10.1 Instantaneous Power 10.2 Average and Reactive Power 10.3 The RMS Value and Power Calculations 10.4 Complex Power

Transcription

1 SINUSOIDAL STEADY-STATE STATE POWER CALCULATIONS C.T. Pan Instantaneous Powe 10. Aveage and Reactive Powe 10.3 The RMS Value and Powe Calculations 10.4 Complex Powe C.T. Pan

2 10.5 Powe Calculations 10.6 Maximum Powe Tansfe 10.7 Measement of Powe C.T. Pan Instantaneous Powe Evey electic device o equipment has a powe ating. Exceeding the powe ating can do pemanent damage to an appliance. C.T. Pan 4

3 10.1 Instantaneous Powe (1) instantaneous powe 瞬時功率 () aveage (eal) powe 平均功率,( 實功率, 有效功率 ) (3) eactive (imaginay) powe 無效功率,( 虛功率 ) (4) complex powe 複數功率 (5) appaent powe 視在功率 C.T. Pan Instantaneous Powe Assume passive sign convention pt () = vt ()() it watt instantaneous powe C.T. Pan 6

4 10.1 Instantaneous Powe v'( t) = V cos( ωt+ θ ) m i'( t) = I cos( ωt+ θ ) m i v C.T. Pan Instantaneous Powe vt () = V cos( ωt+ θ θ ) m v i it () = I cos( ωt) m C.T. Pan 8

5 10.1 Instantaneous Powe pt () = VI cos( ωt+ θ θ )cos( ωt) m m v i 1 1 QcosAcosB= cos( A B) + cos( A+ B) VI m m VI m m pt () = cos( θv θi) + cos( ωt+ θv θi) Qcos( A+ B) = cosacosb sinasin B VI m m VI m m pt () = cos( θv θi) + cos( θv θi)cos( ωt) VI m m sin( θv θi)sin( ωt) C.T. Pan Aveage and Reactive Powe pt () = P+ Pcosωt Qsinωt VmIm whee P = cos( θv θi), in watt ( 瓦 ) VmIm Q = sin( θv θi), in va ( 乏 ) 1 t + T Q = τ τ T Qis called the eactive powe 0 P p( ) d, called aveage powe t0 C.T. Pan 10

6 10. Aveage and Reactive Powe What is the coesponding elation in phaso domain? If vt () = V cos( ωt + θ ) it () = Im cos( ωt + θi) then V = Vm θv I = I θ 1 P = VmIm cos( θv θi) 1 Re[ * = VI ] * Q = 1 Im[ VI ] C.T. Pan 11 m m It tnsoutthat i v 10.3 The RMS Value and Powe Calculations The effective value, I eff, of a peiodic cent i(t) is the equivalent dc cent that delives the same aveage powe to a esisto as the peiodic cent. C.T. Pan 1

7 10.3 The RMS Value and Powe Calculations 1 = T t + T 0 P i ( τ) Rd t0 τ P= I R eff C.T. Pan The RMS Value and Powe Calculations 1 t0 + T P= P Ieff = i ( ) d T τ τ t0 Theeffectivevalueof apeiodic signal isits oot mean squae( RMS) value. C.T. Pan 14

8 10.3 The RMS Value and Powe Calculations If it () = I cos( ωt+ θ ) m I T m then Ims = cos ( t ) 0 i dt T ω + θ Im = Thus, wecandefinethems phaso of it () as Im I ms = θ C.T. Pan 15 i 10.4 Complex Powe VI m m Q P= cos( θv θi) 1 Re[ * = VI ] * V I = Re[ ] * = Re[ V ms Ims ] C.T. Pan 16

9 10.4 Complex Powe Similaly VI m m Q= sin( θ θ ) VI m m Similaly Q= sin( θv θi) 1 uvv* = Im[ VI ] uv v * * 1 V I = = Im[ VI] uuuvuuuv * = I [ V I ] 1 Im[ ] m ms ms * * V I ms ms = Im[ ] * = Im[ V ms I ms ] Then, wecandefinethecomplex powe uv uuuvuuuv S = P+ jq= V I Then, wecandefinethecomplex powe * S = P+ jq= V ms Ims C.T. Pan 17 v i 10.4 Complex Powe Apowetiangle uv S = P + Q uv S = P+ jq uv S : appaent powe inva P: eal powe, in w Q: eactive powe, in va ( volt ampeeeactive) C.T. Pan 18

10 10.4 Complex Powe Powe factodefinition θ v pf = cos( θ θ) 1 θ i v = sin( θ θ) v i i is called the powe facto angle Reactive factodefinition f C.T. Pan Complex Powe Fom complex ohm' s law V ms = ZIms = ( R+ jx) Ims * S = V ms Ims = ZIms = RIms + jx Ims = P+ jq C.T. Pan 0

11 10.4 Complex Powe pf = cos θ, P pf = S θ=θv θi Q is a mease of the enegy exchange between the soce and the eactive pat of the load. Reactive powe epesents a lossless intechange between the load and the soce. C.T. Pan Complex Powe V V Z = = θ θ I I m uv V Vm Z = v = v θv θii I Im m = R+ jx uv Q P= S cos( θv θi) jxuv uv = S pf S = R+ Q P = S cos( θ θ ) = S pf S Q = 0, esistive load, pf = 1.0 Q > 0, v inductive i load, lagging pf Q< 0, capacitive load, leading pf Q= 0, esistive load, pf = 1.0 Q> 0, inductive load, lagging pf Q< 0, capacitive load, leading pf C.T. Pan

12 10.4 Complex Powe S θ v -θ i P jq Lagging pf means that cent lags voltage. Leading pf means that cent leads voltage. C.T. Pan 3 V ms 10.5 Powe Calculations = S V ms I * ms it () = 10cos( wt+ 30 ) A vt () = 100cos( wt+ 30 ) V 100 V ms = 30 V 10 Ims = 30 A = VA V ms = j0 VA = = I ms R R P = 500 W, Q = 0va pf = RI ms = cos0 = 1.0 C.T. Pan 4

13 Time 10.5 Powe Calculations domain pt () = vtit ()() = 100cos( wt+ 30) 10cos( wt + 30) = 500(1+ cos wt) VA C.T. Pan Powe Calculations V ms = jxi ms it () = 10 cos( t 45 ) vt () = 100 cos( t+ 45 ) * S = V ms Ims = = = 0+ j500va P= 0, Q= 500va pf = cos90 = 0, lagging C.T. Pan 6

14 10.5 Powe Calculations Time domain pt () = v'( ti )'( t) = 100cos( t + 90) 10cost = 500[cos90+ cos(t + 90)] VA = 500sint = Qsin, t Q = 500va C.T. Pan Powe Calculations it () = 10 cos(t+ 60 ) A vt () = 100 cos(t 30 ) V * S = V ms Ims = VA = VA= P+ jq P= 0, Q= 500va pf = 0, leading C.T. Pan 8

15 10.5 Powe Calculations Time domain i'( t) = 10 cost v'( t) = 100cos(t 90 ) pt () = v'( ti )'( t) = 1000cos(t 90)cost = 500[cos( 90) + cos(4t 90)] =+ 500sin4t = Qsin4t = ( 500)sin4t Q= 500va, leading C.T. Pan Powe Calculations Example. Given the load voltage and cent o v( t) = 60cos( ωt 10 ) V o i() t = 1.5cos( ωt + 50 ) A Find (a) S& S (b) P and Q (c) pf and Z L Solution ( ) 60 o a V ms = 10 V 1.5 o I ms = 50 A * S = V ms I ms 60 o 1.5 = = VA C.T. Pan o 30 o

16 10.5 Powe Calculations o o ( b) S= 45cos( 60 ) + j45sin( 60 ) =.5 j38.97 P =.5 W Q= VAR VA o () c pf = cos( 60 ) = 0.5 leading o V o ZL = = = Ω o I C.T. Pan 31 V 10.5 Powe Calculations Consevation of AC Powe I I 1 u I Z 1 Z Assume ms phasos The complex powe supplied by the soce * * S = VI = V( I1+ I) * * = VI1 + VI = S1+ S C.T. Pan 3

17 Similaly V 10.5 Powe Calculations I Z 1 Z V 1 V u Assume ms phasos The complex powe supplied by the soce * * S = VI = ( V1+ V ) I * * = V1I + V I = S1+ S C.T. Pan Powe Calculations The complex ( eal, o eactive ) powe of the soce equals the espective sum of the complex ( eal, o eactive ) powes of the individual loads. C.T. Pan 34

18 10.5 Powe Calculations Most industial loads ae inductive and ae opeated at a low lagging powe facto. I Z θ S θ θ θ V pf 1 Q = cos θ, θ = tan P V θ V I = Z θ C.T. Pan 35 I 10.5 Powe Calculations This will sacificing some geneato eal powe output capability to povide the eactive powe. Also, fom the uses view point, to educe enegy cost due to the penalty of low pf, it is woth to incease the pf. Solution: fo inductive loads, Q L > 0 add paallel capacito, Q C < 0 without influencing the aveage powe P. I L V u L C.T. Pan 36

19 V I I L 10.5 Powe Calculations I C then oiginal load S P Q V I * 1 = = 1 = Z C.T. Pan 37 V θ 1 V = Z θ pf1 = cosθ1 Suppose afte pf coection pf = cosθ u S = P u + jq u = S cosθ + js sinθ = P + jq 1 θ 1 u S 1 u S but θ 10.5 Powe Calculations P = S cosθ = P S = 1 P 1 cosθ P1 then Q = sinθ = P tanθ cosθ SC = S S = P tanθ P tanθ * * V V V ω c C C * Z 1 C * j S = V I = V = = = jv ω c ( j ) ω c P (tan θ tan θ ) = V ω c c 1 1 = P 1 1 C.T. Pan 38 (tan θ tan θ ) ω V

20 10.5 Powe Calculations Example : u voltage soce : VS = 10 V ( ms) ω = 60 π ad / s electical load :absobs 4KW at a lagging pf of 0.8 Find C necessay to aise the pf to 0.95 Solution : 1 0 pf = 0.8, θ = cos 0.8= u S P 4kw 1 1 = = = cosθ VA u = = = 0 Q1 S1 sinθ sin VAR C.T. Pan Powe Calculations when pf = θ = cos 0.95= u 4kw 4000w S = = = VA cosθ 0.95 u Q = S sinθ = VAR Q=Q Q = = VAR 1 V Q C = = V ω c 1 ω c c = = 310.5µ F π C.T. Pan 40

21 10.6 Maximum Powe Tansfe Given a linea two-teminal ac cicuit, N, with load impedance Z L Fom Thevenin stheoem V TH C.T. Pan Maximum Powe Tansfe Z = R + jx Let TH TH TH Then Z = R + jx L L L VTH VTH I= = ZTH+ ZL ( RTH+ RL) + jx ( TH+ XL) 1 1 VTH RL P= I RL = ( ) ( R + R ) + ( X + X ) TH L TH L Necessay condition fo maximum output P P = 0...(A) R L P X = 0...(B) C.T. Pan 4 L

22 10.6 Maximum Powe Tansfe Fom (B) : XL= XTH Fom (A) : R = R + ( X + X ) L TH TH L Hence, fo maximum output powe Z = R + jx = R jx = Z * L L L TH TH TH P 1 V TH V = RTH = ( ) 8R max RTH TH TH amplitude phaso C.T. Pan Maximum Powe Tansfe In case the load is pely eal Then, Z = R + j0, X = 0 L L R = R + X = Z L TH TH TH L C.T. Pan 44

23 10.6 Maximum Powe Tansfe Example 1: Detemine Z L to get maximum aveage powe output P Z L amplitude phaso 10 0 o V Z = j5ω+ 4 Ω //(8Ω j6 Ω) V TH TH =.933+j4.467 Ω 8 j6 = = 4 + (8 j6) V C.T. Pan Maximum Powe Tansfe Fom maximum powe tansfe theoem * ZL ZTH.933 j4.467 P Z L = = Ω V TH = = 8 R TH.368 W C.T. Pan 46

24 10.6 Maximum Powe Tansfe Example : Find R L such that it will absob maximum aveage powe o V C.T. Pan Maximum Powe Tansfe Z TH = (40 j30)//0 j = j.35 Ω j0 0 0 VTH = j0+ 40 j30 = (Amplitude phaso!!!) o V Fom maximum aveage powe tansfe theoem RL = ZTH = = 4.5Ω V TH 0 I = A Z TH + R = L 1 Pmax = I R L 1 = (1.8) (4.5) = 39.9 W C.T. Pan 48

25 10.7 Measement of Powe The wattmete is the instument fo measing the aveage powe. u I L u I L V u L V u L C.T. Pan Measement of Powe An electomagnetic type wattmete consists of two coils. the cent coil (C.C) : in seies with load the voltage coil (V.C) : in paallel with load The mechanical inetia of the moving pat poduces an equilibium deflection angle that is popotional to the aveage value of the poduct v () t i (), t ie.. α 1 T t t T C.T. Pan 50 L v () t i () tdt = P L L L

26 ± 10.7 Measement of Powe The sign maked on the teminals of C.C. and V.C. is fo passive sign convention to ense upscale deflection. Example :Find the eading of W o V ms C.T. Pan Measement of Powe Solution : I L = = A (1+ j10) + (8 j6) 0+ j4 u 150 (8 j6) VL = I L (8 j6) = V 0+ j4 u * 150 (8 j6) 150 S = VL I L = 0+ j4 0 j4 =43.7 j34.6 VA P = 43.7 W C.T. Pan 5

27 Summay Objective 1 : Undestand the ac powe concepts of P, Q, S, S. Objective : Know the ms phasos and be able to calculate vaious ac powes and powe facto. Objective 3 : Undestand the condition fo maximum eal powe tansfe and be able to calculate the maximum eal powe. C.T. Pan 53 Summay Objective 4 : Undestand the pf coection technique and be able to find the desied compensation. Objective 5 : Know how to use a wattmete to mease the aveage powe of an ac load. C.T. Pan 54

28 Summay Chapte Poblems : Due within one week. C.T. Pan 55