BEF BEF Chapter 2. Outline BASIC PRINCIPLES 09/10/2013. Introduction. Phasor Representation. Complex Power Triangle.

Transcription

1 BEF 5503 BEF 5503 Chapter BASC PRNCPLES Outline ntroduction Phasor Representation Coplex Power Triangle Power Factor Coplex Power in AC Single Phase Circuits Coplex Power in Balanced Three-Phase Circuits One-line Diagra pedance Diagra Per Unit Analysis 1

2 ntroduction Power Concepts Per Unit Analysis Basic Principles of Power Systes Coplex Power Power Representations ntroduction R C L Power Syste Loads

3 ntroduction Tie Doain versus Frequency Doain Tie Doain: X M cost Frequency Doain: X X M ntroduction Sinusoids Period: T Tie necessary to go through one cycle Frequency: f = 1/T Cycles per second (Hz) Angular frequency (rads/sec): = f Aplitude: V M 3

4 ntroduction What is the aplitude, period, frequency, and angular (radian) frequency of this sinusoid? ntroduction Coplex Nuber aginary axis y x Real axis x is the real part y is the iaginary part z is the agnitude is the phase Polar Coordinates: A = z q Rectangular Coordinates: A = x + jy x z cos z x y y z sin y tan 1 x 4

5 ntroduction There is a good chance that your calculator will convert fro rectangular to polar, and fro polar to rectangular. Convert to polar: 3 + j4 and -3 - j4 Convert to rectangular: 45 & - 45 ntroduction Eleent Tie doain Phasor doain Resistance v = ir V = R Reactance v = L (di/dt) V = jl = jx Capacitance i = C (dv/dt) = jcv = jbv 5

6 ntroduction Phasor Representation The AC voltages and currents appearing in power systes can be represented by phasors, a concept useful in obtaining analytical solutions to one-phase and three-phase syste design. Phasor is generally defined as a transfor of sinusoidal functions fro the tie doain into the coplex-nuber doain. A phasor is a coplex nuber that contains the agnitude and phase angle inforation of a sinusoidal function. 6

7 Phasor Representation The Euler s identity that relates the exponential function to the trigonoetric functions is given as follows: j cos Re e e j cos j sin j sin e A sinusoidal voltage quantity v(t), with angular speed of ω rad/s, axiu voltage of V, and phase displaceent of θ 1 canbewrittenas: jt t Re V e v( t) V cos 1 1 j V e e j Re t 1 Phasor Representation Since the frequency is fixed in power studies, factor e jωt is equals to 1 and v(t) can be rewritten as: j1 v( t) V cost V e or 1 V 1 The instantaneous voltages and currents is not particularly concerned in power studies but ore with their RMS agnitudes and phase angles. V V j1 e V 1 j e Voltage Current Polar for 7

8 Phasor Representation The representation of RMS voltage and RMS current phasors in rectangular (or Cartesian) for are as follows: V VRMS cos1 j sin1 RMS cos j sin Coplex Power Triangle The coplex power (S in kva) is defined as a coplex nuber with the real part represents active power (P in kw) and the iaginary part represents reactive power (Q in kvar). S P jq Q L or Q C 8

9 Coplex Power Triangle S P Q L Power triangle for inductive loads P S Q C Power triangle for capacitive loads Coplex Power Triangle Exaple.1: Draw the power triangle for the following circuit: 9

10 Coplex Power Triangle Exaple.: Draw the power triangle for the following circuit: Power Factor Most of the loads in power syste are inductive type. For exaple: otors require REACTVE power (Q) to set up the agnetic field, and ACTVE power (P) to produce the useful work (shaft Horse Power). (Power Factor Angle) ACTVE power (P) REACTVE power (Q) 10

11 Power Factor Power Factor is a easure of how efficiently electrical power is consued. 100% of the energy burned is being used to ove the runner fro AtoB. Say, =30, only 87% of the energy burned is being used to ove the runner in the horizontal direction of B, and so extra energy will be required to achieve the sae objective. Power Factor Power Factor is the ratio of Active Power (P) to Apparent/ Total Power (S): (Power Factor Angle) ACTVE power (P) REACTVE power (Q) Power Factor Q tan P P(kW) S(kVA) cos Lagging (nductive Loads) Leading (Capacitive Loads) 11

12 Power Factor pact of poor power factor: Poor PF causes ore power losses Wasted Power Power Factor Exaple.3: An installation supplies the following loads: 10 kw at unity power factor; 15 kva at 0.8 p.f. lagging; 4 kvar leading. Calculate the total kw, kvar, the overall kva and power factor. [Ans. kw, 5 kvar (lag),.56 kva, (lag)] 1

13 Coplex Power in AC Single Phase Circuits To study steady-state behavior of circuits, consider the following sinusoidal voltage and current: v( t) V cost i t ) cos t Note that the current lags the ( voltage by an angle. The instantaneous power is defined as: p ( t ) vt i ( t ) p( t) V p V cos( t) cos( t ) cos cos t 1 ( ) cos cos t cos cos Coplex Power in AC Single Phase Circuits 13

14 Coplex Power in AC Single Phase Circuits V p( t) cos cos t A ore useful quantity is the average power that is being delivered. By averaging the instantaneous power over a specified tie-period, typically for one cycle, the average power, P, becoes: 0 V P cos cost V cos RMS RMS V cos V V ; RMS RMS Coplex Power in AC Single Phase Circuits Lagging power factor eans the current lags the voltage by an angle. Leading power factor eans the current leads the voltage by an angle. Lagging power factor Leading power factor 14

15 Coplex Power in AC Single Phase Circuits t is coon in electric power studies to set the voltage angle as the angular reference. The coplex power or the apparent power, S is defined as the product of voltage ties the conjugate of current, S V * V S V cos jv sin P V cos (W) Q V sin (VAr) S P jq POWER FACTOR cos costan 1 Q P P P Q P S Coplex Power in AC Single Phase Circuits Exaple.4: Consider the following 50 Hz based circuit with the given paraeters: 15

16 Coplex Power in AC Single Phase Circuits Exaple.4 (Cont.): Find the following: a) the source current, b) the active, reactive, and apparent power into the circuit, c) the power factor of the circuit. Coplex Power in AC Single Phase Circuits Solution.4: a) source current V 1000V 1 R jl 0.5 j rad / sec.110 V S 3 H 3 jc1000v j314.16rad / sec F b) power flows A A A A A 1 S V * c) power factor 1000V A j4573.1va P 700.5W Q VAr 3.4 PF cos(3.4) 0.84 lagging VA 16

17 Coplex Power in Balanced Three-Phase Circuits The ajor assuption of all the electric power presently used is generated, transitted, and distributed using balanced three-phase voltage systes. Three-phase operation is preferable to single-phase because a three-phase syste is ore efficient than a singlephase syste, and the flow of power is constant. Coplex Power in Balanced Three-Phase Circuits A balanced three-phase voltage syste is coposed of three single phase voltage sources having the sae agnitude and frequency but tie-displaced fro one another by 10 of a cycle as shown in the phasor diagras below: 17

18 Coplex Power in Balanced Three-Phase Circuits i A (t) V A (t) + V A =V0 Z Z = Z - V A (t)= V sint i B (t) V B (t) + V B =V-10 Z Z = Z - V B (t)= V sin(t-10 ) Current in three phases A B C V0 Z V Z V Z i C (t) V C (t) + V C =V-40 Z Z = Z - V C (t)= V sin(t-40 ) Coplex Power in Balanced Three-Phase Circuits A C B A Loads f three loads balanced, N = A + B + C = 0 B N C 18

19 Coplex Power in Balanced Three-Phase Circuits There are two possible connections of loads and sources in three-phase systes: wye-connection delta-connection Coplex Power in Balanced Three-Phase Circuits Wye (Y) and Delta () Connections a a V an n V bn V ab ca ab a a V cn b V bc b V ca V ca bc V bc V ab b c b c c c Wye (Y) Connection Delta () Connection 19

20 Coplex Power in Balanced Three-Phase Circuits Voltages and currents in Y connection: Van V V bn cn 0 V V V V LL = 3 V L = V ab V V an V V bn V V j V 3 1 3V j 3 V 30 Line voltage leads phase voltage by 30 Coplex Power in Balanced Three-Phase Circuits Phasor diagra for voltages in Y connection: 0

21 1 Voltages and currents in connection: Coplex Power in Balanced Three-Phase Circuits a c b c b a j j ca ab a L = 3 V LL = V Line current lags phase current by j Phasor diagra for currents in connection: Coplex Power in Balanced Three-Phase Circuits

22 Coplex Power in Balanced Three-Phase Circuits Power Relationships: Assue that a balanced three-phase voltage source is supplying a balanced load. The three sinusoidal phase voltages and currents can be written as: V V V a ( t) V sint a ( t) sint b( t) V sint 10 b( t) sint 10 b( t) V sint 10 ( t) sint 10 b Coplex Power in Balanced Three-Phase Circuits The three-phase power can be defined as: P3 V cos V cos V cos a a b b n trigonoetric identity, we get the following: P3 V 3cos cost cost 40 cost 40 The suation of the last three ters, in the above equation, is zero. Thus the three-phase power can be obtained as: P 3V cos 3 c c n phase quantities P3 3 V L L cos n line quantities

23 Coplex Power in Balanced Three-Phase Circuits Coplex Power The above analysis can be extended to include the reactive power, Q, or to get the apparent power, S, for a three-phase syste. S * * 3 3V 3 VL L f V V 0 and, then, S 3 3V 3V P 3 jq cos 3 j sin P 3 Q 3 3V 3V cos sin 3V L 3V L L L cos sin Coplex Power in Balanced Three-Phase Circuits Exaple.5: A wye-connected, balanced three-phase load consisting of three ipedances of 1030 is supplied with a balanced set of line-to-neutral voltages: V V an bn V cn 0V0 0V40 0VV 10 3

24 Coplex Power in Balanced Three-Phase Circuits Exaple.5: a) Calculate the phasor currents in each line, b) Calculate the line-to-line phasor voltages and show the corresponding phasor diagra c) Calculate the apparent power, active power, and reactive power supplied to the load Coplex Power in Balanced Three-Phase Circuits Solution.5: a) The phase currents of the loads are obtained as: an bn cn 00V 30A V 10A V A

25 Coplex Power in Balanced Three-Phase Circuits b) The line voltages are obtained as: V ab Or V V V ab bc ca V V 00V 040V 3 030V a b 3 V 3 V 3 V an bn cn Coplex Power in Balanced Three-Phase Circuits Relation between Phase and Line Voltages in a Wye-Connection: 5

26 Coplex Power in Balanced Three-Phase Circuits c) The powers are given by: * S3 3V 3Van P Q * an 00V 30A VA W 760.0VAr j760.0va One-line Diagra R Y B 3-phase syste one-line representation One-line diagra is a siplified single-line circuit diagra of a balanced three-phase electric power syste. t is indicated by a single line with standard apparatus sybols. 6

27 One-line Diagra Apparatus Sybols of One-line Diagra Or Machine or rotating arature Or Two-winding power transforer Or Three-winding power transforer Or Load Power circuit breaker, oil/ liquid Air circuit breaker Cont One-line Diagra Apparatus Sybols of One-line Diagra Busbar Three-phase, three- wire delta connection Transission line Three-phase wye, neutral ungrounded Fuse Current transforer A Three-phase wye, neutral grounded Aeter Or Potential transforer V Volteter 7

28 One-line Diagra One-line Diagra The inforation on a one-line diagra is vary according to the proble at hand and the practice of the particular copany preparing the diagra. Load/ Power Flow Study Transient Stability Study One-line Diagra Advantages of One-line Diagra Siplicity. One phase represents all three phases of the balanced syste. The equivalent circuits of the coponents are replaced by their standard sybols. The copletion of the circuit through the neutral is oitted. 8

29 One-line Diagra Exaple of one-line diagra (39-bus New England syste) One-line Diagra Exaple of one-line diagra (PSCAD/EMTDC) A SMPLE AC SYSTEM RRL #1 # s T FLAT30 Brk1 Vload [oh] C->G Flt1 ent (ka) Curre Voltage (kv) 3 Phase Source Current Phase Load Voltage 9

30 One-line Diagra Exaple of one-line diagra (PSCAD/EMTDC) NDUCTON MACHN NE WND FARM - ES [H] Use this coponent to study a wind gust or a rap and see the response of the control systes. ES Wind Source Mean w 1.0 A B Ctrl Pi Vw Ctrl = 1 CNT * 50.0 P Vw Actual hub speed of achine * * -1 N 3.0 Pole pairs 0.7 A B N/D D Ctrl Ctrl = 1 M wind gen Vw W A six Pole Machine Mechanical speed = W(pu)**pi*f/(pole paris) Pg W S T Beta * -1 The convention for the nduction achine is that input echanical torque is negative. (ie. otoring ode is positive) Wind Turbine MOD Type Beta BETA Wind Turbine Governor MOD Type T P TME TME CNT pedance Diagra pedance and Reactance Diagras pedance (Z = R + jx) diagra is converted fro oneline diagra showing the equivalent circuit of each coponent of the syste. t is needed in order to calculate the perforance of a syste under load conditions (Load flow studies) or upon the occurrence of a short circuit (fault analysis studies). Reactance (jx) diagra is further siplified fro ipedance diagra by oitting all static loads, all resistances, the agnetizing current of each transforer, and the capacitance of the transission line. t is apply to fault calculations only, and not to load flow studies. pedance and reactance diagras soeties called the Positive-sequence diagra. 30

31 pedance Diagra One-line diagra of an electric power syste pedance Diagra pedance diagra corresponding to the one-line diagra 31

32 pedance Diagra Reactance diagra corresponding to the one-line diagra E1 E E3 Generators 1 and Transforer T1 Transission Transforer Line T Gen. 3 Per Unit Analysis Coon quantities used in power syste analysis are voltage (kv), current (ka), voltaperes (kva or MVA), and ipedance (Ω). t is very cubersoe to convert currents to a different voltage level in a power syste having two or ore voltage levels. Per-unit representation is introduced such that the various physical quantities are expressed as a decial fraction or ultiples of base quantities and is defined as: actual quantity Quantity in per - unit base value of quantity 3

33 Per Unit Analysis Exaple.6: For instance, if a base voltage of 75 kv is chosen, actual voltages of 47.5 kv, 75 kv, and kv becoe 0.90, 1.00, and 1.05 per-unit, respectively. For siplicity, per-unit is always written as pu. For single-phase systes: Base current, A base kva 1 base voltage, kv base voltage, V Base ipedance base current, A Base power, kw 1 Base power, MW 1 base kva 1 base MVA 1 LN LN (base voltage, kv Base ipedance base kva 1 LN ) Per Unit Analysis For three-phase systes: Base current, A Base power, kw 3 Base power, MW 3 base MVA 3 3 X base voltage, kv (base voltage, kv Base ipedance base MVA base kva 3 base MVA 3 3 LL ) LL 33

34 Per Unit Analysis Exaple of power flow data in PU (39-bus New England syste) Per Unit Analysis Exaple.7: A three-phase, wye-connected syste is rated at 100 MVA and 13 kv. Express 80 MVA of three-phase apparent power as a per-unit value referred to (a) the three-phase syste MVA as base and (b) the single-phase syste MVA as base. Solution.7: (a) For the three-phase base, Base MVA = 100 MVA = 1 pu Base kv = 13 kv (LL) = 1 pu Per-unit MVA = 80/100 = 0.8 pu (b) For the single-phase base, Base MVA = 1/3 X 100 MVA = MVA = 1 pu Base kv = 13/ 3 = 76.1 kv = 1 pu Per-unit MVA = 1/3 X 80/ = 0.8 pu 34

35 Per Unit Analysis Changing the Base of Per-unit Quantities The ipedance of individual generators and transforers are generally in ters of % or pu quantities based on their own ratings (By anufacturer). For power syste analysis, all ipedances ust be expressed in pu on a coon syste base. Thus,it is necessary to convert the pu ipedances fro one base to another (coon base, for exaple: 100 MVA). Per-unit ipedance of a circuit eleent (actual ipedance, ) X (base MVA) (base voltage, kv) Per Unit Analysis The equation shows that pu ipedance is directly proportional to base MVA and inversely proportional p to the square of the base voltage. Therefore, to change fro old base pu ipedance to new base pu ipedance, the following equation applies: Per - unit Z new per - unit Z old base kv base kv old new base MVA base MVA new old 35

36 Per Unit Analysis Exaple.8: The reactance X of a generator is given as 0.0 pu based on the generator s naeplate rating of 13. kv, 30 MVA. The base for calculations is 13.8 kv, 50 MVA. Find X on this new base. Solution.8: x" pu Per Unit Analysis Exaple.9: A 30 MVA 13.8 kv three-phase generator has a subtransient reactance of 15%. The generator supplies two otors over a transission line having transforers at both ends, as shown in the one-line diagra below. The otors have rated inputs of 0 MVA and 10 MVA, both 1.5 kv with x = 0%. The three-phase transforer T1 is rated at 35 MVA, Y kv with leakage reactance of 10%. Transforer T is coposed of three single-phase transforers each rated at 10 MVA, Y kv with leakage reactance of 10%. Series reactance of the transission line is 80 Ω. Draw the reactance diagra with all reactances arked in per unit. Select the generator rating as base in the generator circuit. 36

37 Per Unit Analysis Solution.9: The three-phase rating of transforer T is 3 X 10 MVA = 30 MVA. ts line-to-line voltage ratio is X 67 = kv. A base of 30 MVA, 13.8 kv in the generator circuit requires a 30 MVA base in all parts of the syste and the following voltage bases: n transission line: 13.8(115/13.) = 10. kv. n otor circuit: 10.(1.5/116) = 13 kv. The reactances of the transforers converted to the proper base are: Transforer T1:X = 0.1 (13./13.8) (30/35) = pu. Transforer T:X =01(15/13) 0.1(1.5/13) = pu. The base ipedance of the transission line is (10. kv) /30 MVA = Ω and the reactance of the line is (80/481.6) = pu. Reactance of otor 1 = 0. (1.5/13) (30/0) = 0.77 pu. Reactance of otor = 0. (1.5/13) (30/10) = pu. Per Unit Analysis Solution.9: Reactance diagra: 37

38 Per Unit Analysis Exaple.10: Prepare an ipedance diagra of the given power syste. Show all the ipedances and source voltages in p.u. on a 100 MVA, 13 kv base in the transission line circuit. The necessary data for this syste are as follows: G1 : 50 MVA, 1. kv, X = 0.15 p.u. G : 0 MVA, 13.8 kv, X = 0.15 p.u. T1 : 80 MVA, 1./ 161 kv, X = 0.10 p.u. T : 40 MVA, 13.8/ 161 kv, X = 0.10 p.u. Load : 50 MVA, 0.80 power factor lagging, operating at 154 kv. Load odeled as a parallel cobination of resistance and inductance. Per Unit Analysis One-line diagra of the given power syste: (40 + j160) (0 + j80) (0 + j80) 38

39 Per Unit Analysis Solution.10: MVA base = 100 MVA. kv base: At transission line: 13 kv At G1 circuit : At G circuit : X13 kv kv. X13 kv kv. Per Unit Analysis pedance in p.u.: X pu(g1) = X pu(g) = X pu(t1) = p ( ) X pu(t) = j0.15 X j pu p.u j0.15 X j p.u j0.10 X j p.u j0.10 X j p.u

40 Per Unit Analysis pedance base at transission line circuit: Z base(tline) = Z pu(tline AB) = (13 kv) MVA 40 j j p.u Z pu(tline AC) = Z pu(tline BC) = 0 j j p.u Per Unit Analysis Load: 50 MVA (0.8 + j0.6) = (40 + j30) MVA R = X L = (154 kv) 40 MW (154 kv) 30 MVar R pu = X Lpu = p.u p.u

41 Per Unit Analysis Sources: E G1 pu = = 1./10.00 = 1. p.u. E G pu = = 13.8/ = 1. pu p.u. Per Unit Analysis pedance diagra: j pu j pu j pu j pu j pu j pu j pu 1. pu pu j4.537 pu 1. pu 41