Calculating The Short Circuit Current And Input Impedance Of Each Node .

Lecture 23 Power Engineering - Egill Benedikt Hreinsson Calculating the short circuit current and input impedance of each node in the power system (by matrix methods) 1

Power Engineering - Egill Benedikt Hreinsson Lecture 23 System Equations Summary The power system is defined by the following matrix equations: The conventional form of the load flow equations I bus Ybus Vbus Vbus Z bus I bus Z bus Ybus 1 2

Power Engineering - Egill Benedikt Hreinsson Lecture 23 System Equations Summary (2) Vbus is the vector of complex voltage phasors on each bus. The elements of Vbus are complex numbers V1 V 2 Vbus V3 # Vn 3

Power Engineering - Egill Benedikt Hreinsson Lecture 23 4 System Equations Summary (2) Ibus is the vector of injected currents into each bus. The elements of Ibus are also complex numbers In normal operation, each injection can be – from a generator ( ) – or into a load (-) In abnormal circumstances and operation an injection can feed into a short circuit I bus I1 I 2 I3 # I n

Power Engineering - Egill Benedikt Hreinsson Lecture 23 A Power System Model for a Short Circuit Study Let us review a previous equation describing the system. There is 1 “injection” of current at bus #3 . This is now the fault current but it could as well be a current from a generator The resulting bus voltage changes are on the left hand side Multiply the 3rd row by the column vector and we get. The z’33 element in the modified Zbus matrix is the input impedance at node #3 V1 0 V z '11 z '12 " z '1n 0 2 z ' z z ' ' " 22 1n V3 21 I3 # % # # # # z ' z 'n 2 " z 'nn Vn n1 0 V3 z '31 0 z '32 0 z '33 I 3 z '34 0 . z '3n 0 V3 z '33 I 3 V3 V3 I3 z '33 zin ,3 5

Power Engineering - Egill Benedikt Hreinsson Lecture 23 6 We get 2 equivalent circuits: I3,fault ΔV1 V3,0 Zf By the previous matrix calculations the power system has been reduced to a single Thevenin equivalent circuit ΔV2 V3,0 z’33 zin,3 The power system I3,fault Zf

Power Engineering - Egill Benedikt Hreinsson Lecture 23 7 Calculation of fault currents The general equation for calculating the fault current from the previous simple circuit is therefore: The most important element is the diagonal element of the modified impedance matrix, z 33 If we have a solid fault zf 0 and the fault current takes its maximum or: – V3,0 is the pre-fault voltage at node 3 – I3,fault is the short circuit current – z 33 is the input impedance at node 3 V3,0 I3,fault z’33 Zf I 3, fault V3,0 z f z '33 I 3, fault V3,0 z '33

Lecture 23 Power Engineering - Egill Benedikt Hreinsson A Short Circuit Calculation Algorithm (A review) 1. Define the power system and its operational conditions 2. Calculate the pre-fault power flow and calculate prefault voltages and currents 3. Calculate the internal impedance (input impedance) and pre-fault voltage (Thevenin equivalent circuit) 4. Calculate the fault currents (i.e. system wide changes of currents) 5. Calculate the post fault currents as the sum of the prefault currents and the fault currents (by the superposition principle) 8

Lecture 23 Power Engineering - Egill Benedikt Hreinsson 9 Fault Current Calculations (A review) Pre-fault currents I0 (load currents) are usually in phase or near in phase with the voltage The fault currents If are primarily inductive, i.e. out of phase with the voltage. Short circuit currents load currents Power system components are reactive Load components are resistive Pre-fault load currents can often be ignored V I0 If Total current I0 If

Lecture 23 Power Engineering - Egill Benedikt Hreinsson Zbus building algorithm Building the matrix “from scratch” in many steps adding one element at a time 10

Lecture 23 Power Engineering - Egill Benedikt Hreinsson Building the Zbus matrix directly It is often difficult to invert the Ybus matrix by traditional matrix inversion methods The Ybus matrix is symmetric and usually large and sparse The Zbus matrix is symmetric and full (and usually large) We often need to change the configuration of the underlying power system, (for instance in contingencies where a single link is added/deleted)) – We need to remove a line or add a transmission line or a link A direct step-by-step matrix building method is advantageous. 11

Power Engineering - Egill Benedikt Hreinsson Lecture 23 12 Zbus building algorithm The Zbus matrix can be built step by step be a direct 1 method rather than calculating Ybus-1 Consider a general n-bus power system We add one link at a time In each step we transform the matrix as follows Z bus,old Z bus,new 2 k 4 3 n J The earth is a special node “reference node”

Power Engineering - Egill Benedikt Hreinsson Lecture 23 13 Zbus building algorithm As an example take a 3-bus system with 5 impedances (links) We build the full matrix in 5 steps and add 1 impedance at a time The Zbus matrix is defined as follows 1 3 z1 z3 z2 1 2 z4 2 3 z5 v1 z11i1 z12i2 z13i3 Vbus Z bus I bus v2 z21i1 z22i2 z23i3 Z bus Ybus v3 z31i1 z32i2 z33i3 -1 J

Power Engineering - Egill Benedikt Hreinsson Lecture 23 14 5 steps in Zbus building algorithm 1st step Zbus is going to be a 3x3 matrix built in 5 steps: 3rd step 2 3 2 1 4th step 2 3 2nd step 3 J 3x3 J J 1x1 2x2 5th step 2 J J 3x3 3 1 1 1 2 3x3 3

Lecture 23 Power Engineering - Egill Benedikt Hreinsson 4 different types of steps 1. Add a new node and from it a new branch to ground 2. Add a new node and from it a new branch to an existing node 3. Add a new branch between 2 existing nodes 4. Add a new branch from an existing node to ground 15

Power Engineering - Egill Benedikt Hreinsson Lecture 23 16 Numerical network example in building Zbus Consider also the following z3 numerical example 2 Assume we have 5 impedances and 3 buses 1 Again, we build the matrix in 5 steps The branches in the numerical example have z1 impedance values: J 2 z4 z3 3 z5 z5 z2 z4 3 1 z1 z2 z1 j 0.15 z2 j 0.075 z3 j 0.1 z4 j 0.1 z5 j 0.1

Power Engineering - Egill Benedikt Hreinsson Lecture 23 The 1st step in the Zbus building algorithm We add branch and inject a current at any node #1 and get a voltage there v z i 1 11 1st step i1 v1 z11i1 z12i2 z13i3 v2 z21i1 z22i2 z23i3 v3 z31i1 z32i2 z33i3 By comparing the above equations we see that the Zbus matrix after this step is a 1x1 (single element): v2 v1 v3 z1 J 1x1 Z bus [ z1 ] 17

Power Engineering - Egill Benedikt Hreinsson Lecture 23 The 1st step in the Zbus building algorithm In our numerical example: z1 j 0.15 , z2 j 0.075 , z3 j 0.1 , z4 j 0.1 and z5 j 0.1 Therefore Z bus [ z1 ] [ j 0.15] 1st step i1 v2 v1 z1 J 1x1 v3 18

Power Engineering - Egill Benedikt Hreinsson Lecture 23 19 The 2nd step in the Zbus building algorithm We now add a branch to the reference node from a node (#2) not previously connected v1 z1i1 0 i2 v2 0 i1 z2 i2 2nd step i1 v1 Vbus Z bus(old) [ z1 ] J 0 i1 z2 i2 Z bus(new) v3 z2 z1 The Zbus will now be a 2x2: v1 z1 v2 0 i2 v2 2x2 Z bus(old) 0 0 z2

Power Engineering - Egill Benedikt Hreinsson Lecture 23 20 The 2nd step in the Zbus building algorithm (2) Assume Zbus,old is a matrix of dimension (k x k) We want to add a node (# k 1) that has not previously been added to our set of nodes. From this node we add the new impedance, znew to the reference node We transform Zbus,old to Zbus,new by adding a row and a column to the Zbus,old (# k 1) Add zeros to all elements of the new row and column except the new diagonal element which is the new impedance, znew The dimensions of the matrix are increased by 1 and are now (k 1) times (k 1) Dimension of Zbus(old) k Z bus,new row # k 1 column # k 1 0 # Z bus ,old 0 0 " 0 znew

Lecture 23 The 2nd step in the Zbus building algorithm Power Engineering - Egill Benedikt Hreinsson In our numerical example we get in step #2: 2nd step v1 i1 v1 z1 0 i1 Vbus v2 0 z2 i2 Z bus(old) [ z1 ] [ j 0.15] Z bus(new) Z bus(old) 0 0 j 0.15 z2 0 i2 v2 z2 z1 J j 0.075 0 v3 2x2 21

Power Engineering - Egill Benedikt Hreinsson Lecture 23 22 The 3rd step in the Zbus building algorithm We now add a branch to a previously connected node (#2) from a new node (#3) , not previously connected We have the following equations: v1 z1 v 0 2 v3 v2 z4i3 v1 z1 v 0 2 v3 0 0 z2 z2 3rd step i1 0 i1 z2 i2 i3 i2 v1 v2 z4 z2 z1 J 0 i1 z2 i2 z2 z4 i3 Zbus,new v3 Z bus,old z12 z22 2x2 z12 z22 z2 z4 i3

Lecture 23 The 3rd step in the Zbus building algorithm (2) Power Engineering - Egill Benedikt Hreinsson Assume Zbus,old is a matrix of dimension (k x k) We want to add a node (# k 1) that has not previously been added to our set of nodes. From this node we add the new impedance, znew to a previously connected node (# j) Dimension of Zbus(old) k We transform Zbus,old to Zbus,new by adding a row (number k 1) and a column (number k 1) to the Zbus,old Duplicate row/column #j in Zbus,old as the new row/column. The new diagonal element is Z bus,old found by adding the new impedance znew to Z bus,new the diagonal element zjj of Zbus,old The dimensions of the matrix are increased by z j1 z j 2 " z jk row # k 1 1 23 column # k 1 z2 j # zkj z jj znew z1 j

Power Engineering - Egill Benedikt Hreinsson Lecture 23 24 The 3rd step in the Zbus building algorithm In our numerical example: z1 j 0.15 , z2 j 0.075 , z3 j 0.1 , z4 j 0.1 and z5 j 0.1 0 i1 v1 z1 0 Therefore: v2 0 z2 z2 i2 v3 0 z2 z2 z4 i3 0 0 i1 j 0.15 0 j 0.075 j 0.075 i2 j 0.075 j 0.175 i3 0 3rd step i1 i2 v1 v2 v3 z4 z2 z1 J 2x2 i3

Power Engineering - Egill Benedikt Hreinsson Lecture 23 The 3rd step in the Zbus building algorithm Therefore, in our numerical example, after the 3rd step: Z bus j 0.15 0 0 0 j 0.075 j 0.075 0 j 0.075 j 0.175 25

Lecture 23 The 4th step in the Zbus building algorithm (1) Power Engineering - Egill Benedikt Hreinsson We now add a branch between 2 4th step previously connected nodes (#1) and (#2) This new impedance, z3 znew in general causes 1 a circular current, iL to flow. The circular current is equivalent to an extra injected current of iL at node #1 and of of -iL at node #2 For the 3 x 3 matrix in the figure, this is equivalent to the following adjustments of the voltage/current equations i v1 z11 v z 2 21 v3 z31 z12 z22 z32 z13 i1 iL z23 i2 iL z33 i3 i2 znew v2 v 3 v1 i z 3 4 i L z2 z1 J 2x2 26

Lecture 23 The 4th step in the Zbus building algorithm (2) Power Engineering - Egill Benedikt Hreinsson In general, assume Zbus,old is a matrix of dimension (k x k) Assume that a new element, znew is added between node #m and node #n in a system that already has k nodes. (In the example, m 1 and n 2) v1 z1,1 The equations v z2,1 2 assuming a # # circular current of vm zm ,1 magnitude iL v z n ,1 n are as follows: # # v z k 1 k 1,1 vk zk ,1 z1,2 " " We can expand the equations as shown on the next slide " z2,2 z1, k 1 z2, k 1 # % zm , k 1 zn , k 1 # zk 1,2 zk ,2 " " zk 1, k 1 " zk , k 1 zm ,2 zn ,2 Z bus,old % z1, k i1 z2, k i2 # # zm , k im iL zn , k in iL # # zk 1, k ik 1 zk , k ik 27

Lecture 23 The 4th step in the Zbus building algorithm (3) We rewrite the former equation We define a new vector, a which represents the difference between columns # m and #n in the Zbus,old matrix Therefore the former equation will be as follows: The voltage drop across the new branch, znew will be: Power Engineering - Egill Benedikt Hreinsson v1 z1,1 v z 2 2,1 # # vk zk ,1 z1,2 " z1,k i1 z1m z1n z2,2 z2,k i2 # iL % # # # zk ,2 " zk ,k ik zkm zkn Vbus Z bus,old I bus a iL vn vm znew iL vm vn znew iL 0 z1m z1n # a # z z km kn 28