Transcript Lecture 3
ECE 476 Renewable Energy Systems Lecture 3: Complex Power, Three-Phase Prof. Tom Overbye Dept. of Electrical and Computer Engineering University of Illinois at Urbana-Champaign [email protected] Announcements • Please read Chapters 2 and 4 • HW 1 is 2.7, 16, 24, 33; due Thursday 9/1 • • • Problem 2.47 moved to HW 2 HW 1 will be turned in (for other homework we may have an in-class quiz) For Problem 2.33 you need to use the PowerWorld Software. You can download the software and cases at the below link; get Version 19 (August 11, 2016) www.powerworld.com/gloveroverbyesarma Power Consumption in Devices Resistors only consume real power 2 PResistor I Resistor R Inductors only consume reactive power 2 Q Inductor I Inductor X L Capacitors only generate reactive power 2 QCapacitor I Capacitor X C QCapacitor VCapacitor XC 1 XC C 2 (Note-some define X C negative) 3 Example First solve basic circuit 400000 V I 4000 Amps 1000 V 400000 (5 j 40) 4000 42000 j16000 44.920.8 kV S V I * 44.9k20.8 4000 17.9820.8 MVA 16.8 j 6.4 MVA 4 Example, cont’d Now add additional reactive power load and resolve Z Load 70.7 pf 0.7 lagging I 564 45 Amps V 59.713.6 kV S 33.758.6 MVA 17.6 j 28.8 MVA 5 Power System Notation Power system components are usually shown as “one-line diagrams.” Previous circuit redrawn 17.6 MW 28.8 Mvar -16.0 MW -16.0 Mvar 59.7 kV 17.6 MW 28.8 Mvar slack 40.0 kV 16.0 MW 16.0 Mvar Transmission lines Generators are are shown as a shown as circles single line Arrows are used to show loads 6 Reactive Compensation Key idea of reactive compensation is to supply reactive power locally. In the previous example this can be done by adding a 16 Mvar capacitor at the load 16.8 MW 16.0 MW 6.4 MVR 0.0 MVR 44.94 kV 16.8 MW 6.4 MVR 40.0 kV 16.0 MW 16.0 MVR 16.0 MVR Compensated circuit is identical to first example with just real power load 7 Reactive Compensation, cont’d • Reactive compensation decreased the line flow from 564 Amps to 400 Amps. This has advantages – – – Lines losses, which are equal to I2 R decrease Lower current allows utility to use small wires, or alternatively, supply more load over the same wires Voltage drop on the line is less • Reactive compensation is used extensively by utilities • Capacitors can be used to “correct” a load’s power factor to an arbitrary value. 8 Power Factor Correction Example Assume we have 100 kVA load with pf=0.8 lagging, and would like to correct the pf to 0.95 lagging S 80 j 60 kVA cos 1 0.8 36.9 PF of 0.95 requires desired cos 1 0.95 18.2 Snew 80 j (60 Qcap ) 60 - Qcap 80 tan18.2 60 Qcap 26.3 kvar Qcap 33.7 kvar 9 Distribution System Capacitors 10 Definition of Reactive Power p (t ) Vmax cos(t V ) I max cos(t I ) 1 p (t ) Vmax I max [cos(V I ) cos(2t V I )] 2 1 Vmax I max cos(2t V I ) 2 1 Vmax I max cos(2t 2 I )cos(V I ) sin(2t 2 I )sin(V I ) 2 P cos(2t 2 I ) Q sin(2t 2 I ) p (t ) P (1 cos(2t 2 I )) Q sin(2t 2 I ) 11 PowerWorld Simulator Overview • Used for power system analysis and visualization – Runs in Windows • Download free 42 bus educational version at – http://www.powerworld.com/gloveroverbyesarma • Image on right shows the problem 2.33 power system (case) Balanced Three-Phase () Systems • A balanced three-phase () system has – – – three voltage sources with equal magnitude, but with an angle shift of 120 equal loads on each phase equal impedance on the lines connecting the generators to the loads • Bulk power systems are almost exclusively 3 • Single-phase is used primarily only in low voltage, low power settings, such as residential and some commercial 13 Balanced 3 -- No Neutral Current I n I a Ib I c V In (10 1 1 Z * * * * S Van I an Vbn I bn Vcn I cn 3 Van I an 14 Advantages of 3 Power • Can transmit more power for same amount of wire (twice as much as single phase) • Torque produced by 3 machines is constant • Three-phase machines use less material for same power rating • Three-phase machines start more easily than single-phase machines 15 Three-Phase - Wye Connection • There are two ways to connect 3 systems – – Wye (Y) Delta () Wye Connection Voltages Van V Vbn V Vcn V 16 Wye Connection Line Voltages Vca Vcn Vab -Vbn Van Vbn Vbc Vab (α = 0 in this case) Van Vbn V (1 1 120 3 V 30 Vbc 3 V 90 Vca 3 V 150 Line-to-line voltages are also balanced 17 Wye Connection, cont’d • Define voltage/current across/through device to be phase voltage/current • Define voltage/current across/through lines to be line voltage/current VLine 3 VPhase 130 3 VPhase e j 6 I Line I Phase S3 * 3 VPhase I Phase 18 Delta Connection For the Delta phase voltages equal line voltages Ica For currents Ia I ab I ca Ic Ib Ibc Iab Ia 3 I ab I b I bc I ab Ic I ca I bc * S3 3 VPhase I Phase 19 Three-Phase Example Assume a -connected load is supplied from a 3 13.8 kV (L-L) source with Z = 10020 Vab 13.80 kV Vbc 13.8 0 kV Vca 13.80 kV 13.80 kV I ab 138 20 amps I bc 138 140 amps I ca 1380 amps 20 Three-Phase Example, cont’d I a I ab I ca 138 20 1380 239 50 amps I b 239 170 amps I c 2390 amps * S 3 Vab I ab 3 13.80kV 138 amps 5.7 MVA 5.37 j1.95 MVA pf cos 20 lagging 21 Delta-Wye Transformation To simplify analysis of balanced 3 systems: 1) Δ-connected loads can be replaced by 1 Y-connected loads with ZY Z 3 2) Δ-connected sources can be replaced by VLine Y-connected sources with Vphase 330 22 Delta-Wye Transformation Proof From the side we get Vab Vca Vab Vca Ia Z Z Z Hence Vab Vca Z Ia 23 Delta-Wye Transformation, cont’d From the Y side we get Vab ZY ( I a I b ) Vca ZY ( I c I a ) Vab Vca ZY (2 I a I b I c ) Since Ia I b I c 0 I a I b I c Hence Vab Vca 3 ZY I a 3 ZY Vab Vca Z Ia Therefore ZY 1 Z 3 24 Three Phase Transmission Line 25 Per Phase Analysis • Per phase analysis allows analysis of balanced 3 systems with the same effort as for a single phase system • Balanced 3 Theorem: For a balanced 3 system with – – All loads and sources Y connected No mutual Inductance between phases 26 Per Phase Analysis, cont’d • Then – – – All neutrals are at the same potential All phases are COMPLETELY decoupled All system values are the same sequence as sources. The sequence order we’ve been using (phase b lags phase a and phase c lags phase a) is known as “positive” sequence; later in the course we’ll discuss negative and zero sequence systems. 27