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# Review of basic concepts (kuliah ke 3)

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### Review of basic concepts (kuliah ke 3)

1. 1. Review of Basic Concepts Pekik Argo Dahono
2. 2. Average and RMS Concepts• Periodic signals x(t ) = x(t + T )• Average value 1 t o +T x= T to∫ x(t )dt• RMS value 1 t o +T 2 X = T to∫ x (t )dt
3. 3. Single-Phase Power Concept• Sinusoidal voltage and current v = 2V cos(ωt ) i = 2 I cos(ωt − φ )• Instantaneous power p = vi = VI 444+ cos(2ω3] + VI 44 sin (2ωt ) 1 cos φ [12444t ) 1sin φ244 3 resistive part reactive part
4. 4. Single-Phase Power Concept• Active or Average Power t o +T ∫ 1 P= pdt =VI cos φ T to• Reactive Power Q = VI sin φ• Apparent Power S = VI
5. 5. Single-Phase Power Concept• Power triangle S = P +Q 2 2 2• Power factor P PF = = cos φ S
6. 6. Single-Phase Power Concept E 2 = (V + ΔV )2 + δV 2 I P + jQ E R jX Load V δV 2 << (V + ΔV )2 Thus ΔV ≈ E-V = RI cos φ + XI sin φ RP + XQ XQ = ≈ E V V δV Losses = RI 2 = R(S / V )2 δ φ VI ΔV [ = R (P / V )2 + (Q / V )2 ]
7. 7. Balanced Three-Phase Power• Voltage and current va = 2V cos(ωt ) ( π vb = 2V cos ωt − 23 ) vc = 2V cos(ωt + 23 ) π ia = 2 I cos(ωt − φ ) ( π ib = 2 I cos ωt − 23 − φ ) ( π ic = 2 I cos ωt + 23 − φ )
8. 8. Balanced Three-Phase Power• Instantaneous power p = va ia + vb ib + vc ic p = 3VI cos φ• Instantaneous power is a constant that is equal to average power
9. 9. Balanced Three-Phase Power• Reactive power is defined as Q = 3VI sin φ• Apparent power is defined as S = 3VI
10. 10. Three-Phase Power System Rs I Rs 3I Rs I Rs Rs I Rs Ro Ro Ro Ro / 3 Rs RsLosses = 3RI 2 Losses = 18RI 2 PF = 1.0 PF = ?
11. 11. Three-Phase Power System Rs Rs Rs − j1 j1 1 Rs PF = ?
12. 12. Three-Phase Four-Wire SystemsSeries Losses Consideration : Three - Phase Three - Wire System (ΔP = rs I a + I b + I c + I n 2 2 2 2 ) (unbalanced) Ie = (I 2 ) + Ib + Ic / 3 2 2 aΔP = 3rs I e 2 (V ) (balanced) I a + Ib + Ic + I n 2 2 2 2 Ve = 2 an + Vbn + Vcn / 3 2 2Thus, I e = 3Shunt Losses Consideration : = (V 2 ab + Vbc + Vca 2 2 )/ 9 Vao + Vbo + Vco + Vno 2 2 2 2ΔP = (unbalanced) Rsh Ve2ΔP = 3 (balanced) Rsh Vao + Vbo + Vco + Vno 2 2 2 2Thus, Ve = 3or Ve = (V 2 an 2 2 2 2 2 ) + Vbn + Vcn + Vab + Vbc + Vca / 12Apparent Power S e = 3Ve I ePower Factor = P / S e
13. 13. Fourier Series Representation• Fourier series ∞ x(t ) = co + 2 ∑C n =1 n sin (nωt + θ n )• Average value x = co• RMS Value ∞ X = co + 2 ∑n =1 2 Cn
14. 14. Total Harmonic Distortion (THD)• Voltage signal ∞ v= 2 ∑V n =1 n sin (nωt + θ n )• Total Harmonic Distortion 1/ 2 ⎛ ∞ ⎞ ⎜ ∑ ⎜ Vn2 ⎟ ⎟ THD = ⎝ n=2 ⎠ V1
15. 15. Power concept under nonsinusoidal waveforms ∞• Voltage v = Vo + 2 ∑V n =1 n cos(nωt + α n ) ∞• Current i = Io + 2 ∑I n =1 n cos(nωt + β n )• Instantaneous power p = vi
16. 16. Power concept under nonsinusoidal waveforms• Average power ∞ P = Vo I o + ∑V I n =1 n n cos(α n − β n )• Apparent power S = Vrms I rms• Power factor ∞ P Vo I o + ∑V I n =1 n n cos(α n − β n ) PF = = S Vrms I rms
17. 17. Example• Voltage v = 1 + 210 cos(100πt ) + 2 2 cos(300πt )• Current ( ) i = 2 5 cos 100πt − 60 − 2 cos(300πt ) o• Average power ( ) P = 50 cos 60 o + 2 cos(π ) = 25• RMS Voltage and Current V = 12 + 10 2 + 2 2 = 105 I = 5 2 + 12 = 26• Power factor P 25 PF = = = 0,478 S 105 6
18. 18. Sinusoidal voltage caseAverage power : P = V1 I1 cos(α1 − β1 )Power factor : P I I1 PF = = 1 cos(α1 − β1 ) = cos φ1 S I rms ⎡ 2 ∞ 2⎤ 1/ 2 ⎢ I1 + ⎢ ∑In ⎥ ⎥ ⎣ ⎦ φ1 = α1 − β1 n=2where :Relationship between power factor and THD: cos φ1 PF = 1 + THD 2
19. 19. Transformer and inductor• Inductor is used to store temporary energy and also used to smoothing the current• Transformer is used for voltage conversion, galvanic isolation, and also used to store temporary energy.• Transformer and inductor are the heaviest component in power electronics system.• In power electronics applications, transformer sometimes has to operate with both dc and ac voltages or currents.
20. 20. Electromagnetism Hukum Ampere : ∫ H.dl = ℑ = NI At Hlc = NII Φ H= NI At/m lc A μNI B = μH = Wb/m 2N lilit lc μA Φ = BA = NI Wb lc μAN 2 λ = NΦ = I lc λ μAN 2 L= = I lc
21. 21. Airgap Influence ℑ = NI = H c lc + H g g Bc Bg ⎛ l g ⎞ NI = lc + g = Φ⎜ c + ⎟ μ μo ⎜ μAc μ o Ag ⎟ ⎝ ⎠ Ac ≈ Ag = AI Φ μ = μ r μo NI Φ=N lilit g lc g + μr μo A μo A N 2I λ = NΦ = lc g + μr μo A μo A N2 L= ≈ μ o AN 2 / g lc g + μr μo A μo A
22. 22. Transformeri1 i2v1 N1 N2 v2
23. 23. Ideal Transformer i1 i2 • • v1 N1 = =− i2v1 N1 N 2 v2 v2 N 2 i1 Ideal transformer neither dissipates nor stores energy
24. 24. Practical Transformer i1 i2 Ll1 • • Ll 2v1 Lm N1 N2 v2
25. 25. Symmetrical ComponentsAny unbalanced three-phase quantities can be composed into three symmetrical components:- Positive sequence components- Negative sequence components- Zero sequence components Ib2 I c1 I ao = I bo = I co I a1 I a2 I b1 I c2
26. 26. Symmetrical Components⎡Vao ⎤ ⎡1 1 1 ⎤ ⎡Va ⎤ ⎡Va ⎤ ⎡ 1 1 1⎤ ⎡Va1 ⎤⎢V ⎥ = 1 ⎢1 a ⎢V ⎥ = ⎢a 2 1⎥ ⎢Va 2 ⎥⎢ a1 ⎥ 3 ⎢ a 2 ⎥ ⎢Vb ⎥ ⎥⎢ ⎥ ⎢ b⎥ ⎢ a ⎥⎢ ⎥⎢Va 2 ⎥ ⎢1 a 2 a ⎥ ⎢Vc ⎥ ⎢Vc ⎥ ⎢ a ⎣ ⎦ ⎣ a2 1⎥ ⎢Vao ⎥ ⎦⎣ ⎦⎣ ⎦ ⎣ ⎦⎣ ⎦ 2π ja= e 3 I co I c2 I c1 Ia Ic I ao Ib I a1 I a2 I bo I b1 I b2
27. 27. Balanced nonsinusoidal quantities Let assume: ∞ ∞ ∞ ∑ [( )] ∑ [( )] ∑ π Vn cos n ωt + 23va = Vn cos[n(ωt )] vb = Vn cos n ωt − 2π 3 vc = n =1 n =1 n =1 For n=1:va1 = V1 cos(ωt ) (π vb1 = V1 cos ωt − 23 ) π vc1 = V1 cos ωt + 23 ( ) For n=2:va 2 = V2 cos(2ωt ) ( π vb 2 = V2 cos 2ωt + 23 ) π vc 2 = V2 cos 2ωt − 23 ( ) For n=3: va 3 = V3 cos(3ωt ) vb3 = V3 cos(3ωt ) vc3 = V3 cos(3ωt )
28. 28. Balanced nonsinusoidal quantitiesFor n=3k-2, The harmonics are similar to positive sequence quantities.For n=3k-1, the harmonics are similar to negative sequence quantities.For n=3k, the harmonics are similar to zero quantities.
29. 29. Symmetrical components• Symmetrical components theory can be applied to both steady-state phasor quantities and instantaneous quantities.• Symmetrical components theory can be derived also by using linear algebra and treated as variable transformation
30. 30. Voltage and Current across the inductor diL 1 to + t vL = L iL = ∫ vL dt + iL (to ) dt L toSteady-state:iL (t ) = iL (t + T ) vL (t ) = vL (t + T )Thus t +T ∫t vL dt = 0Average voltage across the inductor under steady-state condition is zero.
31. 31. Voltage and current across the capacitor dvC 1 to + t iC = C vC = ∫ iC dt + vC (to ) dt C toSteady-state:iC (t ) = iC (t + T ) vC (t ) = vC (t + T )Thus t +T ∫t iC dt = 0Average current through the capacitor under steady-state conditions is zero.
32. 32. Batasan Topologi• Sumber tegangan hanya boleh diparalel jika sama besar• Sumber arus hanya boleh diseri jika sama besar• Sumber tegangan tidak boleh dihubungsingkat• Sumber arus tidak boleh dibuka• Sumber tegangan bisa berupa kapasitor• Sumber arus bisa berupa induktor
33. 33. Hubungan Berikut Harus Dihindari ++
34. 34. DualitasSumber tegangan Sumber arusHubungan paralel Hubungan seriInduktor KapasitorResistor KonduktorReverse conducting switch Reverse blocking switchVariabel tegangan Variabel arus
35. 35. Contoh Dualitas R L C + Vs is Ip vp G C L dis 1 t dv p1 tVs = Ris + L + ∫ is dt + vC (0) I p = Gv p + C + ∫ v p dt + iL (0) dt C 0 dt L 0
36. 36. Contoh Dualitas+ +
37. 37. Penggunaan Komputer• PSIM, MATLAB, EMTP, PSPICE, etc.• Switching Concept• Averaging Concept
38. 38. Switching Concept ii io vo=SwEd S1 R Vo(ω)=Sw(ω)EdEd S2 vo Io(ω)=Vo(ω)/Z(ω) L Ii(ω)=Sw(ω)Io(ω)Sw=1 IF S1 ON and S2 OFFSw=0 IF S1 OFF and S2 ON
39. 39. Switching Concept ii S1 io R Ed S 2 vo IF vref > car THEN S w = 1 ELSE S w = 0 L dio = (S w E d − Rio ) / L dt ii = S wiovref +car −
40. 40. Averaging Concept 0 < t < TON ii S1 io dio Rio + L = Ed dt R TON < t < Ts S 2 vo dio Ed Rio + L =0 dt L Averaging dio Rio (TON + TOFF ) + L (TON + TOFF ) = Ed TON dtvref Divided by Ts results in + dio T Rio + L = E d ON = vocar − dt Ts
41. 41. D-Q Transformf qdo = Kf abc f abc = K -1f qdo(f qdo )T = [ f q fd fo ] ⎡ cos θ sin θ 1⎤(f abc ) T = [ fa fc ] ⎢ (π K = ⎢cos θ − 23 -1 ) ( π sin θ − 23 ) ⎥ 1⎥ fb ( ⎢cos θ + 2π ) ( π sin θ + 23 ) 1⎥ ⎡cosθ ( π cos θ - 23 ) ( π) cos θ + 23 ⎤ ⎣ 3 ⎦ 2⎢K = ⎢ sinθ 3⎢ 1 ( π sin θ - 23 ) ( ) π ⎥ sin θ + 23 ⎥ 1 1 ⎥ ⎣ 2 2 2 ⎦ tθ = ∫ ω (ς )dς + θ o 0
42. 42. D-Q Transformfb fq θ fa fd fc
43. 43. Exampleia = 2 I cos ω s t ( 23π )ib = 2 I cos ω s t +ic = 2 I cos(ω s t − 23 ) πiq = 2 2I 3 [ ( π) ( cos ω s t cos ωt + cos ω s t + 23 cos ωt + 2π 3 )+ cos(ω s t − 23π )cos(ωt − 23π )]iq = 2 I cos(ω s − ω )tid = 2 2I 3 [ ( ) ( π π ) ( π ) ( π sin ω s t sin ωt + sin ω s t + 23 sin ωt + 23 + sin ω s t − 23 sin ωt − 23 )]id = 2 I sin (ω s − ω )tio = 0
44. 44. D-Q TransformPower InvariancePabc = va ia + vb ib + vc ic 3 (Pabc = Pqdo = vq iq + vd id + 2vo io 2 )Instantaneous Reactive Power : 3 (q = v q id − v d iq 2 )
45. 45. DQ Transform of Stationary Elements v abc = pλabc [ ] [ ] v qdo = Kp K -1λabc = Kp K -1 λqdo + KK -1 pλqdo ⎡ − sin θ cos θ 0⎤v abc = ri abc [ ] p K -1 ⎢ ( π = ω ⎢ − sin θ − 23 ) ( π cos θ − 23 ) ⎥ 0⎥v qdo = KrK i qdo-1 ⎣ ( ⎢− sin θ + 2π 3 ) ( π cos θ + 23 ) 0⎥ ⎦ ThusKrK -1 = r ⎡ 0 1 0⎤ [ ] Kp K -1 = ω ⎢ − 1 0 0⎥ ⎢ ⎥ ⎢ 0 0 0⎥ ⎣ ⎦
46. 46. DQ Transform of Stationary Elementsv qdo = ωλdq + pλqdo(λdq ) = [λd T − λq ]vq = ωλd + pλqvd = −ωλq + pλdv o = pλ o
47. 47. DQ Transform of Stationary Elements iq R L + ia R L vq ωLida id L R ib n vd ωLiqb + ic io R Lc vo
48. 48. DQ Transform• If the speed of reference frame ω is equal to the supply frequency, it is called synchronous reference frame.• If the speed of reference frame is zero, it is called stationary reference frame.• If the speed of reference frame is not equal to the supply frequency, it is called asynchronous reference frame.
49. 49. Space Vector N N e jθ + e − jθna = cos θ = 2 2 2 ( π j θ − 23 ) + e − j (θ − 23π )nb = N 2 ( π cos θ − 23 =)2 N e 2 ( π j θ + 23 ) + e − j (θ + 23π )nc = N 2 ( π cos θ + 23 = ) N e 2 2Fa = na ia + nb ib + nc ic N jθ ⎛ π − j 23 j 23 ⎞ π N − jθ ⎛ π j 23 − j 23 ⎞ π= e ⎜ ia + ib e + ic e ⎟ + e ⎜ ia + ib e + ic e ⎟ 4 ⎝ ⎠ 4 ⎝ ⎠ 3 N jθ r 3 N − jθ r *= e i+ e i 8 8r 2⎡ π − j 23 j 23 ⎤ πi = ia + ib e + ic e 3⎢ ⎣ ⎥ ⎦is called space vector of current.The same definitions apply to voltage and current.
50. 50. Example of Space Vectoria = 2 I cos ω s t = 2 2 [ I e jω s t + e − jω s t ] (ib = 2 I cos ω s t + 2π ) = 2 I ⎡e j (ω s t + 23π ) + e − j (ω s t + 23π ) ⎤ 3 2 ⎢ ⎣ ⎥ ⎦ 2 ⎡ j (ω s t − 23π ) − j (ω s t − 23π ) ⎤ ( πic = 2 I cos ω s t − 23 = ) 2 I e ⎢ ⎣ +e ⎥ ⎦ri = 3 2 ⎨ ⎩ [ ] ⎢ ⎣ j (ω t + 2π ) I ⎧ e j ω s t + e − jω s t + ⎡ e s 3 + e − j (ω s t + 23 ) ⎤ − j 23 π ⎥ ⎦ e π ⎢ ⎣ j (ω t − 2π ) + ⎡e s 3 + e − j (ω s t − 23 ) ⎤ j 23 ⎫ π ⎥ ⎦ e ⎬ π ⎭ri = 2 Ie jω s t
51. 51. Transformation of Space Vector r ω r − j ωt x = xe Previous example rω j (ω s −ω )t i = 2 Ie It should be noted : r xa = Re[x ] r ⎡ xe j 23π ⎤ xb = Re ⎢ ⎣ ⎥ ⎦ r ⎡ xe − j 23π ⎤ xc = Re ⎢ ⎣ ⎥ ⎦
52. 52. Numerical Methods To Solve Differential Equations Let f ( x, t ); x (t o ) = xo ; h Euler Method : x(t o + h ) = x(t o ) + hf ( xo , t o ) Fourth - Order Runge - Kutta Method : k1 = f (xo , t o ) ⎛ k1 h⎞ k 2 = f ⎜ xo + , t o + ⎟ ⎝ 2 2⎠ ⎛ k h⎞ k 3 = f ⎜ xo + 2 , t o + ⎟ ⎝ 2 2⎠ k 4 = f ( xo + k 3 , t o + h ) h x(t o + h ) = x(t o ) + (k1 + 2k 2 + 2k3 + k 4 ) 6
53. 53. Example di = 10 − 5idti0 = 0h = 0 .1i (0.1) = 0 + 0.1× (10 − 5 × 0 ) = 1i (0.2 ) = 1 + 0.1× (10 − 5 × 1) = 1.5i (0.3) = 1.5 + 0.1× (10 − 5 × 1.5) = 1.75i (0.4 ) = 1.75 + 0.1× (10 − 5 × 1.75) = 1.875
54. 54. Numerical IntegrationWe have N data for the function y in the interval from a to b : b−aLet h = N b N −1 ∑∫a ydt = h n=0 yi (the simplest rule) b h⎡ N −1 ⎤∫ ∑ ydt = ⎢ y0 + y N + 2 yi ⎥ 2⎢ n =1 ⎥a ⎣ ⎦
55. 55. ExampleT 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9i 1 2 3 1 2 3 1 2 3 1 h = 0.1 N = 10 N −1 ∫0idt = h ∑ i = 1.9 0.9 0 N −1 ∑ i 2 = 4.3 0.9 2 ∫0 i dt = h 0