Ee 1404 Power System Lab Manual

DHANALAKSHMI COLLEGE OF ENGINEERINGDEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING EE 1404 – POWER SYSTEM SIMULATION LABORATORY LAB MANUAL/ OBSERVATION 2009 – 2010 ODD SEMESTER NAME : REG : BRANCH : PREPARED BY V.BALAJI, M.Tech, (Ph.D) Asst.Professor /EEE PREPARED BY V.BALAJI, M.Tech, (Ph.D) Asst.Professor /EEE Page 1 LIST OF EXPERIMENTS 1. COMPUTATION OF PARAMETERS AND MODELLINGOF TRANSMISSION LINES. 2. FORMATION OF ADMITTANCE MATRICES. 3. FORMATION OF IMPEDANCE MATRICES. 4. SOLUTION OF POWER FLOW USING GAUSS-SEIDEL METHOD. 5. SHORT CIRCUIT ANALYSIS. 6. SOLUTION OF POWER FLOW USING NEWTON-RAPHSON METHOD. 7. LOAD – FREQUENCY DYNAMICS OF SINGLE AREA POWER SYSTEMS. 8. LOAD – FREQUENCY DYNAMICS OF TWO AREA POWER SYSTEMS. 9. TRANSIENT AND SMALL SIGNAL STABILITY ANALYSIS – SINGLE MACHINE INFINITE BUS SYSTEM. 10. ECONOMIC DISPATCH IN POWER SYSTEMS PREPARED BY V.BALAJI, M.Tech, (Ph.D) Asst.Professor /EEE Page 2 I CYCLE 1. COMPUTATION OF PARAMETERS AND MODELLINGOF TRANSMISSION LINES. 2. FORMATION OF ADMITTANCE MATRICES. 3. FORMATION OF IMPEDANCE MATRICES. 4. SOLUTION OF POWER FLOW USING GAUSS-SEIDEL METHOD. 5. SHORT CIRCUIT ANALYSIS. II CYCLE 6. SOLUTION OF POWER FLOW USING NEWTON-RAPHSON METHOD. 7. LOAD – FREQUENCY DYNAMICS OF SINGLE AREA POWER SYSTEMS. 8. LOAD – FREQUENCY DYNAMICS OF TWO AREA POWER SYSTEMS. 9. TRANSIENT AND SMALL SIGNAL STABILITY ANALYSIS – SINGLE MACHINE INFINITE BUS SYSTEM. 10. ECONOMIC DISPATCH IN POWER SYSTEMS PREPARED BY V.BALAJI, M.Tech, (Ph.D) Asst.Professor /EEE Page 3 Professor /EEE Page 4 .Tech.CONTENTS S.NO Date Name of the Experiment Page NO Marks Remarks out of 10 /Signature PREPARED BY V. M.BALAJI. (Ph.D) Asst. M.NO Date Name of the Experiment Page NO Marks Remarks out of 10 /Signature PREPARED BY V. (Ph.D) Asst.S.Tech.Professor /EEE Page 5 .BALAJI. capacitance and conductance. (Ph.COMPUTATION OF PARAMETERS AND MODELLING OF TRANSMISSION LINES Expt. Three phase – Asymmetrical Transposed GMD = geometric mean of the three distance of the symmetrically place conductors = 3√DABDBCDCA PREPARED BY V.Tech.3 THEORY : Transmission line has four parameters – resistance.No : Date : AIM : (i) To determine the positive sequence line parameters L and C per phase per kilometre of a three phase single and double circuit transmission lines for different conductor arrangements.2 ln (Dm / Ds) Where. SOFTWARE REQUIRED: MATLAB 5. The inductance and capacitance are due to the effect of magnetic and electric fields around the conductor. r = radius of conductor III. inductance. Dm = geometric mean distance (GMD) Ds = geometric mean radius (GMR) I. M. (ii) To understand modeling and performance of medium lines.D) Asst. Single phase 2 wire system GMD = D GMR = re-1/4 = r′ Where.BALAJI. Inductance: The general formula: L = 0.Professor /EEE Page 6 . the inductances and capacitances can be evaluated using the formula. Three phase – symmetrical spacing GMD = D GMR = re-1/4 = r′ Where. r = radius of conductor II. The resistance of the conductor is best determined from the manufactures data. Tech.GMR = re-1/4 = r′ Where.Professor /EEE Page 7 ..D) Asst. are bundle conductor. DSADSB and DSC are GMR of each phase group and given by DSA = 4√(Dsb Da1a2)2 = [Dsb Da1a2]1/2 DSB = 4√(Dsb Db1b2)2 = [Dsb Db1b2]1/2 DSC = 4√(Dsb Dc1c2)2 = [Dsb Dc1c2]1/2 where. a2……. are bundle conductor PREPARED BY V.. GMRL is equivalent geometric mean radius and is given by GMRL = (DSADSBDSC)1/3 where. Ds is the GMR of each subconductor and d is bundle spacing Three phase – Double circuit transposed: The inductance per phase in milli henries per km is L = 0. is given by Lx = 0. Dsb = ra1’= rb1= ra’2 = rb’2 = rc’2 if a1. M.Dan )……. GMD = mn √ (Daa′ Dab′)…….2 ln (GMD / GMRL) mH/km where.Dnn) where. a2….(DnaDnb……. r′a = ra e(-1/ 4) Bundle Conductors: The GMR of bundle conductor is normally calculated GMR for two sub conductor c = (Ds * d)1/2 GMR for three sub conductor Dsb =(Ds * d2)1/3 GMR for four sub conductor Dsb = 1.Dnm′) GMR = n2√ (Daa Dab…….(Dna′……. (Ph.2 ln (GMD/GMR) where. r = radius of conductors Composite conductor lines The inductance of composite conductor X.09 (Ds * d3)1/4 where. Dsb =GMR of bundle conductor if conductor a1.BALAJI. (Ph.Professor /EEE Page 8 . DBC&DCA are GMD between each phase group A-B. 47/7 Bobolink conductor per phase with flat horizontal spacing of 11m between phases a and b and between phases b and c.Tech.000-cmil. GMD is the “Geometric mean distance” which is same as that defined for inductance under various cases. B-C.1793 cm and a GMR of 0.D) Asst. 1. EXERCISES: 1 A three-phase transposed line composed of one ACSR.GMD is the equivalent GMD per phase” & is given by GMD = [DAB * DBC * DCA]1/3 where. PROCEDURE: 1. (b) Verify the results using the MATLAB program. 5. (a) Determine the inductance and capacitance per phase per kilometer of the above two lines.New – M – File 3. The conductors have a diameter of 3.BALAJI. Type and save the program in the editor window. but it is to be operated at a higher voltage and therefore the phase spacing is increased to 14m as measured from the center of the bundles.439 cm. View the results. The line is to be replaced by a three-conductor bundle of ACSR 477. The spacing between the conductors in the bundle is 45 cm. 26/7 Hawk conductors having the same cross sectional area of aluminum as the single-conductor line. C-A which are given by DAB = [Da1b1 * Da1b2 * Da2b1 * Da2b2]1/4 DBC = [Db1c1 * Db1c2 * Db2c1 * Db2c2]1/4 DCA = [Dc1a1 * Dc2a1 * Dc2a1 * Dc2a2]1/4 Capacitance A general formula for evaluating capacitance per phase in micro farad per km of a transmission line is given by C = 0.8839 cm. PREPARED BY V. 4. The conductors have a diameter of 2. M. Enter the command window of the MATLAB.43. 2. DAB.000 cmil. Execute the program by either pressing Tools – Run.625 cm and a GMR of 1. The new line will also have a flat horizontal configuration. Create a new M – file by selecting File .0556/ ln (GMD/GMR) µF/km Where. L = 0.2*log(GMD/GMRL) C = 0.D) Asst.0556/log(GMD/GMRC) MANUAL CALCULATIONS : PREPARED BY V.Professor /EEE Page 9 .Tech. GMRL.PROGRAM : [GMD. M.BALAJI. (Ph. GMRC] = gmd. (Ph.BALAJI.D) Asst.Tech.PREPARED BY V. M.Professor /EEE Page 10 . B=Z. is=((C*vrph)+(D*ir)). ir=irold*(pfr-(j*k)).Professor /EEE Page 11 .732*vr*10^3*. bc=input('value of bc').BALAJI.D constants .7788*rad). M. vs=((A*vrph)+(B*ir)). angle(is). Z=rnew+i*(XL). L=2*10^(-7)*log(deq/newrad).sending end voltage. r=input('resistance/ph/km'). pr=input('receving end power in mw'). f=input('frequency').C.2. receiving end voltage and regulation. rnew=r*l.732. D=A. u=angle(is). k=sin(acos(pfr)).6m transposed. D=input('diameter in m'). f=angle(vs).8 pf lagging . vr=input('receving end voltage in kv'). XL=2*pi*f*L*l*1000. (Ph.0. Y=i*(2*pi*f*C*l*1000). L C rnew A B C abs(vs) abs(is) angle(vs)*180/pi angle(is)*180/pi PFS eff reg PREPARED BY V. deq=(ab*bc*ca)^(1/3).current .Find A. eff=((pr*10^6)/(3*abs(vs)*abs(is)*PFS))*100. newrad=(0.854*10^-12)/log(deq/rad). angle(vs). transmission efficiency .Tech. l=input('length of the line in km'). PROGRAM : ab=input('value of ab'). ca=input('value of ca').8). reg=(((abs(vs)/abs(A))-abs(vrph))/abs(vrph))*100. vrph=(vr*10^3)/1.D) Asst. rad=D/2. A three phase overhead line 200km long R = 0.B.power factor and power when the line is delivering full load of 50MW at 132kV . irold=(pr*10^6)/(1. pfr=input('receving end powerfactor').16 ohm/km and Conductor diameter of 2cm with spacing 4.5. C=Y*(1+(Y*Z)/4). PFS=cos(f-u). A=1+((Y*Z)/2). C=(2*pi*8. Professor /EEE Page 12 .Tech.MANUAL CALCULATIONS: PREPARED BY V.D) Asst. (Ph. M.BALAJI. (Ph.RESULT : PREPARED BY V.D) Asst.Tech. M.BALAJI.Professor /EEE Page 13 . The equivalent circuit of Tap changing transformers is included while forming Y-bus matrix. yid ydi ……… ydd where. 3. Every transmission line should be represented by π. Enter the command window of the MATLAB.equivalent. (ii) Check the results obtained in using MATLAB. The off diagonal elements are unaffected.New – M – File Type and save the program in the editor window.Professor /EEE Page 14 . 4. Yii = Self admittance Ydi = Transfer admittance PROCEDURE: 1. I. Create a new M – file by selecting File . Shunt impedances are added to diagonal element corresponding to the buses at which these are connected. PREPARED BY V. In most of the power system studies it is required to form y.bus matrix of the system by considering certain power system parameters depending upon the type of analysis. 2.No : Date : AIM: To determine the admittance matrices for the given power system network. SOFTWARE REQUIRED: MATLAB THEORY: Bus admittance is often used in power system studies.BALAJI. View the results. Y-bus may be formed by inspection method only if there is no mutual coupling between the lines. M. FORMATION OF Y-BUS MATRIX Generalised Y-bus = yii ……….FORMATION OF BUS ADMITTANCE MATRICES Expt. Execute the program by either pressing Tools – Run. (Ph.D) Asst. 5.. EXERCISE: (i) Determine the Y bus matrix and Z bus matrix for the power system network shown in fig.Tech. D) Asst.2 3 4 0 0.Tech.0 0 2 0 0. M. Y = ybus(z) MANUAL CALCULATIONS: PREPARED BY V. (Ph.4 1 3 0 0.Professor /EEE Page 15 .8 1 2 0 0.BALAJI.PROGRAM : z = [0 1 0 1.08].2 2 3 0 0. M.Professor /EEE Page 16 .D) Asst. (Ph.PREPARED BY V.BALAJI.Tech. Professor /EEE Page 17 .Tech. (Ph.RESULT: PREPARED BY V. M.D) Asst.BALAJI. Create a new M – file by selecting File . In one method we can form the bus admittance matrix and than taking its inverse to get the bus impedance matrix. 2. View the results. The bus impedance matrix are very useful in fault analysis.BALAJI.New – M – File 3.No: Date : AIM : To determine the bus impedance matrices for the given power system network. (ii) Check the results obtained in using MATLAB. In another method the bus impedance matrix can be directly formed from the reactance diagram and this method requires the knowledge of the modifications of existing bus impedance matrix due to addition of new bus or addition of a new line (or impedance) between existing buses.Tech. PREPARED BY V. EXERCISE: (i) Determine the and Z bus matrix for the power system network shown in fig.D) Asst. SOFTWARE REQUIRED: MATLAB 5. Execute the program by either pressing Tools – Run.Professor /EEE Page 18 . M. Type and save the program in the editor Window 4. PROCEDURE: 1.FORMATION OF BUS IMPEDANCE MATRICES Expt. 5. The bus impedance matrix can be determined by two methods. (Ph.3 THEORY: FORMATION OF Z-BUS MATRIX In bus impedance matrix the elements on the main diagonal are called driving point impedance and the off-diagonal elements are called the transfer impedance of the buses or nodes. Enter the command window of the MATLAB. D) Asst.BALAJI. Zbus = inv(Y) Vbus = Zbus*Ibus MANUAL CALCULATIONS: PREPARED BY V.2 2 3 0 0. 0]. Y = ybus(z) Ibus = [-j*1. 0. (Ph.PROGRAM : z = [0 1 0 1.25.Tech.2 3 4 0 0.1.Professor /EEE Page 19 .0 0 2 0 0.4 1 3 0 0. -j*1.08].8 1 2 0 0. M. Professor /EEE Page 20 . M. (Ph.D) Asst.PREPARED BY V.BALAJI.Tech. D) Asst.Professor /EEE Page 21 . M. (Ph.BALAJI.Tech.RESULT: PREPARED BY V. SOLUTION OF POWER FLOW USING GAUSS-SEIDEL METHOD Expt. 5.Professor /EEE Page 22 .Tech. (Ph. View the results.New – M – File 3. Type and save the program in the editor Window 4.∑ Vq k q=p+1 The reactive power of bus-p is given by p-1 n q=p QPk+1 =(-1) x Im (Vpk)* ∑ Ypq Vqk+1 + ∑ Ypq Vqk q=1 PROCEDURE: 1.No : Date : AIM : To understand. PREPARED BY V. Execute the program by either pressing Tools – Run. in particular. the mathematical formulation of power flow model in complex form and a simple method of solving power flow problems of small sized system using GaussSeidel iterative algorithm SOFTWARE REQUIRED: MATLAB 5.BALAJI. The non-linear load flow equation is given by 1 Vp k+1 Pp – j Qp (Vp ) k * p-1 n k+1 = Ypp . M.3 THEORY: The GAUSS – SEIDEL method is an iterative algorithm for solving a set of non-linear load flow equations.D) Asst. Enter the command window of the MATLAB.∑ Ypq Vq q=1 . Create a new M – file by selecting File . 2. 3))). Program %Gauss Sedial clc. y=zeros(bus.The magnitude at bus 1 is adjusted to 1.BALAJI. Find the slack bus real and reactive power.4).max(data(:.p)+data(q.2)).EXERCISE : The figure shows the single line diagram of a simple 3 buses power system with generator at bus 1. Determine the phasor values of the voltage at the load buses 2 and 3.Tech.3)==p) y(p.Professor /EEE Page 23 .2)==p|data(q. if(data(q.p)=y(p. (Ph. Line impedance are marked in pu.05pu. for p=1:bus. The base value is 100kVA.1)). bus=max(max(data(:. The scheduled loads at buses 2 and 3 are marked on the diagram. data=[1 1 2 10-j*20 2 1 3 10-j*30 3 2 3 16-j*32] elements=max(data(:. Verify the result using MATLAB.D) Asst. end end end PREPARED BY V. for q=1:elements. M.bus). The line charging susceptances are neglected. for q=1:bus. fprintf ('%g'.Tech. g2=(((p3-j*q3)/conj(v3))+(-y(1. while(abs(dx1)&abs(dx2)>=0.4)). b1=input('enter q2 in MVAR:').05.v3]). v3=1+j*0. iter=0.Professor /EEE Page 24 . dx2=g2-v3. pu=input('enter the base value in MVA'). if (p~=q) for r=1:elements if((data(r. p2=(a1/pu). iter=iter+1.3)*v3))/y(2.2).3)*g1))/y(3. end MANUAL CALCULATION PREPARED BY V. q2=(b1/pu). M. dx1=g1-v2. dx2=1+j*0. end end end end end a1=input('enter p2 in MW:'). v1=1.D) Asst.3)==p)) y(p. b2=input('enter q3 in MVAR'). g1=(((p2-j*q2)/conj(v2))+(-y(1. v3=v3+dx2. v2=v2+dx1. disp('iter v2 v3'). q3=(b2/pu).2)==p&data(r. dx1=1+j*0.3)*v1)+(-y(2.for p=1:bus.iter).00001)&iter<7.q)=-(data(r. p3=(a2/pu). v2=1+j*0.3).3)==q)|(data(r. a2=input('enter p3 in MW:'). (Ph.BALAJI.2)==q&data(r.disp([v2.2)*v1)+(-y(2. (Ph.BALAJI. M.PREPARED BY V.D) Asst.Professor /EEE Page 25 .Tech. D) Asst.BALAJI.Tech.Professor /EEE Page 26 . (Ph.RESULT PREPARED BY V. M. Bpq = Conductance and Susceptance of admittance Ypq respectively. EXERCISE PREPARED BY V.ep (fq Gpq .Professor /EEE Page 27 .Tech.eq Bpq) Vp2 = ep2 + fp2 where.D) Asst.No : Date : AIM : To determine the power flow analysis using Newton – Raphson method SOFTWARE REQUIRED : MATLAB THEORY : The Newton Raphson method of load flow analysis is an iterative method which approximates the set of non-linear simultaneous equations to a set of linear simultaneous equations using Taylor’s series expansion and the terms are limited to first order approximation.BALAJI.eq Bpq) QP = Σ q=1 fp (eq Gpq + fq Bpq) . The load flow equations for Newton Raphson method are non-linear equations in terms of real and imaginary part of bus voltages. ep = Real part of Vp fp = Imaginary part of Vp Gpq. PP = Σ n q=1 n ep(eq Gpq + fq Bpq) + fp (fq Gpq . M. (Ph.SOLUTION OF POWER FLOW USING NEWTON-RAPHSON METHOD Expt. BALAJI.1)=(abs(v2)*abs(v1)*abs(ybus(2. delp20=gbus(2. t= 0.02 pu.Professor /EEE Page 28 .04+j*0.764 -2.2)-p2. del2=angle(v2).1))*cos((angle(ybus(2.5 1. v2=1+j*0.0 0.941+j*11.764 -2. Consider the 3 bus system each of the 3 line bus a series impedance of 0.3))*sin((angle(ybus(2.3))*sin((angle(ybus(2. delp30=gbus(3.528 -2. v3=1.0 2 0.0].3)))+del3-del2)).941+j*11. q2=-(abs(v2)*abs(v1)*abs(ybus(2.3)))+del3del2)).6 0. M. J(1. del1=angle(v1).882-j*23.941+j*11.2)))+del2-del3)).u and a total shunt admittance of j0.Tech.1))*cos((angle(ybus(3.764 -2.1))*sin((angle(ybus(2.02 + j0.0 0.882-j*23.2)=-(abs(v2)*abs(v3)*abs(ybus(2. PD 1 2 3 2 0 1.1)))+del1-del2))abs(v2)*abs(v2)*abs(ybus(2.001 v1=1.882-j*23. del3=angle(v3).D) Asst.3)=(abs(v1)*abs(ybus(2.0 0. QD 1 0 0.3)-q2. p3=(abs(v3)*abs(v1)*abs(ybus(3.941+j*11.764 5.2))*cos((angle(ybus(3.1)))+del1del2))+(abs(v2)*abs(v3)*abs(ybus(2.2))))+(abs(v2)*abs(v3)* abs(ybus(2.1)))+del1- PREPARED BY V.3))*sin((angle(ybus(2.941+j*11. QG 0.5 generation.5)-gbus(2. ybus = [5.1))*cos((angle(ybus(2.0 0. J(1.3))*cos((angle(ybus(3.0 0. J(1. Verify the result using MATLAB PROGRAM : %NEWTON RAPHSON METHOD clc.08 p.3)))+del3del2)).PG Generation.3))))+(abs(v2)*abs(v3)* abs(ybus(3.528].1.1)))+del1del2))+abs(v2)* abs(v2)*abs(ybus(2.0 3 1.2))*cos((angle(ybus(2.5 0 1 QG3 = ? 2.04 Unspecified V3 = 1.1))*sin((angle(ybus(2.6 Load Real power Reactive Power Voltage Specified V1=1.1)))+del1del3))+abs(v3)*abs(v3)*abs(ybus(3.04+j*0. gbus = [1 2.0 1.528 -2.941+j*11.4)-gbus(3.4)-gbus(2.2))*sin((angle(ybus(2.2))))-(abs(v2)*abs(v3)* abs(ybus(2. The specified quantities at the buses are given below : Bus Real load Reactive demand.764 5.3))+del3-del2)).2)-p3.3))*cos((angle(ybus(2. delq20=gbus(2.5 0.764 -2. (Ph.1)) %abs(v2) for i=1:10 p2=(abs(v2)*abs(v1)*abs(ybus(2.04 demand. %abs(ybus(2. 2)))+del2del3)).2))*sin((angle(ybus(3.2))*cos((angle(ybus(3. del2=A1(1. A1-delA0.delq20].1)))+del1del3))+(abs(v3)*abs(v2)*abs(ybus(3. del3=A1(2.1))*sin((angle(ybus(2. J(3. A1=[del2.3)=(abs(v3)*abs(ybus(3.1)))+del1del2))-2*(abs(v2)*abs(ybus(2.Tech.D) Asst.3))*cos((angle(ybus(2.3) )* cos((angle(ybus(2. A=[del2.b0]+delA1.1)))+del1del2))-(abs(v2)*abs(v3)*abs(ybus(2.2))*sin((angle(ybus(3. if((A1-delA0)<=t) break.3))*cos((angle(ybus(2. J(2.2)=(abs(v2)*abs(v3)*abs(ybus(2.delp30.1).1)=-(abs(v3)*abs(v2)*abs(ybus(3. end A1 MANUAL CALCULATIONS : PREPARED BY V.del3.3)))+del3-del2)).del2))+2*(abs(v2)*abs(ybus(2. (Ph. delA1.abs(v2)]. delA1=inv(J)*delA0.3)=-(abs(v2)*abs(ybus(2.Professor /EEE Page 29 .1).3)))+del2del3)). J(3.3))* sin((angle(ybus(2.2)))+del2-del3)).3)))+del3-del2)).1))*cos((angle(ybus(2. abs(v2)=A1(3.2))*sin((angle(ybus(2. J(3.del3. M.2))))+(abs(v3)*abs(ybus(2.1).2))))(abs(v3)*abs(ybus(2. b0=abs(v2). J(2. end J inv(J).1)=(abs(v2)*abs(v1)*abs(ybus(2. J(2. delA0=[delp20.2)=(abs(v3)*abs(v1)*abs(ybus(3.3)))+del2del3)).1))*sin((angle(ybus(3.BALAJI.2))*cos((angle(ybus(2.2)))+del2del3)). BALAJI.D) Asst. M.Professor /EEE Page 30 .PREPARED BY V. (Ph.Tech. Tech.PREPARED BY V.Professor /EEE Page 31 . (Ph.D) Asst.BALAJI. M. Professor /EEE Page 32 .Tech.RESULT: PREPARED BY V.BALAJI. (Ph.D) Asst. M. Tech.SHORT CIRCUIT ANALYSIS Expt.3 THEORY : Symmetrical Fault : II. (Ph.Professor /EEE .No : Date : AIM : To become familiar with modelling and analysis of power systems under faulted condition and to compute the fault level. M. both symmetric and unsymmetric. post-fault voltages and currents for different types of faults. Three phase fault : From the thevenin’s equivalent circuit Vth ′′ Fault current If = Z th Where Vth = Thevenin’s Voltage Z th = Thevenin’s Impedance Unsymmetrical Fault : Single line to ground fault : Fault current If = Ia = 3Ia1 Ia1 = Ea Z1+Z2+Z0 Line to line fault: Fault current If = Ia1(a2 – a ) Ea Ia1 = Z1+Z2 Double Line to ground fault : Fault current If = 2 Ia0 +( Ia1+ Ia2) (a2 + a ) Ea Ia1 = Z1 + Z0Z2 Page 33 PREPARED BY V.D) Asst.BALAJI. PROGRAM REQUIRED: MATLAB 5. Enter the command window of the MATLAB. EXERCISE : PREPARED BY V. negative and zero phase sequence currents Z1 .New – M – File 3. Create a new M – file by selecting File .Professor /EEE Page 34 . Execute the program by either pressing Tools – Run. Ia2 and Ia0 are positive.Tech. Type and save the program. (Ph. 4. negative and zero phase sequence impedances PROCEDURE: 1. M.Ia1) * Z0 Z0 + Z2 Ia0 = − ( Ia1 − Ia2) Fault MVA = √3 * If * Vpu where. Ia1.D) Asst.BALAJI.Z0 + Z2 Ia2 = (. 2.Z2 and Zo are positive. View the results. 10 0.1 per unit.15 3 0 0. M. Zbus2) llfault(zdata1. The neutral of each generator is grounded through a current limiting reactor of 0.Zbus1) lgfault(zdata0.05 0.1 per unit. The system data expressed in per unit on a common 100 MVA base is tabulated below.35 0. Determine the fault current for the following faults.10 0.25]. (Ph. (c) A line to line fault at bus3 through a fault impedance Zf = j0.25 2 0 0.15 0.25 X0 0.125 0.10 0. zdata0 = [0 1 0 0.10 1 2 0 0.15 0. zdata1.35 2 3 0 0.30 1 3 0 0.30 0.25 X2 0. zdata2. Zbus2) PREPARED BY V. Zbus0.10 0.15 0. Zbus1.Tech. symfault(zdata1. (b) A single line to ground fault at bus3 through a fault impedance Zf = j0.Professor /EEE Page 35 .D) Asst.10 0. Zbus1.25/3 per unit on a 100MVA base. Zbus0. zdata1.25 2 0 0.05 0. zdata2.15 0.40 0 2 0 0. (a) A balanced three phase fault at bus 3 through a fault impedance Zf = j0.125 0.10 0.15 0. Zbus1 = zbuild(zdata1) Zbus0 = zbuild(zdata0) Zbus2 = Zbus1. zdata2.7125]. PROGRAM : zdata1 = [0 0 1 1 2 1 0 0.125 3 0 0. The generators are running on no load at their rated voltage and rated frequency with their emfs in phase.15 0. (d) A double line to ground fault at bus3 through a fault impedance Zf = j0. Item G1 G2 T1 T2 L12 L13 L23 Base MVA 100 100 100 100 100 100 100 Voltage Rating kV 20 20 20/220 20/220 220 220 220 X1 0. Zbus1.The one line diagram of a simple power system is shown in figure.BALAJI. Zbus2) dlgfault(zdata0.1 per unit.7125 Verify the result using MATLAB program.1 per unit. zdata2 = zdata1. BALAJI.MANUAL CALCULATIONS: PREPARED BY V.Tech. (Ph. M.D) Asst.Professor /EEE Page 36 . Professor /EEE Page 37 . M.BALAJI.PREPARED BY V.D) Asst.Tech. (Ph. (Ph. M.D) Asst.PREPARED BY V.BALAJI.Tech.Professor /EEE Page 38 . D) Asst. M.Tech.BALAJI.RESULT PREPARED BY V. (Ph.Professor /EEE Page 39 . 5.D) Asst. 2. After forming the block diagram . Restoration of frequency to nominal value requires secondary control action which adjust the load . This function is referred to as load – frequency control(LFC).LOAD – FREQUENCY DYNAMICS OF SINGLE AREA POWER SYSTEMS Expt . Pick up the blocks from the simulink library browser and form a block diagram. PREPARED BY V. A change in system load cases a change in the speed of all rotating masses ( Turbine – generator rotor systems) of the system leading to change in system frequency. This is one of the foremost requirements in proving quality power supply. No : Date AIM : To become familiar with modelling and analysis of the frequency and tie-line flow dynamics of a power system without and with load frequency controllers (LFC) and to design better controllers for getting better responses.New – Model 3. : PROCEDURE : 1. The speed change form synchronous speed initiates the governor control (primary control) action result in all the participating generator – turbine units taking up the change in load. Double click the scope and view the result. M.Tech.BALAJI.reference set points of selected ( regulating) generator – turbine units. save the block diagram.Professor /EEE Page 40 . Create a new Model by selecting File . (Ph. 4. stabilizing system frequency. The primary objectives of automatic generation control (AGC) are to regulate system frequency to the set nominal value and also to regulate the net interchange of each areas to the scheduled value by adjusting the outputs of the regulating units. THEORY : Active power control is one of the important control actions to be perform to be normal operation of the system to match the system generation with the continuously changing system load in order to maintain the constancy of system frequency to a fine tolerance level. Enter the command window of the MATLAB. Tech. (b) Use MATLAB to obtain the root locus plot. Use MATLAB to obtain the time domain performance specifications and the frequency deviation step response.5sec Governor time constant τg = 0. PREPARED BY V. A sudden load change of 50MW( PL = 0.2 per unit) occurs. (c) The governor speed regulation is set to R = 0.D = 0.8 (a) Use the Routh – Hurwitz array to find the range of R for control system stability.D) Asst. (Ph.e. (i) (ii) Find the steady state frequency deviation in Hz.2sec Generator inertia constant H = 5sec Governor speed regulation = R per unit The load varies by 0.EXERCISE: 1.BALAJI.An isolated power station has the following parameters Turbine time constant τT = 0.Professor /EEE Page 41 .The turbine rated output is 250MW at nominal frequency of 60Hz.8 percent for a 1 percent change in frequency . M.05 per unit. i. D) Asst.MANUAL CALCULATIONS: PREPARED BY V. (Ph. M.Tech.Professor /EEE Page 42 .BALAJI. (Ph. M.BALAJI.Professor /EEE Page 43 .PREPARED BY V.Tech.D) Asst. Professor /EEE Page 44 . M.D) Asst.RESULT: PREPARED BY V. (Ph.BALAJI.Tech. Professor /EEE Page 45 . Restoration of frequency to nominal value requires secondary control action which adjust the load . After forming the block diagram .New – Model 3.BALAJI. save the block diagram.D) Asst. 4. M.reference set points of selected ( regulating) generator – turbine units.LOAD – FREQUENCY DYNAMICS OF TWO AREA POWER SYSTEMS Expt . This is one of the foremost requirements in proving quality power supply. A change in system load cases a change in the speed of all rotating masses ( Turbine – generator rotor systems) of the system leading to change in system frequency. : PROCEDURE: 1. PREPARED BY V. 5.Tech. Create a new Model by selecting File . No : Date AIM : To become familiar with modelling and analysis of the frequency and tie-line flow dynamics of a two area power system without and with load frequency controllers (LFC) and to design better controllers for getting better responses. The speed change form synchronous speed initiates the governor control (primary control) action result in all the participating generator – turbine units taking up the change in load. Double click the scope and view the result. THEORY: Active power control is one of the important control actions to be perform to be normal operation of the system to match the system generation with the continuously changing system load in order to maintain the constancy of system frequency to a fine tolerance level. (Ph. Pick up the blocks from the simulink library browser and form a block diagram. The primary objectives of automatic generation control (AGC) are to regulate system frequency to the set nominal value and also to regulate the net interchange of each areas to the scheduled value by adjusting the outputs of the regulating units. 2. stabilizing system frequency. This function is referred to as load – frequency control(LFC). Enter the command window of the MATLAB. 6sec The units are operating in parallel at the nominal frequency of 60Hz. SIMULINK BLOCK DIAGRAM : PREPARED BY V.Professor /EEE Page 46 . (b) Construct the SIMULINK block diagram and obtain the frequency deviation response for the condition in part(a).9 H2=4 1000MVA τg1 = 0.Tech.0625 D2=0. A load change of 187.load coeff.D) Asst.5sec 2 R2=0. Inertia Constant Base Power Governor Time Constant Turbine Time Constant 1 R1=0.5 MW occurs in area1. A two area system connected by a tie line has the following parameters on a 1000MVA common base Area Speed Regulation Frequency –sens.2sec τT1 =0.BALAJI.u.EXERCISE : 1. (a) Dertermine the new steady state frequency and the change in the tie-line flow.6 H1=5 1000MVA τg1 = 0.05 D1=0.3sec τT1 =0. The synchronizing power coefficient is computed from the initial operating condition and is given to be Ps = 2 p. M. (Ph. D) Asst.BALAJI. (Ph. M.MANUAL CALCULATION: PREPARED BY V.Professor /EEE Page 47 .Tech. PREPARED BY V.BALAJI, M.Tech, (Ph.D) Asst.Professor /EEE Page 48 PREPARED BY V.BALAJI, M.Tech, (Ph.D) Asst.Professor /EEE Page 49 RESULT: PREPARED BY V.BALAJI, M.Tech, (Ph.D) Asst.Professor /EEE Page 50 loss of load. current and power flows are steady.jQe PREPARED BY V. The system may come back to a steady state condition maintaining synchronism or it may break into subsystems or one or more machines may pull out of synchronism. The generating units run a synchronous speed and system frequency. Small signal stability: When a power system is under steady state. M. (ii) Rotor oscillations of increasing magnitude due to lack of sufficient damping torque.. voltage. (i) Steady increase in rotor angle due to lack of synchronising torque. types of controllers etc. (Ph. In the former case the system is said to be stable and in the later case it is said to be unstable.f)) S* Stator Current It = = Et* Voltage behind transient condition E1 = Et + j Xd1It Et* Pe .3 THEORY : Stability : Stability problem is concerned with the behaviour of power system when it is subjected to disturbance and is classified into small signal stability problem if the disturbances are small and transient stability problem when the disturbances are large. Transient stability: When a power system is under steady state. FORMULA : Reactive power Qe = sin(cos-1(p.Professor /EEE Page 51 . the transmission system strength. loss of generation etc. the system may be subjected to small disturbances such as variation in load and generation.TRANSIENT AND SMALL SIGNAL STABILITY ANALYSIS – SINGLE MACHINE INFINITE BUS SYSTEM Expt. normal operating condition.No : Date : AIM : To become familiar with various aspects of the transient and small signal stability analysis of Single-Machine-Infinite Bus (SMIB) system PROGRAM REQUIRED : MATLAB 5. the load plus transmission loss equals to the generation in the system.BALAJI. When a large disturbance such as three phase fault. change in mechanical toque etc. Instability that may result from small disturbance may be of two forms.Tech.. The nature of system response to small disturbance depends on the operating conditions. occurs the power balance is upset and the generating units rotors experience either acceleration or deceleration.D) Asst. change in field voltage. ∠ EB Prefault Operation: X = j Xd1+ jXtr + X1 X2 X1 + X2 E1 x EB Power Pe = X δo = sin Pe * X -1 sinδo E1 * EB During Fault Condition: Pe = PEii = 0 Find out X from the equivalent circuit during fault condition Post fault Condition: Find out X from the equivalent circuit during post fault condition E1 x EB Power Pe = X δmax = π .Voltage of infinite bus EB = Et . X3 = X1 + X2 Angular separation between E1 and EB δo = ∠ E1 . (Ph.P2max sinδo PREPARED BY V.BALAJI.P2maxcosδo δ δ δ Cosδcr = P3max .D) Asst.δo Pm Pe = sinδmax Critical Clearing Angle: Pm(δmax .Professor /EEE Page 52 .Tech.δo ) + P3maxcosδmax .j( X3 + Xtr )It X1 X2 where. M. The generator is delivering real power Pe = 0. Determine the critical clearing angle and the critical fault clearing time. X1 = 0. X3) For b) Pm = 0. both lines are intact. EXERCISE : 1. E. 4. X1.Tech.δo) tcr = √ πfo Pm Sec PROCEDURE : 1. X1. M. X3) PREPARED BY V.8.Professor /EEE Page 53 . X3 = 0.17.17. a) A temporary three-phase fault occurs at the sending end of the line at point F. X3 = 0. Create a new M – file by selecting File . 2.D) Asst. V. (Ph. Type and save the program.8. E.65. Execute the program by either pressing Tools – Run 5. X2. V = 1.Critical Clearing Time: 2H (δcr .0.8 per unit and Q = 0. X2 = inf. A 60Hz synchronous generator having inertia constant H = 5 MJ/MVA and a direct axis transient reactance Xd1 = 0. E = 1. . Reactances are marked on the diagram on a common system base.65.When the fault is cleared. eacfault(Pm.New – M – File 3.3 per unit is connected to an infinite bus through a purely reactive circuit as shown in figure. X2 = 1. X2. PROGRAM : Pm = 0.074 per unit to the infinite bus at a voltage of V = 1 per unit. Enter the command window of the MATLAB. b) Verify the result using MATLAB program. E = 1. eacfault(Pm.8. V = 1. View the results. V.BALAJI. X1 = 0.65.0.8. MANUAL CALCULATION: PREPARED BY V.BALAJI.D) Asst.Professor /EEE Page 54 . (Ph.Tech. M. D) Asst.Tech. (Ph.Professor /EEE Page 55 .BALAJI.RESULT: PREPARED BY V. M. 2.Tech.BALAJI.No : Date AIM : To understand the fundamentals of economic dispatch and solve the problem using classical method with and without line losses.Professor /EEE Page 56 .D) Asst.NS Out of these NS set of generation schedules.ECONOMIC DISPATCH IN POWER SYSTEMS Expt. their production cost functions. ………………PGN(K) }. (Ph. To determine : The set of generation schedules. PGi . : PROGRAM REQUIRED : MATLAB 5. M. i = 1.2………N Which minimize the total production cost. PG2(k) . This economic dispatch problem is mathematically stated as an optimization problem. the system operator has to choose the set of schedules.……. which is essentially the sum of the production cost of all the generating units. FT = Σ Fi (PGi ) i=1 N (1) (2) and satisfies the power balance constraint φ= Σ PGi –PD = 0 i=1 N (3) and the operating limits PREPARED BY V. with negligible transmission loss and with N number of spinning thermal generating units the total system load PD at a particular interval can be met by different sets of generation schedules {PG1(k) .3 THEORY : Mathematical Model for Economic Dispatch of Thermal Units Without Transmission Loss: Statement of Economic Dispatch Problem In a power system. Min . Given : The number of available generating units N. which minimize the system operating cost.. their operating limits and the system load PD. k = 1. λ .max ____(9) (10) for ≤ λ for PGi = PGi. . M.max The units production cost function is usually approximated by quadratic function Fi (PGi) = ai PG2i + bi PGi + ci .N ∂L / ∂λ = 0 = Σ PGi – PD i=1 N (7) (8) The solution to ED problem can be obtained by solving simultaneously the necessary conditions (7) and (8) which state that the economic generation schedules not only satisfy the system power balance equation (8) but also demand that the incremental cost rates of all the units be equal be equal to λ which can be interpreted as “incremental cost of received power”.min ≤ PGi ≤ PGi. L as Min : L (PG1 …….Professor /EEE Page 57 ..λ [Σ PGi – PD] i=1 i=1 N N (6) The necessary conditions for the existence of solution to (6) are given by ∂L / ∂PGi = 0 = dFi (PGi) / dPGi . i=1.BALAJI.Tech.mi ≥λ for Economic Schedule PGi = (λ -bi)/ 2ai . By omitting the inequality constraints (4) tentatively.N PREPARED BY V. the reduce ED problem (1).…….D) Asst.PGN .. i = 1.min ≤ PGi ≤ PGi. (Ph.2.…….(2) and (3) may be restated as an unconstrained optimization problem by augmenting the objective function (1) with the constraint φ multiplied by LaGrange multiplier. . .N where ai . . λ to obtained the LaGrange function. bi and ci are constants Necessary conditions for the existence of solution to ED problem (4) (5) The ED problem given by the equations (1) to (4).PGi.2…………….max ` PGi = PGi. 2. When the inequality constraints(4) are included in the ED problem the necessary condition (7) gets modified as dFi (PGi) / dPGi = λ PGi. λ) = Σ Fi(PGi) . i = 1. 4. Execute the program by either pressing Tools – Run.009]. (Ph.Neglecting line losses and generator limits. 0. Enter the command window of the MATLAB. EXERCISE : 1.New – M – File 3. C2 = 400 + 5. M. while abs(DelP) >= 0.The fuel cost functions for three thermal plants in $/h are given by C1 = 500 + 5.Tech.006. PD is 800MW...006 P22 . 200]. 5. beta = [5.004 P12 .BALAJI. C3 = 200 +5.Incremental fuel cost λ= PD + Σ( bi/2ai ) / Σ (1/2ai) i=1 i=1 N N (11) PROCEDURE : 1. 5.001 PREPARED BY V. DelP = 10. Verify the result using MATLAB program.004. 400. Type and save the program.D) Asst.8 P3 + 0. fprintf(' ') disp(['Lamda P1 P2 P3 DP'. PD = 800.3. View the results. 5. gamma = [0.5.5 P2 + 0.8]. find the optimal dispatch and the total cost in $/h by analytical method. PROGRAM : alpha = [500. 2. ' grad Delamda']) iter = 0.Professor /EEE Page 58 . 0.009 P3 . lamda = input('Enter estimated value of Lamda = '). Create a new M – file by selecting File . 2 P1 in MW P2 in MW P3 in MW The total load .3 P1 + 0. (Ph.P(1). P = (lamda .DelP.Delamda]) lamda = lamda + Delamda.sum(P). Delamda = DelP/J.D) Asst.*P.Tech.P(3).*P + gamma.Professor /EEE Page 59 .iter = iter + 1.1). M. disp([lamda./(2*gamma).P(2). end totalcost = sum(alpha + beta.^2) MANUAL CALCULATION: PREPARED BY V. J = sum(ones(length(gamma)./(2*gamma)). DelP = PD .beta).BALAJI.J. PROGRAM : cost = [500 5. The fuel cost functions for three thermal plants in $/h are given by C1 = 500 + 5.D) Asst.009 P32 .3 P1 + 0. Generation limits: 200 ≤ P1 ≤ 450 MW 150 ≤ P2 ≤ 350 MW 100 ≤ P3 ≤ 225 MW Find the optimal dispatch and the total cost in $/h by analytical method. C3 = 200 +5. (Ph.8 P3 + 0. PD is 975MW.004 P12 . dispatch gencost P1 in MW P2 in MW P3 in MW MANUAL CALCULATION: PREPARED BY V.2.5 0.5 P2 + 0.009]. mwlimits = [200 450 150 350 100 225]. Pdt = 975. Verify the result using MATLAB program. M.Tech.004 400 5.BALAJI.006 200 5.006 P22 . The total load .3 0. C2 = 400 + 5.Professor /EEE Page 60 .8 0. PREPARED BY V. (Ph.D) Asst.BALAJI. M.Professor /EEE Page 61 .Tech. BALAJI.D) Asst.Professor /EEE Page 62 .PREPARED BY V. (Ph.Tech. M. Tech.RESULT : PREPARED BY V.BALAJI.D) Asst. M. (Ph.Professor /EEE Page 63 . Documents Similar To Ee 1404 Power System Lab ManualSkip carouselcarousel previouscarousel nextMatlab ProgramBus Admittance Impedance. Matrix Algorithm. 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