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Lecture 1

Lecture 1. Balanced Three-phase Systems From Network to Single-phase Equivalent Circuit. Power plant. Transformer. Switching station. Line. Load. Three-phase system. Single-line diagram. Equivalent single-phase circuit. Balanced Three-phase Systems.

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Lecture 1

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  1. Lecture 1

  2. Balanced Three-phase SystemsFrom Network to Single-phase Equivalent Circuit Power plant Transformer Switching station Line Load Three-phase system Single-line diagram Equivalent single-phase circuit

  3. Balanced Three-phase Systems • A three-phase system can be analyzed by means of a single-phase equivalent when: • The source voltages are balanced or symmetrical • The electrical parameters of the system are symmetrical • The loads are balanced • Three-phase quantities can be determined from single phase voltages and currents when symmetry is assumed between phases.

  4. Balanced and Unbalanced faults • Balanced Cases • three-phase fault • (symmetrical) load flow • Unbalanced Cases • Single line to ground fault • Line to line fault • Double line to ground fault • (unsymmetrical load flow)

  5. Analyzing unbalanced system using Fortescue’s Theorem • Unbalanced faults in power systems require a phase by phase solution method or other techniques. • One of the most useful techniques to deal with unbalanced networks is the “symmetrical component” method, developed in 1918 by C.L. Fortescue.

  6. Symmetrical Components • Reasons for use of symmetrical components: • Unbalanced systems are difficult to handle • -> several independent balanced systems are easier to handle than one unbalanced system. • Transformation of one unbalanced 3-phase system into 3 balanced 3-phase systems. • -> for each system only one phase has to be considered

  7. Analyzing unbalanced system using Fortescue’s Theorem • Any three unbalanced set of voltages or currents can be resolved into three balanced systems of voltages or currents, referred to as the system symmetrical components, defined as follows: • Positive Sequence components: three phasors of equal magnitude displaced 120 degrees from each other following the positive sequence • Negative Sequence components: three phasors of equal magnitude displaced 120 degrees of each other following the negative sequence • ZeroSequencecomponents: three parallel phasors having equal magnitude and angle For a 3-ph system: 3 unbalanced phasors can be resolved into 3 balanced systems of 3 phasors each

  8. Let Va, Vb, Vc be the Phase voltages • According to Fortescue, these can be transformed into • Positive-seq. voltages: Va1, Vb1, Vc1 • Negative-seq. voltages: Va2, Vb2, Vc2 • zero-sequence voltages: Va0, Vb0, Vc0 Thus, Va = Va1 + Va2 + Va0 Vb = Vb1 + Vb2 + Vb0 Vc = Vc1 + Vc2 + Vc0

  9. Vc1 Va1 Va2 Va0 Va1 Vb1 Vc Va Vb Vb2 Va2 Vc2 Va0 Vb0 Vc0

  10. The ‘a’ operator a = 1<1200 = -0.5 + j 0.866 a I rotates I by 1200 a2 = 1<2400 = -0.5 – j 0.866 a3 = 1<3600 = 1<00 = 1 + j 0 1 + a + a2 = 0 a -a2 1 -1 -a a2

  11. From figure previous figures Vb1 = a2Va1 Vc1 = a Va1 Vb2 = a Va2 Vc2 = a2 Va2 Vb0 = Va0 Vc0 = Va0 sub. In Eq. (Slide 8) we get: Thus, Va = Va0 + Va1 + Va2 Vb = Va0 + a2Va1 + a Va2 Vc = Va0 + a Va1 + a2Va2

  12. Matrix Relations Let And Inverse of A is

  13. Matrix Relations Similarly currents can be obtained using their symmetrical components

  14. Matrix Relations • Vp= A Vs; Vs = A-1Vp • Va0 = 1/3 (Va + Vb + Vc) • Va1 = 1/3 (Va + aVb + a2Vc) • Va2 = 1/3 (Va + a2Vb + aVc)

  15. Matrix Relations

  16. Numerical Example • The line currents in a 3-ph 4 –wire system are Ia = 100<300 ; Ib = 50<3000 ; Ic = 30<1800. Find the symmetrical components and the neutral current. • Solution : • Ia0 = 1/3(Ia + Ib + Ic) = 27.29 < 4.70 A • Ia1 = 1/3(Ia + a Ib + a2Ic) = 57.98 < 43.30 A • Ia2 = 1/3(Ia + a2 Ib + a Ic) = 18.96 < 24.90 A • In = Ia + Ib + Ic = 3 Ia0 = 81.87 <4.70 A

  17. Numerical Example 2. The sequence component voltages of phase voltages of a 3-ph system are: Va0 = 100 <00 V; Va1 = 223.6 < -26.60 V ; Va2 = 100 <1800 V. Determine the phase voltages. Solution: Va = Va0 + Va1 + Va2 = 223.6 <-26.60 V Vb = Va0 + a2Va1 + a Va2 = 213 < -99.90 V Vc = Va0 + a Va1 + a2 Va2 = 338.6 < 66.20 V

  18. Numerical Example 3. The two seq. components and the corresponding phase voltage of a 3-ph system are Va0 =1<-600 V; Va1=2<00 V ; & Va = 3 <00 V. Determine the other phase voltages. Solution: Va = Va0 + Va1 + Va2  Va2 = Va – Va0 – Va1 = 1 <600 V Vb = Va0 + a2Va1 + a Va2 = 3 < -1200 V Vc = Va0 + a Va1 + a2 Va2 = 0 V

  19. Numerical Example 4. Determine the sequence components if Ia =10<600 A; Ib = 10<-600 A ; & Ic = 10 <1800 A. Solution: Ia0 = 1/3 (Ia + Ib + Ic) = 0 A Ia1 = 1/3 (Ia + a Ib + a2Ic) = 10<600 A Ia2 = 1/3 (Ia + a2 Ib + a Ic) = 0 A Thus, If the phasors are balanced, Two Sequence components will be zero.

  20. Ia Ia Ib

  21. Numerical Example 5. Determine the sequence components if Va = 100 <300 V; Vb = 100 <1500 V ; and Vc = 100 <-900 V. Solution: Va0 = 1/3(Va + Vb + Vc) = 0 V Va1 = 1/3(Va + a Vb + a2Vc) = 0 V Va2 = 1/3(Va + a2 Vb + a Vc) = 100<300 V Observation: If the phasors are balanced, Two sequence components will be zero.

  22. Three phase power in symmetrical components • S = VpTIp* = [A Vs]T[A Is]* = VsT AT A* Is* = 3 VsTIs* = 3Va0 Ia0* + 3Va1 Ia1* + 3Va2 Ia2* note that AT = A

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