Create Presentation
Download Presentation

Download Presentation
## Dave Angell Idaho Power 21st Annual Hands-On Relay School

- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -

**Symmetrical Components IAn Introduction to Power System**Fault Analysis Using Symmetrical Components Dave Angell Idaho Power 21st Annual Hands-On Relay School**Basic Course Topics**• Terminology • Phasors • Equations • Fault Analysis Examples • Calculations**Symmetrical Component Phasors**• The unbalanced three phase system can be transformed into three balanced phasors. • Positive Sequence • Negative Sequence • Zero Sequence**Positive Phase Sequence**• Each have the same magnitude. • Each positive sequence voltage or current quantity is displaced 120° from one another.**Positive Phase Sequence**• The positive sequence quantities have a-b-c, counter clock-wise, phase rotation.**Negative Phase Sequence**• Each have the same magnitude. • Each negative sequence voltage or current quantity is displaced 120° from one another.**Negative Phase Sequence**• The negative sequence quantities have a-c-b, counter clock-wise, phase rotation.**Zero Phase Sequence**• Each zero sequence quantity has the same magnitude. • All three phasors with no angular displacement between them, all in phase.**Symmetrical Components Equations**• Each phase quantity is equal to the sum of its symmetrical phasors. • Va = Va0 + Va1+Va2 • Vb = Vb0 + Vb1 +Vb2 • Vc = Vc0 + Vc1 +Vc2 • The common form of the equations are written in a-phase terms.**The a Operator**• Used to shift the a-phase terms to coincide with the b and c-phase • Shorthand to indicate 120° rotation. • Similar to the j operator of 90°.**Rotation of the a Operator**• 120° counter clock-wise rotation. • A vector multiplied by 1 /120° results in the same magnitude rotated 120°.**Rotation of the a2 Operator**• 240° counter clock-wise rotation. • A vector multiplied by 1 /240° results in the same magnitude rotated 240°.**B-Phase Zero Sequence**• We replace the Vb sequence terms by Va sequence terms shifted by the a operator. • Vb0 = Va0**B-Phase Positive Sequence**• We replace the Vb sequence terms by Va sequence terms shifted by the a operator • Vb1 = a2Va1**B-Phase Negative Sequence**• We replace the Vb sequence terms by Va sequence terms shifted by the a operator • Vb2 = aVa2**C-Phase Zero Sequence**• We replace the Vc sequence terms by Va sequence terms shifted by the a operator. • Vc0 = Va0**C-Phase Positive Sequence**• We replace the Vc sequence terms by Va sequence terms shifted by the a operator • Vc1 = aVa1**C-Phase Negative Sequence**• We replace the Vc sequence terms by Va sequence terms shifted by the a operator • Vc2 = a2Va2**What have we produced?**• Va = Va0+ Va1 + Va2 • Vb = Va0 + a2Va1 + aVa2 • Vc = Va0 + aVa1 + a2Va2**Symmetrical Components Equations**• Analysis • To find out of the amount of the components • Synthesis • The combining of the component elements into a single, unified entity**Symmetrical Components Synthesis Equations**• Va = Va0+ Va1 + Va2 • Vb = Va0 + a2Va1 + aVa2 • Vc = Va0 + aVa1 + a2Va2**Symmetrical Components Analysis Equations**• Va0 = 1/3 ( Va + Vb + Vc) • Va1= 1/3(Va + aVb + a2Vc) • Va2= 1/3 (Va + a2Vb + aVc)**Symmetrical Components Analysis Equations - 1/3 ??**• Where does the 1/3 come from? • Va1= 1/3(Va + aVb + a2Vc) • Va = Va0 + Va1 + Va2 • When balanced**Symmetrical Components Analysis Equations - 1/3 ??**• Va1= 1/3(Va + aVb + a2Vc) • Adding the phases**Symmetrical Components Analysis Equations - 1/3 ??**• Va1= 1/3(Va + aVb + a2Vc) • Adding the phases yields**Symmetrical Components Analysis Equations - 1/3 ??**• Va1= 1/3(Va + aVb + a2Vc) • Adding the phases yields 3 Va. • Divide by the 3 and now Va = Va1**Example VectorsAn Unbalanced Voltage**• Va = 13.4 /0° • Vb = 59.6 /-104° • Vc = 59.6 /104°**Analysis Results in These Sequence Quantities**• Va0 = -5.4 • Va1 = 42.9 • Va2 = -24.1**The Synthesis Equation Results in the Original Unbalanced**Voltage**Three phase fault**Positive Phase to phase fault Positive Negative Phase to ground fault Positive Negative Zero Symmetrical Components Present During Shunt Faults**Symmetrical Component Review of Faults Types**• Let’s return to the example fault reports and view the sequence quantities present**Single Line to Ground Fault**• Voltage • Negative and zero sequence 180 out of phase with positive sequence • Current • All sequence are in phase