1 / 38

Reflection & Refraction

Reflection & Refraction. At Discontinuity (change of Z 0 ): 1-Adjustment to keep the proportionality of V and I 2-in form of initiation of 2 new waves The new waves : Reflected & Transmitted Satisfying portionality & continuity

bunme
Télécharger la présentation

Reflection & Refraction

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Reflection & Refraction • At Discontinuity (change of Z0): 1-Adjustment to keep the proportionality of V and I 2-in form of initiation of 2 new waves • The new waves : Reflected & Transmitted • Satisfying portionality & continuity • The energy conservation Auto. Satisfied • α: reflection coeff. β: refraction coeff. • α=(ZB-ZA)/(ZB+ZA), β=2ZB/(ZB+ZA)

  2. Energy Conservation • Assuming ZA>ZB, I1=V1/ZA, I2=-V2/ZA I3=V3/ZB, V1+V2=V3,I1+I2=I3 • Solving for V2,V3 : V2=[ZB-ZA]/[ZA+ZB] V1 V3=2ZB/[ZA+ZB] V1 I1V1=V1/ZA V2/ZA +V3/ZB=V1/ZA{[(ZA-ZB)/(ZA+ZB)]+4ZAZB/(ZA+ZB)} =V1/ZA

  3. Traveling on multiple joint • i.e. a line connected to n other lines I3B=V3B/ZB, I3C=V3C/ZC, …. I3N=V3N/ZN I2A=-V2A/ZA • For Continuity of Voltage: V1A+V2A=V3B=V3C=….=V3N I1A+I2A=I3B+I3C+…+I3N • These sufficient for Analysis

  4. Line Termination • Open CCT voltage coeff.s :α=1, β=2 • Sh. CCT Voltage coeff.s : α=-1,β=0 • A real surge: V=V0(e^-αt - e^-βt) • For a C termination: ZB(s)=1/C1s • a=[(1/C1s)-ZA]/[1/(C1s)+ZA], • b=2/C1s/[1/(C1s)+ZA] • v2(s)=av1(s)

  5. Time response of C termination • if α=1/(C1ZA): v2(s)=V1/s{[(1/C1s –ZA)/(1/C1s+ZA)]} • V2(t)=V1(1-2e^-αt),V3(t)=V1(2-2e^-αt) • Interpretation of V3 response: 1- At step application, sh. CCT. : O/P zero 2- Finally open CCT.: O/P 2V1 • Similarly For a termination Inductance L1 : • v2(s)=v1/s [(L1s-ZA)/(L1s+ZA)] • Assuming 1/β=L1/ZA  • v2(s)=v1{1/(s+β)-β/[s(s+β)]}, v3=v1[2/(s+β)]

  6. Time response of L termination • V2(t)=-V1(1-2e^-βt) • V3(t)=2V1e^-βt • Application of Thevenin theorem to: calculation of refl. & refr. at Termination • Steps : to calculate current in ZB 1- branch 1st removed,& V0 across it 2-all sources sh.& replaced by int. Imp.s 3-looking to its terminals x,y ; ZA determined • I=V0/(ZA+ZB),VB=IZB=V0ZB/(ZA+ZB)

  7. Attenuation and Distortion • rate of Electrical energy supplied: 1/2CV ν watts, dissipated rate: GVν both ~ V result in an exp. Form voltage wave: V0exp(-G/C t) • current wave supplies: 1/2LI, dissipate IR ; both ~ I :  I=I0 exp(-R/L t) • However to preserve the relation of V/I=Z0 • requires: R/L=G/C or R/G=L/C=Z0=V/I • says: IR=VG rate. loss. LR=rate. loss. Line Leak. • In power trans. res. losses>> leakance losses

  8. Switching Operations and Transmission Lines • Source Impedance • Voltage on Line: V(0)/s x Z0/(Ls+Z0) • V=V(0)[1-exp(- Z0t/L)] • complicated source • The source impedance shown • When study energization of single line

  9. Closing Resistor • stiff source impress 100% voltage on line • closing resistor reduce percentage impressed by factor: Z0/(Z0+R) • S2 close 1st , S1 short some time later • Comparison of reclosing transient voltage

  10. Lattice Diagram • Example of Line: Voltage: at instant t, and at point M Add incident & reflected up to that instant • A general Method • voltage&current at any location vs time

  11. Example(Lattice Diagram Appl.) • A sys. of O/H line & Cable O/H parameters: Zc=270Ω,T=100μs Cable parameters: Zc’=30Ω,T’=50μs • Unit step, Zs=0 • C open CCT • VB, IB ?

  12. Refl. & Refr coefficients • αA=-1 , βA=0 • for B junction if O/H ~1, cable~2 αB1-2=(30-270)/300=-0.8 βB1-2=600/300=0.2 αB2-1=0.8 βB2-1=540/300=1.8 • αC=1 β=2 • Consider the Lattice Diagram

  13. Lattice Diagram of Example • T=2T’ t=0 eB=0 t=T eB=1-.8=.2 t=2T eB=1-.8+.36=.56 t=3T eB=.56+.8-.352=1 t=4T eB=1-.36+.288+.51 =1.454 t=5T eB=1.454+.352+.285-.28=1.8

  14. Voltage Variation at B • The voltage at B 1-rising continuously 2-increasing to 2 pu 3- since C open • What current is expected? • Any possible response?

  15. Current Refl. & Refr. coefficients • αA=1 βA=0 • αB12=0.8 βB12=1.8 • αB21=-.8 βB21=0.2 • αC=-1 βc=0 • draw a similar Lattice Diagram

  16. Lattice Diagram for Current • t=0 iB=0 • t=T iB=(1+.8)/Zc=1.8/Zc • t=2T iB=1.44/Zc • t=3T iB=2.59/Zc • T=4T iB=1.425/Zc • T=5T iB=1.774/Zc • Next the IB curve

  17. Variation of IB • variation different • no similarity • there is some similarity in single line propagation • Method capable of application in a software • High memory size

  18. Characteristic Method • Wave Equations: L ∂i/dt+Ri+∂e/∂x=0 (1) C ∂e/dt+Ge+∂i/∂x=0 (2) Difference of a function of 2 variables: de=∂e/∂t dt+ ∂e/∂x dx di=∂i/∂t dt + ∂i/∂x dx From these if ∂e/∂x,∂i/∂t as follows ∂e/∂x=[de-∂e/∂t dt]/dx ∂i/∂t=[di-∂i/∂x dx]/dt be substituted in EQ 1: L{ [di-∂i/∂x dx]/dt} +RI+ {[de-∂e/∂t dt]/dx}=0 (3) ∂e/∂t=- 1/C ∂i/∂x – G/C e [from (2) substituted in (3)]

  19. Reforming the Equations • Ldi/dt+Ri+de/dx+G/C e dt/dx+ (-Ldx/dt+1/C dt/dx) ∂i/∂x=0 (4) • term in ()=0 to cancel the partial derivatives; then • 2 resultant ODEs: Ldi/dt+Ri+de/dx+1/C Ge dt/dx=0 (5) (dx/dt)=1/LC or in form of: • LdI dx/dt+Ridx+de+1/C Ge dt/dx=0 dx/dt=+(-)1/√LC (6)

  20. Solution based on Characteristic Method • if dx/dt=1/√LC: √L/C di+(RI+√L/C Ge)dx+de=0 (7) • If dx/dt=-1/√LC: -√L/C di +(Ri-√L/C Ge)dx+de=0 (8) • The characteristics are straight Lines called Forward & Backward • e & i are found from above EQs

  21. Finding lossless line solution • dx/dt=1/√LC=v, de=-√L/C di=-Zc di (9) • dx/dt=-1/√LC=-v de=√L/C di=Zc di (10) • 1st method employed by Bergeron 1928 in Hydraulic • Application to single phase transmission line

  22. Integration of ODEs 7 & 8 • integrating EQ set (7): e=-Zci+c1 (9), x=vt+c2 (10) • where c1 & c2 are constants: found from initial conditions • X=0 line terminal, if point (d, t) satisfy EQ (10) then satisfy EQ(9) and: e(d,t)=-Zc i(d,t)+c1 , d=vt+c2 (11) • Similarly for point (0,t’): e(0,t’)=-Zc i(0,t’)+c1, 0=vt’+c2 (12) • Subtracting EQs 11 & 12 respectively: • e(d,t)-e(0,t’)=-Zc[i(d,t)-i(0,t’)], d=v(t-t’)

  23. Solution Continued • t’=t-d/v=t-τ, where: τ=d/v • e(d,t)-e(0,t-τ)=-Zc [i(d,t)-i(0,t-τ)] (13) and: • e(0,t)-e(d,t-τ)=Zc[i(0,t)-i(d,t-τ)] (14) • Rearranging (13)&(14): e(d,t)=-Zc i(d,t)+[e(0,t-τ)+Zc i(0,t-τ)] (15) e(0,t)= Zc i(0,t)+[e(d,t-τ)- Zc i(d,t-τ)] (16) • Defining, 2 terms in right brackets as History dependent voltage sources; Ef(0,t-τ)=-[e(0,t-τ)+Zc i(0,t-τ)] Eb(d,t-τ)=-[e(d,t-τ)-Zc i(d,t-τ)]

  24. Lossless line Equivalent CCTs • Substituting in (15)&(16) e(d,t)=-Zc i(d,t)-Ef(0,t-τ) (17) e(0,t)=Zc i(0,t)–Eb(d,t-τ) (18) • Equiv. CCT. ,  • The Norton Eq. CCT more useful

  25. Line Norton Eq. CCT. • rewriting (17)&(18): i(d,t)=-1/Zc e(d,t)-If(0,t-τ) (19) i(0,t)=1/Zc e(0,t) + Ib(d,t-τ)(20) If & Ib Hist. depend. Cur. Sources: If(0,t-τ)=-1/Zc e(0,t-τ)-i(0,t-τ) Ib(d,t-τ)=-1/Zc e(d,t-τ)–i(d,t-τ) Simple H.D.S. evaluation: Ef(0,t)=-[2e(0,t)+Eb(d,t-τ)] Eb(d,t-τ)=-[2e(d,t)-Ef(0,t-τ)]

  26. Eq. CCT. Of Lumped Elements • Inductance: • ea-eb=L(dia,b/dt) • Trapezoidal Rule: ia,b(t)-ia,b(t-∆t)= 1/L∫(ea-eb)dt= 1/L{[ea(t)-eb(t)]+ [ea(t-∆t)+ eb(t-∆t)]}/2 . ∆t • ia,b(t)=∆t/2L [ea(t)- eb(t)] +Ia,b(t-∆t) • Ia,b(t-∆t)=ia,b(t-∆t)+ ∆t/2L[ea(t-∆t)-eb(t-∆t)]

  27. Eq. CCT for Lumped Capacitor • Similar derivation: • ia,b(t)=2C/∆t[ea(t)-eb(t)]+Ia,b(t-∆t) • Where: Ia,b(t-∆t)=-ia,b(t-∆t)-2C/∆t [ea(t-∆t)-eb(t-∆t)] • all in form of: algebraic EQs

  28. Distributed Line Model in 3ph network • for a 3ph lossless line in general: [-∂eph/∂x]=[L][∂iph/∂t] [-∂iph/∂x]=[C][∂eph/∂t] • wave EQs similarly for 3ph is: [∂eph/∂x]=[L][C][∂eph/∂t] [∂iph/∂x]=[C][L][∂iph/∂t] [L],[C] inductance & capacitance matrices of 3 ph line with mutuals

  29. Similarity Transformation • to solve the complexity of EQs • instead of 3ph Domain, Modal Domain solved for 3 independent voltages • Results of Modal Domain Transferred to 3ph • [eph]=[M][eM] and [iph]=[N][iM] • [∂eM/∂x]= [M]-[L][C][M][∂eM/∂t]=[Λ][∂eM/∂t] • [∂iM/∂x]= [N]-[L][C][N][∂iM/∂t]=[Λ][∂iM/∂t]

  30. Similarity Transformation • [Λ] is diagonal matrix • Diagonal elements are eigen values of: [L][C] or [C][L] • EQ of λn is independent of other modes: ∂eM/∂x=λn ∂eM/∂t λn≈LC of single phase • A case where: [M]=[N] is shown  • Vn=√1/λn,τn=l/vn • Zn=vn.λn

  31. Bergeron EQs for 3ph network • Eq. Modal Domains of 3ph. • i1a-2a(t)= -1/Za e1a(t)-Ifa(t-τa) • i1b-2b(t)= -1/Zb e1b(t)-Ifb(t-τb) • i1c-2c(t)= -1/Zce1c(t)-Ifc(t-τc)

  32. The 3ph Eq CCT Equations • In matrix form: [iM(1-2)]=-[Λ/]-[eM1]-[IMf(t-τ)] [IMf(t-τ)]=-{[Λ/][eM2(t-τ)]-[iM(2-1)(t-τ)]} Then : [N]-[i1-2(t)]=[Λ/]-[M][e1(t)]+[IMf(t-τ)] Or: [i1-2(t)]=[G][e1(t)] + [I] Where: [G]=[N][Λ/]-[M]- [I]=-[N]{[Λ/]-[eM2(t-τ)]+[iM(1-2)(t-τ)]}

  33. 1st Mid Term Exam • Question 1: • Xc1=20/20=20Ω, Xc2=40Ω C1=1/(3.14x20)=159.1μF,C2=79.6μF • Vp=20√2/√3 Ceq=C1C2/[C1+C2] • Z0=√[(40x238.68)/12661]=0.868Ω

  34. Q1 continued • dI/dt+1/τs dI/dt +I/T=0 • i(s)=(s+1/Ts)/[s+s/Ts+1/T]I(0)+I’(0)/[s+s/Ts+1/T] • I(0)=0, I’(0)=Vc(0)/L • i(s)=Vc(0)/L x 1/[s+s/Ts+1/T] • i(s)=Vc(0)/L x 1/[s+1/T] undamped • I(t)=Vc(0)/Z0 sinω0t • Ip=Vp/0.868=18.81 KA • Ip%=13.5/18.81=0.715fig4.4:λ=2.0 • λ=Z0/R=0.868/R=2  R=0.434 Ω

  35. Q1 solution • Vcf=Vpx159.1/[159.1+79.6]=10.88KV • in undamp, C2swing to 21.76 KV • with damping: C1V1=C1V1(0)-C2V2 V1=V1(0)-C2/C1V2 V1=IR+L dI/dt +V2, I=C2dV2/dt dV2/dt+L/R dV2/dt+V2/LC=V1(0)/LC2 V2(0)=V’2(0)=0 V2(s)=V1(0)/T 1/[s(s+s/Ts+1/T)] xC1/[C1+C2] 2x15/21.76=1.38fig 4.7: λ=1.8, R=Z0/λ R=Z0/λ=0.868/1.8=0.482Ω

  36. Question 2 • 30/[20√3]=0.866 KA • I XL /[20/√3]=0.12 • XL=0.12x20/√3/0.868=1.6 Ω • L=1.6/314.15=5.1 mH • R=0.05x20√3/0.868=0.666Ω

  37. Q2 continued • Z=√0.666+1.6=1.73Ω • Φ=tan-1.6/0.666=tan-2.4=67.38◦ • Z0=√0.0051/(1.2x10^-8)=651.92Ω • λ=651.92/0.666=978.8 • TRV :Almost undamped:2Vp=2x16.33=32.66 KV • t=Π√LC=3.1415√0.0051x1.2x10^-8=24.47 μs • k=20/32.66=0.6125 fig 4.7:η=1.2

  38. Q2 continued • R/651.9=1.2 R=782.3Ω • RRRV=32.65/24.57=1.327 KV/μs • t’=3.6  t=3.6x 7.82=28.15μs • RRRV=20/28.15=0.71 KV/μs

More Related