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Transcript

ECE 476 POWER SYSTEM ANALYSIS Lecture 3 Three Phase, Power System Operation Professor Tom Overbye Department of Electrical and Computer Engineering Reading and Homework • • For lecture 3 please be reading Chapters 1 and 2 For lectures 4 through 6 please be reading Chapter 4 – • we will not be covering sections 4.7, 4.11, and 4.12 in detail HW 1 is 2.7, 12, 21, 26; due Thursday 9/4 1 Balanced 3 Phase () Systems A balanced 3 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 2 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 3 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 constrant Three phase machines use less material for same power rating Three phase machines start more easily than single phase machines 4 Three Phase - Wye Connection There are two ways to connect 3 systems Wye (Y) Delta () Wye Connection Voltages Van V Vbn V Vcn V 5 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 6 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 7 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 8 Three Phase Example Assume a -connected load is supplied from a 3 13.8 kV (L-L) source with Z = 10020W Vab 13.80 kV Vbc 13.8 0 kV Vca 13.80 kV 13.80 kV I ab 138 20 amps W I bc 138 140 amps I ca 1380 amps 9 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 10 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 11 Delta-Wye Transformation Proof From the side we get Vab Vca Vab Vca Ia Z Z Z Hence Vab Vca Z Ia 12 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 13 Three Phase Transmission Line 14 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 15 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. 16 Per Phase Analysis Procedure To do per phase analysis 1. Convert all load/sources to equivalent Y’s 2. Solve phase “a” independent of the other phases 3. Total system power S = 3 Va Ia* 4. If desired, phase “b” and “c” values can be determined by inspection (i.e., ±120° degree phase shifts) 5. If necessary, go back to original circuit to determine line-line values or internal values. 17 Per Phase Example Assume a 3, Y-connected generator with Van = 10 volts supplies a -connected load with Z = -jW through a transmission line with impedance of j0.1W per phase. The load is also connected to a -connected generator with Va”b” = 10 through a second transmission line which also has an impedance of j0.1W per phase. Find 1. The load voltage Va’b’ 2. The total power supplied by each generator, SY and S 18 Per Phase Example, cont’d First convert the delta load and source to equivalent Y values and draw just the "a" phase circuit 19 Per Phase Example, cont’d To solve the circuit, write the KCL equation at a' 1 ' ' ' (Va 10)(10 j ) Va (3 j ) (Va j 3 20 Per Phase Example, cont’d To solve the circuit, write the KCL equation at a' 1 ' ' ' (Va 10)(10 j ) Va (3 j ) (Va j 3 10 (10 j 60) Va' (10 j 3 j 10 j ) 3 Va' 0.9 volts Vb' 0.9 volts Vc' 0.9 volts ' Vab 1.56 volts 21 Per Phase Example, cont’d * ' Va Va Sygen 3Va I a* Va 5.1 j 3.5 VA j 0.1 " " Va Sgen 3Va ' * Va 5.1 j 4.7 VA j 0.1 22 Power System Operations Overview Goal is to provide an intuitive feel for power system operation Emphasis will be on the impact of the transmission system Introduce basic power flow concepts through small system examples 23 Power System Basics All power systems have three major components: Generation, Load and Transmission/Distribution. Generation: Creates electric power. Load: Consumes electric power. Transmission/Distribution: Transmits electric power from generation to load. – – Lines/transformers operating at voltages above 100 kV are usually called the transmission system. The transmission system is usually networked. Lines/transformers operating at voltages below 100 kV are usually called the distribution system (radial). 24 Small PowerWorld Simulator Case Load with green arrows indicating amount of MW flow Bus 2 20 MW -4 MVR Bus 1 1.00 PU 204 MW 102 MVR 1.00 PU 106 MW 0 MVR 150 MW AGC ON 116 MVR AVR ON -14 MW -34 MW 10 MVR 4 MVR 34 MW -10 MVR Home Area Used to control output of generator -20 MW 4 MVR 100 MW Note the power balance at each bus 14 MW -4 MVR 1.00 PU Bus 3 102 MW 51 MVR 150 MW AGC ON 37 MVR AVR ON Direction of arrow is used to indicate direction of real power (MW) flow 25 Power Balance Constraints Power flow refers to how the power is moving through the system. At all times in the simulation the total power flowing into any bus MUST be zero! This is know as Kirchhoff’s law. And it can not be repealed or modified. Power is lost in the transmission system. 26 Basic Power Control Opening a circuit breaker causes the power flow to instantaneously(nearly) change. No other way to directly control power flow in a transmission line. By changing generation we can indirectly change this flow. 27 Transmission Line Limits Power flow in transmission line is limited by heating considerations. Losses (I2 R) can heat up the line, causing it to sag. Each line has a limit; Simulator does not allow you to continually exceed this limit. Many utilities use winter/summer limits. 28 Overloaded Transmission Line 29 Interconnected Operation Power systems are interconnected across large distances. For example most of North America east of the Rockies is one system, with most of Texas and Quebec being major exceptions Individual utilities only own and operate a small portion of the system, which is referred to an operating area (or an area). 30 Operating Areas Transmission lines that join two areas are known as tie-lines. The net power out of an area is the sum of the flow on its tie-lines. The flow out of an area is equal to total gen - total load - total losses = tie-flow 31 Area Control Error (ACE) The area control error is the difference between the actual flow out of an area, and the scheduled flow. Ideally the ACE should always be zero. Because the load is constantly changing, each utility must constantly change its generation to “chase” the ACE. 32 Automatic Generation Control Most utilities use automatic generation control (AGC) to automatically change their generation to keep their ACE close to zero. Usually the utility control center calculates ACE based upon tie-line flows; then the AGC module sends control signals out to the generators every couple seconds. 33 Three Bus Case on AGC Bus 2 -40 MW 8 MVR 40 MW -8 MVR Bus 1 1.00 PU 266 MW 133 MVR 1.00 PU 101 MW 5 MVR 150 MW AGC ON 166 MVR AVR ON -39 MW -77 MW 25 MVR 12 MVR 78 MW -21 MVR Home Area Generation is automatically changed to match change in load 100 MW 39 MW -11 MVR Bus 3 1.00 PU 133 MW 67 MVR 250 MW AGC ON 34 MVR AVR ON Net tie flow is close to zero 34 Generator Costs There are many fixed and variable costs associated with power system operation. The major variable cost is associated with generation. Cost to generate a MWh can vary widely. For some types of units (such as hydro and nuclear) it is difficult to quantify. For thermal units it is much easier. These costs will be discussed later in the course. 35 Economic Dispatch Economic dispatch (ED) determines the least cost dispatch of generation for an area. For a lossless system, the ED occurs when all the generators have equal marginal costs. IC1(PG,1) = IC2(PG,2) = … = ICm(PG,m) 36 Power Transactions Power transactions are contracts between areas to do power transactions. Contracts can be for any amount of time at any price for any amount of power. Scheduled power transactions are implemented by modifying the area ACE: ACE = Pactual,tie-flow - Psched 37 100 MW Transaction Bus 2 8 MW -2 MVR -8 MW 2 MVR Bus 1 1.00 PU 225 MW 113 MVR 1.00 PU 0 MW 32 MVR 150 MW AGC ON 138 MVR AVR ON -92 MW -84 MW 27 MVR 30 MVR 85 MW -23 MVR Home Area 93 MW -25 MVR Bus 3 Scheduled Transactions 100.0 MW Scheduled 100 MW Transaction from Left to Right 100 MW 1.00 PU 113 MW 56 MVR 291 MW AGC ON 8 MVR AVR ON Net tie-line flow is now 100 MW 38 Security Constrained ED Transmission constraints often limit system economics. Such limits required a constrained dispatch in order to maintain system security. In three bus case the generation at bus 3 must be constrained to avoid overloading the line from bus 2 to bus 3. 39 Security Constrained Dispatch Bus 2 -22 MW 4 MVR 22 MW -4 MVR Bus 1 1.00 PU 357 MW 179 MVR 1.00 PU 0 MW 37 MVR 100% 194 MW OFF AGC -142 MW 49 MVR 232 MVR AVR ON 145 MW 100% -37 MVR Home Area Bus 3 Scheduled Transactions 100.0 MW -122 MW 41 MVR 100 MW 124 MW -33 MVR 1.00 PU 179 MW 89 MVR 448 MW AGC ON 19 MVR AVR ON Dispatch is no longer optimal due to need to keep line from bus 2 to bus 3 from overloading 40 Multi-Area Operation If Areas have direct interconnections, then they may directly transact up to the capacity of their tie-lines. Actual power flows through the entire network according to the impedance of the transmission lines. Flow through other areas is known as “parallel path” or “loop flows.” 41 Seven Bus Case: One-line System has three areas 44 MW -42 MW -31 MW 0.99 PU 3 1.05 PU 1 106 MW -37 MW AGC ON 62 MW 79 MW 2 40 MW 20 MVR Top Area Cost 8029 $/MWH 1.00 PU -32 MW Case Hourly Cost 16933 $/MWH 32 MW 80 MW 30 MVR 4 110 MW 40 MVR 38 MW -61 MW 1.04 PU 31 MW -77 MW 5 -39 MW 40 MW 94 MW AGC ON Area top has five buses -14 MW 1.01 PU 130 MW 40 MVR 168 MW AGC ON -40 MW 1.04 PU 6 Area left has one bus 20 MW -20 MW 40 MW 1.04 PU 20 MW 200 MW 0 MVR Left Area Cost 4189 $/MWH 200 MW AGC ON -20 MW 7 200 MW Right Area Cost 0 MVR 4715 $/MWH 201 MW AGC ON Area right has one bus 42 Seven Bus Case: Area View Top 40.1 MW 0.0 MW Area Losses 7.09 MW -40.1 MW 0.0 MW System has 40 MW of “Loop Flow” Left Area Losses 0.33 MW Right 40.1 MW 0.0 MW Actual flow between areas Scheduled flow Area Losses 0.65 MW Loop flow can result in higher losses 43 Seven Bus - Loop Flow? Top 4.8 MW 0.0 MW -4.8 MW 0.0 MW Left Area Losses -0.00 MW 100 MW Transaction between Left and Right Area Losses 9.44 MW Right 104.8 MW 100.0 MW Note that Top’s Losses have increased from 7.09MW to 9.44 MW Area Losses 4.34 MW Transaction has actually decreased the loop flow 44 Pricing Electricity Cost to supply electricity to bus is called the locational marginal price (LMP) Presently some electric makets post LMPs on the web In an ideal electricity market with no transmission limitations the LMPs are equal Transmission constraints can segment a market, resulting in differing LMP Determination of LMPs requires the solution on an Optimal Power Flow (OPF) 45 3 BUS LMPS - OVERLOAD IGNORED Bus 2 Gen 2’s cost is $12 per MWh 60 MW 60 MW Bus 1 10.00 $/MWh 0 MW 10.00 $/MWh 120 MW 120% 180 MW 0 MW Gen 1’s cost is $10 per MWh 60 MW Total Cost 1800 $/hr 120% 120 MW 60 MW 10.00 $/MWh Bus 3 180 MW 0 MW Line from Bus 1 to Bus 3 is over-loaded; all buses have same marginal cost 46 LINE OVERLOAD ENFORCED Bus 2 20 MW 20 MW Bus 1 10.00 $/MWh 60 MW 12.00 $/MWh 100 MW 80% 100% 120 MW 0 MW 80 MW Total Cost 1921 $/hr 80% 100% 100 MW 80 MW 14.01 $/MWh Bus 3 180 MW 0 MW Line from 1 to 3 is no longer overloaded, but now the marginal cost of electricity at 3 is $14 / MWh 47