Power system model Olof Samuelsson EIEN15 Electric Power Systems L2 1 Outline • Previously: Models for lines, generator, power electronic converter, transformer • Single line diagram • Per unit • Bus admittance matrix • Thévenin equivalent • Bus impedance matrix EIEN15 Electric Power Systems L2 2 Single line diagram (Sw enlinjeschema) Three-phase circuit Generator Transformer Line Transformer Load All three phases equal = symmetry single phase represents all three: Single-phase circuit Don’t draw impedances. Nodes are busbars in substations: Single-Line Diagram or One-Line Diagram EIEN15 Electric Power Systems L1 3 Bus (bar) (Sw samlingsskena, skena) • Circuit diagram: Threephase node • One line diagram: bus bar • Reality – Aluminum bus bars on porcelain support EIEN15 Electric Power Systems L1 4 Single line diagram in PowerWorld • Specific to PowerWorld – Pie charts and animated arrows visualize line flows – “Dog bone” rotors in generators EIEN15 Electric Power Systems L1 5 Example 2.3 PW Many voltage levels • Transmission with EHV • Subtransmission with Medium Voltage • Distribution with MV • Distribution with LV Low Voltage • Transform loads across transformer? • Per unit normalization a better way – Normalizes all values – Eliminates ideal transformers – HOW? EIEN15 Electric Power Systems L2 6 Per unit base values and normalization • Theoretically – Any two of S, V, I and Z • Practically – Choose system MVA base and voltage base at one voltage level • Transformer turns ratio gives voltage base on other side Ex. Vbase1=11 kV at 130/10 kV transformerVbase2=(130/10)11 = 143 kV • In each voltage zone: Ibase=Sbase/(√3Vbase) and Zbase =Vbase2/Sbase • Sbase is used for S, P and Q, similarly Zbase is used for Z, R and X • Normalize each S, V, I, Z to corresponding base value EIEN15 Electric Power Systems L2 7 Example: Per unit on system base Xeq 10kV/130kV N1 N2 • Nameplate data: 50 MVA, 130kV/10kV, Xeq=0.2@10kV • Use rated MVA and kV values as base values • Determine p.u. value of X on both sides EIEN15 Electric Power Systems L2 8 Per unit transformer model • p.u. eliminates ideal transformers – One % value on name plate Vbase1 Vbase2 – Simple p.u. model only a Zeq! • System model with many voltage levels – Define system MVA base – Define voltage base at each voltage level (voltage zone) – Per unit removes ideal transformers – Only transformer impedances remain SGO Example 3.4 EIEN15 Electric Power Systems L2 9 Changing per unit base • Generator and transformer data often on its own (rated) Sbase • When changing base, actual value is invariant: Zp.u.new·Zbasenew=Zactual,= Zp.u.old·Zbaseold Vp.u.new·Vbasenew=Vactual,kv= Vp.u.old·Vbaseold Ip.u.new·Ibasenew=Iactual,kA= Ip.u.old·Ibaseold Sp.u.new·Sbasenew=Sactual,MVA= Sp.u.old·Sbaseold EIEN15 Electric Power Systems L2 10 Example – Per unit on component base Xd Xd Xd • Two generators in a power plant: Each unit: 20 MVA, 20 kV, Xd=1.1p.u.@ component base • Combine the two generators to one large: • Determine Xd on the new component base EIEN15 Electric Power Systems L2 11 Example: Single line diagram • Draw single line diagram for circuit with this per phase circuit diagram – use busbars, lines, loads and generators EIEN15 Electric Power Systems L1 12 Setting up a network model • Collect all impedances in one model: • Assume voltages and impedances known • Kirchhoff’s current law KCL I1 I10 I12 I13 I 2 I 20 I 21 I 24 I 3 I 30 I 31 I 34 I 4 I 40 I 42 I 43 • Ohm’s law with admittance I j 0 y j 0V j I ij yij (Vi V j ) • Combine the above equations to matrix notation I=YV EIEN15 Electric Power Systems L1 13 Bus admittance matrix Ybus • Admittance Y=1/Z • Ohm’s law for networks • Matrix equation I= YbusVbus – Nodal current balances • Current vector I – Injection from sources (generators) – Negative injection from loads not included in Ybus EIEN15 Electric Power Systems L1 14 Setting up Ybus Element ii by inspection Sum of all admittances connected to bus i Element ij by inspection –(admittance connecting buses i and j) Element ij from measurements Voltage source at node j Voltage sources at nodes≠j set to zero Current into bus i is Ybus,ijVj EIEN15 Electric Power Systems L1 15 Ybus properties • With reference and no shunts – Row and column sums zero – Ybus not invertible • Reference bus removed – Dimensions N-1 x N-1 • Sparse (many elements are zero) and symmetric • A compact network model – Contains both series and shunt impedances EIEN15 Electric Power Systems L1 16 SGO Problem 2.38 p 85 • Set up the bus admittance matrix for this system • First replace generator Thévenin equivalents with Norton equivalents (current source + admittance in parallel) EIEN15 Electric Power Systems L1 17 Simplify network model • If only buses with current injection are of interest, a new Ybus with only those buses can be computed • Reorder and partition vectors, rearrange matrix – Current vector in nonzero and zero elements – Voltage vector in buses to keep and to skip I Y12 Y12 Vkeep V 0 Y Y 21 22 skip I Y12Vkeep Y12Vskip 0 Y21Vkeep Y22Vskip I Y12Vkeep Y12Vskip 1 Vskip Y22 Y21Vkeep 1 1 I Y12Vkeep Y12Y22 Y21Vkeep Y12 Y12Y22 Y21 Vkeep Reduced Ybus • Typically only generator buses are kept EIEN15 Electric Power Systems L2 18 Example Matrix reduction • y10= y20= y30= y40=1 • y12= y13= y24= y34=2 • Bus 3 has no current injection • Eliminate it and find new Ybus 0 V1 I1 5 2 2 I 2 V 5 0 2 2 2 I 3 2 0 5 2 V3 I 0 2 2 5 V4 4 EIEN15 Electric Power Systems L2 19 Bus impedance matrix Zbus • Vbus = Zbus I • If Ybus is invertible: Zbus = Ybus-1 • Zbus by inspection difficult • Eliminating a bus very easy – Just remove corresponding row and column – Previous example EIEN15 Electric Power Systems L2 20 Setting up Zbus Element ij of Zbus from measurement – 1 p.u. current source at node j – Current sources at nodes≠j to zero – Voltage at bus i is Zbus,ij EIEN15 Electric Power Systems L2 21 (Per phase) Thévenin equivalent ZTH Represents passive network Also for entire power system VTH ~ VTH no-load voltage (Sw tomgångsspänning) ZTH short-circuit impedance (Sw kortslutningsimpedans) » Equivalent Z of network » What is measured at terminals with all V sources set to zero EIEN15 Electric Power Systems L4 22 Short-circuit current – Z=0 connected at terminals ZTH VTH ~ – Short-circuit current (Sw kortslutningsström) ISC » ISC=VTH/(√3ZTH)≈1/(√3ZTH) p.u. (VTH line-line voltage) » Determines circuit breaker rating (Sw märkström för effektbrytare) – ISC limited by ZTH » X still gives small Vdrop » Important advantage of AC » Extra X may be inserted EIEN15 Electric Power Systems L4 23 Short-circuit power ZTH – Short-circuit power in MVA (Sw kortslutningseffekt) VTH ~ ISC » Also ”short-circuit capacity” » Also ”fault level” – SSC=√3VTHISC p.u. – SSC=VTH2/ZTH ≈1/ZTH p.u. – SSC not useful power – Note: Voltage before short-circuit times current during short-circuit EIEN15 Electric Power Systems L4 24 Network strength • ZLOAD >> ZTH SLOAD << SSC ZTH – small voltage drop across ZTH – load voltage insensitive of load VTH ~ SLOAD – strong, urban load • ZLOAD not >> ZTH SLOAD not << SSC/2 – load voltage sensitive to load – weak, rural load • SLOAD > SSC/2 – impossible EIEN15 Electric Power Systems L4 25 Weak network ZTH ~ VTH • Nearby motor load + V M – – Starting current = peak in Im – Dip in feeding voltage Im – Voltage recovers • Not everywhere V – Common in rural (weak) networks – Uncommon in urban (strong) networks Start t EIEN15 Electric Power Systems L4 26 Example: ZTH at different voltage levels All transformer x to same base by multiplying by 100 MVA/Sbase,old • 400/130 kV, x=0.1 p.u. @ 750 MVA – 0.013 p.u. @ 100 MVA base • 130/20 kV, x=0.1 p.u. @ 40 MVA – 0.25 p.u. @ 100 MVA base – 18.75 x 0.013 p.u. • 20/0.4 kV, x=0.1 p.u. @ 0.8 MVA – 12.5 p.u. @ 100 MVA base – 50 x 0.25 p.u. and 937.5 x 0.013 p.u. • The last transformer dominates ZTH j0.013 ZTH400 VTH ~ 400 kV j0.25 130 kV j12.5 20 kV 0.4 kV Example: ZTH and SSC in network 3a. At 3 kV busbar a) » X3kV=(X300MVA+X6MVA)//X2MVA 3b. After transformers 3 kV » X3kV +(X0,5MVA//X0,6MVA) Increase XSC, decrease SSC » Go through transformer/line b) Decrease XSC, increase SSC » Add generator » Add parallel transformer/line EIEN15 Electric Power Systems L4 28 ZTH is easy to get from Zbus Element ii of Zbus – Short-circuit impedance ZTH at bus i • Conditions – Zbus has neutral as reference – Generators have internal impedance – Loads can be included in Zbus • Practical for large systems EIEN15 Electric Power Systems L4 29 Summary • Single line diagram for overview • Per unit eliminates ideal transformers • We can build a network model! – Bus admittance matrix the starting point – Per unit is the key – Many voltage levels not a problem • Thévenin equivalent of system “behind” a bus EIEN15 Electric Power Systems L2 30
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