Power system model. Olof Samuelsson. EIEN15 Electric Power Systems L2 1
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1 Power system model Olof Samuelsson 1
2 Outline Previously: Models for lines, generator, power electronic converter, transformer Single line diagram Per unit Bus admittance matrix Bus impedance matrix Thévenin equivalent 2
3 Consumption is traditionally called load In a national system model one load may represent a city! V 1. No voltage dependence - constant P and Q True for motors and power electronics 2. Voltage dependent - constant Z True for heating 3. Load with recovery time constant Fast for motors, slow for thermostats t P t EIEN15 Electric Power Systems L1 3
4 Transforming Z across transformer Impedances are transformed as the turns ratio squared ' 2 ' 2 ' 2 Z N N I N N U N N I U I U Z = = = = EIEN15 Electric Power Systems L1 4
5 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 5
6 Bus (bar) (Sw samlingsskena, skena) Reality: bus bars Aluminum pipes on porcelain support Circuit diagram: Threephase node One line diagram: bus bar 6
7 Single line diagram in PowerWorld Specific to PowerWorld Pie charts and animated arrows visualize line flows Dog bone rotors in generators 7 Example 2.3 PW
8 Example: Single line diagram Draw single line diagram for circuit with this per phase circuit diagram use busbars, lines, loads and generators 8
9 Many voltage levels Transmission with EHV Subtransmission with Distribution with MV Distribution with LV Medium Voltage Transform loads across transformer? Per unit normalization a better way Normalizes all values Eliminates ideal transformers HOW? 9
10 Per unit base values for normalization 1. Choose system MVA base and voltage base at one voltage level 2. Transformer turns ratio gives voltage base on other side Ex. V base1 =11 kv at 130/10 kv transformer V base2 =(130/10)11 = 143 kv 3. In each voltage zone: I base =S base /( 3V base ) and Z base =V base2 /S base 4. Normalize each S, V, I, Z to corresponding base value S base is used for S, P and Q, similarly Z base is used for Z, R and X 10
11 Example: Per unit on system base X eq 10kV/130kV N 1 N 2 Example: Kraftringen Östra Mottagningsstationen Lund Nameplate data: 50 MVA, 130kV/10kV, X eq Use rated MVA and kv values as base values Determine p.u. value of X eq on both sides 11
12 Per unit transformer model p.u. eliminates ideal transformers One % value on name plate Simple p.u. model only a Z eq! S base V base1 I base1 I pu1 S base V base2 I base2 I pu2 =I pu1 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
13 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,ω = 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 13
14 Example Per unit on component base 20 MVA, 20 kv X d =1.1p.u.@20 MVA X d X d 20 MVA, 20 kv X d =1.1p.u.@20 MVA X d? MVA, 20 kv X d =?? p.u.@? MVA 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 14
15 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 15
16 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 generators (>0) and loads (<0) not in Y bus Reference bus (often neutral/ground) removed Gives N-1 x N-1 matrix that is invertible 16
17 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 17
18 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) 18
19 Simplify network model Remove bus(es) with no current injection are of interest Reorder and partition vectors, rearrange matrix Current vector in nonzero and zero elements Voltage vector in buses to keep and to skip I Y Y V I = Y keep = 0 Y Y V skip 11V keep Y12Vskip I = Y11V 0 = Y21V keep + Y22Vskip Vskip = Y I = Y 11 V keep Y 12 Y 1 22 Y 21 V keep 1 ( Y11 Y12Y22 Y21) V keep = Reduced Y + Y Removing bus 3 of four in Matlab: keep=[1 2 4], skip=3 and Y12=Y(keep,skip) etc bus keep 1 22 Y V V keep skip 19
20 Bus impedance matrix Z bus V bus = Z bus I, where Z bus = Y -1 bus if Y bus invertible Z bus by inspection difficult 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 Eliminating a bus very easy Just remove corresponding row and column 20
21 Element ii of Z bus is a Thévenin impedance Diagonal element ii of Z bus Thévenin 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 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 22
23 Z TH gives 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» In Z TH X>>R for line, transf, synch gen» X limits current» If needed, extra X may be inserted» R gives losses and V drop from active power 23
24 Z TH gives 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., Voltage before short-circuit times current during short-circuit S SC =V TH2 /Z TH 1/Z TH p.u. S SC not useful power 24
25 Z TH gives network strength Z TH Z LOAD >> Z TH S LOAD << S SC small voltage drop across Z TH load voltage insensitive of load V TH ~ S 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 Max S impedance matching 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 V Start t 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 27
28 Summary Large systems are often visualized using Per unit on common base ideal transformer Per unit on component base makes parameters Many voltage levels can be managed with per unit on base A matrix can represent an entire network with components such as Seen from one bus a can represent an entire system It is easy to set up the matrix by inspection, while its inverse the holds Thévenin impedances (where?) 28
29
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