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)

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

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

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)

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)]

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 ZB1- branch 1st removed,& V0 across it2-all sources sh.& replaced by int. Imp.s3-looking to its terminals x,y ; ZA determined I=V0/(ZA+ZB),VB=IZB=V0ZB/(ZA+ZB)

Attenuation and Distortion rate of Electrical energy supplied: 1/2CV ν watts, dissipated rate: GVν both ~ V result in an exp. Form voltage wave: V0exp(-G/C t) current wave supplies: 1/2LI, dissipate IR ; 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: IR=VG rate. loss. LR=rate. loss. Line Leak. In power trans. res. losses>> leakance losses

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

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

Lattice Diagram Example of Line:Voltage: at instant t, and at point MAdd incident &

reflected up to that instant

A general Method voltage&current at

any location vs time

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 ?

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

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

eB=1.454+.352+.285-.28=1.8

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?

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

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

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

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)]

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)

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

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

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’)

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)]

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

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)]

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)]

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

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

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]

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

1 1 1

1 -2 1

1 1 -2

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/Zc e1c(t)-Ifc(t-τc)

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)]}

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

Q1 continued dI/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=0.715fig4.4:λ=2.0 λ=Z0/R=0.868/R=2 R=0.434 Ω

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/dtdV2/dt+L/R dV2/dt+V2/LC=V1(0)/LC2V2(0)=V’2(0)=0V2(s)=V1(0)/T 1/[s(s+s/Ts+1/T)] xC1/[C1+C2]2x15/21.76=1.38fig 4.7: λ=1.8, R=Z0/λR=Z0/λ=0.868/1.8=0.482Ω

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

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

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