ECE 41/51 Electric Energy Systems Power Systems Analysis I Basic Princiles Instructor: Kai Sun Fall 014 1
Outline Power in a 1-hase AC circuit Comlex ower Balanced 3-hase circuit
Single Phase AC System it ( ) = I cos( ωt+ θ ) m i + vt ( ) = Vm cos( ωt+ θv) - Z Load 3
Phasor Reresentation vt ( ) = V cos( ωt+ θ ) = V cos( ωt+ θ ) m v v it ( ) = I cos( ωt+ θ ) = I cos( ωt+ θ ) m i i V= V θ v I= I θ i RMS hasors of v(t) and i(t) Reference A hasor is a comlex number that carries the amlitude and hase angle information of a sinusoidal function Phasors reresent sinusoidal signals of a common frequency (ω) as vectors in the comlex lane w.r.t. a chosen reference signal. Phasor reresentation is a maing from the time domain to comlex number domain. 4
Instantaneous ower delivered to the load: Using trigonometric identity 1 cos Acos B = cos( A- B) + cos( A+ B) θ = θ θ v V = V /, I = I / m [ ] i Imedance angle >0 for inductive load and <0 for caacitive load Root-mean-square (RMS) values m t () = vtit ()() = VI cos( ωt+ θ )cos( ωt+ θ ) m m v i 1 = VmIm[ cos( θv θi) + cos( ωt + θv + θi) ] 1 = VmIm{ cos( θv θi) + cos[( ωt + θv) ( θv θi)] } 1 = VmIm{ cosθ + cos[( ωt + θv) θ] } 1 = VmIm[ cosθ + cos ( ωt + θv) cosθ + sin ( ωt + θv) sinθ ] ( t) = V I cos θ[1 + cos ( ωt + θv)] + V I sinθsin ( ωt + θv) () t () t R X 5
Examle.1 Assume vt ( ) = 100cosωt Z = 1.5 60 Ω Calculate the RMS current hasor: 1 100 0 1 I = = 80 60 o 1.5 60 it ( ) = 80cos( ωt 60 ) 100 v(t)=v m cos ωt, i(t)=i m cos(ωt -60) 6000 (t)=v(t) i(t) 50 4000 0 000-50 0-100 0 90 180 70 360 ωt, degree 4000 3000 000 R (t) Eq..6-000 0 90 180 70 360 ωt, degree 4000 000 0 X (t) Eq..8 1000-000 0 0 90 180 70 360-4000 ωt, degree 0 90 180 70 360 ωt, degree 6
( t) = V I cos θ[1 + cos ( ωt + θv)] + V I sinθsin ( ωt + θv) () t () t R X Observations on R (t): Oscillating at frequency ω (twice of the source frequency) Changing between 0 and V I cosθ (always ositive) Average value is V I cosθ It is the energy flow into the circuit Observations on X (t): Oscillating at frequency ω (twice of the source frequency) Changing between - V I sinθ and V I sinθ (either ositive or negative) Average value is zero It is the energy borrowed & returned by the circuit. It does no useful work in the load but takes some line caacity 7
Real and Reactive Powers ( t) = V I cos θ[1 + cos ( ωt + θv)] + V I sinθsin ( ωt + θv) V I Aarent ower def P = V I cosθ Real ower or active ower (average ower) Unit ~ watt or W Power factor (PF): cosθ= cos(θ v -θ i ) () t () t Unit ~ volt amere or VA (kva or MVA) R Lagging or leading when θ =θ v -θ i >0 or <0 X Q P def Q = V I sinθ Reactive ower Unit ~ var (volt-amere reactive). Some eole use Var, VAR or VAr Q>0 or <0 when θ=θ v -θ i >0 or <0 8
Characteristics of instantaneous ower (t) ( t) = V I cos θ[1 + cos ( ωt+ θv)] + V I sinθsin ( ωt+ θv) For a ure resistor load (thermal load), () t () t R = P[1 + cos ( ωt+ θ )] + Qsin ( ωt+ θ ) θ=θ v -θ i =0, PF=1 (unity PF) P= V I, so all electric energy becomes thermal energy For a ure inductive load, v θ=θ v -θ i =90, PF=0 P=0, so electric energy = constant (no transformation to other forms) (t) oscillates between the source and the magnetic field associated with the inductive load For a ure caacitive load, θ=θ v -θ i =-90, PF=0 P=0, so electric energy = constant (no transformation to other forms) (t) oscillates between the source and the electric field associated with the caacitive load v X 9
Examle.1 Z = 1.5 60 Ω vt ( ) = 100cosωt o it ( ) = 80cos( ωt 60 ) RMS hasors: V 100 80 = 0, I = 60 o o ( t) = V I cos θ[1 + cos ωt] + V I sinθsin ωt = 4000cos 60 [1 + cos ωt] + 4000sin 60 sin ωt () () () () R t X t R t X t P Q 100 v(t)=v m cos ωt, i(t)=i m cos(ωt -60) 6000 (t)=v(t) i(t) 50 4000 0-50 000 0 P -100 0 90 180 70 360 ωt, degree 4000 R (t) Eq..6-000 0 90 180 70 360 ωt, degree 4000 X (t) Eq..8 3000 000 1000 P -000 0 0 90 180 70 360-4000 ωt, degree 0 90 180 70 360 ωt, degree 000 0 Borrowing Returning Q Q 10
What does absorbing or generating reactive ower mean? Reactive ower with a circuit element does not lead to real energy consumtion Starting from the time of v(t)=v m, An inductive element (lagging PF) first absorbs and then returns the same amount of electric energy. Such a attern is considered absorbing reactive ower, and it hels reduce oscillations of ower A caacitive element (leading PF) first generates and then absorbs the same amount of electric energy. Such a attern is considered generating reactive ower, and it hels maintain voltage For a real-world ower system with long-distance transmission, low voltage at receiving ends is usually a big reliability concern. Shunt/series caacitors can be added as sources of reactive ower or var 11
Comlex Power def S = P + jq = V I cosθ + jv I sinθ = V I θ = V I θ θ = VI v i = S θ = P + Q θ Reference S is the aarent ower θ=tan -1 (Q/P) When θ>0 (θ i <θ v, i.e. lagging PF), Q>0 i.e. absorbing Q (inductive load) When θ<0 (θ i >θ v, i.e. leading PF), Q<0 i.e. generating Q (caacitive load) 1
PF=cosθ=P/ S (lagging/leading is told by +/- sign of Q) P= S cosθ= S PF = Q= = S sin θ S 1 PF, if lagging S sin θ S 1 PF, if leading If the load imedance is Z, i.e.v=zi, then S = VI = ZII = Z I = R I +jx I = P + jq P=R I Q=X I Z R X The load imedance angle θ is also called ower angle (ϕ in some literature) Aarent ower S I indicates heating and is used as a rating unit of ower equiment 13
Some useful equations: VV V S = VI = = = V Y Z Z Z = V S Y = S V If Z is urely resistive P = V R If Z is urely reactive Q V V V = = or X ωl 1/ ωc 14
Comlex Power Balance S = VI = V[ I + I + I ] = VI + VI + VI = S + S + S 1 3 1 3 1 3 P + jq = P + jq + P + jq + P + jq 1 1 3 3 Due to the conservation of energy, the total comlex ower delivered to the loads in arallel is the sum of comlex owers delivered to each (by KCL). In other words, a balance must be maintained all the time for both real ower and reactive ower 15
Examle. V = 100 0 V, Z = 60 + j0 Ω, Z = 6 + j1 Ω, Z = 30 j30ω 3 1 Find the ower absorbed by each load and the total comlex ower I I I 1 3 V 100 0 100 = = = = 0 + j0 A Z1 60 + j0 60 V 100 0 00 00(1 j) = = = = = 40 j80 A Z 6 + j1 1+ j 5 V 100 0 40 40(1 + j) = = = = = 0 + j0 A Z 30 j30 1 j 3 S = VI = 100 0 (0 j0) = 4, 000 W+ j0 var 1 1 3 3 S = VI = 100 0 (40 + j80) = 48, 000 W+ j96, 000 var S = VI = 100 0 (0 j0) = 4, 000 W j4, 000 var S = S1 + S + S3 = 96,000 W+ j7,000 var 16
Other aroaches I = I + I + I = (0 + j0) + (40 j80) + (0 + j0) 1 3 = 80 j60 = 100 36.87 A S (100 0 )(100 36.87 ) 10,000 36.87 VA = VI = = = 96,000 W + j7,000 var S S S V (100) = = = 4,000 W + j0 var 60 1 Z1 V (100) = = = 48,000 W + j96,000 var 6 j1 Z V (100) = = = 4,000 W j4,000 var 30 + j30 3 Z3 17
Theorem of Conservation of Comlex Power For a network sulied by indeendent sources all at the same frequency (all voltages and currents are assumed to be sinusoids), the sum of the comlex ower sulied by the indeendent sources equals the sum of the comlex ower received by all the other branches of the network For a single source with elements in series or arallel, it is roved by Kirchhoff s voltage or current law (KVL/KCL) For a general case, it is roved by Tellegen s theorem. Alication of the theorem: a art of the network can be relaced by an equivalent indeendent source 18
Some examles on Bergen and Vittal s book 19
0
Assume V 1 = V = V 3 1
Power Factor Correction P= V I cosθ= S PF When PF<1, current I needs to increase by 1/PF to deliver the same P as that with PF=1. Thus, the major loads of the system are exected to have close to unity PFs. That is not a roblem for residential and small commercial customers. However, industrial loads are inductive with low lagging PFs, so caacitors may be installed to imrove PFs
Examle.3 P+jQ (a) (b) Find P, Q, I and PF at the source Find caacitance C of the shunt caacitor to imrove the overall PF to 0.8 lagging C Solution: (a) I I 1 00 0 = = 0 A 100 00 0 = = 4 j8 A 10 + j0 S = VI = 00 0 ( j0) = 400 W+ j0 var 1 1 S = VI = 00 0 (4 + j8) = 800 W + j1600 var S = P + jq = 100 + j1600 = 000 53.13 VA S 000 53.13 I = = = 10 53.13 A V 00 0 PF = cos(53.13 ) = 0.6 lagging (b) New ower angle θ 1 = cos (0.8) = 36.87 Q = P tanθ = 100 tan(36.87 ) = 900 var Q Z c c = 1600 900 = 700 var c 1 V (00) = = = = j57.14 Ω jωc S j700 6 10 C = = 46.4 μf 377(57.14) 3
New aarent ower and current S = 100 + j900 = 1500 36.87 000 1500 S 1500 36.87 I = = = 7.5 36.87 V 00 0 10 7.5 Learn Examle.4 (high-voltage) High-voltage caacitor bank (150kV - 75MVAR) (Source: wikiedia.org) 4
Question True or False? A load sulied by a voltage source has 0.7 ower factor lagging. A shunt caacitor connected across the load will necessarily imrove the ower factor of the load side. 5
Comlex Power Flow V1 = V1 δ1 V = V δ P 1 +jq 1 P 1 +jq 1 I 1 = V δ V δ 1 1 Z γ V1 V = δ1 γ δ γ Z Z V1 V S1 = VI 1 1 = V1 δ 1[ 1 ] Z γ δ Z γ δ V1 V1 V = γ γ + δ1 δ Z Z P V V V = cosγ cos( γ + δ δ ) Z Z 1 1 1 1 Q V V V = sin γ sin( γ + δ δ ) Z Z 1 1 1 1 If R=0, i.e. Z =X and γ=90 o P V V = sin( δ δ ) X 1 1 1 V1 Q1 = [ V1 V cos( δ1 δ )] X 6
P V V = sin( δ δ ) X 1 1 1 V1 Q1 = [ V1 V cos( δ1 δ )] X Since R=0, there are no transmission line losses, and P 1 =-P 1 In a normal ower systems, V i 100% and δ 1 -δ is small (e.g. 5~30 o ). P 1 sin(δ 1 -δ ) δ 1 -δ Real ower flow is governed mainly by the angle difference of the terminal voltages. For examle, if V 1 leads V, i.e. δ 1 -δ >0, the real ower flows from node 1 to node. Theoretical maximum ower transfer (i.e. static transmission caacity): V V P1 max = sin 90 = X V V X 1 1 Q 1 V 1 - V cos(δ 1 -δ ) V 1 - V Reactive ower flow is determined by the magnitude difference of terminal voltages 7
Examle.5 Determine the real and reactive owers sulied or received by each source and the ower loss in the line (i.e. consumtion of Z) S 1 =P 1 +jq 1 S 1 =P 1 +jq 1 I 1 V = 10 5 V, V = 100 0 V, Z = 1 + j7 Ω 1 I I 1 1 10 5 100 0 = = 3.135 110.0 A 1+ j7 100 0 10 5 = = 3.135 69.98 A 1+ j7 1000 800 600 P 1 400 S = VI = 376. 105.0 = 97.5 W + j363.3 var 1 1 1 S = VI = 313.5 69.98 = 107.3 W j94.5 var 1 1 1 P, Watts 00 0 P L S = S1 + S1 = 9.8 W + j68.8 var = P + jq L L L -00-400 L 1 (1)(3.135) 9.8 W= 1 1 P = RI = = P + P L 1 (7)(3.135) 68.8 var = 1 1 Q = XI = = Q + Q -600-800 P 1-1000 -40-30 -0-10 0 10 0 30 Source #1 Voltage Phase Angle, δ 1 8 (angle difference)
Balanced Three-Phase Circuits A balanced source: three sinusoidal voltages are generated having the same amlitude but dislaced in hase by 10 o Instantaneous ower delivered to the external loads is constant Positive hase sequence Negative hase sequence 9
Balanced Y-connected Loads EAn = E α A EBn = E α A 10 ECn = E α A 40 VAn = EAn ZG Ia Van = VAn ZLIa Let V an be the reference: Phase voltages (line-to-neutral voltages): V an = V 0 V V 10 bn = V V 40 cn = Line voltages (line-to-line voltages): V = V V = V (1 0 1 10 ) = 3 V 30 ab an bn V = V V = V (1 10 1 40 ) = 3 V 90 bc bn cn V = V V = V (1 40 1 0 ) = 3 V 150 ca cn an V = 3V L V bc 30
I I a I b c Van Van 0 = = = I θ Z Z θ Vbn = = I 10 θ Z Vcn = = I 40 θ Z I L = I On the neutral line (return line) I n =I a +I b +I c =0 The neutral line may not actually exist (one line er hase) The balanced 3-hase ower system roblems can be solved on a er-hase basis, e.g. for Phase A only. The other two hases carry identical currents excet for hase shifts. 31
Balanced -connected Loads V L = V Let I ab be the reference: I I ab bc I ca = I 0 = I 10 = I 40 I = I I = I (1 0 1 40 ) = 3 I 30 a ab ca I = I I = I (1 10 1 0 ) = 3 I 150 b bc ab I = I I = I (1 40 1 10 ) = 3 I 90 c ca bc I = 3 L I 3
-Y Transformation I a = f Y (V an, Z Y ) =V an /Z Y Z Z n I a = f (V an, Z ) Z I a V V V + V = + = Z Z Z ab ac ab ac I a = 3V an Z V ca V ab +V ac Comared to : I a = V Z an Y Z Y = Z 3 V + V = 3 V 30 + 3 V 30 ab ac an an = 3 V ( 3/ + j/+ 3/ j/) = 3V an an 33
Balanced Three-Phase Power v = V cos( ωt+ θ ) an v v = V cos( ωt+ θ 10 ) bn v v = V cos( ωt+ θ 40 ) cn v For balanced loads i = I cos( ωt+ θ ) an i i = I cos( ωt+ θ 10 ) bn i i = I cos( ωt+ θ 40 ) cn i Total instantaneous ower: = 3 vani + a vbni + φ b vcnic = V I cos( ωt+ θ )cos( ωt+ θ ) v i + V I cos( ωt+ θ 10 ) cos( ωt+ θ 10 ) v i + V I cos( ωt+ θ 40 ) cos( ωt+ θ 40 ) v i 3 φ = V I [cos( θv θi) + cos( ωt+ θv + θi)] + V I [cos( θ θ ) + cos(ωt+ θ + θ 40 )] v i v i + V I [cos( θ θ ) + cos(ωt+ θ + θ 480 )] v i v i Zero Total instantaneous ower is constant: = 3 V I cosθ = P 3φ 3φ Where does reactive ower go? 34
Extend the concet of comlex ower to 3-hase systems P3 = 3 V I cosθ φ Q3 φ = 3 V I sinθ S = P + jq = V I 3φ 3φ 3φ 3 Exressed in terms of line voltage V L and line current I L Y-connection: V = V / 3 I = IL -connection: V L = V I = I / 3 L L P3 φ = 3 VL IL cosθ Q3 = 3 V I sin φ L L θ S = 3φ 3VI L L Do we have? Note: θ is the angle between the hase voltage and the hase current for the same hase (e.g. Phase A) 35
Examle.7 (a) (b) (c) (d) The current, real ower and reactive owers drawn from the suly The line voltage at the combined loads The current er hase in each load The total real and reactive owers in each load and the line 36
V =110-j0 37