04-Electric Power. ECEGR 452 Renewable Energy Systems

Transcription

1 04-Electric Power ECEGR 452 Renewable Energy Systems

2 Overview Review of Electric Circuits Phasor Representation Electrical Power Power Factor Dr. Louie 2

3 Introduction Majority of the electrical energy produced by renewable resources is ultimately transmitted and consumed within the interconnected power system Integration of renewable resources into the power system is a challenge We need to understand basics of electric power Dr. Louie 3

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5 Review of Electric Circuits Voltage in AC circuits: v(t) = V max cos(wt + q V ) V max : voltage amplitude (Volts) w: frequency (rad/sec) q V : phase angle of the voltage (rad) Current in AC circuits: i(t) = I max cos(wt + q i ) I max : amplitude (Amperes) q i : phase angle of the voltage (rad) Dr. Louie 5

6 Review of Electric Circuits amplitude phase angle t v2( t) 1. 5cos 377t 45 v1( t) 1. 0 cos Dr. Louie 6

7 Review of Electric Circuits Conversion of radians (q rad ) to degrees (q deg ) q deg q rad 180 Conversion of degrees to radians q rad q deg 180 Dr. Louie 7

8 Review of Electric Circuits Frequency in North American power systems is 60 Hz f: frequency (Hertz) w = 2f ~ 377 rad/sec Other parts of the world 50 Hz is common We assume 60 Hz unless otherwise noted Voltage waveform is set as a reference, so q V = 0 o Dr. Louie 8

9 Phasor Transform Shorthand for writing sinusoidal functions Used for steady-state calculations Contains amplitude and phase angle information Assumed that frequency is known Relies on Euler s Identity v(t) v cos wt q max v Phasor Transform V max 2 q v Note: division by square root of 2 is used in power system analysis ( effective phasor ) Dr. Louie 9

10 Example Write the phasor representation of v 1(t) 1. 41cos 377t 0 Write the phasor representation of v 2(t) cos 377t 45 Write the phasor representation of v 3(t) 1. 41cos t 0 Dr. Louie 10

11 Example write the phasor representation of v 1(t) 1. 41cos 377t 0 write the phasor representation of v 2(t) cos 377t 45 write the phasor representation of v 3(t) 1. 41cos t 0 solution V1 10 solution V solution V3 10 Dr. Louie 11

12 Notation Lecture slides use bold uppercase variables (e.g. V, I) for phasors and other vectors Capital letters (e.g. V, I) or absolute values of phasors ( V, I ) are used to indicate the magnitude of the phasor V Vq Lowercase variables (e.g. v, i) are preferred to represent scalars not associated with phasors and vectors Notable exceptions P, Q for real and imaginary power Dr. Louie 12

13 Phasor Transform We use the effective phasor because P v i cos( ) rms rms So we can then write Unless otherwise specified, assume that voltages and currents are given in RMS and all phasors are effective phasors Also note: P VIcos( ) j90 o o o e cos( 90 ) jsin( 90 ) jsin( 90 ) j where j is the imaginary operator j 1 Dr. Louie 13

14 Phasor Transform Different expressions of a voltage phasor v max 2 Current: i max 2 Impedance: Define: q q q q j v V rms v = V v V v Ve q q q q q j I I rms I = I I I I Ie q j Z qz ZqZ Ze q Z q q (remember this!) v i I v Dr. Louie 14

15 Phasors Phasors have a direct geometric interpretation Polar form V1 145 q V 1 45 o V2 30 Horizontal is reference Dr. Louie 15

16 Phasors Another way of specifying phasors is in rectangular form Let the Y-axis be the imaginary (j) axis Let the X-axis be the real axis Resolving into real and imaginary components V V j j0 imag V1 145 V2 30 real Dr. Louie 16

17 Phasors imag j real j 1 90 Dr. Louie 17

18 Addition of Phasors Addition and subtraction of phasors are simple using rectangular form Simply add/subtract the real values and add/subtract the imaginary values V3 V1 V j Addition is tip to tail Subtraction is tail to tip imag V 2 V 1 V 3 real Dr. Louie 18

19 V V Multiplication of Phasors Multiplication and division are easier in polar form For multiplication: multiply magnitudes, add angles For division: divide magnitudes, subtract angles V V ( 145 )( 30 ) 345 add angles multiply magnitudes V 4 imag real Dr. Louie 19

20 Example If A. B. C. D. 0 V , what is jv? Dr. Louie 20

21 Example If A. B. C. D. 0 V , what is jv? j V ( 190 )( 110 ) 1100 Dr. Louie 21

22 Phasors Another way of specifying phasors is in rectangular form Let the Y-axis be the imaginary (j) axis Let the X-axis be the real axis Resolving into real and imaginary components V V j j0 imag V1 145 V2 30 real Dr. Louie 22

23 For resistors i(t) i cos( wt q ) max v(t) i(t)r v(t) Ri cos( wt q ) max Phasor Analysis of Resistors Transforming into phasor form: i i V V RIe RI jq i Voltage and current are in phase i(t) + v(t) - R Dr. Louie 23

24 For inductors i(t) i cos( wt q ) max di v(t) L dt v(t) Lwi sin( wt q ) Lwi o cos( wt q 90 ) Transforming into phasor form: i max i max i o v(t) Lwi cos( wt q 90 ) V V Phasor Analysis of Inductors w max jqi L Ie e j90 jqi jlwie jlw o I j90 using e j + v(t) L - i i(t) Dr. Louie 24

25 Phasor Analysis V jwli Define X L = wl (inductive reactance) Therefore V = jx L I I lags V by 90 deg. A similar derivation for capacitors yields X C = 1/(wC) (capacitive reactance) V = -jx C I I leads V by 90 deg. Dr. Louie 25

26 Phasor Analysis We can rewrite Ohm s Law to include complex impedances V = IZ Z: complex impedance (Ohms) Z = R + jx L - jx C (if in series) 1/Z = 1/R + j/x L - j/x C (if in parallel) Z will have a magnitude and phase associated with it Z Zq z Dr. Louie 26

27 Complex Power P is also known as Real Power, Active Power, Average Power P V I cos( ) P s Re VI * * is the complex conjugate operator, it denotes a change in sign of the imaginary part Conjugation is needed so that the difference in phase between voltage and current is considered, rather than their sum Dr. Louie 27

28 Complex Power Let S be the complex power defined as S VI then Let Q be the reactive power defined as Then * * VI S P Re Re Q Q S And therefore: S Im Im P VI jq * Q is also known as imaginary power S is also known as apparent power Dr. Louie 28

29 Complex Power Technically, units of S, Q and P are watts* To avoid confusion, alternate units are used in practice S: Volt-Amps (VA) Q: Volt-Amps Reactive (VAR) Inductors, capacitors consume/supply reactive power, Q S and Q are defined values a meaningful physical interpretation is elusive *See C. Gross On VA's, VAR's, and Other Traditions in Electric Power Engineering Dr. Louie 29

30 Power Triangle Relationships between S, P and Q can be shown graphically S = P + jq imaginary Note: P, Q are scalars representing the real and imaginary parts of S S P Q real Dr. Louie 30

31 Power Triangle Consider P = V I cos() Since S = VI* = VI = V I Then P = S cos() P is the projection of S onto the real axis S Q P Dr. Louie 31

32 Complex Power A similar result can be found for Q P = S cos() Q = S sin() Q is the projection of S onto the imaginary axis Imaginary axis S P Real axis Q Dr. Louie 32

33 Complex Power Cheat Sheet P Re{ VI } 2 =Re{ IZI } I Re{ Z} 2 P I R P V I cos 2 P I Z cos Q Im{ VI } 2 Q I X Q V I sin 2 Q I Z sin S P cos jq Q 1 tan ( ) P P P Q 2 2 Dr. Louie 33

34 Power Factor Power factor is non-negative cos() = cos( ) Need to distinguish between and S Q P -Q Dr. Louie 34

35 Power Factor For example let q v = 0 o Case 1: q i = 30 o Capacitive circuit PF = Case 2: q i = -30 o Inductive circuit PF = Same power factor Dr. Louie 35

36 Leading/Lagging Power Factor Must describe the PF value along with whether the current leads or lags voltage Lagging: current lags voltage (inductive) Leading: current leads voltage (capacitive) Useful mnemonic: ELI the ICE man Inductive, lagging, positive S V Q I P capacitive, leading, negative I V P S Q Dr. Louie 36

37 Why are Inductive Circuits Lagging? Recall V = jx L I I = V/(jX L ) = -jv/x L S = VI * I V therefore I * = jv * /X L S = jvv * /X L = j V 2 /X L I * S V angle is always +90 degrees for purely inductive circuits Dr. Louie 37

38 Leading/Lagging Power Factor Similar result for capacitive circuits S = -j V 2 /X C Note: the presence of resistance does not affect whether or not a circuit is leading or lagging, but it does affect the magnitude of the power factor circuits with L and C (or L, C and R) must be analyzed before leading or lagging can be determined Dr. Louie 38

39 Phasor Analysis Find the current out of the source, the power out of the source, and power consumed by the resistor assuming: V s = 120 Volts at 60 Hz L = 0.01 Henry R = 10 Ohms I L V s R Dr. Louie 39

40 Phasor Analysis V s = 120 Volts (RMS) at 60 Hz L = 0.01 Henry jx L = jwl = j(60 x 2)x.01 = j3.77 R = 10 Ohms I L V s R Dr. Louie 40

41 Phasor Analysis Z ( 3. 77j 10) tan 10 Z V s IZ 1200 I A phasor diagram I V s Dr. Louie 41

42 Power from the source Phasor Analysis P V I cos( ) ( 120)( )cos( ) 1. 26kW s Power consumed by the load resistor 2 2 P I R kW The inductor does not consume any power P I L V s R Dr. Louie 42

43 Summary P, Q, S related by power triangle S Q P P has a physical interpretation, Q and S do not S is a vector, Q and P are scalars Resistors associated with P; inductors/capacitors associated with Q Dr. Louie 43