Power system model. Olof Samuelsson. EIEN15 Electric Power Systems L2

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1 Power system model Olof Samuelsson EIEN15 Electric Power Systems L2 1

2 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

3 Single line diagram (Sw enlinjeschema) Three-phase circuit Generator Transformer Line Transformer Load All three phases equal = symmetry single phase represents all three: Don t draw impedances. Nodes are busbars in substations: Single-phase circuit Single-Line Diagram or One-Line Diagram EIEN15 Electric Power Systems L1 3

4 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

5 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

6 Many voltage levels Transmission with EHV Subtransmission with Distribution with MV Distribution with LV Medium Voltage 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

7 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. V base1 =11 kv at 130/10 kv transformerv base2 =(130/10)11 = 143 kv In each voltage zone: I base =S base /( 3V base ) and Z base =V base2 /S base S base is used for S, P and Q, similarly Z base is used for Z, R and X Normalize each S, V, I, Z to corresponding base value EIEN15 Electric Power Systems L2 7

8 Example: Per unit on system base X eq 10kV/130kV N 1 N 2 Nameplate data: 50 MVA, 130kV/10kV, X eq =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

9 Per unit transformer model p.u. eliminates ideal transformers One % value on name plate Simple p.u. model only a Z eq! V base1 V base2 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

10 Changing per unit base Generator and transformer data often on its own (rated) S base When changing base, actual value is invariant: Z p.u.new Z basenew =Z actual, = Z p.u.old Z baseold V p.u.new V basenew =V actual,kv = V p.u.old V baseold I p.u.new I basenew =I actual,ka = I p.u.old I baseold S p.u.new S basenew =S actual,mva = S p.u.old S baseold EIEN15 Electric Power Systems L2 10

11 Example Per unit on component base X d X d X d Two generators in a power plant: Each unit: 20 MVA, 20 kv, X d =1.1p.u.@ component base Combine the two generators to one large: Determine X d on the new component base EIEN15 Electric Power Systems L2 11

12 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

13 Setting up a network model Collect all impedances in one model: Assume voltages and impedances known Kirchhoff s current law KCL I I I I I I I I I I I I I4 I40 I42 I43 Ohm s law with admittance I I j0 ij y y ij j0 V ( V i j V j ) Combine the above equations to matrix notation I=YV EIEN15 Electric Power Systems L1 13

14 Bus admittance matrix Y bus Admittance Y=1/Z Ohm s law for networks Matrix equation I= Y bus V bus Nodal current balances Current vector I Injection from sources (generators) Negative injection from loads not included in Y bus EIEN15 Electric Power Systems L1 14

15 Setting up Y bus 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 Y bus,ij V j EIEN15 Electric Power Systems L1 15

16 Y bus properties With reference and no shunts Row and column sums zero Y bus 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

17 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

18 Simplify network model If only buses with current injection are of interest, a new Y bus 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 I Y12Vkeep Y12Vskip I Y12V 0 Y21 Y22 Vskip 0 Y21V keep Y22V skip V Y I Y 12 V keep Y 12 Y 1 22 Y 21 V keep Typically only generator buses are kept 1 12 Y Y ReducedY Y Y bus 21 V keep skip keep 1 22 Y Y V V skip keep EIEN15 Electric Power Systems L2 18

19 Example Matrix reduction y 10 = y 20 = y 30 = y 40 =1 y 12 = y 13 = y 24 = y 34 =2 Bus 3 has no current injection Eliminate it and find new Ybus 19 EIEN15 Electric Power Systems L V V V V I I I I

20 Bus impedance matrix Z bus V bus = Z bus I If Y bus is invertible: Z bus = Y -1 bus Z bus by inspection difficult Eliminating a bus very easy Just remove corresponding row and column Previous example EIEN15 Electric Power Systems L2 20

21 Setting up Z bus Element ij of Z bus from measurement 1 p.u. current source at node j Current sources at nodes j to zero Voltage at bus i is Z bus,ij EIEN15 Electric Power Systems L2 21

22 (Per phase) Thévenin equivalent V TH ~ Z TH Represents passive network Also for entire power system V TH no-load voltage (Sw tomgångsspänning) Z TH 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

23 Short-circuit current Z=0 connected at terminals Z TH Short-circuit current (Sw kortslutningsström)»i SC =V TH /( 3Z TH ) 1/( 3Z TH ) p.u. (V TH line-line voltage) V TH ~ I SC» Determines circuit breaker rating (Sw märkström för effektbrytare) I SC limited by Z TH» X still gives small V drop» Important advantage of AC» Extra X may be inserted EIEN15 Electric Power Systems L4 23

24 Short-circuit power Z TH Short-circuit power in MVA (Sw kortslutningseffekt) V TH ~ I SC» Also short-circuit capacity» Also fault level S SC = 3V TH I SC p.u. S SC =V TH2 /Z TH 1/Z TH p.u. S SC not useful power Note: Voltage before short-circuit times current during short-circuit EIEN15 Electric Power Systems L4 24

25 Network strength Z TH Z LOAD >> Z TH S LOAD << S SC V TH ~ S LOAD small voltage drop across Z TH load voltage insensitive of load strong, urban load Z LOAD not >> Z TH S LOAD not << S SC /2 load voltage sensitive to load weak, rural load S LOAD > S SC /2 impossible EIEN15 Electric Power Systems L4 25

26 Weak network Z TH Nearby motor load ~ V TH + V M I m Starting current = peak in I m Dip in feeding voltage Voltage recovers Not everywhere V Common in rural (weak) networks Uncommon in urban (strong) networks Start t EIEN15 Electric Power Systems L4 26

27 Example: Z TH at different voltage levels All transformer x to same base by multiplying by 100 MVA/S base,old 400/130 kv, x= MVA MVA base 130/20 kv, x= MVA MVA base x p.u. 20/0.4 kv, x= MVA MVA base 50 x 0.25 p.u. and x p.u. The last transformer dominates Z TH Z TH400 j0.013 j0.25 j12.5 V TH ~ 400 kv 130 kv 20 kv 0.4 kv

28 Example: Z TH and S SC in network 3a. At 3 kv busbar 3 kv b) a)»x 3kV =(X 300MVA +X 6MVA )//X 2MVA 3b. After transformers»x 3kV +(X 0,5MVA //X 0,6MVA ) Increase X SC, decrease S SC» Go through transformer/line Decrease X SC, increase S SC» Add generator» Add parallel transformer/line EIEN15 Electric Power Systems L4 28

29 Z TH is easy to get from Z bus Element ii of Z bus Short-circuit impedance Z TH at bus i Conditions Z bus has neutral as reference Generators have internal impedance Loads can be included in Z bus Practical for large systems EIEN15 Electric Power Systems L4 29

30 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

Power system model. Olof Samuelsson. EIEN15 Electric Power Systems L2 1

Power system model. Olof Samuelsson. EIEN15 Electric Power Systems L2 1 Power system model Olof Samuelsson 1 Outline Previously: Models for lines, generator, power electronic converter, transformer Single line diagram Per unit Bus admittance matrix Bus impedance matrix Thévenin