Three-phase analysis of unbalanced systems

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1 Exercises of course ELEC Electric power systems analysis and operation Three-phase analysis of unbalanced systems Thierry Van Cutsem t.vancutsem@ulg.ac.be March 205 / 26

2 Objectives Three-phase analysis of unbalanced systems It is common practice to have the lines, cables, transformers and generators modelled by their positive, negative and zero sequence equivalent circuits (the latter are assumed decoupled) for each of those components, we consider how to pass from the symmetric-component model (+,, o) to the three-phase model (a, b, c) for other components, we derive the (a, b, c) model directly all these individual models are assembled to build the three-phase bus admittance matrix Y and the vector of three-phase injected currents Ī of the linear model: Ī = Y V which is solved to obtain the vector V of three-phase bus voltages for each component, a MATLAB script is available and its use is described. 2 / 26

3 Remarks Three-phase analysis of unbalanced systems All voltages refer to their local grounds as in the Theory, the neutrals are eliminated thus, for an N-bus system, Y is a 3N 3N matrix Y and Ī are built for the balanced system; imbalances resulting from faults, loads, lines, etc. will be added by modifying Y and Ī some models require values from an initial power flow computation of the balanced system the per unit system uses the single-phase base power (see Chapter 5) however, the base impedances are the same as with a three-phase base power: three-phase base power S B3 = 3V B I B3 = 3U B I B3 single-phase base power S B = V B I B I B3 = S B3 = S B3 3V B 3UB Z B3 = V B I B3 = 3V 2 B S B3 = U2 B S B3 I B = S B V B Z B = V B I B = V 2 B S B 3 / 26

4 Load Three-phase analysis of unbalanced systems Load Star-connected load Input data: complex power consumed in one phase: P + jq (in pu) magnitude of the phase-to-neutral voltage under which this power is consumed: V (in pu) impedance between neutral and ground: z n (in pu) y = P jq V 2 y n = z n 4 / 26

5 Load Ī a = y( V a V n ) () Ī b = y( V b V n ) (2) Ī c = y( V c V n ) (3) Ī a + Īb + Īc = y n Vn (4) Adding (, 2 and 3) and introducing the result in (4) yields: V n = y V a + y V b + y V c y tot with y tot = 3y + y n (5) Introducing (5) into (,2,3) and arranging the results in matrix form: Ī ā y y 2 y tot y 2 y tot y 2 y tot V a I = y 2 y tot y y 2 y tot y 2 y tot V b bīc y 2 y tot y2 y tot y y 2 V c y tot } {{ } contribution to matrix Y Question. Determine the admittance matrix when z n = 0 ( solid grounding)? 5 / 26

6 Load Delta-connected load Input data: complex power consumed in each branch: P + jq (in pu) magnitude of the phase-to-phase voltage under which those powers are consumed: U (in pu) y = P jq U 2 6 / 26

7 Load Ī a = y( V a V b ) + y( V a V c ) Ī b = y( V b V a ) + y( V b V c ) Ī c = y( V c V a ) + y( V c V b ) which can be rewritten in matrix form as: Ī ā 2y y y I = y 2y y bīc y y 2y } {{ } contribution to matrix Y V a V b V c Remarks this admittance matrix is singular because the voltage reference node is not located in the circuit ( indefinite admittance matrix ) this is not a problem since it will be combined with other admittance matrices to form Y. 7 / 26

8 Load Matlab script Yload.m 8 / 26

9 Balanced Norton equivalent Balanced Norton equivalent Input data: complex power produced by one of the three phases: P + jq (in pu) Thévenin impedance: z th (in pu) complex (phase-to-neutral) voltage of phase a: Va (in pu) Ē a = V a + z th P jq V a Ē b = a 2 Ē a Ē c = aēa 9 / 26

10 Balanced Norton equivalent Ī a = ( V a Ē a )/z th Ī b = ( V b Ē b )/z th Ī c = ( V c Ē c )/z th which can be rewritten in matrix form as: Ī ā /z th 0 0 I = 0 /z th 0 bīc 0 0 /z th } {{ } contribution to matrix Y V a V b V c Ē a /z th Ē b /z th } Ē c /z th {{ } contrib. to vector Ī 0 / 26

11 Balanced Norton equivalent Matlab script Norton equ.m / 26

12 Generator Three-phase analysis of unbalanced systems Generator Input data: positive, negative and zero-sequence impedances: z +, z and z o (in pu) impedance between neutral and ground: z n (in pu) complex power produced by one of the three phases: P + jq (in pu) complex (phase-to-neutral) voltage of phase a: V a (in pu) 2 / 26

13 Generator To identify Īnor,a, Īnor,b and Īnor,c we assume initial balanced operating conditions: Hence: V = V o = 0 V+ = V a Ī = Īo = 0 Ī + = Īa = P jq V a Ē + = V + z + Ī + = V a + z + P jq V a Contribution to vector Ī : Ī nor,a Ī nor,b Ī nor,c = Ē + /z + a 2 Ē + /z + a Ē + /z + Contribution to admittance matrix Y : with: YF = YT = TYF T /z /z /(z o + 3z n ) and T = a 2 a a a 2 3 / 26

14 Generator Matlab script Norton gen.m 4 / 26

15 Line or cable Three-phase analysis of unbalanced systems Line or cable Full three-phase symmetry is assumed. Input data: positive-sequence series impedance: z + = r + + jx + (in pu) zero-sequence series s: z o = r o + jx o (in pu) positive-sequence half shunt susceptance: b + (in pu) zero-sequence half shunt susceptance: b o (in pu) 5 / 26

16 Line or cable It has been shown (see theory): z + = z s z m z o = z s + 2z m from which one obtains: z s = 2z + + z o z m = z o z z s z m z m and the impedance matrix: Z = z m z s z m z m z m z s The voltage-current relations of the series part are: V A V g I A V B V g V C V g = Z I B I C + and I A I B I C I A I B I C = = Z I A I B I C V A V g V B V g V C V g = Z V A V g V B V g V C V g Z V A V g V B V g V C V g V A V g V B V g V C V g Z V A V g V B V g V C V g (6) (7) 6 / 26

17 Line or cable It has been shown (see theory): b + = b s b m b o = b s + bz m from which one obtains: and the admittance matrix: b s = 2b + + b o b m = b o b b s b m b m j B = j b m b s b m b m b m b s For the shunt part on the left, the voltage-current relations are: I a V a V g I A I b I c = j B V b V g V c V g + I B I C (8) Similarly, for the shunt part on the right: I a V a V g I b I c = j B V b V g V c V g + I A I B I C (9) 7 / 26

18 Line or cable Taking into account that: V a = V A V b = V B V c = V C V a = V A V b = V B V c = V C Eqs. (6, 7, 8, 9) can be combined into: I a I b I j B + Z Z c = I a I b Z j B + Z I c } {{ } 6 6 admittance matrix Y V a V g V b V g V c V g V a V g V b V g V c V g 8 / 26

19 Line or cable Matlab script Yline.m 9 / 26

20 Transformer Three-phase analysis of unbalanced systems Transformer Input data: positive-sequence series impedance: R + jx (in pu) positive-sequence magnetizing reactance: X m (in pu) zero-sequence series impedance: R o + jx o (in pu) zero-sequence magnetizing reactance: X mo (in pu) impedance neutral - ground on primary and secondary sides: z n, z n2 (in pu) complex transformer ratio n not used in some transformers 20 / 26

21 Transformer Admittance matrix of the direct-sequence equivalent two-port (Ī + ) abc = ( V + ) abc + jx m (Ī+) a b c = n R + jx [( V + ) abc n ] R + jx ( V + ) a b c ] [ n ( V + ) a b c ( V + ) abc or in matrix form: [ (Ī+ ) abc (Ī+) a b c ] = [ jx m + R+jX n R+jX n R+jX n 2 R+jX } {{ } posit.-sequ. admittance matrix ] [ ( V + ) abc ( V + ) a b c ] 2 / 26

22 Transformer Admittance matrix of the negative-sequence equivalent two-port Replacing n by n in the direct-sequence admittance matrix: [ ] [ ( Ī ) abc jx m + R+jX n ] [ = R+jX ( V ) abc (Ī ) a b c n ( V ) a R+jX n 2 R+jX b c } {{ } negative-sequ. admittance matrix ] Admittance matrix of the zero-sequence equivalent two-port. Transformer of type Ynd* [ (Īo ) abc (Īo) a b c ] [ = jx mo + R o+jx o+3z n } {{ } zero-sequ. admit. matrix ] [ ( V o ) abc ( V o ) a b c ] 22 / 26

23 2. Transformer of type Ynyn0 Transformer [ ( Ī o ) abc (Ī o ) a b c ] = [ jx mo + R o+jx o+3z n+3z n2/n 2 n R o+jx o+3z n+3z n2/n 2 n R o+jx o+3z n+3z n2/n 2 n 2 R o+jx o+3z n+3z n2/n 2 } {{ } zero-sequ. admittance matrix ] [ ( Vo ) abc ( V o ) a b c ] 3. Transformer of any other type [ ( Ī o ) abc (Ī o ) a b c ] = [ jx mo 0 ] 0 0 } {{ } zero-sequ. admit. matrix [ ( Vo ) abc ( V o ) a b c ] 23 / 26

24 Transformer Assembling the admittance matrices of all three sequences (Ī + ) abc ( V + ) abc (Ī ) abc ( V ) abc (Ī o ) abc (Ī+) a = ( V o ) abc YF b c ( V + ) a b c (Ī ) a b c ( V ) a b c (Ī o ) a b c ( V o ) a b c with where YF = jx m + R+jX 0 0 n R+jX jx m + R+jX 0 0 n R+jX y 0 0 y 2 n R+jX n n 2 R+jX 0 0 R+jX 0 0 n 2 R+jX y y 22 [ ] y y 2 is the zero-sequence admittance matrix previously derived. y 2 y / 26

25 Getting back to the a, b, c, a, b, c phases Transformer Using the inverse Fortescue transformation: or Ī ā I bīc Ī a Ī b Ī c = T T Ī ā I bīc Ī a Ī b Ī c T 0 0 T = YF YF T T V a V b V c V a V b V c T 0 0 T } {{ } the sought admittance matrix. Yeah! V a V b V c V a V b V c 25 / 26

26 Transformer Matlab script Ytrfo.m 26 / 26

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