AC Power. by Prof. Dr. Osman SEVAİOĞLU Electrical and Electronics Engineering Department
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1 by Prof. Dr. Osman SEVAİOĞLU Electrical and Electronics Engineering Department EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 1
2 Voltage Waveform Consider the following AC circuit driven by a source with voltage waveform; V(t) 25,0 20,0 V(t) = V max cos wt V max 15,0 Phasor representation of this voltage waveform will then be V max 0 o 10,0 5,0 0,0-5, ,0 + I(t) -15,0-20,0 V(t) = V max cos wt Load = R + j X = Z Z = R 2 + X 2, = Tan -1 (X / R) EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 2
3 METU Phase Shift in Voltage Waveform Consider now, the following AC circuit driven by a a source with an AC voltage waveform; V(t) = V max cos ( wt - v ) Phasor representation of the shifted voltage waveform will then be V(t) V max 25,0 20,0 15,0 v / w = t 1 t 1 = 3 msec V max - v 10,0 5,0 I(t) 0,0-5, ,0 V(t) = V max cos ( wt - v ) Load = R + j X -15,0-20,0 = Z Z = R 2 + X 2, = Tan -1 (X / R) EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 3
4 Current Waveform Relation Between Voltage and Current Phasors + I(t) = I max cos ( wt - I ) Load = R + j X = Z V max v = I max - I V(t) V(t) 25,0 20,0 15,0 10,0 5,0 0,0-5,0-10,0-15,0-20,0 v / w = t 1 t 1 = 3 msec V max Z I max I(t) 25,0 20,0 15,0 10,0 5,0 0,0-5,0-10,0-15,0-20,0 Z = R 2 + X 2, = Tan -1 (X / R) = v - I = Tan -1 ( X / R ) I / w EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 4
5 METU Current Waveform Phasor representation of this current will then be I(t) 25,0 20,0 I / w I max - I I max 15,0 Waveform representation of this current will then be + I(t) = I max cos ( wt - I ) I(t) = I max cos ( wt - I ) 10,0 5,0 0,0-5,0-10,0-15,0-20, V(t) Load = R + j X = Z Z = R 2 + X 2, I = Tan -1 ( X / R ) EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 5
6 Voltage and Current Waveforms Please note that the angle difference v - I depends only on the resistance R and reactance X of the load = v - I = Tan -1 ( X / R ) V 25,0 /w =( v - I )/w V max = ^ I 20,0 15,0 10,0 V(t) 5,0 0,0-5,0-10, V(t) I(t) Load -15,0-20,0 EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 6
7 - Power Expression Voltage and current waveforms are V(t) = V max cos wt I(t) = I max cos ( wt - ) Power waveform will then be S(t) = V(t) x I(t) * = V max cos wt x I max cos ( wt + ) cos (a + b) = cosa cos b - sin a sin b + cos (a - b) = cosa cos b + sin a sin b cos (a + b) + cos (a - b) = 2 cosa cos b cosa cos b = ½ [ cos (a + b) + cos (a-b) ] Please note that; - sign here has been altered due to the conjugation operation in the definition of power: V max V(t) S = V x I* / w =( v - I ) / w 25,0 20,0 15,0 10,0 5,0 0,0-5,0-10,0-15,0-20, V I I * V ^ V cos ( wt ) ^ I cos ( wt - ) ^ I cos ( wt + ) ^ V cos ( wt ) EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 7
8 - Power Expression Power waveform is the product of the voltage and conjugate of the current waveforms S(t) = V max I max cos wt x cos ( wt + ) cosa x cos b = ½ [ cos (a + b) + cos (a - b) ] S(t) = ½ V max I max [ cos ( wt + wt + ) + cos ( wt - wt - ) ] = ½ V max I max [ cos ( 2wt + ) + cos ] = (V max / 2 ) (I max / 2) [ cos ( 2wt + )+ cos ] =V rms I rms [ cos ( 2wt + ) + cos ] V rms = V max / 2 = v - I = Tan -1 ( X/R) V(t) Vmax*coswt I(t) Imax*cos(wt+theta) S(t) V(t) * I(t) EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 8 V(t) S(t) I(t)*
![cos ] =V rms I rms [ cos ( 2wt + ) + cos ] V rms = V max / 2 = v - I = Tan -1 ( X/R) 500 400 300 200 100 0-100 -200-300 -400 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17](/docs-images/44/21019897/images/page_8.jpg "18 19 20 V(t) Vmax*coswt I(t) Imax*cos(wt+theta) S(t) V(t) * I(t) EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O.")
9 Decomposition of Power Expression Power expression may be decomposed into two components Sinusoidal (AC) Term DC Term =S(t) avg / w S(t) = V rms I rms cos ( 2wt + ) + V rms I rms cos Sinusoidal (AC) Term Constant (DC) Term S(t) V(t) * I(t) Constant (DC) Term EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 9
10 Average Power Expression Average Power Average is zero Sum of these areas is zero S(t) = V rms I rms [ cos ( 2wt + ) + cos ] S (t) avg = (1 / T ) S(t) dt 0 S(t) avg = (1/T) P(t) dt T = (1/T) V rms I rms [ cos ( 2wt + ) + cos ] dt = (1/T) V rms I rms cos ( 2wt + ) dt + (1/T) V rms I rms cos dt Sinusoidal (AC) Term S(t) V(t) * I(t) Constant (DC) Term + zero Constant (DC) Term EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 10
11 Average Power Expression Average Power S(t) avg = (1/T) P(t) dt = (1/T) V rms I rms cos dt = (1/T) V rms I rms cos dt = (1/T) T V rms I rms cos = V rms I rms cos T dt = T V(t) I(t) DC Term = S(t) avg Load / w S (t) avg = V rms I rms cos EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 11
12 Example Calculate the average and instantaneous powers dissipated in the load shown below Parameters: Question V rms = 100 Volts Z load = 3 + j 4 Ohms = o Ohms + V(t) = V max cos wt = 2 V rms cos wt = cos wt = cos wt I(t) Load Z load = o EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 12 0 V(t) S(t) V(t) Vmax*coswt I(t) Imax*cos(wt+theta) S(t) V(t) * I(t) S(t) avg I(t)*
13 Example Solution 500 S(t) avg 400 S(t) V max = 100 x 2 = Volts I = V max 0 o Z = o o V(t) I(t)* = o Amp 100 I rms = I max / 2 = 28.8 / 2 = 20 Amp + I(t) V(t) = V max cos wt = 2 V rms cos wt = cos wt = cos wt Load Z load = o V(t) Vmax*coswt I(t) Imax*cos(wt+theta) S(t) V(t) * I(t) EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 13
14 Example Solution S(t) S(t) avg S(t) avg = V rms I rms cos = 100 x 20 cos = 1200 Watt V(t) I(t)* S(t) = V(t) I(t) = 2 x100 cos wt x 2 x 20 cos (wt o ) = 4000 cos wt x cos (wt o ) V(t) = Vmax cos wt = 2 V rms cos wt = cos wt = cos wt I(t) Load Z load = o V(t) Vmax*coswt I(t) Imax*cos(wt+theta) S(t) V(t) * I(t) Please note that; - sign here has been changed to + due to the conjugate operation in the definition of power: S = V x I* EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 14
15 Example Solution S(t) S(t) avg S(t) = 4000 cos wt x cos (wt o ) or by using the following identity; V(t) I(t)* cosa cos b = ½ [ cos (a + b) + cos (a - b) ] 100 S(t) = 2000 [cos (2wt o ) + cos o ] + I(t) V(t) = V max cos wt Load = 2 V rms cos wt = cos wt = cos wt Z load = o V(t) Vmax*coswt I(t) Imax*cos(wt+theta) S(t) V(t) * I(t) EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 15
![13 o ) + cos 53.13 o ] + I(t) V(t) = V max cos wt Load = 2 V rms cos wt = 100 2 cos wt = 144.4 cos wt Z load = 5 53.](/docs-images/44/21019897/images/page_15.jpg "13 o 0-100 -200-300 -400 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 V(t) Vmax*coswt I(t) Imax*cos(wt+theta) S(t) V(t) * I(t) EE")
16 METU Complex Power Active Power Expression + I(t) Load Expanding the first term in the power expression; S(t) = V rms I rms [ cos ( 2wt + ) + cos ] = V rms I rms (cos2wt cos - sin2wt sin + cos ) V(t) 500 cos ( a + b ) = cosa cos b - sin a sin b 400 S(t) S(t) avg and recombining the cosine terms; = V rms I rms [ cos ( 1 + cos2wt ) - sin 2wt sin ] V(t) I(t)* P(t) = Active power Q(t) = Reactive power EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page V(t) Vmax*coswt I(t) Imax*cos(wt+theta) S(t) V(t) * I(t)
![I rms [ cos ( 1 + cos2wt ) - sin 2wt sin ] 300 200 100 V(t) I(t)* P(t) = Active power Q(t) = Reactive power EE 209 Fundamentals of Electrical and](/docs-images/44/21019897/images/page_16.jpg "Electronics Engineering, Prof. Dr. O.")
17 Complex Power S(t) = V rms I rms [ cos ( 1 + cos2wt ) - sin2wt sin ] = V rms I rms cos ( 1 + cos2wt ) - V rms I rms sin sin2wt 500 ¼ Period = 2.5 msec P(t) avg = V rms I rms cos ( 1 + cos2wt ) - Q max sin2wt 400 S(t) P(t) =Active power P(t) Q max = V rms I rms sin 100 Q(t) = Reactive power EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page Q(t) S(t) V(t) * I(t) P(t) Vrms Irms*cos(Theta)*(1+cos(2*wt)) Q(t) -Vrms Irms*SIN(Theta)*SIN(2*wt)
18 METU Active and Reactive Powers Summary S(t) = V rms I rms cos ( 1 + cos2wt ) - V rms I rms sin sin2wt ¼ Period = 2.5 msec P(t) avg 350 Active power P(t) avg = V rms I rms cos 300 Reactive power Q(t) = - Q max sin 2wt P(t) Marmara, Unimar Power Plant 470 MW Q(t) P(t) Vrms Irms*cos(Theta)*(1+cos(2*wt)) Q(t) -Vrms Irms*SIN(Theta)*SIN(2*wt) EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 18
19 METU Active and Reactive Power Waveforms Power (P, Q) 350 ¼ Period = 2.5 msec P(t) avg + V(t) Load P(t) Full period = 10 msec ¼ period = 2.5 msec ¼ period = 360 o / 4 = 90 o P(t) = Active power Q(t) =Reactive power Q(t) -150 P(t) Vrms Irms*cos(Theta)*(1+cos(2*wt)) -200 Q(t) -Vrms Irms*SIN(Theta)*SIN(2*wt) EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 19
20 METU Active and Reactive Power Waveforms One revolution = 10 msec Hence, 2.5 msec = 360 o / 4 = 90 o ¼ Period = 2.5 msec P(t) avg Please note that Q leads P by 90 o P(t) Q P(t) = Active power msec 90 o Q(t) =Reactive power P -100 Q(t) -150 P(t) Vrms Irms*cos(Theta)*(1+cos(2*wt)) -200 Q(t) -Vrms Irms*SIN(Theta)*SIN(2*wt) EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 20
21 Active Power Waveform P(t) = V rms I rms cos ( 1 + cos2wt ) = V rms I rms cos + V rms I rms cos cos2wt P(t) = Active Power (Watt) 350 P(t) avg = 150 W 300 P (t) avg P(t) sinusoidal 250 Active Power (P) V(t) Load EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 21
22 Reactive Power Waveform Q(t) = V rms I rms sin x sin2wt = Q max sin 2wt Q(t), Reactive Power (VAR) Q(t) ¼ Period = 2.5 msec Q(t) sinusoidal Q max 50 + Reactive Power (Q) V(t) -100 Load Q(t) -Vrms Irms*SIN(Theta)*SIN(2*wt) EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 22
23 Reactive Power (Q) Reactive Power Waveform (During the first 5 mseconds) Q(t), Reactive Power (VAR) 200,0 + Load 150,0 100,0 V(t) 50,0 0,0-50, Q max -100,0-150,0-200,0 ¼ Period = 2.5 msec EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 23
24 Reactive Power (Q) Reactive Power Waveform (During the next 5 mseconds) Q(t), Reactive Power (VAR) 200,0 150,0 + V(t) Load 100,0 50,0 Q max 0,0-50,0-100, ¼ Period = 2.5 msec -150,0-200,0 EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 24
25 Phosors Representation of Active and Reactive Powers Mean of P(t) is called active power Peak of Q(t) is called reactive power P(t) = Active Power (Watt) P(t) avg = 150 W Q max = 160 VAR 0 200,0 150,0 100,0 50,0 0,0-50,0-100,0-150,0-200, Q(t), Reactive Power (VAR) EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 25
26 Phosors Representation of Active and Reactive Powers Voltage and Current Angle = 90 o, Time for ¼ revolution = 20/4 =5 msec Angular speed: w = 2 f = 2 x 3.14 x 50 = 314 rad/sec Time for one revolution = 1/f = 1/ 50 = sec = 20 msec Hence, Angle for ¼ revolution = 360 o / 4 = 90 o Time for ¼ revolution = 20 msec / 4 = 5 msec Active and Reactive Power Angular speed: w = 2 w = 4 f = 2 x314 = 628 rad/sec Time for one revolution = 10 msec Time for ¼ revolution = 10 msec / 4 = 2.5 msec EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 26
27 Phosors Representation of Active and Reactive Powers Power (P, Q, S) S = V x I * + Load V(t) 150 S(t) = V(t) x I(t) S(t) I * o 100 V(t) I(t) o 50 I V EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 27
28 Phosors Representation of Active and Reactive Powers Total power S = V rms I rms S = P 2 + Q 2 Q S Active power P = V rms I rms cos Reactive power Q = V rms I rms sin Power (P, Q, S) Q = S sin = o P S^ V(t) + Load P = S cos S(t) = V(t) x I(t) EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 28
29 Phosors Representation of Active and Reactive Powers S - Total power (k) VA (kva) P- Active power (k) Watt (kw) Q S Q - Reactive power (k) VAR (kvar) Please note that this angle depends only on the resistance R and reactance X of the load, i.e. Q = S sin = o S^ = Tan -1 X / R = Tan -1 Q / P P = S cos P EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 29
30 Basic Conversions Polar Representation P = S cos, Q = S sin Rectangular Representation S S = P 2 + Q 2, = Tan -1 (Q / P) P + j Q Please note that this angle depends only on the resistance R and reactance X of the load Q =S sin Q S S^ = Tan -1 X / R = Tan -1 Q / P = o P P =S cos X / R = Q / P i.e. if X = 0 Q = 0 EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 30
31 Active Reactive Powers (in the first 5 mseconds) Q Reactive Power (kvar) S Total (Complex) Power (kva) Wagon P Active Power (kw) Railway EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 31
32 Active Reactive Powers (in the next 5 mseconds) Railway Wagon P Active Power (kw) Q Reactive Power (kvar) S Total (Complex) Power (kva) EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 32
33 METU Active Reactive Powers Reactive Power (kvar) Total Power (kva) P(t) average > 0 P Active Power (kw) Wagon Q Railway S Wagon P Active Power (kw) , Q Reactive Power (kvar) EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 33
34 Active-Reactive Powers Polar Representation P = S cos, Q = S sin Rectangular Representation S S = P 2 + Q 2, = Tan -1 (Q / P) P + j Q Reactive Power (kvar) Total Power (kva) Q, Reactive Power (kvar) S, Total Power (kva) Q S Wagon Railway P Active Power (kw) P, Active Power (kw) EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 34
35 Active-Reactive Powers Total power S = V rms I rms S = P 2 + Q 2 Q, Reactive Power (kvar) S, Total (Complex) Power (kva) Active power P = V rms I rms cos Reactive power Reactive Power (kvar) Q = V rms I rms sin Total Power (kva) P, Active Power (kw) Wagon Q Railway S P Active Power (kw) S, Total Power (kva) EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 35
36 Power Meters Analog Digital Power Analyzer Clamp Type Current Transformers EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 36
37 Reactive Power Compensation Definition Reactive power compensation is partial or full cancellation of the reactive component of complex power by introducing a negative (compensation) component Q, Reactive Power (kvar) S, Total Power (kva) Q new Reactive Power (kvar) S new, Total Power (kva) Q comp., Reactive Power Compensation (kvar) P, Active Power (kw) new P, Active Power (kw) EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 37
38 Power Factor Definition Cosine of the angle between S and P is called Power Factor of the load Power Factor = p.f. = cos = cos ( Tan -1 Q / P ) Please note that reducing Q means reducing the angle, and hence increasing power factor Hence, reactive power compensation is sometimes called as Power Factor Correction, i.e. correcting power factor to a value near unity Q, Reactive Power (kvar) Q comp., Reactive Power Compensation (kvar) Q new Reactive Power (kvar) new S, Total Power (kva) P, Active Power (kw) S new, Total Power (kva) P, Active Power (kw) EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 38
39 Full or Partial Compensation Partial Compensation Partial compensation is the case, where Power Factor is raised to a value not reaching unity Power Factor new = p.f. new = cos new = cos ( Tan -1 Q new / P ) Full Compensation Full compensation is the case, where Power Factor is unity Power Factor new = p.f. new = cos new = cos ( Tan -1 0 / P ) = 1 Q, Reactive Power (kvar) Q new Reactive Power (kvar) Q new = 0 new new = 0 S new = P S, Total Power (kva) P, Active Power (kw) S new, Total Power (kva) P, Active Power (kw) P, Active Power (kw) EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 39
40 Charging Principle Applied by TEDAS If Q / P > 1/3 Then, reactive power is charged If Q / P < 1/3 Then, reactive power is free Please note that means; Q / P > 1/3 Tan -1 (1/3) = o cos o = Q, Reactive Power (kvar) S, Total Power (kva) P, Active Power (kw) EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 40
41 Why Partial Compensation is Preferred? Partial Compensation Due to economic reasons, full compensation is rarely implemented Reactive power needed to raise p.f. from 0.95 to 1 is the same as that for raising p.f. from 0.83 to 0.95 (not worthwhile) Power Factor 1,00 0,95 Reactive power needed to raise p.f. from 0.83 to 0.95 Reactive power needed to raise p.f. from 0.95 to unity P + j Q new P + j Q 0,90 0,85 0,80 0,75 0,70 + _ j Q comp P = 3000 kw p.f. = 0.65 = o (lag) Q = 3507 kvar 0,65 0,0 0,5 1,0 1,5 2,0 2,5 3,0 3,5 Q comp (MVAR) Q comp = Q = Q Q new EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 41
42 Application of Reactive Power Compensation Reduction of Equipment Loading Transformers Lines Cables are priced with respect to the power rating (kva) Prices of these equipments on the other hand, are determined by the cross-section ( mm 2 ) of the equipment Cross Section Current Capacity ( mm 2 ) ( Amp ) Cross section (size) of the cable (mm 2 ) EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 42
43 Application of Reactive Power Compensation Reduction of Equipment Loading Cable roller Hence, the power rating S (kva) of a cable is merely determined by the cross section, which must be minimized in order to reduce the investment to be made for the cable EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 43
44 Application of Reactive Power Compensation Reduction of Equipment Loading Hence, the power rating S (kva) of a cable is merely determined by the cross section, which must be minimized in order to reduce the investment to be made for the cable Q = S sin Q = o S S^ Hence, S (kva) must be minimized P Power (S) P = S cos V(t) + S(t) = V(t) x I(t) Load EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 44
45 Alternative Ways of Reducing S (kva) Alternative Ways of Reducing S (kva) a) Reducing the overall loading; P + jq (kw + j kvar) (Overall comsumption) Unreasonable, since the active consumption P is determined by the needs of the comsumer, who consumes electricity a) Reducing only Q (kvar) Possible, reasonable Please note that active power does not change after compensation Q, Reactive Power (kvar) Q c, Compensation (kvar) S, Total Power (kva) P, Active Power (kw) Q new Reactive Power (kvar) new S new, Total Power (kva) P, Active Power (kw) EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 45
46 METU Example Question The factory shown on the RHS draws a load at 6300 V nominal voltage P + j Q = 120 kw + j 140 kvar Calculate the amount of reactive power needed in order to raise the power factor of the factory to 0.95 (Lagging) TEDAŞ 6300 V (rms) Mains 3 km O/H line P + j Q new P + j Q + j Q comp P + j Q = 120 kw + j 140 kvar _ Q comp = Q = Q Q new EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 46
47 Example Answer Uncompensated (Given) Case Tan = Q / P = 140 / 120 = = Tan = o p.f.: cos = 0.65 (lagging) Compensated Case cos new = 0.95, new = cos = o Tan new = Tan o = Tan new = Q new / P Q new = x P = kvar Q comp = Q = Q Q new = = kvar Q =140 kvar Q comp., Reactive Power Compensation (kvar) Q new = (kvar) new S Total Power (kva) P = 120 kw S new, Total Power (kva) P = 120 kw EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 47
48 METU Example Question Now, for the previous problem, calculate the reduction in line losses as a result of this compensation by assuming that line impedance is R + j X = 10 + j 20 Ohms TEDAŞ 6300 V Mains 3 km O/H line R + jx = 10 + j20 Ohms P + j Q new P + j Q + j Q comp P + j Q = 120 kw + j 140 kvar _ Q comp = Q = Q Qnew EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 48
49 S = V I* I = S / V = / 6300 = x 1000 / 6300 = Amp Example S new = V I new * I new = / 6300 = x 1000 / 6300 = Amp P loss = R I 2 = 10 x = Watts P loss-new = R I 2 new = 10 x = Watts P loss = = Watts TEDAŞ 6300 V Mains 3 km O/H line R + jx = 10 + j20 Ohms P + j Q new P + j Q + j Qcomp P + j Q = 120 kw + j 140 kvar _ Q comp = Q = Q Qnew EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 49
50 Example Now, calculate the return rate of the investment to be made for the compensator, by assuming that the retail price of electricity is 16 Cents/kWh and price of capacitor is USD/kVAR P loss = = Watts Investment = kvar * USD/kVAR = 18546,11 USD Saving = / 1000 * 16 Cent/kWh/ 100 = USD / hour Return Rate = Investment / Saving = / = hours = days = 1.58 years EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 50
51 METU Example TEDAŞ 6300 V Mains Question Now, for the previous problem, determine the minimum cross section of the line for the alternative cases, when line is compansated and uncompensated I new I initial = Amp = Amp 3 km O/H line R + jx = 10 + j20 Ohms P + j Q new + _ P + j Q j Q comp The cheaper alternative P + j Q = 120 kw + j 140 kvar Q comp = Q = Q Qnew Cross Section Current Capacity ( mm 2 ) ( Amp ) , , , , , , , , , , , , ,0 EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 51
52 Question Question Now, for the previous problem, calculate the shunt capacitance in Farads needed for the amount of compensation found above Q comp = Q = Q Q new = = kvar Q comp = V I comp * I comp = Q comp / V = VA /6300 V = Amp X comp = V / I comp = 6300 / = Ohms X comp = 1 / ( jwc ) Q new = Q + Q comp C = 1 / ( jwx ) = 1 / ( x ) = mf Q comp < 0 TEDAŞ 6300 V Mains 3 km O/H line R + jx = 10 + j 20 Ohms P + j Q new P + j Q + _ j Qcomp I comp P + j Q = 120 kw + j 140 kvar Q comp = Q = Q Qnew EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 52
53 Medium Voltage Capacitor Banks Shunt connection of large capacity capacitors in power systems EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 53
54 Installation of MV Capacitor Banks EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 54
55 Electronic Capacitors in a Motherboard EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 55
56 Another Example Early in the history or electricity, Thomas Edison's General Electric company was distributing DC electricity at 110 volts in the United States. Then Nikola Tesla the devised a system of three-phase AC electricity at 240 volts. Threephase meant that three alternating currents slightly out of phase were combined in order to even out the great variations in voltage occurring in AC electricity. He had calculated that 60 cycles per second or 60Hz was the most effective frequency. Tesla later compromised to reduce the voltage to 110 volts for safety reasons. EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 56
57 Another Example Europe goes to 50 Hz: With the backing of the Westinghouse Company, Tesla's AC system became the standard in the United States. Meanwhile, the German company AEG started generating electricity and became a virtual monopoly in Europe. They decided to use 50 Hz instead of 60 Hz to better fit their metric standards, but they kept the voltage at 110 V. Unfortunately, 50 Hz AC has greater losses and is not as efficient as 60 HZ. EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 57
58 Another Example Due to the slower speed, 50Hz electrical generators are 20 % less effective than 60Hz generators. Electrical transmission at 50 Hz is about % less efficient. 50Hz transformers require larger windings and 50 Hz electric motors are less efficient than those meant to run at 60Hz. They are more costly to make to handle the electrical losses and the extra heat generated at the lower frequency. Europe goes to 220 V Europe stayed at 110 V AC until the 1950s, just after World War II. They then switched over to 220 V for better efficiency in electrical transmission. Great Britain not only switched to 220 V, but they also changed from 60Hz to 50 Hz to follow the European lead. Since many people did not yet have electrical appliances in Europe after the war, the change-over was not that expensive for them. U.S. stays at 110 V, 60Hz EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 58
59 Another Example The United States also considered converting to 220 V for home use but felt it would be too costly, due to all the 110 V electrical appliances people had. A compromise was made in the U.S. in that 220 V would come into the house where it would be split to 110 V to power most appliances. Certain household appliances such as the electric stove and electric clothes dryer would be powered at 220 V. EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 59
60 Did everybody follow this part carefully? EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 60
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