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Lecture 1

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Balanced Three-phase Systems From Network to Single-phase Equivalent Circuit Power plant TransformerSwitching station LineLoad Three-phase system Single-line diagram Equivalent single-phase circuit

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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.

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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)

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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.

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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

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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 Zero Sequence components: 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

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Let V a, V b, V c be the Phase voltages According to Fortescue, these can be transformed into 1.Positive-seq. voltages: V a1, V b1, V c1 2.Negative-seq. voltages: V a2, V b2, V c2 3.zero-sequence voltages: V a0, V b0, V c0 Thus, V a = V a1 + V a2 + V a0 V b = V b1 + V b2 + V b0 V c = V c1 + V c2 + V c0

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V a1 V b1 V c1 V a2 V c2 V b2 V a1 V a2 V a0 VaVa VcVc VbVb V a0 V b0 V c0

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The ‘a’ operator a = 1<120 0 = -0.5 + j 0.866 a I rotates I by 120 0 a 2 = 1<240 0 = -0.5 – j 0.866 a 3 = 1<360 0 = 1<0 0 = 1 + j 0 1 + a + a 2 = 0 1 a2a2 a -a -a 2

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From figure previous figures V b1 = a 2 V a1 V c1 = a V a1 V b2 = a V a2 V c2 = a 2 V a2 V b0 = V a0 V c0 = V a0 sub. In Eq. (Slide 8) we get: Thus,V a = V a0 + V a1 + V a2 V b = V a0 + a 2 V a1 + a V a2 V c = V a0 + a V a1 + a 2 V a2

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Matrix Relations Let And Inverse of A is

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Matrix Relations Similarly currents can be obtained using their symmetrical components

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Matrix Relations V p = A V s ; V s = A -1 V p V a0 = 1/3 (V a + V b + V c ) V a1 = 1/3 (V a + aV b + a 2 V c ) V a2 = 1/3 (V a + a 2 V b + aV c )

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Matrix Relations

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Numerical Example 1.The line currents in a 3-ph 4 –wire system are I a = 100<30 0 ; I b = 50<300 0 ; I c = 30<180 0. Find the symmetrical components and the neutral current. Solution : I a0 = 1/3(I a + I b + I c ) = 27.29 < 4.7 0 A I a1 = 1/3(I a + a I b + a 2 I c ) = 57.98 < 43.3 0 A I a2 = 1/3(I a + a 2 I b + a I c ) = 18.96 < 24.9 0 A I n = I a + I b + I c = 3 I a0 = 81.87 <4.7 0 A

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Numerical Example 2. The sequence component voltages of phase voltages of a 3-ph system are: V a0 = 100 <0 0 V; V a1 = 223.6 < -26.6 0 V ; V a2 = 100 <180 0 V. Determine the phase voltages. Solution: V a = V a0 + V a1 + V a2 = 223.6 <-26.6 0 V V b = V a0 + a 2 V a1 + a V a2 = 213 < -99.9 0 V V c = V a0 + a V a1 + a 2 V a2 = 338.6 < 66.2 0 V

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Numerical Example 3. The two seq. components and the corresponding phase voltage of a 3-ph system are V a0 =1<-60 0 V; V a1 =2<0 0 V ; & V a = 3 <0 0 V. Determine the other phase voltages. Solution: V a = V a0 + V a1 + V a2 V a2 = V a – V a0 – V a1 = 1 <60 0 V V b = V a0 + a 2 V a1 + a V a2 = 3 < -120 0 V V c = V a0 + a V a1 + a 2 V a2 = 0 V

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Numerical Example 4. Determine the sequence components if I a =10<60 0 A; I b = 10<-60 0 A ; & I c = 10 <180 0 A. Solution: I a0 = 1/3 (I a + I b + I c ) = 0 A I a1 = 1/3 (I a + a I b + a 2 I c ) = 10<60 0 A I a2 = 1/3 (I a + a 2 I b + a I c ) = 0 A Thus, If the phasors are balanced, Two Sequence components will be zero.

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IaIa IbIb IaIa

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Numerical Example 5. Determine the sequence components if V a = 100 <30 0 V; V b = 100 <150 0 V ; and V c = 100 <-90 0 V. Solution: Va0 = 1/3(Va + Vb + Vc) = 0 V Va1 = 1/3(Va + a Vb + a 2 Vc) = 0 V Va2 = 1/3(Va + a 2 Vb + a Vc) = 100<30 0 V Observation: If the phasors are balanced, Two sequence components will be zero.

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Three phase power in symmetrical components S = V p T I p * = [A V s ] T [A I s ]* = V s T A T A* I s * = 3 V s T I s * = 3V a0 I a0 * + 3V a1 I a1 * + 3V a2 I a2 * note that A T = A

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