lecture13 - EE 740 Professor Ali Keyhani Lecture #13...

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Unformatted text preview: EE 740 Professor Ali Keyhani Lecture #13 Symmetrical Components Operator Positive Sequence Voltage (1 = ‘+’) Vc1 Va1=V1 Vb1 Va = Va1 = V∠ = V1∠ 0° 0° Vb = Vb1 = V1∠ 240° Vc = Vc1 = V1∠ 120° Let a = 1∠ 120° a2 = 1∠ 240° Va1 = V1 Va1 Vb1 Vc1 Vb1 = a2V1 1 a2 a Vc1 = aV1 = V1 Negative Sequence Voltage (2 = ‘-‘) Vb2 Va2=V2 Vc2 Va2 = V2 Vb2 = aV2 Vc2 = a2V2 Va2 Vb2 Vc2 1 a a2 = V2 Zero Sequence Voltage (“0”) Va0 = V0 Vb0 = V0 Vc0 = V0 We will show that a set of unbalanced voltages Va,Vb,Vc can be written as: Va = Va0 + Va1 + Va2 ⇒ Va = V0 + V1 + V2 Vb = Vb0 + Vb1 + Vb2 ⇒ Vb = V0 + a2V1 + aV2 Vc = Vc0 + Vc1 + Vc2 ⇒ Vc = V0 + aV1 + a2V2 Compact Form: Va Vb Vc Vabc T -1 = 1 1 1 a2 1 a = T = -1 1 a a2 V012 T V0 V1 V2 Vabc = = T 1/3 -1 T V012 V012 T -1 Vabc 1 1 1 1 a a2 1 a2 a V0 ∴ = 1/3(Va + Vb + Vc) V1 = 1/3(Va + aVb + a2Vc) V2 = 1/3(Va + a2Vb + aVc) Similarly: Iabc I012 = = T T I012 -1 Iabc Problem: Consider a balanced, Y-connected, 460 Volt generator. Compute the positive, negative and zero sequence voltages. Vcn Van 460V Vbn n • Van = 460∠0°/√3 = 265.9∠0° Vbn = 265.9∠240° = 265.9a2 Vcn = 265.9∠120° = 265.9a V0 V1 V2 1/3 = 1 1 1 1 a a2 1 a2 a Van Vbn Vcn V0 = 1/3(265.9 + 265.9a2 + 265.9a) = 265.9/3(1 + a + a2) = 0 ⇒ V0 = 0 Show that (1 + a + a2) = 0 V1 = 1/3(265.9 + 265.9a3 + 365.9a3) Recall a3 = 1∠0° V1 = 265.9∠0° V2 = 1/3[265.9 + 265.9a4 + 265.9 a2] a4 = a = 1∠120° V2 = 1/3[(265.9) (1 + a + a2)] = 0 Conclusions: For balanced three - phase sources we have: • Only positive sequence voltages exist. • Zero sequence voltages ‘Do Not Exist’. • Negative sequence voltages ‘Do Not Exist’. Problem: voltages: Consider a 3-φunbalanced set of Va = 4∠ 0° Vb = 3∠ -90° Vc = 8∠ 143.1° Vc 8 143.1° 4 3 Vb Va abc Determine V0, V1 and V2 Solution: V0 = 1/3(Va + Vb + Vc) = 1/3[4 - j3 + 8(- .8 + J.6)] = -.8 + j.6 = 1∠ 143.1° Va0 = Vb0 = Vc0 = V0 = 1∠ 143.1° V1 = 1/3(Va + aVb + a2Vc) V1 = 1/3[4∠ +(1∠ 0° 120° )(3∠ 90° )+(1∠ 240° )(8∠ 143.1° )] V1 = 1/3[13.96 + j4.64] = 4.9∠ 18.4° Va1 = V1 = 4.9∠ 18.4° Vb1 = V1∠ 240° = 4.9∠ 258.4° Vc1 = V1∠ 120° = 4.9∠ 138.4° V2 = 1/3[Va + a2Vb + aVc] V2 = 2.15∠ -86.2° Va2 = V2 = 2.15∠ -86.2° Vb2 = V2∠ 120° = 2.15∠ 33.8° Vc2 = V2∠ -120° = 2.15∠ -206.2° Check: Va = Va0 + Va1 + Va2 = 1∠ 143.1° + 4.9∠ 18.4° + 2.15∠ -86.2° = 4∠ 0° Vb = Vb0 + Vb1 + Vb2 = 1∠ 143.1° + 4.9∠ 258.4° + 2.15∠ 33.8° = 3∠ -90° Vc = Vc0 + Vc1 + Vc2 = 1∠ 143.1° + 4.9∠ 138.4° + 2.15∠ -206.2° = 8∠ 143.1° Zero Sequence Sequence Va0 Vb0 Vc0 Positive 1∠ 143.1° Vc1 4.9 Va1 18.4° Vb1 Negative Sequence Vb2 Vc2 2.15 Va2 -86.2° Original Unbalanced Three Phase System Vc 8 143.1° 4 -90° 3 Vb Va Problem 4: Solution: Balanced 3φ voltages are applied to a balanced 3φ load. Compute I0, I1 and I2. Assume the load is purely resistive and positive phase sequence. Ic Vc Va Vb Ib I0 I+ I= 1/3 1 1 1 1 1 a a2 a2 a Ia Ib Ic Ia RL=10Ω RL RL I0 = 1/3[Ia + Ib + Ic] I+ = 1/3[Ia + aIb + a2Ic] I- = 1/3[Ia + a2Ib + aIc] Balanced Voltages and Balanced Loads Va Ia (+) VL RL (-)&(0) Assume: Va = V∠ = 10∠ 0° 0° Vb = Va∠ 240° = 10∠ 240° Sequence Vc = Va∠ 120° = 10∠ 120° Ia = Van/RL = V/RL∠ = I∠ 0° 0° Ib = V/ RL ∠ 240° = I∠ 240° = a2Ia Ic = V/ RL = I∠ 120° = aIa (+) Phase I0 = 1/3[Ia + Ib + Ic] but (Ia + Ib + Ic)=In=0 for balanced three phase systems I1 = 1/3[Ia + a*a2Ia + a2*aIa] = 1/3Ia[1 + 2] = Ia I2 = 1/3[Ia + a2Ib + aIc] Note: Ib = a2Ia, Ic = aIa I2 = 1/3[Ia + a4Ia + a2Ia] = 1 + a2 + a4 = 0 + j0 I2 = 0 Conclusions: When a balanced set of three phase voltages is applied to a balanced three phase system, only positive sequence currents can flow. That is: •Only Positive Sequence Currents Exist. •Zero Sequence Currents “Do Not Exist”. •Negative Sequence Currents “Do Not Exist”. ...
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