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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

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Reflection & Refraction

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  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

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