Lecture on Three-Phase Power Systems - Lab II | ECEL 302, Lab Reports of Electrical and Electronics Engineering

Download Lecture on Three-Phase Power Systems - Lab II | ECEL 302 and more Lab Reports Electrical and Electronics Engineering in PDF only on Docsity! D. Niebur Lab II: Lecture on Three-Phase Power Systems 1 This lecture and its figures are based on chapter 2.5-2.7 from the book by J. D. Glover and M.S. Sarma, Power System Analysis and Design, 3rd Edition, Brooks/Cole 2002, ISBN 0-534-95367-0. 2.5 Balanced Three-Phase Circuits Balanced Y-Connected Loads Definition A Y-connected load is balanced if all phase impedances are equal: Za  Zb  Zc  Zy Balanced Line-to-Neutral Voltages Definition A line-to-neutral source voltage is balanced if all line-to-neutral voltages have the same magnitude: |Ean |  |Ebn |  |Ecn |  |E| and their phase angles have a differences of 120°: ∠Ebn  ∠Ean − 120° ∠Ecn  ∠Ebn − 120° ∠Ean  ∠Ecn − 120° D. Niebur Lab II: Lecture on Three-Phase Power Systems 2 Definition The voltages are said to be in positive sequence of Ean leads Ebn by 120° and Ebn leads Ecn by 120° Example of a balanced source in positive sequence: Ean  10∠0°V Ebn  10∠ − 120°  10∠240°V Ecn  10∠ − 240°  10∠120°V So far we have looked at balanced line-to-neutral voltages. Balanced Line-to-Line Voltages We use Euler’s formula to convert between polar and cartesian coordinates: ∠  ej  cos  j sin We now introduce the same concept for line to line voltages: Eab  Ean − Ebn  Ean − Ean∠240°  Ean1 − ∠240°  Ean1 − ej240°   Ean1 − cos240° − j sin240°  Ean 1  12  3 2 j  3 Ean 3 2  1 2 j  3 Eancos30°  j sin30°  3 Ean∠30° D. Niebur Lab II: Lecture on Three-Phase Power Systems 5 Ia  Iab − Ica  Iab − Iab∠120°  Iab1 − cos120° − j sin120°  Iab 1 − 12 − j 1 2 3  3 Iab 3 2 − j 1 2  3 Iabcos30° − j sin30°  3 Iab∠ − 30° Similar we obtain Ib  Ibc∠ − 30° Ic  Ica∠ − 30° Property: For a balanced Δ-connected load the line currents are 3 bigger than the Δ −load currents and they lag the Δ −connected load currents by 30°. Δ −Y Conversion for Balanced Loads In Δ : Ia  3 Iab∠ − 30°  3 EabZΔ ∠ − 30° In Y : Ia  EanZy  Eab 3 1 Zy ∠ − 30°  Eab 3 1 Zy  3 EabZΔ  Zy  13 ZΔ D. Niebur Lab II: Lecture on Three-Phase Power Systems 6 Example Balanced Δ and Y loads Given: Balanced source Eab  480V Balanced Δ −connected load ZΔ  30∠40° Series line impedance per phase ZL  1∠85° Goal: Find the Δ −connected load currents IAB, IBC and ICA as well as the voltages across the loads EAB, EBC and ECA Solution: 1. We choose the angle of Eab as a reference  Eab  480∠0°V and compute Ean  Eab 3 ∠ − 30°V  480 3 ∠ − 30°V  277. 1∠ − 30°V 2. We convert the Δ connected load to a y −connected load  Zy  ZΔ3  30 3 ∠40°  10∠40° D. Niebur Lab II: Lecture on Three-Phase Power Systems 7 3. We combine the line impedance and the load impedance: Z  ZL  Zy  1∠85°  10∠40°  10. 73∠43. 78° 4. We draw the equivalent single-phase line-to-neutral diagram for phase a : 5. We compute the balanced line currents for phase a Ia  IA  EanZL  Zy  277. 1∠ − 30°10. 73∠43. 78°  25. 83∠ − 73. 78°A 6. We compute the balanced line currents for phase b and c by moving Ia by ∓120° : Ib  IB  Ia∠ − 120°  25. 83∠ − 193. 78°A Ic  IC  Ia∠120°  25. 83∠46. 22°A 7. We compute the balanced Δ −load currents IAB  Ia 3 ∠30°  25. 83 3 ∠ − 43. 78°A  14. 91∠ − 43. 78°A IBC  Ib 3 ∠30°  25. 83 3 ∠ − 163. 78°A  14. 91∠ − 163. 78°A ICA  Ic 3 ∠30°  25. 83 3 ∠76. 22°A  14. 91∠76. 22°A 8. We compute the balanced load voltages D. Niebur Lab II: Lecture on Three-Phase Power Systems 10 S3  Sa  Sb  Sc  3Sa The 3-phase active power is computed as P3  ReS3   3ReSa   3|Van ||Ia | cos −  We note that p3t  3|Van ||Ia | cos −   P3 Property: So the 3-phase instantenous power is equal to the 3-phase active power. For power calculations in 3-phase systems it does not matter whether we compute instantenous power in the time domain or active power phasor in the frequency domain! Further, the 3-phase reactive power is computed as Q3  ImS3   3ImSa   3|Van ||Ia | sin −  Finally, the 3-phase apparent power is computed as |S3 |  3|Van ||Ia | Instantenous and Complex Power for Balanced Y and Balanced Δ Impedance Loads We can express the voltages as line-to-neutral voltages or line-to-line voltages: For line-to-neutral voltages and y −connected loads we obtain as above p 3t  3|Van ||Ia | cos −   P3 Q3  3|Van ||Ia | sin −  S3  3|Van ||Ia |cos −   j sin −  |S3 |  3|Van ||Ia | For balanced line-to-line voltages and balanced y −connected load currents Ia, Ib and Ic we note that |Vab |  3 |Van | So p 3t  3|Van ||Ia | cos −   P3  3 3 Vab |Ia | cos −   3 Vab |Ia | cos −  Q3  3 Vab |Ia | sin −  S3  3 Vab |Ia |cos −   j sin −  |S3 |  3 |Vab ||Ia | For line-to-line voltages and Δ −connected load currents Iab, Ibc and Ica we note that D. Niebur Lab II: Lecture on Three-Phase Power Systems 11 |Vab |  3 |Van | |Iab |  1 3 |Ia | and therefore p 3t  3|Van ||Ia | cos −   P3  3 3 3 Vab |Iab | cos −   3|Vab ||Iab | cos −  Q3  3Vab |Iab | sin −  S3  3Vab |Iab |cos −   j sin −  |S3 |  |Vab ||Iab | Remark: For balanced systems, it is advantageous to convert the system from a Δ −connected 3-phase system into a y −connected 3-phase system. We then compute all quantities in terms of line-to-neutral voltages, line currents and single phase power for phase a. Only in the final step we compute the quantities for phase b and c and, if necessary, compute the line-to-line voltages and the Δ −load currents. The three-phase power at the load is always the triple of the single phase power at the load. 2.7 Advantages of Balanced Three-Phase versus Single-Phase Systems 1. The transmitted instantenous power is constant the mechanical power at the generator is constant; the mechanical shaft torque at the generator is constant; reduced shift vibration and reduced noise! 2. Because we need only one neutral wire (which carries current only in the unbalanced case) less wires are necessary than for three equivalent single-phase circuits. 3. Smaller-size neutral wire is necessary, since it carries currents only under unbalanced conditions. 4. Less wires result in less power losses ( I2R