one of such algorithms. The Newton-Raphson method of load flow analysis is an iterative method which approximate the set of non-linear simultaneous equations to a set of linear simultaneous equations. The load flow equations for Newton-Raphson method are non-linear equations in terms of real and imaginary part of bus voltages. n Pp = Σ {ep (eq Gpq +fq Bpq)+fp(fqGpq-eq Bpq) q=1 n Qp= Σ{fp (eqGpq+fqBpq)-ep(fqGpq-eqBpq)} q=1 |Vp|2=ep2 +fp2 Where ep = Real part of Vp(Voltage of bus –p) fq = Imaginary part of Vp(Voltage of bus –p) Gpq, Bpq = Conductance and Susceptance of admittance Ypq ALGORITHM: Department of Electrical & Electronics Engineering Power System Simulation Laboratory Step1 : Read the data such as line data ,bus data,specified powers at the generator buses and tolerance for convergence. Step2: Compute Y-bus matrix. Step3: Initialise all the bus voltages. Step4: Iter=1, Bus count=1. Step5: Check for slack bus. if it is a slack bus then go to step 13 otherwise go to next step. Step 6: Calculate the real and reactive power of bus P using the following equation. n Ppk = Σ {epk (eqk Gpq +fqk Bpq)+fpk(fq kGpq-eqk Bpq) q=1 n Qpk= Σ{fpk (eq kGpq+fq kBpq)-ep k(fq kGpq-eq kBpq)} q=1 Step 7: Calculate the change in real power. Change in real power Δkp = Pp spec - Pkp Step 8 : Check for Generator bus. if it is a Generator bus go to step9 otherwise go to step 12. Step 9: Check for reactive power limit violation of Generator bus. Step 10: if the calculated reative power is within specified limit then consider this bus as a generator Bus. Step 11: if the reactive power limit is violated, then treat this bus as a load bus. if Qkp< Qp min then Qp spec = Qp min if Qkp> Qp max then Qp spec = Qp max Step 12: Calculate the change in reactive power for load bus. Step 13: Repeat step 5 to 12 until all residues (change in P , Q and V ) are calculated. For this the bus Department of Electrical & Electronics Engineering Power System Simulation Laboratory Increment the bus count by 1 and go to Step 5 until the bus count is n. Step 14: Determine the largest of the absolute value of the residue (ΔE ) Step 15: Compare ΔE and convergence. If ΔE < convergence then go to step 20 else go to next step. Step 16: Determine the values of Jacobian matrix Step 17: Calculate the increment in real and reactive part of voltages. Step 18: Calculate the new bus voltages. epk+1=epk +Δepk ; p=1,2,3,……… n except slack bus fpk+1=fpk +Δfpk ; p=1,2,3,……… n except slack bus |Vpk+1|=√(epk+1)2 +(fpk+1)2 and δpk+1 =tan-1 (epk+1∕ fpk+1) Therefore , Vpk+1=| Vpk+1| < δpk+1 Step 19: Advance the iteration count, i.e., k=k+1 and go to step 4. Step 20: Calculate the line flows. Department of Electrical & Electronics Engineering Power System Simulation Laboratory FLOWCHART: START Read line data, bus data, Pp, Qp, Vp spec, Qp max, Qp min Compute Y-bus matrix by inspection method Initialize all the bus voltages suitably Calculate ep°&fp° for p=1 z Set Iter count, Iter=1,Set bus count p=1 No Check for Set bus count,p=1 slack bus d Calculate Ppk Qpk & ΔPk f No Check for b generator bus Yes c Department of Electrical & Electronics Engineering Power System Simulation Laboratory c Yes Check if Set Qpspec=Qpmin Qpk < Qpmin No Yes Set Qp,spec=Qp,max Check if Qpk > Qpmax No b Calculate ΔPk Calculate|ΔVp|2 f Advance bus count p=p+1 No Check if d p>n Yes 1 Department of Electrical & Electronics Engineering Power System Simulation Laboratory 1 Determine the largest absolute value of residue, ΔE Yes Check if ΔE<є No Determine the elements of Calculate the line flows and slack bus power the jacobian matrix(J) and slack bus power STOP Calculate the voltage increment Δepk and Δfpk Calculate |Vpk+1| and δpk+1 No Check for generator bus Yes epk+1=|Vp spec|cosδpk+1 fpk+1=|Vp spec|sinδpk+1 Advance iteration count,k=k+1 z Department of Electrical & Electronics Engineering Power System Simulation Laboratory ONE LINE DIAGRAM: G 1 4 North Lake 3 Main 2 6 1 4 7 3 5 South 2 Elm 5 G DATA FOR SAMPLE SYSTEM IMPEDANCE AND LINE CHARGING FOR THE SAMPLE SYSTEM: Bus code Impedance Line From - To (R+jX) Charging B/2 Department of Electrical & Electronics Engineering Power System Simulation Laboratory 1-2 0.02+j0.06 0.0+j 0.030 1-3 0.08+j0.24 0.0+ j0.025 2-3 0.06+j0.08 0.0+ j0.02 2-4 0.06+j0.08 0.0+ j0.02 2-5 0.04+j0.12 0.0+ j0.015 3-4 0.01+j0.03 0.0+ j0.010 4-5 0.08+j0.24 0.0+ j0.025 GENERATION ,LOADS AND BUS VOLTAGES FOR SAMPLE SYSTEM: Bus Bus Generation Generation Load Load No Voltage MW MVAR MW MVAR 1 1.06+j0.0 0 0 0 0 2 1.00+j0.0 40 30 20 10 3 1.00+j0.0 0 0 45 15 4 1.00+j0.0 0 0 40 5 5 1.00+j0.0 0 0 60 10 PROCEDURE TO ENTER THE DATA FOR PERFORMING STUDIES USING MI- POWER Mi-Power-Database-Configuration Open Power system Network Editor.Select Menu option Database Configure.Configure Database dialog is popped up.Click Browse button. Open dialog box is popped up,where you are going to browse the desired directory and specify the name of the database to be associated with the single line Department of Electrical & Electronics Engineering Power System Simulation Laboratory diagram.Click open button after entering the desired database name.Configure Database dialog will appear with path chosen.Click Ok button on the Configure Database dialog. Uncheck the Power System Libraries and Standard Relay Libraries.For this example these Standard libraries are not needed ,because all the data is given on pu for power system libraries (liketransformer ,line ,generator ),and relay libraries are required only for relay co-ordination studies.If Libraries are selected, standard libraries will be loaded along with database.Click Electrical Information tab.Since the impedances are given on 100MVA base, check the pu status.Enter the Base MVA and Base frequency.Click on Breaker Ratings button to give Breaker ratings.Click OK button to create the database to return to Network Editor. Bus Base Voltage Configuration In the network editor,Configure the base voltages for the single line diagram.Select Menu option Configure Base Voltage.If necessary change the Base Voltages ,colour,Bus width and click OK. Procedure to Draw First Element-Bus Click on Bus icon provided on power system tool bar.Draw a bus and a dialog appears prompting to give the Bus ID and Bus Name.Click OK. Database manager with corresponding Bus Data form will appear. Modify the Area number,Zone number and Contigency Weightage data if it is other than the default values.If this data is not furnished,keep the default values.Usually the minimum and maximum voltage ratings are ±5% of the rated voltage.If these ratings are other than this modify these fields.Otherwise keep the default values. Bus description field can be effectively used if the bus name gives more tha 8 characters.If Bus name is more than 8 characters ,then a short name is given in the bus name field and the Bus description field can be used to abbreviate the Bus name.For example let us say the bus name is Northeast, then bus name can be given as NE and the Bus description field can be North East. After entering data click Save which invokes Network Editor .Follow the same procedure for remaining buses. Following table gives data for other buses Bus Bus Nominal Number Name Voltage(kV) Department of Electrical & Electronics Engineering Power System Simulation Laboratory 2 South 220 3 Lake 220 4 Main 220 5 Elm 220 Procedure to Draw Transmission Line Click on Transmission Line icon provided on power system tool bar. To draw the line click in between two buses and to connect to the from bus double clicking Left Mouse Button on the From Bus and join it to another bus by double clicking the mouse button on the To Bus. Element ID dialog will appear. Enter Element ID number and click OK. Database manager with corresponding Line\Cable Data form will be open. Enter the details of that line. Enter Structure Ref No. as 1 and click on Transmission Line Library>>button. Line & cable Library form will appear.Enter Transmission line library data . After entering data Save and Close. Line\Cable Data form will appear.Click Save, which invokes Network Editor to update next element.Data for elements given in the following table. TRANSMISSION LINE ELEMENT DATA: Department of Electrical & Electronics Engineering Power System Simulation Laboratory Line From To No.Of Structure No Bus Bus circuits Ref.No 1 1 2 1 1 2 1 3 1 2 3 2 3 1 3 4 2 4 1 3 5 2 5 1 4 6 3 4 1 5 7 4 5 1 2 TRANSMISSION LINE LIBRARY DATA: Structure Structure Ref Resistance Reactance Line Thermal Rating Ref No Name charging B/2 1 Line 1-2 0.02 0.06 0.03 100 2 Line 1- 0.08 0.24 0.025 100 3&4-5 3 Line 2- 0.06 0.18 0.02 100 3&2-4 4 Line 2-5 0.04 0.12 0.015 100 5 Line 3-4 0.01 0.03 0.01 100 Department of Electrical & Electronics Engineering Power System Simulation Laboratory Generator 1 Element Data Manufacturer Ref No 1 No.of units parallel 1 Specified Voltage 233.200 Derated MVA 100 Scheduled Power 80 Real Power Min 0 Real Power Max 80 Reactive Power Min 0 Reactive Power Max 60 Since the specified voltage is given as 1.06pu,click Compute Volt button and give 1.06 value. Voltage will be calculated and appear in the specified voltage field. Since generator at bus 1 is mentionas slack bus,only specified voltage will have importance. Enter Manufacturer Ref. No as 1 and click on Generator Library button.Generator library form will appear Generator 1 Library Data MVA Rating 100 MW Rating 80 KV Rating 220 Manufacturer Gen1 Name After entering data Save and close.In Generator Data form click Save.Network Editor screen will be Department of Electrical & Electronics Engineering Power System Simulation Laboratory invoked.Similarly connect generator 2 at bus 2.Enter its details as given in the following table. Procedure to Draw Generator Click on Generator icon provided on power system tool bar.Connect it to bus 1 by clicking the Left Mouse button on Bus 1.The Element ID dialog will appear.Enter ID number and click OK.Database with corresponding Generator Data form will appear.Enter all the details. Generator 2 Element Data Manufacturer Ref No 2 No.of units parallel 1 Specified Voltage 220 Derated MVA 50 Scheduled Power 40 Real Power Min 0 Real Power Max 40 Reactive Power Min 30 Reactive Power Max 30 Generator 1 Library Data MVA Rating 100 MW Rating 80 KV Rating 220 Manufacturer Gen1 Name Procedure to Enter Load Data: Department of Electrical & Electronics Engineering Power System Simulation Laboratory Click on Load icon provided on power system tool bar.Connect Load 1 at BUS 2 by clicking the Left Mouse button on Bus 2.Element ID dialog will appear.Give Id number as 1 and say OK.Load Data form will appear .Enter load details.Then click Save button,which invokes Network Editor.Enter the Load detail as given in the following table. Load Bus MW MVAR No No 1 2 20 10 2 5 60 10 3 3 45 15 4 4 40 5 Solve Load Flow Analysis Select Menu option Solve Load Flow Analysis.When Study Info button is clicked,following dialog will open.Select Newton-Raphson method and enter acceleration factor as1.4&P-Tolerance and Q-Tolerance as 0.01.Click OK. Execute Load Flow Analysis and click on Report in Load Flow Analysis dialog to view Report. OUTPUT: Department of Electrical & Electronics Engineering Power System Simulation Laboratory ------------------------------------------------------------------------------- Date and Time : Mon Mar 15 10:25:39 2010 ------------------------------------------------------------------------------- LOAD FLOW BY NEWTON RAPHSON METHOD CASE NO : 1 CONTINGENCY : 0 SCHEDULE NO : 0 CONTINGENCY NAME : Basecase RATING CONSIDERED : NOMINAL ------------------------------------------------------------------------------- VERSION NUMBER : 6.1 LARGEST BUS NUMBER USED : 5 ACTUAL NUMBER OF BUSES : 5 NUMBER OF 2 WIND. TRANSFORMERS : 0 NUMBER OF 3 WIND. TRANSFORMERS : 0 NUMBER OF TRANSMISSION LINES : 7 NUMBER OF SERIES REACTORS : 0 NUMBER OF SERIES CAPACITORS : 0 NUMBER OF CIRCUIT BREAKERS : 0 NUMBER OF SHUNT REACTORS : 0 NUMBER OF SHUNT CAPACITORS : 0 NUMBER OF SHUNT IMPEDANCES : 0 NUMBER OF GENERATORS : 2 NUMBER OF LOADS : 4 NUMBER OF LOAD CHARACTERISTICS : 0 NUMBER OF UNDER FREQUENCY RELAY: 0 NUMBER OF GEN CAPABILITY CURVES: 0 NUMBER OF FILTERS : 0 NUMBER OF TIE LINE SCHEDULES : 0 NUMBER OF CONVERTORS : 0 NUMBER OF DC LINKS : 0 ------------------------------------------------------------------------------- LOAD FLOW WITH NEWTON RAPHSON METHOD : 6 NUMBER OF ZONES : 1 PRINT OPTION : 3 - BOTH DATA AND RESULTS PRINT PLOT OPTION : 1 - PLOTTING WITH PU VOLTAGE NO FREQUENCY DEPENDENT LOAD FLOW, CONTROL OPTION: 0 BASE MVA : 100.000000 NOMINAL SYSTEM FREQUENCY (Hzs) : 60.000000 FREQUENCY DEVIATION (Hzs) : 0.000000 FLOWS IN MW AND MVAR, OPTION : 0 SLACK BUS : 1 TRANSFORMER TAP CONTROL OPTION : 0 Q CHECKING LIMIT (ENABLED) : 0 REAL POWER TOLERANCE (PU) : 0.00010 REACTIVE POWER TOLERANCE (PU) : 0.00010 MAXIMUM NUMBER OF ITERATIONS : 15 BUS VOLTAGE BELOW WHICH LOAD MODEL IS CHANGED : 0.75000 CIRCUIT BREAKER RESISTANCE (PU) : 0.00000 CIRCUIT BREAKER REACTANCE (PU) : 0.00010 Department of Electrical & Electronics Engineering Power System Simulation Laboratory TRANSFORMER R/X RATIO : 0.05000 ------------------------------------------------------------------------------ ANNUAL PERCENTAGE INTEREST CHARGES : 15.000 ANNUAL PERCENT OPERATION & MAINTENANCE CHARGES : 4.000 LIFE OF EQUIPMENT IN YEARS : 20.000 ENERGY UNIT CHARGE (KWHOUR) : 2.500 Rs LOSS LOAD FACTOR : 0.300 COST PER MVAR IN LAKHS : 5.000 Rs ------------------------------------------------------------------------------- ZONE WISE MULTIPLICATION FACTORS ZONE P LOAD Q LOAD P GEN Q GEN SH REACT SH CAP C LOAD ---- -------- -------- -------- -------- -------- -------- -------- 0 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1 1.000 1.000 1.000 1.000 1.000 1.000 1.000 BUS DATA BUS NO. AREA ZONE BUS KV VMIN-PU VMAX-PU NAME ------- ---- ---- -------- -------- -------- -------- 1 1 1 220.000 0.950 1.050 Bus1 2 1 1 220.000 0.950 1.050 Bus2 3 1 1 220.000 0.950 1.050 Bus3 4 1 1 220.000 0.950 1.050 Bus4 5 1 1 220.000 0.950 1.050 Bus5 ------------------------------------------------------------------------------- TRANSMISSION LINE DATA STA CKT FROM FROM TO TO LINE PARAMETER RATING KMS NODE NAME* NODE NAME* R(P.U) X(P.U.) B/2(P.U.) MVA --- --- ---- -------- ---- -------- --------- --------- --------- ------ ----- 3 1 1 Bus1 2 Bus2 0.02000 0.06000 0.03000 100 1.0 0 1 1 Bus1 3 Bus3 0.08000 0.24000 0.02500 100 1.0 3 1 4 Bus4 5 Bus5 0.08000 0.24000 0.02500 100 1.0 3 1 2 Bus2 3 Bus3 0.06000 0.18000 0.02000 100 1.0 3 1 2 Bus2 4 Bus4 0.06000 0.18000 0.02000 100 1.0 3 1 2 Bus2 5 Bus5 0.04000 0.12000 0.01500 100 1.0 Department of Electrical & Electronics Engineering Power System Simulation Laboratory 3 1 3 Bus3 4 Bus4 0.01000 0.03000 0.01000 100 1.0 ------------------------------------------------------------------------------- ------------------------------------------------------------------------------- TOTAL LINE CHARGING SUSCEPTANCE : 0.29000 TOTAL LINE CHARGING MVAR AT 1 PU VOLTAGE : 29.000 NUMBER OF LINES OPENED ON BOTH THE ENDS : 1 TOTAL LINE CHARGING SUSCEPTANCE OF EXISTING LINES : 0.24000 TOTAL LINE CHARGING MVAR AT 1 PU VOLTAGE OF EXISTING LINES : 24.000 ------------------------------------------------------------------------------- TOTAL CAPACITIVE SUSCEPTANCE : 0.00000 pu - 0.000 MVAR TOTAL INDUCTIVE SUSCEPTANCE : 0.00000 pu - 0.000 MVAR ------------------------------------------------------------------------------- GENERATOR DATA SL.NO* FROM FROM REAL Q-MIN Q-MAX V-SPEC CAP. MVA STAT NODE NAME* POWER(MW) MVAR MVAR P.U. CURV RATING ------ ---- -------- --------- --------- --------- --------- ---- ------- ---- 1 1 Bus1 40.0000 30.0000 30.0000 1.0000 0 50.00 3 2 2 Bus2 40.0000 30.0000 30.0000 1.0000 0 50.00 3 ------------------------------------------------------------------------------- LOAD DATA SLNO FROM FROM REAL REACTIVE COMP COMPENSATING MVAR VALUE CHAR F/V * NODE NAME* MW MVAR MVAR MIN MAX STEP NO NO STAT ---- ---- -------- -------- -------- -------- ------- ------- ------- ---- ---- 1 2 Bus2 60.000 10.000 0.000 0.000 0.000 0.000 0 0 3 0 2 5 Bus5 45.000 15.000 0.000 0.000 0.000 0.000 0 0 3 0 3 3 Bus3 45.000 15.000 0.000 0.000 0.000 0.000 0 0 3 0 4 4 Bus4 40.000 5.000 0.000 0.000 0.000 0.000 0 0 3 0 ------------------------------------------------------------------------------- INCLUDING OUT OF SERVICE VALUES TOTAL SPECIFIED MW GENERATION : 80.00000 TOTAL MIN MVAR LIMIT OF GENERATOR : 60.00000 Department of Electrical & Electronics Engineering Power System Simulation Laboratory TOTAL MAX MVAR LIMIT OF GENERATOR : 60.00000 TOTAL SPECIFIED MW LOAD : 190.00000 reduced 190.00000 TOTAL SPECIFIED MVAR LOAD : 45.00000 reduced 45.00000 TOTAL SPECIFIED MVAR COMPENSATION : 0.00000 reduced 0.00000 ------------------------------------------------------------------------------- IN SERVICE VALUES TOTAL SPECIFIED MW GENERATION : 80.00000 TOTAL MIN MVAR LIMIT OF GENERATOR : 60.00000 TOTAL MAX MVAR LIMIT OF GENERATOR : 60.00000 TOTAL SPECIFIED MW LOAD : 190.00000 reduced 190.00000 TOTAL SPECIFIED MVAR LOAD : 45.00000 reduced 45.00000 TOTAL SPECIFIED MVAR COMPENSATION : 0.00000 reduced 0.00000 ------------------------------------------------------------------------------- GENERATOR DATA FOR FREQUENCY DEPENDENT LOAD FLOW SLNO* FROM FROM P-RATE P-MIN P-MAX %DROOP PARTICI BIAS NODE NAME* MW MW MW FACTOR SETTING C0 C1 C2 ------ ---- -------- -------- --------- --------- --------- --------- --------- 1 1 Bus1 40.000 0.0000 40.0000 4.0000 0.0000 0.0000 0.0000 0.0000 0.0000 2 2 Bus2 40.000 0.0000 40.0000 4.0000 0.0000 0.0000 0.0000 0.0000 0.0000 ------------------------------------------------------------------------------- Slack bus angle (degrees) : 0.00 ------------------------------------------------------------------------------- ------------------------------------------------------------------------------- Iteration count 0 maxp 0.450000 maxq 0.285001 Iteration count 1 maxp 0.043225 maxq 0.015055 Iteration count 2 maxp 0.298492 maxq 1.111672 Iteration count 3 maxp 0.011113 maxq 0.041154 Iteration count 4 maxp 0.000015 maxq 0.000114 Iteration count 5 maxp 0.000000 maxq 0.000001 ------------------------------------------------------------------------------- BUS VOLTAGES AND POWERS NODE FROM V-MAG ANGLE MW MVAR MW MVAR MVAR Department of Electrical & Electronics Engineering Power System Simulation Laboratory NO. NAME P.U. DEGREE GEN GEN LOAD LOAD COMP ---- -------- ------ ------ -------- -------- -------- -------- -------- 1 Bus1 1.0000 0.00 158.767 19.952 0.000 0.000 0.000 < 2 Bus2 0.9588 -5.43 40.000 30.000 60.000 10.000 0.000 > 3 Bus3 0.9154 -9.90 0.000 0.000 45.000 15.000 0.000 @ 4 Bus4 0.9176 -9.84 0.000 0.000 40.000 5.000 0.000 @ 5 Bus5 0.9206 -9.02 0.000 0.000 45.000 15.000 0.000 @ ------------------------------------------------------------------------------- NUMBER OF BUSES EXCEEDING MINIMUM VOLTAGE LIMIT (@ mark) : 3 NUMBER OF BUSES EXCEEDING MAXIMUM VOLTAGE LIMIT (# mark) : 0 NUMBER OF GENERATORS EXCEEDING MINIMUM Q LIMIT (< mark) : 1 NUMBER OF GENERATORS EXCEEDING MAXIMUM Q LIMIT (> mark) : 1 ------------------------------------------------------------------------------- LINE FLOWS AND LINE LOSSES SLNO CS FROM FROM TO TO FORWARD LOSS % NODE NAME NODE NAME MW MVAR MW MVAR LOADING ---- -- ---- -------- ---- -------- -------- -------- -------- -------- ------- 1 1 1 Bus1 2 Bus2 158.767 19.952 5.1467 9.6825 160.0! 2 1 1 Bus1 3 Bus3 LINE IS OPEN 3 1 4 Bus4 5 Bus5 -4.863 -1.597 0.0227 -4.1560 6.0& 4 1 2 Bus2 3 Bus3 41.614 8.889 1.2054 0.1019 44.4^ 5 1 2 Bus2 4 Bus4 40.888 7.891 1.1530 -0.0637 43.4^ 6 1 2 Bus2 5 Bus5 51.119 13.491 1.2333 1.0496 55.1$ 7 1 3 Bus3 4 Bus4 -4.592 -6.214 0.0060 -1.6621 8.4& ------------------------------------------------------------------------------- ! NUMBER OF LINES LOADED BEYOND 125% : 1 @ NUMBER OF LINES LOADED BETWEEN 100% AND 125% : 0 # NUMBER OF LINES LOADED BETWEEN 75% AND 100% : 0 $ NUMBER OF LINES LOADED BETWEEN 50% AND 75% : 1 ^ NUMBER OF LINES LOADED BETWEEN 25% AND 50% : 2 & NUMBER OF LINES LOADED BETWEEN 1% AND 25% : 2 * NUMBER OF LINES LOADED BETWEEN 0% AND 1% : 0 ------------------------------------------------------------------------------- Department of Electrical & Electronics Engineering Power System Simulation Laboratory NEW SYSTEM FREQUENCY FOR ISLAND 1 : 60.000000 Hzs ------------------------------------------------------------------------------- Summary of results TOTAL REAL POWER GENERATION : 198.767 MW TOTAL REAL POWER DRAWAL -ve g : 0.000 MW TOTAL REACT. POWER GENERATION : 49.953 MVAR GENERATION pf : 0.970 TOTAL SHUNT REACTOR INJECTION : 0.000 MW TOTAL SHUNT REACTOR INJECTION : 0.000 MVAR TOTAL SHUNT CAPACIT.INJECTION : 0.000 MW TOTAL SHUNT CAPACIT.INJECTION : 0.000 MVAR TOTAL REAL POWER LOAD : 190.000 MW TOTAL REAL POWER INJECT,-ve L : 0.000 MW TOTAL REACTIVE POWER LOAD : 45.000 MVAR LOAD pf : 0.973 TOTAL COMPENSATION AT LOADS : 0.000 MVAR TOTAL HVDC REACTIVE POWER : 0.000 MVAR TOTAL REAL POWER LOSS (AC+DC) : 8.767104 MW ( 8.767104+ 0.000000) PERCENTAGE REAL LOSS (AC+DC) : 4.411 TOTAL REACTIVE POWER LOSS : 4.952199 MVAR ------------------------------------------------------------------------------- Zone wise distribution Description Zone # 1 ---------------- ---------- MW generation 198.7669 MVAR generation 49.9526 ------------------------------------------------------------------------------- LINE FLOWS AND LINE LOSSES SLNO CS FROM FROM TO TO FORWARD LOSS % NODE NAME NODE NAME MW MVAR MW MVAR LOADING Department of Electrical & Electronics Engineering Power System Simulation Laboratory ---- -- ---- -------- ---- -------- -------- -------- -------- -------- ------- 1 1 1 Bus1 2 Bus2 158.767 19.952 5.1467 9.6825 160.0! 2 1 1 Bus1 3 Bus3 LINE IS OPEN 3 1 4 Bus4 5 Bus5 -4.863 -1.597 0.0227 -4.1560 6.0& 4 1 2 Bus2 3 Bus3 41.614 8.889 1.2054 0.1019 44.4^ 5 1 2 Bus2 4 Bus4 40.888 7.891 1.1530 -0.0637 43.4^ 6 1 2 Bus2 5 Bus5 51.119 13.491 1.2333 1.0496 55.1$ 7 1 3 Bus3 4 Bus4 -4.592 -6.214 0.0060 -1.6621 8.4& ------------------------------------------------------------------------------- ! NUMBER OF LINES LOADED BEYOND 125% : 1 @ NUMBER OF LINES LOADED BETWEEN 100% AND 125% : 0 # NUMBER OF LINES LOADED BETWEEN 75% AND 100% : 0 $ NUMBER OF LINES LOADED BETWEEN 50% AND 75% : 1 ^ NUMBER OF LINES LOADED BETWEEN 25% AND 50% : 2 & NUMBER OF LINES LOADED BETWEEN 1% AND 25% : 2 * NUMBER OF LINES LOADED BETWEEN 0% AND 1% : 0 ------------------------------------------------------------------------------- NEW SYSTEM FREQUENCY FOR ISLAND 1 : 60.000000 Hzs ------------------------------------------------------------------------------- Summary of results TOTAL REAL POWER GENERATION : 198.767 MW TOTAL REAL POWER DRAWAL -ve g : 0.000 MW TOTAL REACT. POWER GENERATION : 49.953 MVAR GENERATION pf : 0.970 TOTAL SHUNT REACTOR INJECTION : 0.000 MW TOTAL SHUNT REACTOR INJECTION : 0.000 MVAR TOTAL SHUNT CAPACIT.INJECTION : 0.000 MW TOTAL SHUNT CAPACIT.INJECTION : 0.000 MVAR TOTAL REAL POWER LOAD : 190.000 MW TOTAL REAL POWER INJECT,-ve L : 0.000 MW TOTAL REACTIVE POWER LOAD : 45.000 MVAR LOAD pf : 0.973 TOTAL COMPENSATION AT LOADS : 0.000 MVAR TOTAL HVDC REACTIVE POWER : 0.000 MVAR Department of Electrical & Electronics Engineering Power System Simulation Laboratory TOTAL REAL POWER LOSS (AC+DC) : 8.767104 MW ( 8.767104+ 0.000000) PERCENTAGE REAL LOSS (AC+DC) : 4.411 TOTAL REACTIVE POWER LOSS : 4.952199 MVAR ------------------------------------------------------------------------------- Zone wise distribution Description Zone # 1 ---------------- ---------- MW generation 198.7669 MVAR generation 49.9526 Department of Electrical & Electronics Engineering Power System Simulation Laboratory Exp No. 6 Date: 08.02.2010 DC LOAD FLOW ANALYSIS USING MATLAB AIM : To determine DC power flow among various buses and DC power generated by slack bus using MATLAB. THEORY: The Newton power flow is the most robust power flow algorithm used in practice. The one draw back is the fact that the term in the Jacobian matrix must be recalculated during each iteration and then the entire set of linear equations, Must also be resolved during each iterations. Such situation where a lot of computations have to be made a linear approximation of the load flow problem can be made to save computation time, this is called the DC load flow. The DC Load flow is principally different from the decoupled load flow. In the DC power flow, the non-linear load flow equations are linearized Department of Electrical & Electronics Engineering Power System Simulation Laboratory to ease the calculation and to speed up the computation, of the unknown voltages, this means that the actual model of the power system is altered, then it affects the final solution of the load flow. In the decoupled load flow, the non linear load flow equation are solved iteratively and approximation are made to the Jacobin matrix only, therefore only the speed of the convergence is affected, but the final result remains the same. The DC load flow can also be applied to find the fairly good approximation of the unknown voltages, that can be used as initial values in a Newton-Raphson / decoupled load flow solution or calculations. The decoupled power flow results in The above equations can be written in matrix form as Department of Electrical & Electronics Engineering Power System Simulation Laboratory Simplifying eqns (1) and (2) using ΔP - Δθ relationship, Assume rik << Xik this changes - Bik to Eliminate all shunt reactions to ground. Eliminate all shunts to ground, which arise from auto transformer. Simplifying the Δ θ – Δ|E1| relationship of eqn (2) Omit all effects from phase shift transformer The resulting equations are Department of Electrical & Electronics Engineering Power System Simulation Laboratory Where the terms in the matrices are Assuming a branch from i to k. In DC power flow, a further simplification of the power flow algorithm involves simply dropping the Q- V (eqn (4) altogether. This results in a completely linear, non iterative , power flow algorithm. Assume that |Ei| = 1.0pu. Eqn (3) becomes Department of Electrical & Electronics Engineering Power System Simulation Laboratory In dc power flow is only good for calculating MW flows on transmission lines and transformers. It gives no indication of what happen to voltage magnitudes, or MVAR or MVA flows. The power flowing on each line using the DC power flow is then PROBLEM : Determine the dc power flow among various buses and dc power generated by slack bus using MATLAB. For the following network. Fig 1 Department of Electrical & Electronics Engineering Power System Simulation Laboratory Susceptance Matrix Susceptance Matrix after deleting row and column. Department of Electrical & Electronics Engineering Power System Simulation Laboratory Slack Bus Power = 35MW Fig 2. Slack Bus Power = 35MW Department of Electrical & Electronics Engineering Power System Simulation Laboratory ALGORITHM: Step 1 : Read the number of buses. Step 2 : Read the reactance value between the buses. Step 3 : Formulate the C matrix by inversing the reactance values. Step 4 : read the reference bus number as i. Step 5 : Formulate the B matrix by eliminating ith row and column of C matrix Step 6 : Calculate [θ] = [B]-1 [P] Step 7 : Evaluate Pik = (1/Xik) (θi -θk) Pi = Σ Pk k = buses connected to „i‟ Department of Electrical & Electronics Engineering Power System Simulation Laboratory FLOW CHART: Department of Electrical & Electronics Engineering Power System Simulation Laboratory DC LOAD FLOW: clear all n=input('Enter no. of buses ='); x=zeros(n,n); x for i=1:n for j =1:n if (i<j) fprintf('Enter the values of X(%d,%d):',i,j); x(i,j)=input(''); x(j,i)=x(i,j); if (x(i,j)~=0) B(i,j)=(-1)/x(i,j); B(j,i)=(-1)/x(j,i); end end end end fprintf('the reactance matrix is \n'); Department of Electrical & Electronics Engineering Power System Simulation Laboratory sum=0; for i=1:n for j=1:n sum=sum+B(i,j); end B(i,i)=-1*sum; sum=0; end fprintf('The susceptance matrix is \n'); B p=input('Enter the reference bus = '); B(p,:)=[]; B(:,p)=[]; fprintf('The susceptance matrix after deleting row n column is \n'); B F=zeros(n,1); for i=1:n if (i~=p) fprintf('Enter the value of power(%d):',i); F(i)=input(''); Department of Electrical & Electronics Engineering Power System Simulation Laboratory end end F(p,:)=[]; fprintf('The power at the given bus = ') F H=inv(B); J=H*F; J W=zeros(n,1); for j=1:n-1 W(j,1)=J(j,1); end for i=1:p-1 T(i,1)=W(i,1); end T(p,1)=0; for i=p+1:n T(i,1)=W(i-1,1); end T Department of Electrical & Electronics Engineering Power System Simulation Laboratory E=zeros(n,1); for i=1:n for j=1:n if(i~=j && x(i,j)~=0) E(i,j)=(1/x(i,j))*(T(i,1)-T(j,1)); end end end E add=0; for i=1:n for j=1:n add = add+E(i,j); end E(i,i)=add; add=0; E fprintf('Slack bus power is E(%d,%d)=',p,p) E(p,p) Department of Electrical & Electronics Engineering Power System Simulation Laboratory RESULT: Enter no. of buses =3 x= 0 0 0 0 0 0 0 0 0 Enter the values of X(1,2):.2 Enter the values of X(1,3):.4 Enter the values of X(2,3):.25 the reactance matrix is x= 0 0.2000 0.4000 0.2000 0 0.2500 0.4000 0.2500 0 Department of Electrical & Electronics Engineering Power System Simulation Laboratory The Susceptance matrix is B= 7.5000 -5.0000 -2.5000 -5.0000 9.0000 -4.0000 -2.5000 -4.0000 6.5000 Enter the reference bus = 3 The Susceptance matrix after deleting row n column is B= 7.5000 -5.0000 -5.0000 9.0000 Enter the value of power(1):65 Enter the value of power(2):-100 The power at the given bus = F= 65 -100 Department of Electrical & Electronics Engineering Power System Simulation Laboratory J= 2.0000 -10.0000 T= 2.0000 -10.0000 0 E= 0 60.0000 5.0000 -60.0000 0 -40.0000 -5.0000 40.0000 0 Department of Electrical & Electronics Engineering Power System Simulation Laboratory The power flow E= 65.0000 60.0000 5.0000 -60.0000 -100.0000 -40.0000 -5.0000 40.0000 35.0000 Slack bus power is E(3,3)= ans = 35 Department of Electrical & Electronics Engineering Power System Simulation Laboratory Exp No: 7 a) Date: 24 .2.2010 SIMULATION AND ANALYSIS OF MAGNETIC CIRCUITS USING SIMULINK Injection of third harmonic voltage using three coupled windings AIM: To inject third or higher order harmonic voltage using three coupled windings to a circuit THEORY: Coupled windings or mutual inductance Description The Mutual Inductance block can be used to model two- or three-windings inductances with equal mutual coupling, or to model a generalized multi-windings mutual inductance with balanced or unbalanced mutual coupling. If two- or three-windings inductances model with equal mutual coupling is used, specify the self-resistance and inductance of each winding plus the mutual resistance and inductance. The electrical model for this block in this case is given below: Fig(1) Department of Electrical & Electronics Engineering Power System Simulation Laboratory If a general mutual inductance model is used, specify the number of self windings (not just limited to 2 or 3 windings) plus the Resistance and Inductance matrices that define the mutual coupling relationship between the windings (balanced or not). Type of mutual inductance Set to Two or Three windings with equal mutual terms to implement a three-phase mutual inductance with equal mutual coupling between the windings Winding 1 self impedance The self-resistance and inductance for winding 1, in ohms (Ω) and henries (H). Winding 2 self impedance The self-resistance and inductance for winding 2, in ohms (Ω) and henries (H). Three windings Mutual Inductance If selected, implements three coupled windings; otherwise, it implements two coupled windings. Winding 3 self impedance The Winding 3 self impedance parameter is not available if the Three windings Mutual Inductance parameter is not selected. The self-resistance and inductance in ohms (Ω) and henries (H) for winding 3. Mutual impedance The mutual resistance and inductance between windings, in ohms (Ω) and henries (H). The mutual resistance and inductance corresponds to the magnetizing resistance and inductance on the standard transformer circuit diagram. If the mutual resistance and reactance are set to [0 0], the block implements three separate inductances with no mutual coupling. Generalized mutual inductance: Type of mutual inductance Set to Generalized mutual inductance to implement a multi windings mutual inductance with mutual coupling defined by an inductance and a resistance matrix. Department of Electrical & Electronics Engineering Power System Simulation Laboratory Inductance matrix L The inductance matrix, in Henrys, that defines the mutual coupling relationship between the self windings. It must be a N-by-N symmetrical matrix. Resistance matrix R The resistance matrix, in Ohms, that defines the mutual coupling relationship between the self windings. It must be a N-by-N symmetrical matrix. Negative values are allowed for the self- and mutual inductances as long as the self-inductances are different from the mutual inductance. SIMULINK MODEL OF THE CIRCUIT: Fig(2). Simulink model Department of Electrical & Electronics Engineering Power System Simulation Laboratory Fig(3). Output voltage wave form Department of Electrical & Electronics Engineering Power System Simulation Laboratory RESULT: Three coupled winding circuit has been Simulated using Simulink.The third and higher order harmonics voltage was injected and corresponding output waveform was obtained. Department of Electrical & Electronics Engineering Power System Simulation Laboratory Exp No: 7 b) Date: 24.2.2010 LINEAR TRANSFORMER AIM: To simulate a distribution transformer network feeding line-to-neutral and line-to-line loads. THEORY: Distribution Transformer or Linear Transformer Implement two-or three-winding linear transformer 75KVA,14.4KV/120/120,50 Hz , R1=27.648Ω, L1=0.22 H, R2 =R3=55.296Ω, L2=L3=0. Description The linear transformer block model shown consists of three coupled windings wound on the same core. Fig(4). The linear transformer block model The model takes into account the winding resistances (R1 R2 R3) and the leakage inductances (L1 L2 L3), as well as the magnetizing characteristics of the core, which is modeled by a linear (Rm Lm) branch. Department of Electrical & Electronics Engineering Power System Simulation Laboratory Modeling an Ideal Transformer To implement an ideal transformer model, set the winding resistances and inductances to zero, and the magnetization resistance and inductance (Rm Lm) to infinity. Units Specify the units used to enter the parameters of the Linear Transformer Block. Select pu to use per unit. Select SI to use SI units. Changing the Units parameters from pu to SI, or from SI to pu, will automatically convert the parameters displayed in the mask of the block. Nominal Power and Frequency The nominal power rating Pn in volt-amperes (VA) and frequency fn, in hertz (Hz), of the transformer. Winding 1 Parameters The nominal voltage V1, in volts RMS, resistance, in pu or ohms, and leakage inductance, in pu or henries. The pu values are based on the nominal power Pn and on VI.Set the winding resistances and inductances to zero to implement an ideal winding. Winding 2 Parameters The nominal voltage V2, in volts RMS, resistance,in pu or ohms,and leakage inductance,in pu or henries.The pu values are based on the nominal power Pn and on V2.Set the winding resistances and inductances to zero to implement an ideal winding. Three winding transformer If selected, implements a linear transformer with three windings; otherwise, it implements a two-winding transformer. Winding 3 Parameters The Winding 3 parameters parameter is not available if the three windings transformer parameter is not selected. The nominal voltage in volts RMS (Vrms), resistance, in pu or ohms, and leakage inductance,in pu or henries.The pu values are based on the nominal power Pn and on V3.Set the winding resistances and inductances to zero to implement an ideal winding. Magnetization resistance and reactance The resistance and inductance simulating the core active and reactive losses. When selected , the pu values are based on the nominal power Pn and on VI.For example, to specify 0.2% of active and reactive core losses, at nominal voltage, use Rm=500 pu and Lm=500 pu. Rm must have a finite value when the inductance of winding 1 is greater than zero. Department of Electrical & Electronics Engineering Power System Simulation Laboratory SIMULINK MODEL OF THE SYSTEM: Fig(5).simulink model Department of Electrical & Electronics Engineering Power System Simulation Laboratory Fig(6).Input voltage wave form Department of Electrical & Electronics Engineering Power System Simulation Laboratory Fig(7).Input voltage wave form Department of Electrical & Electronics Engineering Power System Simulation Laboratory Fig(8).Output current wave form Department of Electrical & Electronics Engineering Power System Simulation Laboratory Fig(9).Output voltage wave form Department of Electrical & Electronics Engineering Power System Simulation Laboratory RESULT: Linear transformer network feed up line to neutral and line to line loads was simulated and corresponding output waveform are obtained. Department of Electrical & Electronics Engineering Power System Simulation Laboratory Exp No: 8 Date : 04.02.2010 SIMULATION AND MEASUREMENTS OF THREE PHASE CIRCUIT USING SIMULINK AIM: To Explore the powerlib library in the SimPowerSystems Software Learn how to build a simple circuit from the powerlib library Interconnect Simulink blocks with the given circuit The circuit below represents an equivalent power system feeding a 300 km transmission line. The line is compensated by a shunt indicator at its receiving end. To simplify matters, only one of the three phases is represented. The parameters shown in the figure are typical of a 735 kV power system. CIRCUIT TO BE MODELED: Fig 1. Circuit to be modeled Department of Electrical & Electronics Engineering Power System Simulation Laboratory PROCEDURE: Open the SimPowerSystems main library by entering the following command at the MATLAB prompt. powerlib This command displays a Simulink window showing icons of different block libraries. Open these libraries to produce the windows containing the blocks to be copied into the circuit. Each component is represented by a special icon having one or several inputs and outputs corresponding to the different terminals of the component: Open the Electrical Sources library and copy the AC Voltage Source block into the circuit window. Open the AC Voltage Source dialog box by double-clicking the icon and enter the Amplitude, Phase and Frequency parameters according to the given values. Note that the amplitude to be specified for a sinusoidal source is its peak value (424.4e3*sqrt(2) volts in this case). Change the name of this block from AC Voltage Source to Vs. Copy the parallel RLC branch block, which can be found in the Elements library of powerlib, set its parameters and name it Z_eq. The resistance Rs_eq of the circuit can be obtained from the Parallel RLC Branch block. Duplicate the Parallel RLC Branch block. Select R for the Branch Type parameter and the set the R parameter according to the value given. Enter the values as infinity (inf) and zero (0) for L and C. Once the dialog box is closed, notice that the L and C components have disappeared so that the icon now shows a single resistor. Name this block Rs_eq. The model of a line with uniformly distributed R, L and C parameters normally consists of a delay equal to the wave propagation time along the line. This model cannot be simulated as a linear system because a delay corresponds to an infinite number of states. However, a good approximation of the line with a finite number of states can be obtained by cascading several PI circuits, each representing a small section of the line. Department of Electrical & Electronics Engineering Power System Simulation Laboratory A PI section consists of a series R-L branch and two shunt C branches. The model accuracy depends on the number of PI sections used for the model. Copy the PI Section Line block and set its parameters. It is more convenient to use a Series RLC Load block that allows to specify directly inactive and reactive powers absorbed by the shunt reactor. Copy the Series RLC Load block, which can be found in the Elements library of powerlib. Name this block 110 Mvar. Set its parameters as follows: Vn 424.4e3 V fn 60Hz P 110e6/300 w (quality factor = 300) QL 110e6 vars Qe 0 Note that, as no reactive capacitive power is specified, the capacitor disappears on the block icon when the dialog box is closed. A Voltage Measurement block and current measurement block is used to measure the voltage and current at different nodes. This block is found in the Measurement library of powerlib. To observe the voltage and current a display system is needed. This can be any device found in the Simulink Sinks library. Open the Sinks library and copy the Scope block. If the scope were connected directly at the output of the voltage measurement, it would display the voltage in volts. From the Simulation menu, select Start. A Powergui block is automatically added to the model from which the values of voltage and current are obtained. Open the Scope blocks and observe the parameters at different nodes. Department of Electrical & Electronics Engineering Power System Simulation Laboratory Fig 2. Simulation block diagram Department of Electrical & Electronics Engineering Power System Simulation Laboratory Department of Electrical & Electronics Engineering Power System Simulation Laboratory Department of Electrical & Electronics Engineering Power System Simulation Laboratory Department of Electrical & Electronics Engineering Power System Simulation Laboratory Department of Electrical & Electronics Engineering Power System Simulation Laboratory RESULT: Vrms 1 424400<00V Vrms 2 428722.97<-0.120V Irms 1 261.83 <-89.930A Irms 2 284.15<89.710A A 300 km transmission line has been simulated and the corresponding voltage and current has been measured. Department of Electrical & Electronics Engineering Power System Simulation Laboratory Exp No: 9 Date: 03.03.2010 MODELING OF AUTOMATIC GENERATION CONTROL FOR A TWO AREA NETWORK USING SIMULINK AIM: To construct a SIMULINK block diagram of a two area system network and obtain the frequency deviation response and power deviation response. THEORY: As the system load changes continuously the generation is adjusted automatically to restore the frequency to nominal value. This scheme is known as automatic generation control. The operation objectives of the LFC are to maintain reasonably uniform frequency, to divide the load between generators, and to control the tie- line interchange schedules. The change in frequency and tie-line real power are sensed, which is a measure of the change in rotor angle, δ i.e., the rotor error Δδ to be corrected. The error signal, i.e., Δf and ΔPtie, are amplified, mixed, and transformed into a real power command signal ΔPv, which is sent to the prime mover to call for an increment in the torque. The prime mover, therefore, brings change in the generation output by an amount ΔPg which will change the values of Δf and ΔPtie within the specified tolerance. The first step of analysis of design of control system is mathematical modeling of the system. Proper assumptions and approximations are made to linearize the mathematical equations describing the system and transfer function model obtained for the following components. Department of Electrical & Electronics Engineering Power System Simulation Laboratory GENERATOR MODEL: Applying swing equation of a synchronous machine with small perturbation is given by 2H d2Δδ = ΔPm –ΔPe ws dt2 Or in terms of small deviation in speed With speed expressed in per unit, without explicit per unit notation, we have Taking Laplace transform of above equation, we obtain The above relation is shown in block diagram form as below. Fig 1.Generator Model Department of Electrical & Electronics Engineering Power System Simulation Laboratory LOAD MODEL: The load on a power system consists of a variety of electrical devices. For resistive loads, such as lighting and heating loads, the electrical power is independent of frequency. Motor loads are sensitive to changes in frequency. How sensitive it is to frequency depends on the composite of the speed - load characteristics of all the driven devises. The speed - load characteristic of a composite load is approximated by ΔPe = ΔPL + DΔw where ΔPL is the non frequency - sensitive load change, and DΔw is the frequency - sensitive load change. D is expressed as percent change in load divided by percent change in frequency. Fig 2. Load model Department of Electrical & Electronics Engineering Power System Simulation Laboratory PRIME MOVER MODEL: The model for turbine relates changes in mechanical power output Δ Pm to changes in steam valve position ΔPv. Different types of turbine vary widely in characteristics. The simplest prime mover model for non reheat steam turbine can be approximated with a single time constant τT resulting in following transfer function . the block diagram for a sample turbine is show below Fig 3. Prime mover model The time constant τT is in range of 0.2 to 2.0 seconds. Department of Electrical & Electronics Engineering Power System Simulation Laboratory GOVERNOR MODEL: The speed governor mechanism act as comparator whose output ΔPg is the difference between reference set power ΔPref and power as given from the governor speed characteristics. R is the speed regulation of governor (5-6 percent from zero to full load). Or in s – domain ΔPg (s) =ΔPref (s) - ΔΩ (s) The command ΔPg is transformed through the hydraulic amplifier to the steam valve position command ΔPv. Assuming a linear relationship and considering a simple time constant τg, we have the following S-domain relation. Fig 4. Governor model Department of Electrical & Electronics Engineering Power System Simulation Laboratory AGC IN THE TWO AREA SYSTEM: Consider two areas represented by an equivalent generating unit interconnected by a lossless tie line with reactance Xtie. Each area is represented by a voltage source behind an equivalent reactance as in following figure. Fig 5. Reactance diagram During normal operations real power transferred over tie line is given by P12 = |E||E2| Sin δ12 .................................(1) X12 X12 = X1+Xtie + X2 δ12 =δ1-δ2 Equation 1 can be linearised for a small deviation in the tie line flow ΔP12 from nominal value. ie, =Ps δ12............................................... (2) Ps = Slope of the power angle curve at the initial operating angle δ120= δ 10- δ 20 Department of Electrical & Electronics Engineering Power System Simulation Laboratory ........................... (3) = Ps δ12 Tie line power deviation then takes on the form ΔP12 = Ps (Δδ1- Δδ2) ...................................... (4) The tie line flow appears as a load increase in one area and the load decrease in the other area, depending on the direction of flow. The direction of the flow is dictated by phase angle difference if Δδ1>Δδ2 the power flows from area 1 to area 2. A block diagram representation for the two area LFC containing only primary loop is shown in the following figure. Fig 6. Block diagram representation for the two area LFC containing only primary loop Department of Electrical & Electronics Engineering Power System Simulation Laboratory Consider a load change ΔPl1 in area 1. In steady state both areas have same steady state frequency deviation ie, Δw = Δw1 = Δw2 and ΔPm1 - ΔP12 - ΔP11 = ΔwD1 ........................(5) ΔPm2 + ΔP12 = ΔwD2.......................... (6) The change in mechanical power is determined by governor speed characteristics, given by ΔPml = - .................................. (7) ΔPm2 = ......... ............................ (8) Substituting (7) & (8) in (5) & (6) and solving for Δw, we have Where, B1, B2 are frequency bias factors. Department of Electrical & Electronics Engineering Power System Simulation Laboratory ΔP12 = Change in the tie line power is given by PROBLEM: A two - area system connected by a tie line has the following parameters on a 1000 - MVA common base. Table 1 The units are operating in parallel at the nominal frequency of 60Hz. The synchronizing power coefficient is computed from the initial operating condition and is given to be Ps=2.0 per unit. A load change of 187.5 MW occurs in area 1. Construct the SIMULINK block diagram and obtain the frequency deviation response and power deviation step response. Department of Electrical & Electronics Engineering Power System Simulation Laboratory Fig 7.Simulation block diagram Department of Electrical & Electronics Engineering Power System Simulation Laboratory Fig 8. Power deviation response Department of Electrical & Electronics Engineering Power System Simulation Laboratory Fig 9. Power deviation response Department of Electrical & Electronics Engineering Power System Simulation Laboratory Fig 10.Frequency deviation response Department of Electrical & Electronics Engineering Power System Simulation Laboratory RESULT: The block diagram of a two area system network was constructed using Simulink and the frequency deviation response and power deviation response was obtained Department of Electrical & Electronics Engineering Power System Simulation Laboratory Exp No: 10 Date: 5.3.2010 MODELING AND SIMULATION OF NON-CONVENTIONAL ENERGY SYSTEMS USING MATLAB AIM: To study the generic model of Double-Fed Induction Generator (DFIG) Driven by a Wind Turbine. THEORY: 1. Turbine response to a change in wind speed Open the “Wind Speed” step block specifying the wind speed. Initially, wind speed is set at 8m/s, then at t=5s, wind speed increases suddenly at 14m/s. Start simulation and observe the signals on the “Wind Turbine” scope monitoring the wind turbine voltage, current, generated active and reactive powers, DC bus voltage and turbine speed. At t=5s, the generated active power starts increasing smoothly(together with the turbine speed) to reach its rated value of 9 MW in approximately 15s.Over that time frame the turbine speed will have increased from 0.8 pu to 1.21 pu. Initially, the pitch angle of the turbine blades is zero degree and the turbine operating point follows the red curve of the turbine power characteristics up to point D. Then the pitch angle is increased from 0 deg to 0.76 deg in order to limit the mechanical power. Observe also the voltage and the generated reactive power. The reactive power is controlled to maintain a 1 pu voltage. At nominal power, the wind turbine absorbs 0.68 MVar (generated Q = -0.68 MVar) to control voltage at 1pu. If we change the mode of operation to “Var regulation” with the “Generated reactive power Qref” set to zero, we will observe that voltage increases to 1.021 pu when the wind turbine generates its nominal power at unity power factor. 2. Simulation of voltage sag on the 120 kV system Department of Electrical & Electronics Engineering Power System Simulation Laboratory We will now observe the impact of a voltage sag resulting from a remote fault on the 120-kV system. First in the wind speed step block, disable the wind speed step by changing the Final value from 14 to 8 m/s. Then open the 120 kV voltage source menu. In the parameter “Time variation of”, select “Amplitude”. A 0.15 pu voltage drop lasting 0.5s is programmed to occur at t =5s. Make sure that the control mode is still in Var regulation with Qref=0. Start simulation and open the “Grid” scope. Observe the plant voltage and current as well as the motor speed. Note that the wind farm produces 1.87 voltage an d current as well as the motor speed. Note that the wind farm produces 1.87 MW. At t=5s, the voltage falls below 0.9 pu and at t=5.22s, the protection system trips the plant because an under voltage lasting more than 0.2 s has been detected (look at the protection settings and status in the “PLANT” subsystem). The plant current falls to zero and motor speed decreases gradually, while the wind farm continues generating at a power level of 1.87 MW. After the plant has tripped, 1.25 MW of power (P_B25 measured at bus B25) is exported to the grid. Now, change the wind turbine control mode to “Voltage regulation” and repeat the test. We will notice that the plant does not trip anymore. This is because the voltage support provided by the 5 Mvar reactive power generated by the wind turbines during the voltage sag keeps the plant voltage above 0.9 pu protection threshold. The plant voltage during sag is now 0.93 pu. 3. Simulation of a fault on the 25 kV line system. Finally, we will now observe impact of a single phase to ground fault occurring on the 25 kV line at B25 bus. First disable the 120 kV voltage step. Now open the “Fault” block menu and select “Phase A Fault”. Check that the fault is programmed to apply a 9-cycle single phase to ground fault at t =5 s. We should observe that when the wind tubine is in “Voltage regulation” mode, the positive sequence voltage at wind turbine terminals(V1_B575) drops to 0.8 pu during the fault which is above the under voltage protection threshold(0.75 pu for t>0.1 sec). the wind farm therefore stays in service .However, Department of Electrical & Electronics Engineering Power System Simulation Laboratory if the “Var regulation “ mode is used with Q ref =0, the voltage drops under 0.7 pu and the under voltage protection trips the wind point. We can now observe that the turbine speed increases. At t=40s the pitch angle starts to increase in order to limit the speed. PROCEDURE: 1. Open matlab - new-model file and save it 2. Click start button on matlab- start-simulink-library browser-sim power systems 3. Select the required blocks and elements from the library and drag it to the model file 4. Connect the blocks and save it 5. Set simulation time ( 2 sec ) and start simulation 6. Click on the scope so that output is obtained. Department of Electrical & Electronics Engineering Power System Simulation Laboratory Fig No: 1 Model of Double-Fed Induction Generator driven by a Wind Turbine Department of Electrical & Electronics Engineering Power System Simulation Laboratory Fig no. 2 : Scope output curves of Grid Department of Electrical & Electronics Engineering Power System Simulation Laboratory Fig no. 3 : Scope output curves of Wind Turbine Department of Electrical & Electronics Engineering Power System Simulation Laboratory RESULT The generic model of Double-Fed Induction Generator (DFIG) Driven by a Wind Turbine is studied. Department of Electrical & Electronics Engineering

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