Reflection

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
• a reflection coeff. ß refraction coeff.
• a(ZB-ZA)/(ZBZA), ß2ZB/(ZBZA)

2
Energy Conservation

• Assuming ZAgtZB, I1V1/ZA,
  I2-V2/ZA
• I3V3/ZB,
  V1V2V3, I1I2I3
• Solving for V2,V3
• V2ZB-ZA/ZAZB V1
• V32ZB/ZAZB V1
• I1V1V1?/ZA
• V2?/ZA V3?/ZBV1?/ZA(ZA-ZB)/(ZAZB)?4ZAZB
  /(ZAZB)?
• V1?/ZA

3
Traveling on multiple joint

• i.e. a line connected to n other lines
• I3BV3B/ZB, I3CV3C/ZC, . I3NV3N/ZN
• I2A-V2A/ZA
• For Continuity of Voltage
• V1AV2AV3BV3C.V3N
• I1AI2AI3BI3CI3N
• These sufficient for Analysis

4
Line Termination

• Open CCT voltage coeff.s a1, ß2
• Sh. CCT Voltage coeff.s a-1,ß0
• A real surge VV0(e-at - e-ßt)
• For a C termination
• ZB(s)1/C1s
• a(1/C1s)-ZA/1/(C1s)ZA,
• b2/C1s/1/(C1s)ZA
• v2(s)av1(s)

5
Time response of C termination

• if a1/(C1ZA)
• v2(s)V1/s(1/C1s ZA)/(1/C1sZA)
• V2(t)V1(1-2e-at),V3(t)V1(2-2e-at)
• 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)/(L1sZA)
• Assuming 1/ßL1/ZA ?
• v2(s)v11/(sß)-ß/s(sß), v3v12/(sß)

6
Time response of L termination

• V2(t)-V1(1-2e-ßt)
• V3(t)2V1e-ßt
• 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
• IV0/(ZAZB),VBIZBV0ZB/(ZAZB)

7
Attenuation and Distortion

• rate of Electrical energy supplied
• 1/2CV? ? watts, dissipated rate GV??
• both V?
• result in an exp. Form voltage wave
  V0exp(-G/C t)
• current wave supplies
• 1/2LI?, dissipate I?R
• both I? ? II0
  exp(-R/L t)
• However to preserve the relation of V/IZ0
• requires R/LG/C or R/GL/CZ0?V?/I?
• says I?RV?G rate. loss. LRrate. loss. Line
  Leak.
• In power trans. res. lossesgtgt leakance losses

8
Switching Operations and Transmission Lines

• Source Impedance
• Voltage on Line
• V(0)/s x Z0/(LsZ0)
• VV(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/(Z0R)
• 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
• voltagecurrent at any location vs time

11
Example(Lattice Diagram Appl.)

• A sys. of O/H line Cable
• O/H parameters
• Zc270O,T100µs
• Cable parameters
• Zc30O,T50µs
• Unit step, Zs0
• C open CCT
• VB, IB ?

12
Refl. Refr coefficients

• aA-1 , ßA0
• for B junction if O/H 1, cable2
• aB1-2(30-270)/300-0.8
• ßB1-2600/3000.2
• aB2-10.8 ßB2-1540/3001.8
• aC1 ß2
• Consider the Lattice Diagram

13
Lattice Diagram of Example

• T2T
• t0 eB0
• tT eB1-.8.2
• t2T eB1-.8.36.56
• t3T eB.56.8-.3521
• t4T eB1-.36.288.51
• 1.454
• t5T eB1.454.352.285-.281.8

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

• aA1 ßA0
• aB120.8 ßB121.8
• aB21-.8 ßB210.2
• aC-1 ßc0
• draw a similar Lattice Diagram

16
Lattice Diagram for Current

• t0 iB0
• tT iB(1.8)/Zc1.8/Zc
• t2T iB1.44/Zc
• t3T iB2.59/Zc
• T4T iB1.425/Zc
• T5T iB1.774/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/dtRi?e/?x0
  (1)
• C ?e/dtGe?i/?x0
  (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/?xde-?e/?t dt/dx
• ?i/?tdi-?i/?x dx/dt
• be substituted in EQ 1
• L di-?i/?x dx/dt RI de-?e/?t dt/dx0
• 
  (3)
• ?e/?t- 1/C ?i/?x G/C e from (2) substituted
  in (3)

19
Reforming the Equations

• Ldi/dtRide/dxG/C e dt/dx
• (-Ldx/dt1/C dt/dx) ?i/?x0
  (4)
• term in ()0 to cancel the partial derivatives
  then
• 2 resultant ODEs
• Ldi/dtRide/dx1/C Ge dt/dx0 (5)
• (dx/dt)?1/LC
• or in form of
• LdI dx/dtRidxde1/C Ge dt/dx0
• dx/dt(-)1/vLC
  (6)

20
Solution based on Characteristic Method

• if dx/dt1/vLC
• vL/C di(RIvL/C Ge)dxde0 (7)
• If dx/dt-1/vLC
• -vL/C di (Ri-vL/C Ge)dxde0 (8)
• The characteristics are straight Lines
• called Forward Backward
• e i are found from above EQs

21
Finding lossless line solution

• dx/dt1/vLCv,
• de-vL/C di-Zc di (9)
• dx/dt-1/vLC-v
• devL/C diZc 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-Zcic1 (9), xvtc2
  (10)
• where c1 c2 are constants
• found from initial conditions
• X0 line terminal, if point (d, t) satisfy EQ
  (10)
• then satisfy EQ(9) and
• e(d,t)-Zc i(d,t)c1 , dvtc2
  (11)
• Similarly for point (0,t)
• e(0,t)-Zc i(0,t)c1, 0vtc2
  (12)
• Subtracting EQs 11 12 respectively
• e(d,t)-e(0,t)-Zci(d,t)-i(0,t), dv(t-t)
  

23
Solution Continued

• tt-d/vt-t, where td/v
• e(d,t)-e(0,t-t)-Zc i(d,t)-i(0,t-t)
  (13)
• and
• e(0,t)-e(d,t-t)Zci(0,t)-i(d,t-t)
  (14)
• Rearranging (13)(14)
• e(d,t)-Zc i(d,t)e(0,t-t)Zc i(0,t-t)
  (15)
• e(0,t) Zc i(0,t)e(d,t-t)- Zc i(d,t-t)
  (16)
• Defining, 2 terms in right brackets as History
  dependent voltage sources
• Ef(0,t-t)-e(0,t-t)Zc i(0,t-t)
• Eb(d,t-t)-e(d,t-t)-Zc i(d,t-t)

24
Lossless line Equivalent CCTs

• Substituting in (15)(16)
• e(d,t)-Zc i(d,t)-Ef(0,t-t)
• (17)
• e(0,t)Zc i(0,t)Eb(d,t-t)
• (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-t) (19)
• i(0,t)1/Zc e(0,t) Ib(d,t-t)(20)
• If Ib Hist. depend. Cur. Sources
• If(0,t-t)-1/Zc e(0,t-t)-i(0,t-t)
• Ib(d,t-t)-1/Zc e(d,t-t)i(d,t-t)
• Simple H.D.S. evaluation
• Ef(0,t)-2e(0,t)Eb(d,t-t)
• Eb(d,t-t)-2e(d,t)-Ef(0,t-t)

26
Eq. CCT. Of Lumped Elements

• Inductance
• ea-ebL(dia,b/dt)
• Trapezoidal Rule
• ia,b(t)-ia,b(t-?t)
• 1/L?(ea-eb)dt
• 1/Lea(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/2Lea(t-?t)-eb(t-?t)

27
Eq. CCT for Lumped Capacitor

• Similar derivation
• ia,b(t)2C/?tea(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/?xL?iph/?t
• -?iph/?xC?eph/?t
• wave EQs similarly for 3ph is
• ??eph/?x?LC??eph/?t?
• ??iph/?x?CL??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
• ephMeM and iphNiM
• ??eM/?x?
• M-?LCM??eM/?t????eM/?t?
• ??iM/?x?
• N-?LCN??iM/?t????iM/?t?

30
Similarity Transformation

• ? is diagonal matrix
• Diagonal elements are eigen values of
• LC or CL
• EQ of ?n is independent of other modes
• ??eM/?x??n ??eM/?t?
• ?nLC of single phase
• A case where
• MN is shown ?
• Vnv1/?n,tnl/vn
• Znvn.?n

1 1 1
1 -2 1
1 1 -2
31
Bergeron EQs for 3ph network

• Eq. Modal Domains of 3ph.
• i1a-2a(t)
• -1/Za e1a(t)-Ifa(t-ta)
• i1b-2b(t)
• -1/Zb e1b(t)-Ifb(t-tb)
• i1c-2c(t)
• -1/Zc e1c(t)-Ifc(t-tc)

32
The 3ph Eq CCT Equations

• In matrix form
• iM(1-2)-??/?-?eM1-IMf(t-t)
• IMf(t-t)-??/?eM2(t-t)-iM(2-1)(t-t)
  
• Then
• N-?i1-2(t)??/?-?Me1(t)IMf(t-t)
• Or
• i1-2(t)Ge1(t) I
• Where GN??/?-?M-?
• I-N??/?-?eM2(t-t)iM(1-2)(
  t-t)

33
1st Mid Term Exam

• Question 1
• Xc120?/2020O, Xc240O C11/(3.14x20)159.1µF,C2
  79.6µF
• Vp20v2/v3 CeqC1C2/C1C2
• Z0v(40x238.68)/126610.868O

34
Q1 continued

• d?I/dt?1/ts dI/dt I/T?0
• i(s)(s1/Ts)/s?s/Ts1/T?I(0)I(0)/s?s/Ts1/
  T?
• I(0)0, I(0)Vc(0)/L
• i(s)Vc(0)/L x 1/s?s/Ts1/T?
• i(s)Vc(0)/L x 1/s?1/T? undamped
• I(t)Vc(0)/Z0 sin?0t
• IpVp/0.86818.81 KA
• Ip13.5/18.810.715?fig4.4?2.0
• ?Z0/R0.868/R2 ? R0.434 O

35
Q1 solution

• VcfVpx159.1/159.179.610.88KV
• in undamp, C2swing to 21.76 KV
• with damping
• C1V1C1V1(0)-C2V2? V1V1(0)-C2/C1V2
• V1IRL dI/dt V2, IC2dV2/dt
• d?V2/dt?L/R dV2/dtV2/LCV1(0)/LC2
• V2(0)V2(0)0
• V2(s)V1(0)/T? 1/s(s?s/Ts1/T?)
• 
  xC1/C1C2
• 2x15/21.761.38?fig 4.7 ?1.8, RZ0/?
• RZ0/?0.868/1.80.482O

36
Question 2

• 30/20v30.866 KA
• I XL /20/v30.12
• XL0.12x20/v3/0.8681.6 O
• L1.6/314.155.1 mH
• R0.05x20v3/0.8680.666O

37
Q2 continued

• Zv0.666?1.6?1.73O
• Ftan-?1.6/0.666tan-?2.467.38?
• Z0v0.0051/(1.2x10-8)651.92O
• ?651.92/0.666978.8
• TRV Almost undamped2Vp2x16.3332.66 KV
• t?vLC3.1415v0.0051x1.2x10-824.47 µs
• k20/32.660.6125? fig 4.7?1.2

38
Q2 continued

• R/651.91.2? R782.3O
• RRRV32.65/24.571.327 KV/µs
• t3.6 ? t3.6x 7.8228.15µs
• RRRV20/28.150.71 KV/µs